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3 years ago

The velocity of a 150 kg cart changes from 6.0 m/s to 14.0 m/s. What is the magnitude of the impulse that acted on it?

2 answers:

8 0

M = 150 kg.

Final velocity, v = 14 m/s

Initial Velocity, u = 6 m/s

Impulse = m(v - u)

= 150*(14 - 6)

= 150*8 = 1200 kgm/s or 1200 Ns

kow [346]3 years ago

4 0

Answer: The magnitude of impulse acted on the cart will be 1200 kg.m/s

Explanation:

Impulse is defined as the product of mass of the object and the change in its velocity. It is basically the change in momentum of the body.

Mathematically,

$J=m\Delta v$

where,

J = Impulse

m = mass of the object

$\Delta v=v_2-v_1$ = change in the velocity

We are given:

$m=150kg\\v_1=6m/s\\v_2=14m/s\\J=?kg.m/s$

Putting values in above equation, we get:

$J=150kg\times (14-6)m/s=1200kg.m/s$

Hence, the magnitude of impulse acted on the cart will be 1200 kg.m/s

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The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

Circular Motion

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

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. The penalties for a person's second DUI conviction include completion of __________ hours of DUI school.

Kobotan [32]

The penalties for a person's second DUI conviction include completion of 21 hours of DUI school.

4 0

3 years ago

Which planet is smaller than Earth’s core?

makvit [3.9K]

Answer:

The answer is C. Mars. Mars and Mercury are both smaller than Earth's core. Hope this helps you :)

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A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a

sleet_krkn [62]

When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is

∑ F = R - mg = 0

where mg = weight of the mass = (7.00 kg) g = 68.6 N.

It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that

k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m

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Assume a reaction takes 7.5 moles of anhydrous calcium chloride and energy transfer occurs, which we record as 21.2 joules of en

OLga [1]

Answer: 2.83 J/mol

Explanation:

Heat of solution, sometimes interchangeably called enthalpy of solution, is said to be the energy released or absorbed when the solute dissolves in the solvent. A solute is that which can dissolve in a solvent, to form a solution

Given

No of moles of CaCl = 7.5 mol

Total energy used = 21.2 J

Heat of solution = q/n where

q = total energy

n = number of moles

Heat of solution = 21.2 / 7.5

Heat of solution = 2.83 J/mol

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3 years ago

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