Answers
Because Newton as Pressure = Thrust/Area, the answer will be 7.5.
What happens when two kids of different weights balance on a seesaw and each generates an equal amount of torque in the opposing direction?The rotating equivalent of Newton's first law is this. When each child's weight exerts an equal amount of torque about the pivot in opposing directions, two kids can balance on a seesaw (clockwise and counter clockwise).
Can two people of different weights balance on a seesaw?The idea of the seesaw You are probably aware that any two people, regardless of their weights, can balance each other while seated on opposite sides of a seesaw. All that is required is for each individual to contribute an equal moment, which is determined by multiplying the person's weight by the distance from the fulcrum.
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Related Questions
The coefficient of static friction between a rubber tire and dry pavement is about 0.800. Assume that a car’s engine only turns the two rear wheels and that the weight of the car is uniformly distributed over all four wheels.
What limit does the coefficient of static friction place on the time min required for a car to accelerate from rest to 60.0 mph (26.8 m/s)?
Answers
The coefficient of static friction places a limit of approximately 3.42 seconds on the minimum time required for the car to accelerate from rest to 60.0 mph (26.8 m/s).
To find the limit that the coefficient of static friction places on the minimum time required for a car to accelerate from rest to 60.0 mph (26.8 m/s), we need to consider the maximum acceleration the car can achieve due to the friction between the rear tires and the pavement.
The maximum acceleration can be determined using the formula:
a_max = μs * g
where μs is the coefficient of static friction and g is the acceleration due to gravity (approximately 9.8 m/s²).
In this case, since the car's engine only turns the two rear wheels, the maximum acceleration is limited by the friction force between the rear tires and the pavement.
Now, to calculate the minimum time required to accelerate to 60.0 mph (26.8 m/s), we can use the following kinematic equation:
v = u + a * t
where v is the final velocity (26.8 m/s), u is the initial velocity (0 m/s), a is the acceleration, and t is the time.
Rearranging the equation, we have:
t = (v - u) / a
Plugging in the values, we get:
t = (26.8 m/s - 0 m/s) / a_max
t = 26.8 m/s / (μs * g)
Substituting the given value for the coefficient of static friction (μs ≈ 0.800) and the acceleration due to gravity (g ≈ 9.8 m/s²), we can solve for the minimum time required:
t = 26.8 m/s / (0.800 * 9.8 m/s²)
t ≈ 3.42 seconds
Therefore, the coefficient of static friction places a limit of approximately 3.42 seconds on the minimum time required for the car to accelerate from rest to 60.0 mph (26.8 m/s).
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OFFERING 80+ POINTS!!!!!!!!!
Three small spheres, A, B, and C, have charges with magnitudes qA, qB, and qC, respectively.The three spheres are aligned along a straight line, as shown in the figure. At the instant shown, the net force on sphere A is zero.:
1. Which of the following statements must be true of the signs of the charges?
a. Only charges qA and qB have the same sign.
b. Only charges qA and qC have the same sign.
c. Only charges qB and qC have the same sign.
d. Charges qB and qC have different signs.
e. Charges qA, qB, and qC all have the same sign.
2. The ratio qC/qB is...
a. 9/4
b. 1/1
c. 4/9
d. 1/4
e. 1/9
3. Which of the following is true about the sign of charge qA?
a. The sign of charge qA must be the same as that of qB.
b. The sign of charge qA must be the same as that of qC.
c. The sign of charge qA must be the same as that of either qB or qC greater magnitude.
d. The sign of charge qA must be the same as that of either qB or qC lesser magnitude.
e. It is possible that qA could be either positive or negative.
OFFERING A LOT OF POINTS FOR ANSWERS AND EXPLANATIONS TO ALL THREE!!
Answers
1. Charges qB and qC have different signs.
2. The ratio qC/qB is 9/4.
3. The sign of charge qA must be the same as that of either qB or qC lesser magnitude.
What is electric force?Electric force is the attracting or repulsive interaction between any two charged things. Similar to any force, Newton's laws of motion describe how it affects the target body and how it does so.
