This question involves the concept of kinetic energy.
The student's claim is "right".
<h3>Kinetic Energy</h3>
The energy possessed by a body, by the virtue of its motion is called kinetic energy. Mathematically it is given by the following formula:

where,
- K.E = Kinetic energy
- m = mass
- v = velocity
Therefore,
For the paintball:

K.E = 16200 J
For the pellet:

K.E = 16200 J
Hence, both paintball and pellet will have same kinetic energy. The student is right.
Learn more about kinetic energy here:
brainly.com/question/12669551
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Disease
Lack of natural reourses
war
asteroid
Polution.
Mass (m)=55kg
acceleration (a)=9.81 m/s^2, this is the acceleration due to gravity.
initial velocity=0m/s. The skydiver doesn’t start with any speed because she is on the plane or helicopter.
final velocity=16m/s This is the velocity (speed) the skydiver reaches
The equation we use is KE=.5mv^2
Kinetic energy=.5 mass x velocity^2
KE=.5(55kg)(16m/s)^2
KE=.5(55kg)(256m/s)
KE=.5(14080J)
J=Joules
KE=7040J
Kinetic energy is 7040 Joules (J)
Hope this helps
Answer:
The vapor pressure at 60.6°C is 330.89 mmHg
Explanation:
Applying Clausius Clapeyron Equation
![ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%20%3D%20%5Cfrac%7B%5Cdelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
Where;
P₂ is the final vapor pressure of benzene = ?
P₁ is the initial vapor pressure of benzene = 40.1 mmHg
T₂ is the final temperature of benzene = 60.6°C = 333.6 K
T₁ is the initial temperature of benzene = 7.6°C = 280.6 K
ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol
R is gas rate = 8.314 J/mol.k
![ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%20%5Cfrac%7B31%2C000%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B280.6%7D-%20%5Cfrac%7B1%7D%7B333.6%7D%5D%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%280.003564%20-%200.002998%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%20%280.000566%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%202.1104%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%20e%5E%7B2.1104%7D%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%208.2515%5C%5C%5C%5CP_2%20%3D%20%2840.1%2A8.2515%29mmHg%20%3D%20330.89%20mmHg)
Therefore, the vapor pressure at 60.6°C is 330.89 mmHg
Explanation:
Given that,
Mass of Nichrome, m = 0.5 g
The resistance of the wire, R = 0.673 ohms
Resistivity of the nichrome wire, 
Density, 
(A) The length of the wire is given by using the definition of resistance as :
Volume,

Area,
....(1)

(b) Equation (1) becomes :

Hence, this is the required solution.