Can y’all please help me quickkkkk just give me answers it wouldn’t even take a minute from ur life pleaseee

A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball m

77julia77 [94]

Answer:

The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

Explanation:

Given that,

Mass of large ball = 3.0 kg

Mass of steel ball = 1.0 kg

Velocity = 3.0 kg

After collision,

Velocity = 2.0 m/s

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}$

$-3i+2j=3.0\times v_{2}$

$v_{2}=-i+0.66j$

The direction of the momentum

$tan\theta=\dfrac{0.66}{-1}$

$\theta=tan^{-1}\dfrac{0.66}{-1}$

$\theta=-33.42^{\circ}$

The direction of the momentum with respect to east

$\theta=180-33.42=146.58^{\circ}$

Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.

7 0

3 years ago

Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k

Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is $\bf{6.59~m/s}$ .

Part(B): The direction is $\bf{42.7^{0}}$ south to east.

Explanation:

Given:

Mass of Jack, $m_{1} = 59.0~Kg$

Mass of Jill, $m_{2} = 47..0~Kg$

Initial velocity of Jack, $v_{1i} = 8.00~m/s$

Initial velocity of Jill, $v_{2i} = 0$

Final velocity of Jack, $v_{1f} 5.00~m/s$

The final angle made by Jack after collision, $\alpha = 34.0^{0}$

Consider that the final velocity of Jill be $v_{2f}$ and it makes an angle of $\beta$ with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

$~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}$

where, $v_{2f}^{x}$ is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

$v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s$

Conservation of momentum of the system along north direction is given by

$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

where, $v_{2f}^{y}$ is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

$v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s$

Part(B):

The direction is given by

$\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}$

4 0

4 years ago

When light reflects on a mirror, does it undergo a phase change?

Lynna [10]

The light does not undergo a phase change.

Also, the surface of a mirror is a rigid boundary.

5 0

3 years ago

A solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with anangular speed of 3.5 rad/s. Wha

Lynna [10]

Answer:

kE=0.0735 J

Explanation:

Given that

Radius ,R=10 cm = 0.1 m

Mass ,m= 3 kg

Angular speed ,ω = 3.5 rad/s

We know that moment of inertia for solid sphere given as

$I=\dfrac{2}{5}mR^2$

Kinetic energy

$KE=\dfrac{1}{2}I\omega^2$

Now by putting the values

$KE=\dfrac{1}{2}\times \dfrac{2}{5}mR^2\times \omega^2$

$KE=\dfrac{1}{2}\times \dfrac{2}{5}\times 3\times 0.1^2\times 3.5^2\ J$

kE=0.0735 J

Therefore the kinetic energy will be 0.0735 J

3 0

3 years ago

In your own words, define the following terms. Radiation, Convection, and conduction.

Dafna11 [192]

Radiation: transferred energy
Convection: Transfer of heat
Conduction: transfer of electric charge

3 0

3 years ago