1. As net force on qA is zero, charges qB and qC have different signs.
2. The ratio qC/qB is = (3d)²/(2d)² = 9/4.
3. As the net force on qA is zero, the sign of charge qA must be the same as that of either qB or qC lesser magnitude.
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A train brakes from 40
m/s to a stop over a
distance of 100m,what
is the acceleration of
the train?
Answers
Answer:
the acceleration of the train is - 8 m/s² since it is decelerating as the influence of the brake comes to stop
Explanation:
If a spring is stretched 4m from its starting length when 20n of force is applied, then how much work (in joules) is done by the spring when it is stretched 10 m?
Answers
ANSWER:
250 J
STEP-BY-STEP EXPLANATION:
F = 20N is required to stretch the spring by 4 meters
We know that the force is equal to:
\(F=k\cdot x\)We solve for k (spring constant):
\(k=\frac{F}{x}=\frac{20}{4}=5\text{ N/m}\)The work done in stretching the spring is given by the following equation (in this case the stretch is 10 meters:
\(\begin{gathered} W=\frac{1}{2}k\cdot x^2 \\ \text{ Replacing} \\ W=\frac{1}{2}\cdot5\cdot10^2 \\ W=250\text{ J} \end{gathered}\)The work required is 250 joules.
match the words in the left-hand column to the appropriate blank in the sentences in the right-hand column. use each word only once. view available hint(s)for part a resethelp 1. the blankof the hubble space telescope is better for shorter (bluer) wavelengths of light than for longer (redder) wavelengths of light.target 1 of 7 2. the large research observatories on mauna kea use giant blank.target 2 of 7 3. blankseparate the various colors of light, allowing astronomers to determine stellar composition and many other stellar properties.target 3 of 7 4. the twin 10-m keck telescopes can work together to obtain better angular resolution through a technique known as blank.target 4 of 7 5. the chandra x-ray observatory focuses x rays with blankmirrors.target 5 of 7 6. a 10-meter telescope has a larger blankthan a 4-meter telescope.target 6 of 7 7. galileo's telescope designs using lenses were examples of blank.
Answers
In the sentences for the telescope in the right-hand column from the first-hand column, the relevant blank has been filled.
Describe the telescope.
A telescope is the device used to show an enlarged view of a far-off object.
There are different types of telescopes, and each is used for a particular purpose.
The blank that should be filled in correctly from the first column is as follows:
1. Shorter (bluer) wavelengths of light have better angular resolution than longer (redder) wavelengths of light for the Hubble Space Telescope.
2. Huge reflecting telescopes are employed by the numerous research observatories on Mauna Kea.
3. The spectrograph, which divides light into its various colors, enables astronomers to ascertain the stellar composition as well as a number of other stellar characteristics.
4. The identical 10-m Keck
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A 200-lb man carries a 10-lb can of paint up a helical staircase that encircles a silo with radius 30 ft. If the silo is 60 ft high and the man makes exactly two complete revolutions, how much work is done by the man against gravity in climbing to the top
Answers
Answer:
17.07 kJ
Explanation:
The work done against gravity by the man W equals the potential energy change of the man and can of paint, ΔU
W = ΔU = mgΔy where m = mass of man and can of paint = 200 lb + 10 lb = 210 lb = 210 × 1 kg/2.205 lb, g = acceleration due to gravity = 9.8 m/s² and Δy = height of silo = 60 ft = 60 × 1m/3.28 ft
Since W = mgΔy, we substitute the values of the variables into the equation.
So,
W = mgΔy
W = 210 lb × 1 kg/2.205 lb × 9.8 m/s² × 60 ft × 1m/3.28 ft
W = 123480/7.2324 J
W = 17073.2 J
W = 17.0732 kJ
W ≅ 17.07 kJ
Calculate the buoyant force in air on a kilogram of gold (whose density is about 19 grams per cubic centimeter). The density of air is 1.3 ? 10-3 grams per cubic centimeter. Remember that the buoyant force is equal to the weight of a volume of air that is equal to the volume of the object. (Compare with the weight mg of this much gold, which is 9.8 N.)Fbuoyant =???
