You have a light spring which obeys Hooke's law. This spring stretches 2.74 cm vertically when a 2.30-kg object is suspended from it. Determine the following.
(a) force constant of the spring
N/m
(b) distance the spring stretches if you replace the 2.50-kg object with a 1.25-kg object
cm
(c) amount of work an external agent must do to stretch the spring 6.40 cm from its unstretched position
J
Answers
The force constant is 827.4 N/m, the distance is 1.48 cm and the amount of work done is 3.42 J.
To find the force constant of the spring, we use the equation F = kx, where F is the force applied to the spring, k is the force constant, and x is the displacement of the spring. We are given that the mass of the object is 2.30 kg, so the force applied to the spring is equal to the weight of the object, which is given as:
F = mg = (2.30 kg)(9.8 m/s^2) = 22.54 N.
We are also given that the spring stretches 2.74 cm, or 0.0274 m. Plugging these values into the equation above, we get:
22.54 N = k * 0.0274 m
Solving for k, we find that the force constant of the spring is k = 827.4 N/m.
To find the distance the spring stretches when a 1.25-kg object is suspended from it, we use the same equation F = k*x, but with the new values for F and x. The force applied to the spring is now:
F = mg = (1.25 kg)(9.8 m/s^2) = 12.25 N.
Solving for x, we find that the spring stretches:
x = F/k = (12.25 N)/(827.4 N/m) = 0.0148 m = 1.48 cm.
To find the amount of work an external agent must do to stretch the spring 6.40 cm from its un-stretched position, we use the following equation:
W = F×d
where W is the work done, F is the force applied to the spring, and d is the displacement of the spring.
In this case, the force applied to the spring is given as:
F = k*x = (827.4 N/m)(0.0640 m) = 53.29 N,
and the displacement of the spring is d = 0.0640 m. Plugging these values into the equation above, we find that the work done is
W = (53.29 N)(0.0640 m) = 3.42 J.
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Related Questions
What are possible formulas for impulse? Check all that apply.
Answers
Answer:
J = FΔt
J = mΔv
J = Δp
Explanation:
The Impulse Formula:Δp = Change in momentumF = Applied forceΔt = Elapsed timeThe bending of rocks due to the compression of tectonic plates is called
Ofaulting
O folding
subduction
plyometrics
Answers
Answer:
Folding
Explanation:
Sammy Sosa swings at a 0.15 kg baseball and accelerates it at a rate of3.0 x 10^4 m/s2. How much force does Sosa exert on the ball?
Answers
Given:
The mass of the ball is m = 0.15 kg
The acceleration of the ball is
\(a=3\times10^4\text{ m/s}^2\)To find the force.
Explanation:
The force can be calculated by the formula
\(F=\text{ ma}\)On substituting the values, the force will be
\(\begin{gathered} F=0.15\times3\times10^4 \\ =4500\text{ N} \end{gathered}\)Thus, the Sosa exerts 4500 N of force on the ball.
Jason's heartbeat is measured to be 65 beats per minute.
What is the frequency of heartbeats in hertz?
What is the period for each heartbeat in seconds?
Answers
Answer:
To find the frequency of heartbeats in hertz, we need to convert the beats per minute (BPM) to beats per second (BPS), since frequency is measured in hertz (Hz) which is equivalent to cycles per second.
Frequency = BPM / 60
Frequency = 65 / 60
Frequency ≈ 1.083 Hz
To find the period for each heartbeat in seconds, we need to find the reciprocal of the frequency, which gives the time duration for one complete cycle.
Period = 1 / Frequency
Period = 1 / 1.083
Period ≈ 0.922 seconds per beat
A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s2 . It runs out of fuel at the end of 4.00s and continues to coast upward, reaching a maximum height before falling back to earth. (a) Find the rocket’s velocity and position at the end of 4.00s. (b) Find the maximum
Answers
Answer:
a) The velocity and position of the rocket at the end of 4 seconds are 117.6 meters per second and 235.2 meters, respectively, b) The maximum height reached by the rocket is 940.296 meters, c) The rocket crashes on the ground at a velocity of -96.030 meters per second.
