Answer:
We learned in the previous section that temperature is proportional to the average kinetic energy of atoms and molecules in a substance, and that the average internal kinetic energy of a substance is higher when the substance’s temperature is higher.
If two objects at different temperatures are brought in contact with each other, energy is transferred from the hotter object (that is, the object with the greater temperature) to the colder (lower temperature) object, until both objects are at the same temperature. There is no net heat transfer once the temperatures are equal because the amount of heat transferred from one object to the other is the same as the amount of heat returned. One of the major effects of heat transfer is temperature change: Heating increases the temperature while cooling decreases it. Experiments show that the heat transferred to or from a substance depends on three factors—the change in the substance’s temperature, the mass of the substance, and certain physical properties related to the phase of the substance.
The equation for heat transfer Q is
Q = mcΔT,
Explanation:
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Answer:
v = √ 2 G M/ 
Explanation:
To find the escape velocity we can use the concept of mechanical energy, where the initial point is the surface of the earth and the end point is at the maximum distance from the projectile to the Earth.
Initial
Em₀ = K + U₀
Final
= 
The kinetic energy is k = ½ m v²
The gravitational potential energy is U = - G m M / r
r is the distance measured from the center of the Earth
How energy is conserved
Em₀ =
½ mv² - GmM / = -GmM / r
v² = 2 G M (1 / – 1 / r)
v = √ 2GM (1 / – 1 / r)
The escape velocity is that necessary to take the rocket to an infinite distance (r = ∞), whereby 1 /∞ = 0
v = √ 2GM /
The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
To solve this equation, simply plug the values into the equation for calculating kinetic energy.
KE = 1/2mv^2
500 = 1/2(m)(67^2)
500 =2244.5m
m = 500/2244.5 = 0.222 kg.
By definition, acceleration is the change in velocity per change of time. As time passes by, the time increases in value. So, when the acceleration is decreasing while the time is increasing, then that means that the change of velocity is also decreasing with time. So, optimally, the initial velocity and the velocity at any time are very relatively close to each other,
