Answers
Answer:
The answer is 4, the west.
Answer:
The west
Explanation:
hopes this helps you and your questions
Related Questions
1. The kinetic energy of a car is 8 106 J as it travels along a horizontal road. How much work is required to stop the car in 10 s? (A) zero joules (B) 8 105 J (C) 8 107 J (D)8 104 J (E) 8 106 J
Answers
The power to stop the car with kinetic energy of a car is \(8*10^{6} J\) as it travels along a horizontal road is \(8*10^{5} watt\), option B
What is Kinetic energy ?Kinetic energy can be seen as one that is been recorded when an object is able to move from a place , in a broad term we can say this is the energy that can be attributed to that of someone leaving a place and go to another place hence we can see it as the one in the motion.
The definition of energy as the "power to accomplish work" refers to the capacity to apply a force that moves an object. Even if the word is vague, it is clear what energy actually means: it is the force that causes objects to move. The two types can be attributed to the one we know which are kinetic and potential energy.
\(Power \frac{Energy}{time}\)
\(Energy = 8*10^{6} J\)
\(time = 10 s\)
\(Power = \frac{8*10^{6} J}{10}\)
\(power = 8*10^{5} watt\)
Learn more about kinetic energy at:
https://brainly.com/question/8101588
#SPJ1
proper question;
The kinetic energy of a car is 8 × 106 J as it travels along a horizontal road. How much power is required to stop the car in 10 s? (A) zero joules (B) 8 105 J (C) 8 107 J (D)8 104 J (E) 8 106 J
A 0.0400 kg meter stick is placed on a thin rod at the 30.0 cm mark. What is the minimum mass required to be placed on the 0.00 cm mark on the stick to maintain equilibrium?
Answer in kg
Answers
The minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.
To maintain equilibrium, the torques acting on the meter stick must balance each other. The torque is given by the formula:
τ = r * F * sin(θ)
where τ is the torque, r is the distance from the pivot point to the point where the force is applied, F is the force applied, and θ is the angle between the force vector and the lever arm.
In this case, the meter stick is in equilibrium when the torques on both sides of the pivot point cancel each other out. The torque due to the weight of the meter stick itself is acting at the center of mass of the meter stick, which is at the 50.0 cm mark.
Let's denote the mass to be placed on the 0.00 cm mark as M. The torque due to the weight of M can be calculated as:
τ_M = r_M * F_M * sin(θ)
where r_M is the distance from the pivot point to the 0.00 cm mark (which is 30.0 cm), F_M is the weight of M, and θ is the angle between the weight vector and the lever arm.
Since the system is in equilibrium, the torques on both sides of the pivot point must be equal:
τ_M = τ_stick
r_M * F_M * sin(θ) = r_stick * F_stick * sin(θ)
Substituting the given values:
30.0 cm * F_M = 20.0 cm * (0.0400 kg * 9.8 m/s^2)
Solving for F_M:
F_M = (20.0 cm / 30.0 cm) * (0.0400 kg * 9.8 m/s^2)
F_M = 0.0264 kg * 9.8 m/s^2
F_M = 0.25872 N
Finally, we can convert the force into mass using the formula:
F = m * g
0.25872 N = M * 9.8 m/s^2
M = 0.0264 kg
Therefore, the minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.
For more such questions on equilibrium, click on:
https://brainly.com/question/517289
#SPJ8
The current in the wires of a circuit is 60 milliamps. If the resistance of the circuit were doubled (with no change in voltage), then it’s new current would be _____ milliamps
Answers
Answer:30
Explanation:
Current=60 milliamps
Current=(voltage)/(resistance)
60=(voltage)/(resistance)
Doubling the resistance means multiplying both sides by 1/2
60x1/2=(voltage)/(resistance) x 1/2
30=(voltage)/2(resistance)
Therefore the resistance would be 30 milliamp if we double the resistance
A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of vi = 2.60 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e).
The right end of a horizontal spring labeled k is attached to a wall. Five images show five configurations as a block labeled m approaches, compresses, and then moves away from the spring.
In figure a, the block is to the left of the spring, and an arrow above the block points to the right.
