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MrRissso [65]
3 years ago
13

One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

ds, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 25 m/s. The masses of the two objects are 3.0 and 8.0 kg. Determine the final speed of the two-object system after the collision for the case when the large-mass object is the one moving initially and the case when the small-mass object is the one moving initially.
Physics
1 answer:
Lina20 [59]3 years ago
6 0

Large object moving, small object at rest:

v = 18.18m/s

Small object moving, large object rest:

v= 6.81m/s

Explanation:

p = mv

m1•Vi1 + m2•Vi2 = m1•Vf1 + m2•Vf2

First Scenario (large object moving):

For small object,

p=mv

p= 3kg • 0 = 0

For large object,

p= mv

p= 8kg • 25m/s = 200kg•m/s

Combined,

p1+p2 (small object + large object) = 200m/s

After the collision, there is one mass,

new mass is 8kg + 3kg = 11kg

So now , p = mv again. In this case

200m/s = (11kg)•v

v = (200kg•m/s)/(11kg) = 18.18 m/s

Second Scenario (small object moving):

p1v1 = 3kg•25m/s = 75kg•m/s

p2v2 = 8kg•0 = 0

p1v1 + p2v2 = 75kg•m/s

New mass = 8kg+ 3kg = 11kg

p = mv

75kg•m/s = 11kg•v

v = (75kg•m/s)/11kg = 6.81m/s

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You hold a metal block of mass 40 kg above your head at a height of 2 m.
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Answer:

The work done by gravity is 784 J.

Explanation:

Given:

Mass of the block is, m=40\ kg

Height to which it is raised is, h=2\ m

Acceleration due to gravity is, g=9.8\ m/s^2

Now, work done by gravity is equal to the product of force of gravity and the distance moved in the direction of gravity. So,

\textrm{Work by gravity}=F_g\times h

Force of gravity is given as the product of mass and acceleration due to gravity.

\therefore F_g=mg=40\times 9.8=392\ N. Now,

\textrm{Work by gravity}=F_g\times h=392\times 2=784\ J

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As an object moves from point A to point B only two forces act onit: one force is nonconservative and does -30 J of work, the ot
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To solve this problem we will apply the principles of energy conservation. On the one hand we have that the work done by the non-conservative force is equivalent to -30J while the work done by the conservative force is 50J.

This leads to the direct conclusion that the resulting energy is 20J.

The conservative force is linked to the movement caused by the sum of the two energies, therefore there is an increase in kinetic energy. The decrease in the mechanical energy of the system is directly due to the loss given by the non-conservative force, therefore there is a decrease in mechanical energy.

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A light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of
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Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = 85\frac{1000}{3600}=23.61\ m/s

Initial velocity = u = 0

Equation of motion

v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

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An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the
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Answer:

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Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area A= 1.65 m^2

We need to calculate the flux

Using formula of the magnetic flux

\phi=E\cdot A

\phi = EA\cos\theta

Where,

A = area

E = electric field

Put the value into the formula

\phi=2.35\times1.65\times\cos 25^{\circ}

\phi=2.35\times1.65\times0.91

\phi=3.53\ N-m^2/C

Hence, The magnitude of the electric flux is 3.53\ N-m^2/C

8 0
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