Answers
Buoyant force in air on a kilogram of gold is 0.65N
To calculate the buoyant force on the gold, we need to find the volume of air that has the same volume as the gold, and then calculate the weight of that volume of air.
The volume of the gold can be calculated using its density and mass:
Volume of gold = mass of gold / density of gold
Volume of gold = 1000 g / (19 g/cm^3)
Volume of gold = 52.63 cm^3
Now, we need to find the weight of the same volume of air as the gold. The weight of a volume of air is equal to its density times its volume times the acceleration due to gravity.
Weight of air = density of air x volume of air x acceleration due to gravity
Weight of air = (1.3 x 10^-3 g/cm^3) x (52.63 cm^3) x (9.8 m/s^2)
Weight of air = 0.65 N
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10. Refraction occurs...
A. only with light waves
B. at any, unpredictable time
C. when a wave changes speed
D. only at a wave front
E. All of the above
Answers
An object is attached to a trolley with a 0.80 kg mass, which is then pushed into an identical trolley at a speed of 1.1 m / s. The two trolleys couple together and move at a speed of 0.70 m / s after the collision. Calculate the mass of the object.
Answers
The mass of the object is approximately 0.457 kg.
The mass of the object attached to the trolley can be calculated using the principle of conservation of momentum. Since the two trolleys couple together and move as a single system after the collision, the total momentum before and after the collision should be the same. Given the mass of one trolley is 0.80 kg and the initial speed is 1.1 m/s, the momentum before the collision is 0.80 kg * 1.1 m/s = 0.88 kg·m/s. After the collision, the total mass is the sum of the two trolleys, and the final speed is 0.70 m/s.
Using the momentum equation, the mass of the object can be calculated as follows:
Total momentum before collision = Total momentum after collision
0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s
Solving for the mass of the object, we get:
0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s
0.88 kg·m/s = 0.56 kg + 0.70 kg * mass of the object
0.88 kg·m/s - 0.56 kg = 0.70 kg * mass of the object
0.32 kg = 0.70 kg * mass of the object
Dividing both sides by 0.70 kg, we find:
mass of the object = 0.32 kg / 0.70 kg = 0.457 kg
The two trolleys collide and couple together, the total momentum before the collision is equal to the total momentum after the collision according to the principle of conservation of momentum.
The momentum of an object is defined as the product of its mass and velocity. In this case, the mass of one trolley is known (0.80 kg) and the initial speed is given (1.1 m/s), allowing us to calculate the momentum before the collision.
After the collision, the two trolleys move together at a new speed (0.70 m/s). By setting the initial momentum equal to the final momentum and solving for the unknown mass of the object, we can find its value.
In the calculation, we subtract the masses of the two trolleys from the total mass in order to isolate the mass of the object.
Dividing the difference in momentum by the product of the known mass and the new speed, we obtain the mass of the object. In this case, the mass of the object is approximately 0.457 kg.
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Need help please answer
Answers
Answer:
8 + 6 = 14 miles north
Explanation:
3. What is the weight of a 462-kg bar?
Answers
Answer:
F = M g = 462 kg * 9.80 m/s^2 = 4528 Newtons
It somewhat unfortunate that weight in the English System is measured in pounds (mass in slugs) and in the metric system weight is measured in Newtons (mass in kilograms)
A student mixes .075 kg of an unknown substance at 96.5°C with .075 kg of water at 25.0°C. If the final temperature of the system is 31.15°C, what is the specific heat capacity of the substance?
Answers
Answer:
The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.
Explanation:
We notice that the student is mixing a substance with a high temperature and another substance with a low temperature, where the first release heat to the latter one until thermal equilibrium is reached. By the First Law of Thermodynamics and assuming that the entire system has no energy interactions with the surroundings, we get the following model:
\(\Delta U_{x}+\Delta U_{w} = 0\) (1)
Where \(\Delta U_{x}\) and \(\Delta U_{w}\) are the changes in internal energy for the unknown substance and water, measured in joules.