Explanation:
The complete statement is:
A rocket moves straight upward , starting from rest with an acceleration of 29.4 m/s2. it runs out of fuel at the end of 4.00 s and countinues to coast upward , reaching a maximum height before falling back to earth . a) find the rocket's velocity and position at the end of 4.00 s . b) Find the maximum height the rocket reaches. c) find the velocity on the instant before the rocket crashes on the ground.
a) The rocket accelerates uniformly, whose equations of motion are:
\(y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot a \cdot t^{2}\)
\(v = v_{o} + a\cdot t\)
Where:
\(y\) - Final position, measured in meters.
\(y_{o}\) - Initial position, measured in meters.
\(v_{o}\) - Initial velocity, measured in meters per second.
\(t\) - Time, measured in second.
\(a\) - Acceleration, measured in meters per square second.
\(v\) - Final velocity, measured in meters per second.
If we know that \(y_{o} = 0\,m\), \(v_{o} = 0\,\frac{m}{s}\), \(t = 4\,s\) and \(a = 29.4\,\frac{m}{s^{2}}\), the velocity and position of the rocket are, respectively:
\(y = 0\,m+\left(0\,\frac{m}{s} \right)\cdot (4\,s)+\frac{1}{2}\cdot \left(29.4\,\frac{m}{s^{2}} \right) \cdot (4\,s)^{2}\)
\(y = 235.2\,m\)
\(v = 0\,\frac{m}{s} + \left(29.4\,\frac{m}{s^{2}}\right)\cdot (4\,s)\)
\(v = 117.6\,\frac{m}{s}\)
The velocity and position of the rocket at the end of 4 seconds are 117.6 meters per second and 235.2 meters, respectively.
b) Now, the rocket experiments a free-fall motion. The maximum height of the rocket is obtained by equalizing the equation of velocity to zero and evaluating the equation of position later. That is:
\(y = 235.2\,m+\left(117.6\,\frac{m}{s} \right)\cdot t+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}\)
\(v = 117.6\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right)\cdot t\)
Then,
\(0 = 117.6 -9.807\cdot t\)
\(t = 11.991\,s\)
The maximum height reached by the rocket is:
\(y = 235.2+117.6\cdot (11.991)+\frac{1}{2}\cdot (-9.807)\cdot (11.991)^{2}\)
\(y = 940.296\,m\)
The maximum height reached by the rocket is 940.296 meters.
c) The rocket experiments a free-fall motion and is accelerated by gravity until collision happens. The equations of motion below are presented:
\(0\,m = 940.296\,m+\left(0\,\frac{m}{s} \right)\cdot t+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}\)
\(v = 0\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right)\cdot t\)
From the equation of position we get the instant when rocket hits the ground, whose roots are found by Quadratic Formula:
\(t_{1} \approx 9.792\,s\) and \(t_{2} \approx -9.792\,s\)
Only the first root is physically reasonable.
By the second equation we calculate the final velocity:
\(v = 0-9.807\cdot (9.792)\)
\(v = -96.030\,\frac{m}{s}\)
The rocket crashes on the ground at a velocity of -96.030 meters per second.
A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q
Answers
Answer:
0.3 N
Explanation:
Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.
The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N
Coulomb's law equationF = Kq₁q₂ / r²
Where
F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart Data obtained from the question Initial distance apart (r₁) = rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =? How to determine the final forceFrom Coulomb's law,
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ = constant
F₁r₁² = F₂r₂²
With the above formula, we can obtain the final force as follow:
F₁r₁² = F₂r₂²
1.2 × r² = F₂ × (2r)²
1.2r² = F₂ × 4r²
Divide both side by 4r²
F₂ = 1.2r² / 4r²
F₂ = 0.3 N
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When you drop a 0.37 kg apple, Earth exerts
a force on it that accelerates it at 9.8 m/s² to-
ward the earth's surface. According to New-
ton's third law, the apple must exert an equal
but opposite force on Earth.
