Witch statement about polymers is true

if the 50 kg object slows down to a velocity of 1 m/s how much kinetic energy does it have 100j 50j 25j or none

rodikova [14]

$ek = \frac{1}{2}m {v}^{2}$

$ek = \frac{1}{2}(50) {(1)}^{2}$

$ek = \frac{1}{2}(50)$

$ek = 25j$

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I'm not really sure but I do know that it's not 0 because the object is still moving, even if it's only moving at 1 m/s.

6 0

3 years ago

A 0.1 kg beach ball hit me moving forward at 4 m/s. After hitting me, it bounced back moving at the same speed

xenn [34]

0.4 N-s is the "impulse" acted on the "beach ball".

Option: C

Explanation:

Given that,

Mass of the "beach ball" is 0.1 kg.

The speed of the ball hits is 4 m/s.

We know that,

Whenever an object is collide with other object then an impulse is acted on object, this "impulse" causes "change in momentum".

Impulse acted on the beach ball is "mass" times "velocity".

Impulse = mass × velocity

Impulse = 0.1 × 4

Impulse = 0.4 kg m/s

Impulse = 0.4 N-s

Therefore, the "impulse" acted on the ball is 0.4 N-s.

7 0

3 years ago

Can I please have help

NeTakaya

Answer:

it will usually increase

Explanation:

potato

6 0

2 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

3 years ago

Why should a rain gauge be raised 30cm above the ground

Mrrafil [7]

Explanation:

The rain gauge is normally placed in the ground, leaving the top of the funnel above ground – about 30 cm above the ground so that it can collect water into the container or jar. ... Rain gauges should be placed in an open area where there are no buildings, trees, or other obstacles to block the rain.

5 0

3 years ago

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