By definition of internal energy, we expand the equation above now:
\(m_{x}\cdot c_{x}\cdot (T_{o,x}-T_{f,x})+m_{w}\cdot c_{w}\cdot (T_{o,w}-T_{f,w}) = 0\) (2)
Where:
\(m_{x}\), \(m_{w}\) - Masses of the unknown substance and water, measured in kilograms.
\(c_{x}\), \(c_{w}\) - Specific heats of the unknown substance and water, measured in joules per kilogram-degree Celsius.
\(T_{o,x}\), \(T_{f,x}\) - Initial and final temperatures of the unknown substance, measured in degrees Celsius.
\(T_{o,w}\), \(T_{f,w}\) - Initial and final temperatures of water, measured in degrees Celsius.
Then, we clear the specific heat of the unknown substance:
\(c_{x} = \frac{m_{w}\cdot c_{w}\cdot (T_{f,w}-T_{o,w})}{m_{x}\cdot (T_{o,x}-T_{f,x})}\)
If we know that \(m_{w} = m_{x} = 0.075\,kg\), \(c_{w} = 4186\,\frac{J}{kg\cdot^{\circ}C}\), \(T_{f,w} = T_{f,x} = 31.15\,^{\circ}C\), \(T_{o,x} = 96.5\,^{\circ}C\) and \(T_{o,w} = 25\,^{\circ}C\), then the heat capacity of the unknown substance is:
\(c_{x} = \frac{(0.075\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (31.15\,^{\circ}C-25\,^{\circ}C)}{(0.075\,kg)\cdot (96.5\,^{\circ}C-31.15^{\circ}C)}\)
\(c_{x} = 393.939\,\frac{J}{kg\cdot ^{\circ}C}\)
The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.
Calculate forces and accelerations for a horizontal spring system. A 0.350 kg object attached to a spring of force constant 1.30 x 10 N/m is free to move on the frictionless horizontal surface. If the object is released from rest at x = 0.100 m, find the force on it and its acceleration at x = 0.100 m, x-0.0500m, x = 0. x= -0.0500m and x = -0.100 m.
Answers
The force and acceleration alter as the object moves away from its equilibrium position.
How to calculate force and acceleration?The force on the object at any point is given by Hooke's Law:
F = -kx
where F = force, k = force constant of the spring, and x = displacement of the object from its equilibrium position.
The acceleration of the object at any point is given by Newton's Second Law:
a = F/m
where a = acceleration, F = force, and m = mass of the object.
Using these equations, calculate the force and acceleration at each of the specified points:
At x = 0.100 m:
F = -kx = -(1.30 x 10 N/m)(0.100 m) = -0.130 N
a = F/m = (-0.130 N)/(0.350 kg) = -0.371 m/s² (the negative sign indicates that the acceleration is in the opposite direction to the displacement)
At x = 0:
F = -kx = -(1.30 x 10 N/m)(0) = 0 N (the spring is at its equilibrium position)
a = F/m = 0 N/0.350 kg = 0 m/s²
At x = -0.0500 m:
F = -kx = -(1.30 x 10 N/m)(-0.0500 m) = 0.065 N (note that the force is positive because the displacement is negative)
a = F/m = (0.065 N)/(0.350 kg) = 0.186 m/s²
At x = -0.100 m:
F = -kx = -(1.30 x 10 N/m)(-0.100 m) = 0.130 N
a = F/m = (0.130 N)/(0.350 kg) = 0.371 m/s²
So, the force and acceleration change with the displacement of the object from its equilibrium position.
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an astronaut is said to be weightless when he/she travells in a satellite . does it mean that the earthdoes not attract him/her ?
Answers
Answer:
It does attract
Explanation:
But due to the lack of gravitational pull of the earth he/she feels weightless I hope this helps you
What does volume measure name two different units that might be used to measure volume
Answers
Answer:
Volume measures liquid
Explanation:
Volume can be measured in meters and centimeters
8+10÷5(5×4+2)=?
it ıs said that this question was very diffıcult
can you slove?
