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If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 50.5 m/s in the opposite directi

Naya [18.7K]

Answer:

The magnitude of change in momentum of the ball is $97.5 m$ and impulse is also $97.5 m$

Explanation:

Given:

Velocity of a pitched ball $v _{i} = 47$ $\frac{m}{s}$

Velocity of ball after impact $v_{f} = -50.5$ $\frac{m}{s}$

From the formula of change in momentum,

$\Delta P = m (v_{f} -v_{i} )$

Here mass is not given in question,

Mass of ball is $m$

Change in momentum is given by,

$\Delta P = m (-50.5 -47)$

$\Delta P = -97.5 m$

Magnitude of change in momentum is

$\Delta P = 97.5 m$

And impulse is given by

$J = \Delta P$

$J = -97.5 m$

So impulse and

Therefore, the magnitude of change in momentum of the ball is $97.5 m$ and impulse is also $-97.5 m$

6 0

3 years ago

Analysis of the electrocardiogram can be revealed except ________

Lunna [17]

Answer:

when a person is not breathing

Explanation:

The electrocardiogram shows the cardiac action of the heart as a means of the sinusoidal waves. However, the waves have a different structure as they show the pumping phase, breathing and the resting phase of the heart. The waves continues to be displayed as long as there is systolic and diastolic pressure in the heart muscles. When there is no action, such as the cessation of brain activity, action ceases.

3 0

3 years ago

Will mark the brainliest

AysviL [449]

Answer:

The impulse transferred to the nail is 0.01 kg*m/s.

Explanation:

The impulse (J) transferred to the nail can be found using the following equation:

$J = \Delta p = p_{f} - p_{i}$

Where:

$p_{f}$ : is the final momentum

$p_{i}$ : is the initial momentum

The initial momentum is given by:

$p_{i} = m_{1}v_{1_{i}} + m_{2}v_{2_{i}}$

Where 1 is for the hammer and 2 is for the nail.

Since the hammer is moving down (in the negative direction):

$v_{1_{i}} = -10 m/s$

And because the nail is not moving:

$v_{2_{i}}= 0$

$p_{i} = m_{1}v_{1_{i}} = 0.25 kg*(-10 m/s) = -2.5 kg*m/s$

Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:

$p_{f} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Since the hammer and the nail are moving in the negative direction:

$v_{1_{f}}$ = $v_{2_{f}}$ = -9.7 m/s

$p_{f} = -9.7 m/s(7 \cdot 10^{-3} kg + 0.25 kg) = -2.49 kg*m/s$

Finally, the impulse is:

$J = p_{f} - p_{i} = - 2.49 kg*m/s + 2.50 kg*m/s = 0.01 kg*m/s$

Therefore, the impulse transferred to the nail is 0.01 kg*m/s.

I hope it helps you!

7 0

3 years ago

Equation for progressive wave

koban [17]

Answer:

y(x, t) = A Sin(ωt ± kx)

Explanation:

Waves can be classifies as either stationary (standing), or progressive (travelling). A progressive wave is one the is a traveling wave, transferring energy along its path. While a stationary wave seems not to be moving.

The general equation for a progressive wave is;

y(x, t) = A Sin(ωt ± kx)

Where: A is its amplitude, t is the time, k is the wave number.

When the wave travels in the positive x-axis direction, the equation changes to;

y(x, t) = A Sin(ωt - kx)

When it travels in the negative x- axis direction, the equation becomes;

y(x, t) = A Sin(ωt + kx)

NB: ω = 2 $\pi$ f and k = 2 $\pi$ /λ.

6 0

4 years ago

A circuit has a current of 1.2 A. If the voltage decreases to one-third of its original amount while the resistance remains

beks73 [17]

G 3.6A I think so try this answer

7 0

3 years ago