In figure b, the block is just touching the uncompressed spring, and an arrow labeled vector vi above the block points to the right.
In figure c, the block has compressed the spring by a distance d, and a label indicates vector vf = 0.
In figure d, the block is just touching the uncompressed spring, and an arrow labeled vector v above the block points to the left.
In figure e, the block is a distance D away from the spring, and a label indicates vector v = 0.
(a)
Find the distance of compression d (in m).
m
(b)
Find the speed v (in m/s) at the unstretched position when the object is moving to the left (Figure d).
m/s
(c)
Find the distance D (in m) where the object comes to rest.
m
(d)
What If? If the object becomes attached securely to the end of the spring when it makes contact, what is the new value of the distance D (in m) at which the object will come to rest after moving to the left?
m
Answers
Answer:
(a) Approximately \(0.335\; \rm m\).
(b) Approximately \(1.86\; \rm m\cdot s^{-1}\).
(c) Approximately \(0.707\; \rm m\).
(d) Approximately \(0.228\; \rm m\).
Explanation:
\(v_i\) denotes the velocity of the object in the first diagram right before it came into contact with the spring. Let \(m\) denote the mass of the block. Let \(\mu\) denote the constant of kinetic friction between the object and the surface. Let \(g\) denote the constant of gravitational acceleration.Let \(k\) denote the spring constant of this spring.(a)Consider the conversion of energy in this object-spring system.
First diagram: Right before the object came into contact with the spring, the object carries kinetic energy \(\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2\).
Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.
Third diagram: After the velocity of the object becomes zero, it has moved a distance of \(D\) and compressed the spring by the same distance.
Energy lost to friction: \(\underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D\). Elastic potential energy that the spring has gained: \(\displaystyle \frac{1}{2}\,k\, D^2\).The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:
\(\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2\).
Assume that \(g = 9.81\; \rm m \cdot s^{-2}\). In the equation above, all symbols other than \(D\) have known values:
\(m =1.10\; \rm kg\).\(v_i = 2.60\; \rm m \cdot s^{-1}\).\(\mu = 0.250\).\(g = 9.81\; \rm m \cdot s^{-2}\).\(k = 50.0\; \rm N \cdot m^{-1}\).Substitute in the known values to obtain an equation for \(D\) (where the unit of \(D\!\) is \(m\).)
\(3.178 = 2.69775\, D + 25\, D^2\).
\(2.69775\, D + 25\, D^2 + 3.178 = 0\).
Simplify and solve for \(D\). Note that \(D > 0\) because the energy lost to friction should be greater than zero.
\(D \approx 0.335\; \rm m\).
(b)The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:
\(\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J\).
As the object moves to the left, part of that energy will be lost to friction:
\((\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J\).
The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:
\(2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J\).
Calculate the velocity corresponding to that kinetic energy:
\(\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}\).
(c)As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy (\(1.91\; \rm J\)) would be lost to friction.
How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is \(\mu \cdot m \cdot g\).
\(\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m\).
(d)Similar to (a), solving (d) involves another quadratic equation about \(D\).
Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) \(1.91\; \rm J\).
Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.
\(\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2\).
\(25\, D^2 + 2.69775\, D - 1.90811\approx 0\).
Again, \(D > 0\) because the energy lost to friction is greater than zero.
\(D \approx 0.228\; \rm m\).