Answers
Answer:
220 bastanyan sagot ko yawa
Use PEMDAS
5*4=20+2=22 so 8+10/5(22)
10/5=2
8+2=10
10(22) or 10*22 is 220
You are an engineer designing a pinball machine. Local customers have expressed that their ideal velocity for the pinball (mass = 0.1kg) to be launched at 12.6m/s. In general, people compress the pinball machine spring by 0.2m to reach this initial velocity for the pinball.
image
image
Using the information listed in the description, which type of spring do you choose to most closely reach the targeted launch speed of 12.6 m/s? Show your work and explain your answer using what you know of kinetic energy, elastic potential energy, and conservation of energy.
Answers
The spring constant, k, needed to achieve the desired launch speed of 12.6 m/s is 189.75 N/m, which is the spring constant of a linear spring.
Which kind of spring should you use to launch a rocket at 12.6 m/s, your intended launch speed?The type of spring to choose in order to most closely reach the target launch speed of 12.6 m/s is a linear spring. Using what we know of kinetic energy, elastic potential energy, and conservation of energy, we can use the following equations to determine the spring constant, k, needed to achieve the desired launch speed:Kinetic Energy: KE = 0.5 * m * v^2
Elastic Potential Energy: PE = 0.5 * k * x^2
Conservation of Energy: KE + PE = m * g * x
Given that the mass of the pinball is 0.1 kg, the velocity of the pinball is 12.6 m/s, and the compression of the spring is 0.2 m, we can solve for the spring constant, k:KE = 0.5 * 0.1 * 12.6^2 = 7.59 J
PE = 0.5 * k * 0.2^2 = 7.59 J
m * g * x = 7.59 J
k = (7.59 J) / (0.2^2) = 189.75 N/m
Therefore, the spring constant, k, needed to achieve the desired launch speed of 12.6 m/s is 189.75 N/m, which is the spring constant of a linear spring.To learn more about kinetic energy and elastic potential energy refer to:
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sort the properties of metalloids into the correct categories
Answers
Boron
Sillicon
Germanium
Arsenic
Antimony
Tellurium
Pelorium
Metallic characteristics: lustrous, solid at room temperature
In-between characteristics: semiconducting, amphoteric
Nonmetallic: brittle
A student lifts a box of books 2 meters with a force of 45 N. He then carries the box 10 meters to the living room. What is the total amount of work done in this situation?
PLEASE ANSWER FAST
Answers
Answer:
90J
Explanation:
The only time work is being done is when he lifts the box off the ground. Therefore, using the work formula, 2 x 45, you get 90J. Hope this helps someone.
Two hot air balloons with the same mass and amount of helium put inside of them if one is a rigid material and the other expands which one would be the highest?
Answers
Answer:
One is that atmospheric pressure is dramatically reduced at high altitudes, so a helium balloon expands as it rises and eventually explodes. If you inflate a balloon beyond its limits at room temperature, it will break into small pieces up to about ten centimetres long
Explanation:
A 1.8 kg book has been dropped from the top of the football stadium. Its speed is 4.8 m/s when it is 2.9 m above the ground. How high is the stadium?
Answers
Answer:
the height of the stadium is 4 m
Explanation:
The computation of the height of the stadium is shown below:
but before that total mechanic energy should be determined
E = PE + KE
where
PE = mgh
and, KE = 1 ÷2 mv^2
Now
E = mgh + 1 ÷2 mv^2
= (1.8) (9.8) (2.9) + 1 ÷ 2 (1.8) (4.8)^2
= 71.9J
= 72J
Now the height of the stadium is
TE = mgh
72 = (1.8) × (9.8) × h
So, h = 4 m
Hence, the height of the stadium is 4 m
How do you find the range of a data set?
Answers
Answer: You take the biggest number and subtract it from the smallest number.
Explanation: Hope this helps.
Answer:
Subtract the minimum data value from the maximum data value to find the data range.
Explanation:
\(Hope\) \(this\) \(helps!\)
What are civic responsibilities? (2 points)
a
The responsibilities we have at home
b
The responsibilities we have as part of a democratic society
c
Responsibilities that are reserved for state governments
d
Freedoms that are protected by law
Answers
Answer: B
Explanation:
100%
Answer: The answer is b
Explanation: And a brain list please
How do I measure the drag of a paper airplane?