If the mass of the earth 5.98 x 10²4 kg, what
is the magnitude of the earth's acceleration
toward the apple?
Answer in units of m/s².
Answers
The earth is accelerating toward the apple at a rate of 6.2 × 1025 m/s2.
How is this acceleration determined?The apple weighs m = 0.37 kg.
The apple's speed when it approaches the earth's surface is 9.8 meters per second.
Earth's mass, M, is 5.98 × 1024 kg.
Using Newton's Second Law of Motion, we may now:
The strength of the force exerted by Earth on the apple is,
F = ma
⇒ F = 0.37 × 9.8
⇒ F = 3.626 N
We are informed that the apple must exert an equal but opposite force on Earth in accordance with Newton's third law of motion.
Therefore, the force exerted by the apple on Earth will be of the following magnitude:
F = 3.626 N
Let "A" be the acceleration of the earth relative to the apple in m/s2.
Thus,
The following will be used to determine how quickly the earth is moving toward the apple:
F = MA
⇒ 3.626 = [5.98 × 10²⁴] × A
⇒ A = 3.626 / [5.98 × 10²⁴]
⇒ A = 0.606 × 10⁻²⁴
⇒ A = 6.06 × 10⁻²⁵ m/s²
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When two RF signals on the same frequency arrive at a receiver at the exact same time and their peaks and valleys are in alignment, what is true about these signals?
Answers
Answer:
The two RF signals are in phase.
Explanation:
A wave is a disturbance that travels through a medium which transfers energy from one point to another in the medium without causing any permanent displacement of the particles of the medium.
Characteristics of waves include frequency, wavelength, velocity, etc.
Two types of waves are longitudinal and transverse wave. Radio Frequency (RF) signals travel in the form of transverse waves which have regions of maximum and minimum displacements called crests and troughs.
Travelling waves with the same frequency may be said to be in phase or out of phase depending on whether their crests/peaks or troughs/valleys are reached at the same instant of time.
When two RF signals on the same frequency arrive at a receiver at the exact same time and their peaks and valleys are in alignment or in step, they are said to be in phase.
The phase of a wave involves the relationship between the position of the amplitude peaks and valleys of two waveforms.
Emma draws another diagram showing how adding energy affects this wave. How does her new diagram compare to this diagram?
A
Her diagram shows higher wavelengths.
B
Her diagram shows lower wave amplitudes.
С
Her diagram shows lower wave frequencies.
D
Her diagram shows higher wave amplitudes.
HELPPPP MEEE PLEASEEEE 55 POINTS!!!
Answers
What is the speed of light (in m/s) in air? (Enter your answer to at least four significant figures. Assume the speed of light in a vacuum is 2.997 ✕ 108 m/s.) m/s What is the speed of light (in m/s) in polystyrene? m/s
Answers
Answer:
The speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.
Explanation:
To find the speed of light in air and in polystyrene we need to use the following equation:
\( c_{m} = \frac{c}{n} \)
Where:
\(c_{m}\): is the speed of light in the medium
n: is the refractive index of the medium
In air:
\(c_{a} = \frac{c}{n_{a}} = \frac{2.997 \cdot 10^{8} m/s}{1.0003} = 2.996 \cdot 10^{8} m/s\)
In polystyrene:
\(c_{p} = \frac{c}{n_{p}} = \frac{2.997 \cdot 10^{8} m/s}{1.6} = 1.873 \cdot 10^{8} m/s\)
Therefore, the speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.
I hope it helps you!
For the simple harmonic oscillation where k = 19.6
N/m, A = 0.100 m, x = -(0.100 m) cos 8.08t, and v =
(0.808 m/s) sin 8.08t, determine (a) the total energy, (b)
the kinetic and potential energies as a function of time,
(c) the velocity when the mass is 0.050 m from
equilibrium, (d) the kinetic and potential energies at
half amplitude (x = A/2).