The energy transferred between the object and the spring as a closed system, therefore, conserved are;
(a) The distance of compression, d ≈ 0.3354 meters
(b) The speed in the un-stretched position wen the object is sliding to the left, v ≈ 1.8623 m/s
(c) The distance where the object comes to rest, D ≈ 0.7071 m
(d) The distance the object will come to rest attached to the spring, D ≈ 0.2278 m
The reason the above values are correct are as follows;
The known parameters are;
Mass of the object, m₁ = 1.10 kg
Coefficient of friction, μ = 0.250
The initial speed of the object, \(v_i\) = 2.60 m/s
Force constant of the spring, K = 50.0 N/m
Distance the spring is compressed by the object = d
(a) Conservation of energy principle
\(Kinetic \ energy = \dfrac{1}{2} \cdot m\cdot v^2\)
Work done = Force × Distance
Friction force, \(F_f\) = W × μ
Weight, W = m·g
Weight = Mass × Acceleration
Energy transferred by object = Work done by spring + Work done by friction
\(Energy \ transferred \ by \ object = Kinetic \ energy = \dfrac{1}{2} \times 1.10\times 2.60^2 = 3.718\)
Energy transferred by object = 3.718 J
\(Work \ done \ by \ spring = \dfrac{1}{2} \cdot k\cdot x^2\)
\(Work \ by \ spring \ to \ bring \ object \ to \ rest, \ W_{spring} = \dfrac{1}{2} \times 50\times d^2\)
\(W_{spring}\) = 25·d²
Work done by friction, \(W_{friction}\) = 1.10×9.81×0.250×d = 2.69775·d
Therefore;
3.718 = 25·d² + 2.69775·d
25·d² + 2.69775·d - 3.718 = 0
Solving gives
The distance of the compression d ≈ 0.3354 m
(b) The energy given by the spring = 25·d²
The work done by friction, \(W_{friction}\) = 2.69775·d
Kinetic energy given to object = 0.55·v²
0.55·v² = 25·d² - 2.69775·d
0.55·v² = 25×0.3354² - 2.69775×0.3354
∴ v = √(3.4682) = 1.8623
The velocity of the object at the un stretched position, v ≈ 1.8623 m/s
(c) The kinetic energy, K.E. of the object on the way left is given as follows;
K.E. = 0.5 × 1.10 kg × 3.4682 m²/s² = 1.90751 J
The work done by friction before object comes to rest = 2.69775·D
\(D = \dfrac{1.90751 \, J}{2.69775 \, N} \approx 0.7071 \, m\)
The distance where the object comes to rest, D ≈ 0.7071 m
(d) The work done on spring, \(W_{spring}\) = 25·D'²
Work done on friction, \(W_{friction}\) = 2.69775·D'
Kinetic energy of object, K.E. ≈ 1.90751 J
K.E. = \(W_{spring}\) + \(W_{friction}\)
1.90751 ≈ 25·D'² + 2.6775·D'
25·D'² + 2.6775·D' - 1.90751 = 0
Solving with a graphing calculator gives;
D' ≈ 0.2278 m
The new value of the distance D = 0.2278 m
Learn more about the energy conservation principle here:
https://brainly.com/question/928985
The y-position of a damped oscillator as a function of time is shown in the figure.
This function can be described by the y(t) = A0e-btcos(ωt) formula, where A0 is the initial amplitude, b is the damping coefficient and ω is the angular frequency.
1. What is the period of the oscillator? Please, notice that the function goes through a grid intersection point.
2. Determine the damping coefficient. (include the correct units.)
Answers
The period of the oscillator and the damping coefficient.
t= 1.33s b = 0.0426 s^-1What is the period of the oscillator and what determines the damping coefficient.?Generally, the equation for is mathematically given as
If we see cycles between the timestamps t= 0s and t= 4s, we may deduce that it completes three cycles each and every four seconds. Therefore, the amount of time required to complete one cycle of damping will be
t =4/3sec
t= 1.33s
Generally, the equation for amplitude is mathematically given as
\(A = A_0e^{-bt}\)
Therefore
\(3 = 5e^{-12b}\)
\(0.6 = e^{-12b}\)
Therefore
-0.511 = -12b lne
b = 0.0426 s^{-1}
In conclusion, An influence that acts either from inside an oscillatory system or onto it and has the effect of diminishing or stopping the system from oscillating is known as damping. In physical systems, damping is created by processes that dissipate the energy that is stored in the oscillation. These processes are called "damping agents.". Hence the damping coefficient is
b = 0.0426 s^-1
Read more about damping coefficient
https://brainly.com/question/15723320
#SPJ1
A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air. Find the speed at which the water shoots out of the tank
Answers
Explanation: To get the speed of the efflux we will use the Torricelli‘s equation which states that the speed of the efflux through a small circular hole is given as
v = Square Root 2gh
After inserting the given variables one gets the following efflux velocity
v= square root 2 x 9.81 x 14
v= Square Root 276.54 = 16.6 m/s
How would the calculation change if a typical comet in the Oort cloud is only 1 km in diameter?