Answers
Answer:
hmmmmm ill get back later
Explanation:
A PVC pipe has a length of 45.132 centimeters.a. What are the frequencies of the first three harmonics when the pipe is open at both ends? Include units in your answers.b. What are the frequencies of the first three harmonics when the pipe is closed at one end and open at the other? Include units in your answers.
Answers
ANSWERS
a. f₁ = 380 Hz; f₂ = 760 Hz; f₃ = 1140 Hz
b. f₁ = 190 Hz; f₃ = 570 Hz; f₅ = 950 Hz
EXPLANATION
a. For a pipe of length L open at both ends, the frequencies of the first three harmonics are:
\(\begin{cases}f_1=\frac{v}{2L} \\ \\ f_2=2f_1=\frac{v}{L} \\ \\ f_3=3f_1=\frac{3v}{2L}\end{cases}\)Assuming that the speed of the wave is the speed of sound: 343 m/s and knowing that the length of the pipe is L = 45.132 cm = 0.45132 m we can find the frequencies of the first three harmonics:
\(\begin{cases}f_1=\frac{343m/s}{2\cdot0.45132m}\approx380Hz \\ \\ f_2=2f_1=2\cdot380Hz\approx760Hz \\ \\ f_3=3f_1=3\cdot380Hz\approx1140Hz\end{cases}\)b. For a pipe of length L closed at one end and open at the other, the frequencies of the first three harmonics are:
\(\begin{cases}f_1=\frac{v}{4L} \\ \\ f_2=DNE \\ \\ f_3=3f_1=\frac{3v}{4L}\end{cases}\)In a closed pipe, there can only be odd harmonics (1, 3, 5...). Therefore, the second harmonic does not exist and the "third harmonic" would be the 5th,
\(\begin{cases}f_1=\frac{v}{4L} \\ \\ f_3=3f_1=\frac{3v}{4L} \\ \\ f_5=5f_1=\frac{5v}{4L}\end{cases}\)Again, the length of the pipe is 45.132 cm = 0.45132 m, so the first three harmonics are:
\(\begin{cases}f_1=\frac{343m/s}{4\cdot0.45132m}\approx190Hz \\ \\ f_3=3f_1=3\cdot190Hz=570Hz \\ \\ f_5=5f_1=5\cdot190Hz=950Hz\end{cases}\)Two cars collide head-on and stick together.
Car A, with a mass of 2000 kg, was initially
moving at a velocity of 10 m/s to the east. Car
B, with an unknown mass, was initially at rest.
After the collision, both cars move together at
a velocity of 5 m/s to the west. What is the
mass of Car B?
OF
Answers
The mass of Car B is -6000 kg.
To solve this problem, we can apply the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.
Therefore, we can write the equation for the conservation of momentum as:
(mass of Car A * velocity of Car A) + (mass of Car B * velocity of Car B) = (mass of Car A + mass of Car B) * velocity after collision
Let's substitute the given values into the equation:
(2000 kg * 10 m/s) + (mass of Car B * 0 m/s) = (2000 kg + mass of Car B) * (-5 m/s)
Simplifying the equation:
20000 kg*m/s = -5 m/s * (2000 kg + mass of Car B)
Dividing both sides by -5 m/s:
-4000 kg = 2000 kg + mass of Car B
Subtracting 2000 kg from both sides:
mass of Car B = -4000 kg - 2000 kg
mass of Car B = -6000 kg
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Consider the circuit in Figure 5 with e(t) = 12sin(120pit) V. When S1 and S2 are
open, i leads e by 30°. When SI is closed and S2 is open, i lags e by 30°. When S1 and S2 are closed, i has an amplitude 0.5A. What are R, L, and C?
Answers
Based on the information, it should be noted that the resistance R is 0.5 Ω.