Answers
a. Total energy is 0.098 J
b. Potential and Kinetic Energies is 0.032 sin^2(8.08t) J
c. Velocity at x is -0.808 sin(8.08t) m/s
d. Potential and Kinetic Energies at x is 0.016 sin^2(8.08t) J
Step by step explanationWe can use the following formulas for the energy, velocity, and potential and kinetic energies of a simple harmonic oscillator:
Total Energy: E = 1/2 k A^2Velocity: v = -ωA sin(ωt)Potential Energy: U = 1/2 k x^2Kinetic Energy: K = 1/2 m v^2where ω = √(k/m) is the angular frequency.
Given that k = 19.6 N/m, A = 0.100 m, x = -(0.100 m) cos 8.08t, and v = (0.808 m/s) sin 8.08t, we can find the values of E, U, and K as follows:
(a) Total Energy:
E = 1/2 k A^2 = 1/2 * 19.6 * 0.1^2 = 0.098 J
(b) Potential and Kinetic Energies:
U = 1/2 k x^2 = 1/2 * 19.6 * (-0.1 cos(8.08t))^2 = 0.098 cos^2(8.08t) J
K = 1/2 m v^2 = 1/2 * (0.1) * (0.808 sin(8.08t))^2 = 0.032 sin^2(8.08t) J
(c) Velocity at x = 0.050 m:
When x = 0.050 m, cos(8.08t) = -0.5, so we have:
v = -ωA sin(ωt) = -ω(0.1) sin(8.08t) = -0.808 sin(8.08t) m/s
(d) Potential and Kinetic Energies at x = A/2:
When x = A/2 = 0.050 m, cos(8.08t) = -0.5, so we have:
U = 1/2 k x^2 = 1/2 * 19.6 * (0.050)^2 = 0.0245 J
K = 1/2 m v^2 = 1/2 * (0.1) * (0.808 sin(8.08t))^2 = 0.016 sin^2(8.08t) J
Note that the sum of potential and kinetic energies at any point in time is equal to the total energy, which is constant.
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A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.
a. Where should a third charge, Q , be placed, so that all three charges will be in equilibrium? Express your answer in terms of d.
b. What should be its sign, so that all three charges will be in equilibrium?
c. What should be its magnitude, so that all three charges will be in equilibrium? Express your answer in terms of Q.
Answers
Answer:
a) x = ⅔ d , b) the charge must be negative, c) Q
Explanation:
a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing
∑ F = 0
-F₁₂ + F₂₃ = 0
F₁₂ = F₂₃
let's replace the values
k Q Q / r₁₂² = k Q 4Q / r₂₃²
Q² / r₁₂² = 4 Q² / r₂₃²
suppose charge 3 is placed at point x
r₁₂ = x
r₂₃ = d-x
we substitute
1 / x² = 4 / (d-x) 2
1 / x = 2 / (d-x)
x = 2 (x-d)
x = 2x -2d
3x = 2d
x = ⅔ d
b) The sign of the charge must be negative, to have an attractive charge on the two initial charges
c) Q
24. A body A rests on a smooth horizontal table. Two bodies of mass 2 kg and 10 kg hanging freely, are attached to A by strings which pass over smooth pulleys at the edges of the table. The two strings are taut. When the system is released from rest, it accelerates at 2 m/s2 . Find the mass of A.
Answers
The two strings are taut. When the system is released from rest, it accelerates at 2 m/s2 then, Mass of A = 8m/5 kg.
Let the mass of the body A be ‘m’.
The two strings are taut so they exert a tension ‘T’ on body A.
Let ‘a’ be the acceleration produced in the system.
The free body diagram of body A is given below: mA + 2T = mA + ma = mA + m(2)mA + 10T = mA + ma = mA + m(2)
As the two strings are taut, we can say that tension in both strings is equal.
Therefore 2T = 10T or T = 5T As the body A is resting on a smooth horizontal table, there is no friction force acting on the body A.
The net force acting on body A is the force due to tension in the strings. ma = 2T – mg …(1)
As per the given problem, the system is released from rest.
Hence the initial velocity is zero.
Also, we are given that the system accelerates at 2 m/s2.