Answers
When a typical comet in the Oort cloud is only 1 km in diameter, then the mass of the Oort cloud is changes to 4.18×10⁹ kg.
The mass of each comet int he Oort cloud is,
mass = density × Volume
volume of sphere = 4/3(π×r²) where, r = radius of the sphere
From the given,
diameter = 1 km
radius = Diameter / 2 = 1/2(D)
Volume of the comet = 4/3 (π×r²) = 4/3 (π× (1/2 (10)³)²)
= 4/3×(π) ×(1/8×10⁶)
Mass = density × volume, where density = 1000 kg/m³
= 1000 × 4/3×(π) ×(1/8×10⁶)
= 4.18 ×10⁹ kg
When the typical comet in the Oort cloud is only 12 km in diameter the mass of Oort cloud is 4.18×10⁹ kg.
To know more about the Oort cloud :
https://brainly.com/question/23368033
#SPJ1
if a toy car has a centripetal acceleration of 50 m/s2 and was making the turn at 10 m/s. what was his radius
a 2m
ь 500m
C 5m
d.25m
Answers
Answer:
Explanation:
The centripetal acceleration is expressed as;
a = v²/r
a is the acceleration = 50m/s²
v is the velocity = 10m/s
r is the radius
To get the radius
r = v²/a
r = 10²/50
r = 100/50
r = 2m
Hence its radius is 2m
Question 7(Multiple Choice Worth 4 points)
(04.02 LC)
Which of the following substances is formed during photosynthesis?
A Carbon dioxide
B Light
C Sugar
D Water
Answers
Which of the following substances is formed during photosynthesis?
⠀ ⠀A. Carbon dioxide
⠀ ⠀B. Light
⠀ ⠀C. Sugar
⠀ ⠀D. Water
Answer :Sugar is formed during photosynthesis.
So, correct option is C. Sugar.
Explanation :During photosynthesis, Carbon dioxide and water are used to produce sugar with help of light.
*PLEASE HELP***
Which of the following technologies uses electromagnetic radiation with the
shortest wavelength?
O A. Wireless communications transmitter
O B. Infrared sensor on a DVD player
C. Microwave industrial drying equipment
O D. Compact fluorescent lightbulb
Answers
Answer:
A) wireless communications transmitter
Explanation:
Answer:
Compact fluorescent lightbulb
Explanation:
This is because it has the shortest wavelength
Remaining Time: 1 hour, 49 minutes, 34 seconds.
Question Completion Status:
QUESTION 8
If the magnification is calculated to be +0.5 in a particular situation, what does that tell you about the image?
O A. It is bigger by 0.5
OB
It is half the size of the object.
O C. It is twice the size of the object.
O D. It is smaller by 0.5
O E. It tells you nothing about the image.
QUESTION 9
Answers
Explanation:
i think C . it is twice the size of the object
79. An example of an electrical insulator is
a. graphite.
b. glass.
c. aluminum
d. tungsten
Answers
Answer:
B. Glass
Explanation:
An electrical insulator is a substance that does not conduct electricity.
Glass has tightly bounded electrons, that is why it is an insulator of electricity.
Hey there! The answer to your question is below.
The correct answer would be B.GLASS
Glass is a insulator and rubber too
Glass, wood, plastic and more are all insulators
So the answer would be B. Glass
Hope this helps!
By: xBrainly
A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the origin? What would be the force on a +3e charge at this point? Give the x and y components for both the field and force
Answers
Answer:
Let I and j be the unit vector along x and y axis respectively.
Electric field at origin is given by
E= kq1/r1^2 i + kq2/r2^2j
= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)
= (2.88i + 1.44j)*10^-3 N/C
Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3
F= (1.382 i + 0.691 j) *10^-21
Goodluck
Explanation:
A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a table-top.
Figure attached.
When the Ping-Pong ball exits the tube, which path will it follow in the figure?
d
a
c
b
e
Answers
Answer:
e
Explanation:
A 12 V automobile battery is connected to an electric starter motor. The current
through the motor is 250 A.
a) How much power does the motor dissipate?
b) How much energy does the battery deliver to the motor during the 26 s period?
