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N76 [4]
2 years ago
5

A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initiall

y at rest. Which of the following correctly identifies the change in total kinetic energy and the resulting speed of the objects after the collision? Kinetic Energy Speed
(A) Increases 2 m/s 3.2 m/s
(B) Increases Soold 2 m/s
(C) Decreases 3.2 m/s
(D) Decreases
Physics
1 answer:
Svetlanka [38]2 years ago
4 0

Answer: (d)

Explanation:

Given

Mass of object m=2\ kg

Speed of object u=5\ m/s

Mass of object at rest M=3\ kg

Suppose after collision, speed is v

conserving momentum

\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s

Initial kinetic energy

k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J

Final kinetic energy

k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

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two porters are available to carry a long timber wood.out of them one is weak.how to do you make less load to the weak one? writ
marshall27 [118]

Answer:

Please find the answer in the explanation.

Explanation:

Given that two porters are available to carry a long timber wood.out of them one is weak. how do you make less load to the weak one?

We can make the weak one to carry less load through two different ways or means.

First, if we can locate the centre of gravity and centre of mass of the long timbre wood, the week one can carry the other end away from the center of gravity and centre of mass.

Second, the strong porter can carry the long timbre wood almost to the fulcrum and allow the weak one to support at the other end. By doing this, the weak one will only carry light portion of the load.

4 0
3 years ago
An alternating source drives a series RLC circuit with an emf amplitude of 6.04 V, at a phase angle of +30.3°. When the potentia
Vinvika [58]

Answer:

-8.56V

Explanation:

Our values are given by,

e = 6.04 V

Φ = 30.3

VC = 5.32

We can calculate the voltage across the circuit with the emf formula, that is,

e(t) = e* sin(wt)

e(t) = 6.04 * sin(Φ + π)

e(t) = 6.04 * sin(32.5 + 180)

e(t) = -3.245 V

Now, Using Kirchoff Voltage Law,

e(t) - VR- VL - VC = 0

-3.24 - 0 - VL - 5.32 = 0

Finally we have the potential difference across the inductor.

VL = - 8.56 v

5 0
2 years ago
A typical running track is an oval with 74-mm-diameter half circles at each end. A runner going once around the track covers a d
lisabon 2012 [21]

The centripetal acceleration a is 4.32 \times 10^-4 m/s^2.

<u>Explanation:</u>

The speed is constant and computing the speed from the distance and time for one full lap.

Given, distance = 400 mm = 0.4 m,       Time = 100 s.

Computing the v = 0.4 m / 100 s

                         v = 4 \times 10^-3 m/s.

radius of the circular end r = 37 mm = 0.037 m.

            centripetal acceleration a = v^2 / r

                                                        = (4 \times 10^-3)^2 / 0.037

                                                    a = 4.32 \times 10^-4 m/s^2.

6 0
2 years ago
Suppose the glass paperweight has index of refraction n=1.38. a) find the value of θ for which the reflection on the vertical su
zimovet [89]

Answer:

a)θ=71.89°

b)NO

Explanation:

Given that

For glass n= 1.38

We know that for air n'=1

The angle  for total internal reflection θc given as

sin θc=n'/n

By putting the values

sin θc=n'/n

sin θc=1/1.38

θc=46.43°

n'sinθ = n sinθref

sinθref = cosθc

n'sinθ =  n cosθc

1 x sinθ =1.38 x cos 46.43°

θ=71.89°

b)

NO

8 0
3 years ago
How are a concave lens and a convex lens alike? How are they different?
Verdich [7]

ANSWER

A convex lens acts a lot like a concave mirror. ... A concave lens acts a lot like a convex mirror. Both diverge parallel rays away from a focal point, have negative focal lengths, and form only virtual, smaller images.

5 0
3 years ago
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