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Leto [7]
3 years ago
15

How does approaching a temperature of absolute zero affect kinetic energy.?

Physics
1 answer:
Alexus [3.1K]3 years ago
7 0

First of all the kinetic energy is when the particles move in continuous random motion.

If the temperature is high the colliding particles will collide more. and if the temperature is low the colliding particles will collide less.

Low temperature result in low kinetic energy 
High temperature result in high kinetic energy

Absolute zero is the point where where all molecules have no kinetic energy. It is a theoretical value (it has never been reached).

The Kelvin temperature scale is based on absolute zero being the lowest possible temperature that could theoretically be reached. That is why there is no such thing as a negative Kelvin temperature value.

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A 50 kg pitcher throws a baseball with a mass of 0.15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is
Andreas93 [3]

The velocity of the pitcher is <u>0.105 m/s</u> in a direction opposite to the velocity of the ball.

When no external force acts on a system, the total momentum of the system is conserved. The total initial momentum of the system is equal to the total final momentum of the system.

The pitcher and the ball are initially at rest, therefore, the total initial momentum of the system is zero.

Since no external forces act on the system comprising of pitcher and the ball, the total final momentum of the system is also equal to zero.

If the mass of the pitcher is mp and its speed is vp, the mass of the ball is mb and the ball's speed is vb, then the final momentum of the system of pitcher and the ball is given by,

p=m_pv_p+m_bv_b=0

Therefore,

v_p=-\frac{m_b}{m_p} v_p

Substituet 0.15 kg for mb, 50 kg for mp and 35 m/s for vb.

v_p=-\frac{m_b}{m_p} v_p=-\frac{0.15 kg}{50 kg} (35m/s)=-0.105 m/s

The pitcher has a velocity <u> 0.105 m/s</u> opposite to the direction of the velocity of the ball.

8 0
3 years ago
A toy cart at the end of a string 0.70 m long moves in a circle on a table. The cart has a mass of 2.0 kg and the string has a b
luda_lava [24]
Given that the mass of the toy cart is 2.0 kg and and the acceleration is unknown, the normal formula would be a=f/m where a is acceleration, f is force and m is mass but the string's breaking strength is 40n so I think the formula in this case will be f is greater than m*a
40 is greater than 2a
40 is greater than 2a
40/2 is greater than 2a/2
20m/s² is greater than a 
Therefore the maximum speed the toy cart should have should be less than 20m/s²
8 0
3 years ago
A 3500kg suv is driving north at 36m/s what the momentum
lianna [129]
Your answer is

<span>126000</span>
4 0
3 years ago
The initial and final velocities of two blocks experiencing constant acceleration are respectively −7.45 m/s and 14.9 m/s. (a) T
ikadub [295]

Answer:

a) a_{1}=3.7 m/s^{2}

b) a_{2}=3.68 m/s^{2}

Explanation:

a) The displacement of the first object is 22.5 m, so we can use the next equation:

v_{f}^{2}=v_{i}^{2}+2a\Delta x

a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}

a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.5}

a_{1}=3.7 m/s^{2}

positive acceleration.

b) Using the same equation we can find the second value of the acceleration:

a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}

a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.6}

a_{2}=3.68 m/s^{2}

positive acceleration.

I hope it helps you!

8 0
3 years ago
A dogs tail hit a 1.3kg flower vase, knocking it over. the net force on the flower vase over time is shown below
lesantik [10]

Answer: 0.15 m/s

Explanation:

​

​

8 0
3 years ago
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