PLEASE HELP QUICK
Answers
Answer: \(3000\ W,\ 78,000\ J\)
Explanation:
Given
The voltage of the battery is \(V=12\ V\)
Current through the battery is \(I=250\ A\)
(a) Power is the product of voltage and current
Power dissipates is given by
\(P=VI\\P=12\times 250\\P=3000\ W\)
(b) Energy delivered
\(E=P\times t\)
\(E=3000\times 26\\E=78,000\ J\)
Hi
Please help on question asap if the answer is correct I'll rate you five stars a thanks and maybe even brainliest!
When two light bulbs are connected in series, the resistance in the circuit is increase compared to that with one lightbulb. The increased resistance opposes the flow of current so far fewer electrons pass per second , transferring less energy. The lightbulbs are therefore not as bright as In a circuit with same voltage but only one bulb.
However, when Two lightbulbs are connected in parallel, each loop behaves like a separate circuit . The resistance in each branch is the same as if there were just one light bulb in the whole circuit.there is the same current in each branch of the circuit ,so the bulbs Will light up with the same brightness as a single bulb circuit. The energy stored in the battery will decrease twice as quickly and battery will run out faster than I series circuit.
6) explain the advantages and disadvantages of arranging components in series or parallel.
Answers
Advantages of Components Arranged in Series:
1. Increased resistance: Connecting components in series increases the overall resistance in the circuit. This can be advantageous in situations where you want to limit the flow of current or control the amount of power being dissipated.
2. Voltage sharing: In a series circuit, the total voltage of the power supply is divided among the components. This can be useful when you want to ensure that each component receives a specific voltage.
3. Current consistency: The current remains the same throughout the series circuit, which can be beneficial when you need to ensure a consistent current flow through multiple components.
Disadvantages of Components Arranged in Series:
1. Reduced brightness or performance: As mentioned in the question, when light bulbs are connected in series, the increased resistance reduces the flow of current, resulting in dimmer bulbs or reduced performance in other electrical devices.
2. Single point of failure: If one component in a series circuit fails, it can disrupt the entire circuit and cause all components to stop functioning.
Advantages of Components Arranged in Parallel:
1. Brightness and performance: When components such as light bulbs are connected in parallel, each component receives the full voltage, resulting in brighter bulbs and better performance in other devices.
2. Redundancy: In a parallel circuit, if one component fails, the other components can continue to function independently. This provides redundancy and ensures that the failure of one component does not affect the others.
3. Increased current capacity: Parallel arrangement allows the circuit to handle higher currents, which can be useful in situations where high power or heavy loads are required.
Disadvantages of Components Arranged in Parallel:
1. Increased complexity: Parallel circuits require more wiring and connections, which can make the circuit more complex and harder to manage.
2. Higher cost: Parallel circuits may require more components and wiring, resulting in higher costs for materials and installation.
It's important to consider the specific requirements and constraints of the application when deciding whether to arrange components in series or parallel. Each configuration has its advantages and disadvantages, and the choice depends on factors such as desired current flow, voltage distribution, and reliability.I
Although science strives for objectivity by basing conclusions and explanations on data gathered during scientific investigations,
some subjectivity still exists in science, since it is a human endeavor. What beneficial qualities exist in scientific processes,
methods, and knowledge as a result of science being a human endeavor?
A. bias and favoritism
B.
creativity and discovery
C. inaccuracies and errors
D.
all of these
Answers
Answer: creativity and discovery
Explanation: i got it right on study island .
Answer:
CREATIVITY AND DISCOVERY
Explanation:
STUDY ISLAND
The work done in holding a 60 kg object at a height of 2 m above the floor for 20 s is
Answers
Answer:
zero
Explanation:
The work done is zero because the object has no displacement
I need 6 things to save energy or increase energy efficiency in any part of a house
Answers
Here are six things that can help save energy or increase energy efficiency in any part of a house:
Install energy-efficient light bulbs: Replace traditional incandescent light bulbs with energy-efficient LED light bulbs. LED bulbs consume less electricity and last much longer.
