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LiRa [457]
2 years ago
15

Suppose that Salim hits a 100 g fastball moving towards him at a speed of 42 m/s with his bat as shown in the image.the impulse

of the fastball is 8 kg.m/s in the antiparallel direction of the initial velocity. Then the final momentum of the fastball is --------------( Suppose that the direction of the fastball toward the bat is negative and away from the bat is positive)
1.8 kg m/s
0.6 kg m/s
3.8 kg m/s
2.4 kg m/s
Physics
1 answer:
FinnZ [79.3K]2 years ago
3 0

The final momentum of the ball is 3.8 kgm/s.

<h3>Change in momentum of the ball</h3>

The impulse received by the ball is equal to change in momentum of the ball.

J = ΔP

where;

  • J is the impulse
  • ΔP is change in momentum

ΔP = P₂ - P₁

P₂ = ΔP + P₁

<h3>Final momentum of the ball</h3>

The final momentum of the ball is calculated as follows;

P₂ = 8 + (- 0.1 x 42)

P₂ = 8 - 4.2

P₂ = 3.8 kgm/s

Learn more about change in momentum here: brainly.com/question/7538238

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White light falls on a yellow filter,if:
riadik2000 [5.3K]

Answer:

all colours are absorbed except for the colour of the filter.

Explanation:

When white light passes through a coloured filter, all colours are absorbed except for the colour of the filter. For example, an orange filter transmits orange light but absorbs all the other colours. If white light is shone on an orange filter, only the orange wavelengths will be observed by the human eye.

6 0
3 years ago
Earth's lithospheric plates interact at____.
earnstyle [38]

Explanation:

Tectonic plate interactions are classified into three basic types: Divergent boundaries are areas where plates move away from each other, forming either mid-oceanic ridges or rift valleys. These are also known as constructive boundaries. Convergent boundaries are areas where plates move toward each other and collide.

7 0
2 years ago
When a net force of 17.0 newtons is applied to a dictionary placed on a frictionless table, it accelerates by 3.75 meters/second
4vir4ik [10]
Force = (mass) x (acceleration)    (Newton's second law of motion)

Divide both sides of the equation by 'acceleration', and you have

Mass = (force) / (acceleration)

Mass = 17 newtons / 3.75 meters per second-sqrd = 4.533 kilograms (rounded)
8 0
2 years ago
A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur
morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] =  4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\

Substitute the given parameters and solve for the angular speed;

\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\  \ +\  \ 0.35(0.02^2\  + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0
2 years ago
Name three types of energy that exist in a large piece of charcoal on a grill in the sunlight. Explain why the charcoal has each
qaws [65]

Answer:

A large piece of charcoal on a grill in the sunlight (if it's burning) will consist of the following types of energy:

  1. Chemical
  2. Heat and
  3. Light

Explanation:

Charcoal is basically carbon which is produced when wood is heated strongly in the absence of oxygen. From a chemistry point of view, charcoal contains combustible carbon whose chemical formula is C. Sometimes, H_{2}O which is water may be found in it but in very small units.

All matter contains Heat energy. Charcoal is not an exception. As the charcoal burns, the heat energy is produced along with Light energy.

Light comes in many forms such as Infrared rays, Xrays, Visible Spectrum light, etc.

The glow which the coal gives off fall under the visible spectrum of light.

Cheers

8 0
2 years ago
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