What can you say about the speed of the car?

A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc

nalin [4]

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

$m_sv_s + m_rv_r = 0$

where $m_s = 72.8, m_r = 265$ are the mass of the swimmer and raft, respectively. $v_s = 5.21 m/s, v_r$ are the velocities of the swimmer and the raft after the run, respectively. We can solve for $v_r$

$265v_r + 72.8*5.21 = 0$

$v_b = -72.8*5.21/265 = -1.43 m/s$

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

7 0

2 years ago

A(n) 10.1 g bullet is fired into a(n) 2.41 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9.8

fomenos

Answer:

$v = 186.90\,\frac{m}{s}$

Explanation:

The motion of ballistic pendulum is modelled by the appropriate use of the Principle of Energy Conservation:

$\frac{1}{2}\cdot (m_{p}+m_{b})\cdot v^{2} = (m_{p}+m_{b})\cdot g \cdot h$

The final velocity of the system formed by the ballistic pendulum and the bullet is:

$v = \sqrt{2\cdot g\cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.031\,m)}$

$v\approx 0.78\,\frac{m}{s}$

Initial velocity of the bullet can be calculated from the expression derived of the Principle of Momentum:

$(0.0101\,kg)\cdot v = (2.41\,kg + 0.0101\,kg)\cdot (0.78\,\frac{m}{s} )$

$v = 186.90\,\frac{m}{s}$

5 0

3 years ago

A certain nuclear power plant is capable of producing 1.2×10^9 W of electric power. During operation of the reactor, mass is con

alukav5142 [94]

Answer:

0.00016 kg

Explanation:

Given:

Power = P = 1.2 × 10⁹ Watts

Power = work done / Time

efficiency = 0.30

Input power = 1.2 × 10⁹ / 0.30 = 4 × 10⁹ W

Energy = 4 × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules

E = m c² , where c is the speed of light and m is the mass.

⇒ mass = m = E / c² = (1.44 x 10¹³) / (3 × 10⁸ )²

= 0.00016 kg

6 0

3 years ago

Show that the Mass spring system executes simple harmonic motion(SHM)?

murzikaleks [220]

Explanation:

Show that the motion of a mass attached to the end of a spring is SHM

Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed

at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.

If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.

According to "Hook's Law

F = - Kx ---- (1)

Negative sign indicates that the elastic restoring force is opposite to the displacement.

Where K= Spring Constant

If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion

from a to b and then b to a.

According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by

F = ma ---- (2)

Comparing equation (1) & (2)

ma = -kx

Here k/m is constant term, therefore ,

a = - (Constant)x

or

a a -x

This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.

5 0

3 years ago

In which region of the ear does resonance allow the brain to interpret sound answer

lakkis [162]

I'm not too sure what your asking but here are two answers that may help.
The ear drum amplifies the vibrations.
The cochlea changes vibrations into electric signals.

7 0

3 years ago