Use a programmable thermostat: Installing a programmable thermostat can help you control the temperature of your home more efficiently. You can set it to automatically lower the temperature when you're away from home or sleeping.
Air leaks should be sealed with weatherstripping or caulk around windows and doors. By doing so, you can keep your house more pleasant and prevent draughts.
Install ceiling fans: Ceiling fans can help circulate air and keep your home cool in the summer and warm in the winter. They use less energy than air conditioners or heaters.
Upgrade to energy-efficient appliances: When it's time to replace appliances, choose ones with an Energy Star label. These appliances are designed to use less energy and can help you save money on your energy bills.
Use power strips: Plug electronics and appliances into power strips, which can be turned off when not in use. This can prevent phantom energy use, which occurs when electronics and appliances continue to draw power even when turned off.
Learn more about Energy here:
https://brainly.com/question/1932868
#SPJ1
The "problem of perception" is best characterized as?
Answers
Answer:
making sense of a 3-d world from 2-d data
Explanation:
A crate with a mass of 35.0 kg is pushed with a horizontal force of 150 N. The crate moves at a constant speed across a level, rough surface a distance of 5.85 m
Answers
(a) The work done by the 150 N force is 877.5 Joules.
(b) The coefficient of kinetic friction between the crate and the surface is approximately 0.437.
To answer this problem, we must take into account the work done by the applied force as well as the work done by friction.
(a) The applied force's work may be estimated using the following formula:
Work = Force * Distance * cos(theta)
where the force is 150 N and the distance is 5.85 m. Since the force is applied horizontally and the displacement is also horizontal, the angle theta between them is 0 degrees, and the cosine of 0 degrees is 1.
As a result, the applied force's work is:
Work = 150 N * 5.85 m * cos(0) = 877.5 J
So, the work done by the 150 N force is 877.5 Joules.
(b) Frictional work is equal to the force of friction multiplied by the distance. The work done by friction is identical in amount but opposite in direction to the work done by the applied force since the crate travels at a constant speed.
The frictional work may be estimated using the following formula:
Work = Force of Friction * Distance * cos(theta)
The net force applied on the crate is zero since it is travelling at a constant pace. As a result, the friction force must be equal to the applied force, which is 150 N.
Thus, the work done by friction is:
Work = 150 N * 5.85 m * cos(180) = -877.5 J
Since the work done by friction is negative, it indicates that the direction of the frictional force is opposite to the direction of motion.
The coefficient of kinetic friction may be calculated using the following equation:
Friction Force = Kinetic Friction Coefficient * Normal Force
The normal force equals the crate's weight, which may be computed as:
Normal Force = mass * gravity
where the mass is 35.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2.
Normal Force = 35.0 kg * 9.8 m/s^2 = 343 N
Now, we can rearrange the equation for the force of friction to solve for the coefficient of kinetic friction:
Force of Friction = coefficient of kinetic friction * Normal Force
150 N = coefficient of kinetic friction * 343 N
coefficient of kinetic friction = 150 N / 343 N ≈ 0.437
As a result, the kinetic friction coefficient between the container and the surface is roughly 0.437.
In summary, the work done by the 150 N force is 877.5 Joules, and the coefficient of kinetic friction between the crate and the surface is approximately 0.437.
For more such questions on work, click on:
https://brainly.com/question/8119756
#SPJ8
Given the functions f(x)=(1/x-3)+1 and g(x) = (1/1+4)+3
Which statement describes the transformation of the graph of function f onto the graph of function g?
O The graph shifts 2 units right and 7 units down.
O The graph shifts 7 units left and 2 units up.
O
e graph shifts 7 units right and 2 units down.
O The graph shifts 2 units left and 7 units up.
Answers
The statement that describes the transformation of the graph of function f onto the graph of function g is: The graph shifts 2 units right and 7 units down.
To determine the transformation of the graph of function f onto the graph of function g, we compare the two functions f(x) and g(x) and observe the changes in the equations.
The function f(x) = (1/x - 3) + 1 represents a reciprocal function that is shifted vertically 1 unit up and horizontally 3 units to the right. The reciprocal function is reflected about the line y = x.
The function g(x) = (1/(1 + 4)) + 3 simplifies to g(x) = 4 + 3 = 7, which is a constant function representing a horizontal line at y = 7.