Therefore a = 2 m/s2 …(2)
From the equations (1) and (2), we get, m(2) = 2T – mg …(3)⇒ m(2) = 2×5m – mg⇒ 2m = 10m – g⇒ g = 8m/5
Thus, the mass of A is 8m/5 kg.
Answer: Mass of A = 8m/5 kg.
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a shopper pushes a cart 40.0m south down one aisle and then turns 90.0 degrees and moves 15.0m. He then makes another 90.0 degree turn and moves 20.0m. Find the shoppers total displacement. (there could be more than one correct answer)
Answers
Answer:
Explanation:
The displacement is the distnce of the shopper from the starting point.
Sum of movement along the vertical = 40-20 = 20m
Movement along the horizontal (x direction) = 15.0m
Displacement will be gotten using the pythagoras theorem.
d = √20²+ 15²
d = √400+225
d = √625
d = 25.0m
Hence the shoppers total displacement is 25.0m
Displacement is the shortest distance between the initial and final points. The total displacement of the shopper is 25 m.
Displacement :
It is the difference between the initial and final point or the shortest distance between the initial and final point.
Given Here,
A shopper moves south = 40m
Shopper took a turn and move = 15 m
Shopper again take a turn and move = 20 m
So,
The movement of shopper in Y-axis = 40-20 = 20 m
The movement of shopper in X-axis = 15 m
The displacement can be found using the Pythagoras theorem,
\(\bold{ d = \sqrt{ 20^2+ 15^2}}\\\\\bold {d = \sqrt{ 400+225}}\\\\\bold {d = \sqrt{625}}\\\\\bold {d = 25.0m}\)
Therefore, The total displacement of the shopper is 25 m.
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Water is pumped through the hose shown below, from a lower level to an upper level. Compared to the water at point 1, the water at point 2:
a.has less speed and less pressure
b.has greater speed and less pressure
c.has greater speed and greater pressure
d.has less speed and greater pressure
Answers
We can also apply equation of continuity in the horizontal tubes
A1v1 = A2v2
Clearly area of tube 2 is smaller so velocity will be more
Now tube 2 has a greater pressure and height so according to Bernoulli’s theorem tube 2 has lower pressure
The water at tube 2 has higher speed and lower pressure
The correct option is d has less speed and greater pressure.
How water flow?
Water flows through a combination of gravity and pressure. The movement of water is caused by the force of gravity, which causes water to flow downhill from higher to lower elevations. When water is pumped or forced through a system, pressure is added to the water, causing it to flow in the desired direction.
Water flows through a system of pipes, hoses, or channels, depending on the application. The velocity of water flow depends on various factors, including the diameter of the pipe or channel, the amount of water being pumped or released, and the pressure of the system.
In addition to gravity and pressure, other factors can affect the flow of water, including friction, viscosity, and turbulence. Understanding these factors is essential for designing efficient water systems that can deliver water where it is needed, such as in homes, farms, and cities.
Based on the diagram, the water at point 2 has a greater height than point 1, so it has a higher gravitational potential energy. Therefore, the water at point 2 must have a lower kinetic energy than point 1. Since the kinetic energy of water is related to its speed, we can conclude that the water at point 2 has less speed than point 1.
However, since the water is being pumped from a lower level to an upper level, it means that energy is being added to the system, which increases the pressure of the water. Therefore, the water at point 2 has a greater pressure than point 1.
Thus, the correct option is d) has less speed and greater pressure.
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The direction of force of Earth's magnetic field is from the geographic South
Pole to the geographic North Pole. Where is Earth's magnetic north pole?
A. Near Earth's center
B. Near Earth's equator
C. Near Earth's North Pole
D. Near Earth's South Pole
Answers
Answer:d
Explanation:
Answer:
D
Explanation:
Furthermore, the magnetic pole near earth's geographic north pole is actually the south magnetic pole. When it comes to magnets, opposites attract. This fact means that the north end of a magnet in a compass is attracted to the south magnetic pole, which lies close to the geographic north pole.
