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andrew-mc [135]
3 years ago
13

A 20-kg object sitting at rest is struck elastically in a head-on collision with a 10-kg object initially moving at 3.0 m/s. Fin

d the final velocity of the 10-kg object after the collision.

Physics
1 answer:
Sati [7]3 years ago
3 0

Answer:1m/s

Explanation: As the stationary ball is hit by the moving ball ,the two moves together after collision, with a single velocity. The attached photo further explains how the answer is calculated

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Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 7.10 nc, and the other has
k0ka [10]

Answer:

A. F=107.6nN

B. Repulsive

Explanation:

According to coulombs law, the force between two charges is express as

F=(Kq1q2) /r^2

If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.

Note the constant K has a value 9*10^9

Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m

If we substitute values we have

F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)

F=(282.4×10^-9)/2.6244

F=107.6×10^-9N

F=107.6nN

B. Since the charges are both positive, the force is repulsive

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<span>No. the products include He-4, one neutron, and energy.</span>
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A _____ space zone is one that is obstructed in some way
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A closed space zone is one that is obstrutced in some way.

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An arrow is shot into a falling apple. What type of collision is this an example of?
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A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans
alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

                 y = (0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)

Now, we will compare above equation with the standard form of transeverse wave equation,

                 y = A sin(kx + \omega t)

where,    A is the amplitude = 0.09 m

              k is the wave vector = \frac{\pi}{11}

              \omega is the angular frequency = 4\pi

              x is displacement = 1.40 m

              t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave  will be:

                   v(t) = \frac{dy}{dt}

                v(t) = A \omega cos(kx + \omega t)

        v(t) = (0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)

          v(t) = -0.84 m/s

The acceleration of the particle in the location is

            a(t) = \frac{dv}{dt}

           a(t) = -A \omega 2sin(kx + \omega t)

           a(t) = -(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)

           a(t) = -9.49 m/s^{2}

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 m/s^{2} .

(b)  Wavelength of the wave is given as follows.

               \lambda = \frac{2\pi}{k}

              \lambda = (frac{2\pi}{\frac{\pi}{11})&#10;

              \lambda = 22 m

The period of the wave is

             T = \frac{2 \pi}{\omega}

             T = \frac{2 \pi}{4 \pi}

                = 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

                    v = \frac{\lambda}{T}

                       = \frac{22 m}{0.5 s}

                       = 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0
3 years ago
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