By comparing the equations, we can see that the transformation from f(x) to g(x) involves the following changes:
The term 1/x in f(x) is replaced by the constant 1/(1 + 4) in g(x), resulting in a vertical shift of 7 units up.
The term -3 in f(x) is replaced by 3 in g(x), resulting in a vertical shift of 3 units up.
The +1 in f(x) is replaced by +3 in g(x), resulting in an additional vertical shift of 2 units up.
Therefore, the overall transformation is a shift of 2 units to the right and 7 units down.
Hence, the correct statement is: The graph shifts 2 units right and 7 units down.
For more such questions on graph, click on:
https://brainly.com/question/26865
#SPJ8
1. A small metal weight is suspended from a spring
scale as shown in this illustration
Spring Scale
N
1.0
2.0
3.0
4.0- The spring is measure here at 4.0 N
5.0
6.0
7.0
8.0
9.0
10.0
• What is the gravitational force acting on the weight? Explain.
• What is the net force acting on the weight? Explain.
Answers
Answer:
a) W = -4N, b) F_net = 0
Explanation:
The suspended weight of the spring ladder is subjected to two forces: the elastic force of the spring directed downwards and the gravitational force directed downwards.
\(F_{e}\) - W = 0
F_{e} = W
F_{e} = W = 4 N
a) the gravitational force is the weight of the body
W = -4N
b) as the system is in equilibrium, the net force is zero
F_net = 0
The first P-wave of an earthquake travels 5600 kilometers from the epicenter and arrives at a seismic station at 10:05 a.m. At what time did this earthquake occur?
Ahhhhhh I have a Regent's test in 2 hours and I don't know how to solve this type of question! Any help would be appreciated.
Anyone know what the steps to do this are? I dont even need an answer, just how to get to it. Thank you!
Answers
The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.
The p-waves travel with a constant velocity of 7 km/s
The time can be calculated by using the formula
t = d / v
where
T1 = 10:05 a.m
d is the distance they take to travel from the epicenter
v is the speed of the p-waves
On average, the speed of p-waves is
v = 7 km/s
d = 5600 km (given)
Substituting the values in the formula;
t = d / v
t = 5600 ÷ 7
t = 800 seconds
Converting into minutes,
t = 800 ÷ 60
t = 13.3
≈ 13 mins
T1 - 13 mins = T2
10:05 - 13 mins = 9.52 am
It means the earthquake occurred prior 13 minutes, that is at 9.52 am.
Therefore, the earthquake occurred at 9.52 am.
Learn more about earthquakes from the given link.
https://brainly.com/question/25843505
817 cm3 at 80.8 kPa to 101.3 kPa
Answers
The volume of the gas at 101.3 kPa would be approximately 651.25 cm³.
To calculate the change in volume of a gas from an initial pressure to a final pressure, we can use Boyle's law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.
Boyle's law can be expressed as:
P1 * V1 = P2 * V2
Where:
P1 = Initial pressure (80.8 kPa)
V1 = Initial volume (817 cm³)
P2 = Final pressure (101.3 kPa)
V2 = Final volume (to be calculated)
Let's plug in the values into the equation and solve for V2:
80.8 kPa * 817 cm³ = 101.3 kPa * V2
V2 = (80.8 kPa * 817 cm³) / 101.3 kPa
V2 ≈ 651.25 cm³
Therefore, the volume of the gas at 101.3 kPa would be approximately 651.25 cm³.
For more such questions on volume,click on
https://brainly.com/question/14197390
#SPJ8
Two stars orbit near each other in a galaxy far far away.... If the masses
of both stars is doubled and the distance between them is doubled, what
is the new gravitational force going to be?
Answers
If the mass of both of the objects is doubled, then the force of gravity between them is quadrupled and decreased when distance between two objects is doubled.