I need help with this question There is 4 answersa)b)c)d)
Answers
Given,
The velocity of the joggers, v=-3.50 m/s
The mass of Jim, M=100 kg
The mass of Tom, m=59 kg
(a) The kinetic energy of the system is given by,
\(\begin{gathered} E_a=\frac{1}{2}mv^2+\frac{1}{2}Mv^2 \\ =\frac{1}{2}v^2(m+M) \end{gathered}\)On substituting the known values,
\(\begin{gathered} E_a=\frac{1}{2}\times3.50^2(59.0+100) \\ =973.88\text{ J} \end{gathered}\)Thus the kinetic energy of the system is 973.88 J
(b)
The total momentum of the system is given by,
\(\begin{gathered} p_b=mv+Mv \\ =(m+M)v \end{gathered}\)On substituting the known values,
\(\begin{gathered} p_b=(59+100)\times3.50 \\ =556.5\text{ kg}\cdot\text{ m/s} \end{gathered}\)Thus the total momentum of the system is 556.5 kg m/s
(c)
Given that the velocity of Tom is -v
The total kinetic energy of the system is given by,
\(\begin{gathered} E_c=\frac{1}{2}Mv^2+\frac{1}{2}m(-v)^2 \\ =\frac{1}{2}(Mv^2+m(-v)^2) \end{gathered}\)On substituting the known values,
\(\begin{gathered} E_c=\frac{1}{2}(100\times3.50^2+59\times(-3.50)^2) \\ =973.88\text{ J} \end{gathered}\)Thus the total kinetic energy of the system, in this case, is 973.88 J
(d)
The total momentum of the system is given by,
\(\begin{gathered} p_d=Mv+m(-v) \\ =v(M-m) \end{gathered}\)On substituting the known values,
\(\begin{gathered} p_d=3.50(100-59) \\ =143.5\text{ kg}\cdot\text{ m/s} \end{gathered}\)Thus the total momentum of the system, in this case, is 143.5 kg m/s
A space vehicle is coasting at a constant velocity of 22.3 m/s in the y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.203 m/s2 in the x direction. After 56.7 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the y direction.
Answers
Answer:
Explanation:
Initial velocity in y direction Vy = 22.3 m /s
initial acceleration in x direction ax = .203 m /s ²
time of acceleration t = 56.7 s
final velocity in x direction
v = u + a t
Vx = 0 + .203 x 56.7 = 11.51 m /s
Final velocity in y direction will remain same as initial velocity in y direction = 22.3 m /s because there is no acceleration in y direction .
Magnitude of final velocity
= √ ( Vx² + Vy²)
= √ (22.3² + 11.51² )
= √ ( 497.29 + 132.48)
= 25.1 m /s
Direction of final velocity from y direction be Ф
TanФ = Vx / Vy = 11.51 / 22.3 = .516
Ф = 27.3° .
Particles q1 = -8.99 μC, q2 = +5.16 μC, and
q3=-89.9 μC are in a line. Particles q1 and q2 are separated by 0.220 m and particles q2 and q3 are separated by 0.330 m. What is the net force on particle q1?
Answers
Explanation:
, the net force will be felt on particle approximately 15.39 N to the right.
Galileo
o did not believe friction existed
o believed that friction stopped objects in motion
o believed that friction kept objects in motion
О
assumed that in a frictionless environment objects would never move
Answers
Answer:
object would move but it could be difficult to slow down or stop.
Find the Magnitude of the resultant vector (the actual
path of the boat).
The picture is a little blurry, so here are the stats:
Velocity of the boat is 0.75 m/s
Velocity of the river is 1.2 m/s
Answers
The magnitude of the resultant vector, representing the actual path of the boat, is approximately 1.42 m/s.
To find the magnitude of the resultant vector, we need to consider the boat's velocity and the velocity of the river. The boat's velocity is given as 0.75 m/s, and the river's velocity is given as 1.2 m/s.
Since the boat is moving in a river, we can think of the boat's velocity as a combination of two velocities: its own velocity and the velocity of the river. The resultant vector represents the actual path of the boat, considering both velocities.