How to calculate the resistanceWhen S1 and S2 are open, i leads e by 30°. In this case, the circuit consists of only the inductor (L) and the capacitor (C) in series. Therefore, the impedance of the circuit can be written as:
Z = jωL - 1/(jωC)
Since i leads e by 30°, we can express the phasor relationship as:
I = k * e^(j(ωt + θ))
Z = jωL - 1/(jωC) = j(120π)L - 1/(j(120π)C)
Re(Z) = 0
By equating the real parts, we get:
0 = 0 - 1/(120πC)
Let's assume that there is a resistance (R) in series with the inductor and capacitor. The impedance equation becomes:
Z = R + jωL - 1/(jωC)
Z = R + jωL
Im(Z) = ωL > 0
Substituting the angular frequency and rearranging the inequality, we have:
120πL > 0
L > 0
This condition implies that the inductance L must be greater than zero.
When S1 and S2 are closed, i has an amplitude of 0.5 A. In this case, the impedance is:
Z = R + jωL - 1/(jωC)
Since the amplitude of i is given as 0.5 A, we can express the phasor relationship as:
I = 0.5 * e^(j(ωt + θ))
By substituting this phasor relationship into the impedance equation, we can determine the value of R. The real part of the impedance must be equal to R:
Re(Z) = R
Since the amplitude of i is 0.5 A, the real part of the impedance must be equal to 0.5 A: 0.5 = R
Therefore, the resistance R is 0.5 Ω.
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Question 28 of 30The graph below shows the conservation of energy for a skydiver jumping outof a plane and landing safely on the ground. Which energy is represented byline A?сAEnergy (kJ)BDistance (m)A. Potential energyB. Kinetic energy
Answers
Answer:
Explanation:
Potential energy depends on the height of the object above the ground. As the height decreases, the potential energy decreases. Looking at line A, the energy is decreasing as the distance is increasing. Thus, the energy represented by line A is
A. Potential energy
A marble rolling with speed 20cm/s rolls off the edge of a table that is 80cm high. How far horizontally from the table edge does the marble strike the floor
Answers
Answer:
8 cm
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 20 cm/s
Height (h) = 80 cm
Horizontal distance (s) =?
Next, we shall determine the time taken for marble to get to the ground. This can be obtained as follow:
Height (h) = 80 cm
Acceleration due to gravity (g) = 1000 cm/s²
Time (t) =?
t = √(2h/g)
t = √[(2 × 80)/1000]
t = √(160/1000)
T = √0.16
t = 0.4 s
Finally, we shall determine the horizontal distance travelled by the marble. This can be obtained as illustrated below:
Initial velocity (u) = 20 cm/s
Time (t) = 0.4 s
Horizontal distance (s) =?
s = ut
s = 20 × 0.4
s = 8 cm
Thus, the horizontal distance travelled by the marble is 8 cm.
The horizontal distance traveled by the marble is 8 cm.
The given parameters;
speed of the marble, v = 20 cm/sheight of the table, h = 80 cmThe time of motion of the marble is calculated as follows;
\(h = ut + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.8}{9.8} }\\\\t = 0.4 \ s\)
The horizontal distance traveled by the marble is calculated as follows;
\(X = v_0_x t\\\\X = (20 \times 0.4)\\\\X = 8 \ cm\)
Thus, the horizontal distance traveled by the marble is 8 cm.
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car is moving at 40 m/s. At 10 meters the driver spots a deer on the road and instantly steps on the brakes. If the car is 400 kg how much force must the breaks exert to stop the car in time?
Answers
Answer:
32000 N
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Mass (m) of car = 400 Kg
Force (F) =?
Next, we shall determine the acceleration of the the car. This can be obtained as follow:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Acceleration (a) =?
v² = u² + 2as
0² = 40² + (2 × a × 10)
0 = 1600 + 20a
Collect like terms
0 – 1600 = 20a
–1600 = 20a
Divide both side by –1600
a = –1600 / 20
a = –80 m/s²
The negative sign indicate that the car is decelerating i.e coming to rest.
Finally, we shall determine the force needed to stop the car. This can be obtained as follow:
Mass (m) of car = 400 Kg
Acceleration (a) = –80 m/s²
Force (F) =?
F = ma
F = 400 × –80
F = – 32000 N
NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.