What is gravity?All of the matter in the universe is drawn to all of the other bits of matter in the universe by the attractive force of gravity. It is a very significant force on the scale of moons, planets, stars, and galaxies, and it controls a much of the behavior of these objects. Our feet remain firmly planted on the ground thanks to gravity, which also keeps the Moon in orbit around the Earth, the Earth in orbit around the Sun, the Sun in orbit around the center of our Milky Way galaxy, the Milky Way and Andromeda galaxies orbiting their shared center of mass, and so on. Gravity is important to matter.We gravitate toward things that are closest to us and that are the most massive. The Earth's close proximity to the Sun prevents us from zooming off toward the Sun even though the Sun is much more massive than the Earth.Learn more about Gravitational force refer to :
https://brainly.com/question/24783651
#SPJ1
4. Interpret Data The graph below shows the
motion of an elevator. Explain its motion.
Answers
Answer:
Below
Explanation:
0-1 sec descends at constant rate from 10 to 6 m
1-2 sec stops at 6m
2-3 sec descends at constant rate to 2 m
3-4 sec stops at 2 m
4-5 sec descends at another constant rate to 0 m
A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction between the hockey puck and the metal ramp are μs = 0.40 and μk = 0.30, respectively. The puck's initial speed is 3.8 m/s. What speed does it have when it slides back down to its starting point?
Answers
A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction between the hockey puck and the metal ramp are μs = 0.40 and μk = 0.30, respectively. The puck's initial speed is 3.8 m/s, Speed it will have when it slides back down to its starting point is 2.36 m/s
What is Friction?The resistance to motion of one object moving in relation to another is known as friction. It is not regarded as a fundamental force like gravity or electromagnetic, according to the International Journal of Parallel, Emergent and Distributed Systems(opens in new tab). The electromagnetic attraction between charged particles in two contacting surfaces, according to scientists, is what causes it.
using work energy theorem ,
change in kinetic energy = work done by frictional force
\(\frac{1}{2}\) m(\(x^{2}\)-\(y^{2}\)) = μmghcos30°
where mass is m=200g
x is speed with which it slides back
y is speed at top of metal ramp=3.8 m/s
μ is coefficient of kinetic friction=0.3
g is gravity = 9.8m/\(s^{2}\)
h is height to which hockey puck is reached on metal ramp=1.18m
Substituting the values and solving for speed x,
x=2.36m/s
speed it will have when it slides back down to its starting point is 2.36m/s
to know more about friction, visit;
brainly.com/question/28356847
#SPJ1
A box of mass 210 kg is pulled from rest with a string of tension 1300n inclined at 35° to the horizontal. if the box moved with a speed of 10m/s and frictional force between the box and surface is 100 n, calculate the distance covered.
Answers
If A box of mass 210 kg is pulled from rest with a string of tension 1300n inclined at 35° to the horizontal. if the box moved with a speed of 10m/s and the frictional force between the box and surface is 100 n, Then the distance covered by the box is 10.89 meters.
To calculate the distance covered by the box, we need to analyze the forces acting on it and apply the work-energy principle.
Given:
Mass of the box, m = 210 kg
Tension in the string, T = 1300 N
The angle of inclination, θ = 35°
Frictional force, f = 100 N
Initial speed, u = 0 m/s
Final speed, v = 10 m/s
First, let's resolve the tension force into components parallel and perpendicular to the incline. The parallel component of the tension force can be calculated as:
T_parallel = T * cos(θ)
Next, let's calculate the net force acting on the box along the incline. The net force is given by:
Net force = T_parallel - f
Now, using Newton's second law, we can calculate the acceleration (a) of the box:
Net force = m * a
From the given information, we have the final velocity (v), initial velocity (u), and acceleration (a). We can use the following kinematic equation to calculate the distance covered (s):
v^2 = u^2 + 2as
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Now, let's plug in the given values and calculate the distance covered:
T_parallel = 1300 N * cos(35°) ≈ 1067.35 N
Net force = 1067.35 N - 100 N = 967.35 N
a = (967.35 N) / (210 kg) ≈ 4.61 m/s^2
s = (10 m/s)^2 - (0 m/s)^2 / (2 * 4.61 m/s^2) ≈ 10.89 m
Therefore, the distance covered by the box is approximately 10.89 meters.
To learn more about Newton's law of motion click:
brainly.com/question/29775827
#SPJ1
Helpppp please thankss
Answers
Bounce off of the object.
Hope this helps!