To calculate the resultant vector, we can use vector addition. The magnitude of the resultant vector can be found by taking the square root of the sum of the squares of the boat's velocity and the river's velocity. Mathematically, we have:
Resultant magnitude = √(boat velocity^2 + river velocity^2)
Plugging in the given values, we have:
Resultant magnitude = √(0.75^2 + 1.2^2)
= √(0.5625 + 1.44)
= √2.0025
≈ 1.42 m/s
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If the instantaneous voltage at a given moment in the circuit RL is V=VmaxSIN(150), then the instantaneous current at the...... same instant I=Imaxsin
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The instantaneous current at the same moment in the RL circuit can be expressed as I = Imaxsin(150), where Imax represents the maximum current.
1. Given that the instantaneous voltage at a specific moment in the RL circuit is V = Vmaxsin(150).
2. We can express the current at the same moment using Ohm's Law, which states that V = IR, where V is voltage, I is current, and R is resistance.
3. In an RL circuit, the resistance is represented by the symbol R, and it is typically associated with the resistance of the wire or any resistors in the circuit.
4. However, the given equation does not explicitly mention resistance.
5. Since we are considering an RL circuit, it suggests the presence of inductance (L) along with resistance (R).
6. In an RL circuit, the voltage across the inductor (VL) can be expressed as VL = L(di/dt), where L is the inductance and di/dt represents the rate of change of current.
7. At any given instant, the total voltage across the circuit (V) can be expressed as the sum of the voltage across the resistor (VR) and the voltage across the inductor (VL).
8. Therefore, V = VR + VL.
9. Since the given equation represents the instantaneous voltage (V), we can deduce that V = VR.
10. By comparing V = VR with Ohm's Law (V = IR), we can conclude that I = Imaxsin(150), where Imax represents the maximum current.
The specific values of Vmax, Imax, and the phase angle have not been provided in the question, so we are working with the general expression.
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A motorcycle stoop is at a traffic light, when the light turns green, the motorcycle accelerates to a speed of 78 km/h over a distance of 50 m. What is the average acceleration of the motorcycle over this distance?
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The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².
To calculate the average acceleration of the motorcycle, we can use the formula:
Average acceleration = (final velocity - initial velocity) / time
First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.
Converting the final velocity:
Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s
Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:
Average acceleration = (21.67 m/s - 0 m/s) / time
To find the time taken to reach this velocity, we need to use the formula for average speed:
Average speed = total distance/time
Rearranging the formula:
time = total distance / average speed
Plugging in the values:
time = 50 m / 21.67 m/s ≈ 2.31 seconds
Now we can calculate the average acceleration:
Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²
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Sammy is 5 feet and 5.3 inches tall. What is Sammy's height in inches?
Answers
Answer:
\(\boxed{\sf 65.3 \ inches}\)
Explanation:
1 foot = 12 inches
Sammy is 5 feet tall.
5 feet = ? inches
Multiply the feet value by 12 to find in inches.
5 × 12
= 60
Add 5.3 inches to 60 inches.
60 + 5.3
= 65.3
12 (1 Foot) X 5= 60 + 5.3 = 65.3
Theory of uniform accelerated motion lab reports
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The theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.
What is the theory of uniform accelerated motion?The theory of uniform accelerated motion is a scientific statement regarding the acceleration of an object in steady conditions.
This theory (uniform accelerated motion) is fundamentally applied in physic and astrophysics.
In conclusion, the theory of uniform accelerated motion states that the movement of an element is constant in the same acceleration conditions.
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The cost of gas is $1.27 per liter. How much does a gallon of gas costs?
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The cost of gas is $1.27 per liter and the gallon of gas costs $4.80. where 1 gallon is equal to 3.78541 liters.
To convert the cost of gas from dollars per liter to dollars per gallon, The conversion factor between liters and gallons.
1 gallon is equal to 3.78541 liters (the exact conversion factor is 3.785411784).
Given the cost of gas is $1.27 per liter, let's calculate the cost per gallon:
Cost per gallon = Cost per liter × Liters per gallon
Cost per gallon = $1.27 × 3.78541
Cost per gallon = $4.80
So, a gallon of gas costs $4.80.
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Complete the sentence to describe how light behaves as it relates to its nature (waves or particles). light travels as and interacts with matter as .
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Light behaves as it relates to its nature (waves or particles). light travels as and interacts with matter as wave particles.
What is light?Electromagnetic radiation that is within the region of the electromagnetic spectrum that the human eye can see is referred to as light or visible light.
The main characteristics of light include intensity, polarization, frequency or wavelength spectrum, and propagation direction. One of the fundamental constants of nature is its speed in a vacuum, which is 299 792 458 meters per second (m/s).
Light can be categorized according to two theories: the first categorizes light as particles, and the second categorizes light as waves.
The main difference between the wave and particle natures of light is that the former asserts that light can act as an electromagnetic wave while the latter asserts that light is made up of tiny particles known as photons.
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Answer: 1.waves 2.particles
Explanation: just did it!<3
if john mows 11.5 meters of lawn from est to west in 7.1 seconds, what is the velocity of the laawnmower?
Answers
Answer:
11.5m/7.1s=1,619718309859155m/s
Explanation:
Distance devided per time. For km/h multiply it with 3.6
Two 10-cm-diameter charged rings face each other, 18.0 cmcm apart. Both rings are charged to 30.0 nCnC . What is the electric field strength
Answers
Answer:
E=7453.99 V/m
Explanation:
The electric field on the charged is given by
E= Kqx/(r^2 +x^2)^3/2
Where;
K= constant of Coulomb's law
q= magnitude of charge= 30.0×10^-9 C
r= radius of the rings= 5 cm or 0.05m
x= distance between the rings = 18cm = 0.18 m
Substituting values;
E= 9.0×10^9 × 30.0×10^-9 × 0.18 / [(0.05^2 + (0.18)^2]^3/2
E= 48.6/(2.5×10^-3 + 0.0324)^3/2
E= 48.6/(0.0025 + 0.0324)^3/2
E= 48.6/6.52×10^-3
E=7453.99 V/m
A 75-kg trampoline artist jumps upward from the top of a platform with a vertical speed of 6.0 m/s.
Part A: How fast is he going as he lands on the trampoline, 2.0 m below? (Figure 1)
Express your answer with the appropriate units.
Part B: If the trampoline behaves like a spring of spring constant 6.7×104N/m, how far does he depress it?
Express your answer with the appropriate units.
Answers
Answer:
Explanation:
Part A:
Using conservation of energy, we can find the speed of the trampoline artist just before he lands on the trampoline. At the top of the platform, he has gravitational potential energy equal to mgh, where m is his mass, g is the acceleration due to gravity, and h is the height of the platform. At the bottom of his jump, he has kinetic energy equal to (1/2)mv^2, where v is his speed just before he lands on the trampoline. We can equate these two energies and solve for v:
mgh = (1/2)mv^2
v = sqrt(2gh)
where h = 2.0 m and g = 9.81 m/s^2. Plugging in the values, we get:
v = sqrt(2(9.81 m/s^2)(2.0 m)) = 6.26 m/s
Therefore, the trampoline artist is going 6.26 m/s as he lands on the trampoline.
Part B:
The force exerted by the trampoline on the artist is equal to the weight of the artist, which is mg, where g is the acceleration due to gravity. This force causes the trampoline to compress a distance d, which we want to find. From Hooke's law, we know that the force exerted by a spring is equal to its spring constant times its deformation from its equilibrium length. Therefore:
mg = kd
where k is the spring constant of the trampoline. Solving for d, we get:
d = (mg)/k
where m = 75 kg, g = 9.81 m/s^2, and k = 6.7×10^4 N/m. Plugging in the values, we get:
d = (75 kg)(9.81 m/s^2)/(6.7×10^4 N/m) = 0.109 m
Therefore, the trampoline depresses 0.109 m when the artist lands on it.