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We have:
The function of velocity:
For
t = 5 s:
Answer:
The specific heat of a gas may be measured at constant pressure. - is accurate when discussing specific heat.
Explanation:
What will happen if the sample is the
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When a sample of solid, liquid, or gas matter heats up, it expands. When matter gets hot, its particles gain kinetic energy. ... When matter cools down, its particles lose kinetic energy. The decreased kinetic energy lets the particles come closer together. The kinetic theory of matter can be used to explain how solids, liquids and gases are interchangeable as a result of increase or decrease in heat energy. ... If it is cooled the motion of the particles decreases as they lose energy.
Answer:
A factor of 2*4 = 8
Explanation:
F_g = (G*m1*m2)/r^2
where m1 and m2 are the two masses, G is Newton's gravitational constant, and r is the distance between the center of mass of the two objects.
So, if you double m1 and quadruple m2:
m1' = 2*m1
m2' = 4*m2
Then F_g' = (G*m1'*m2')/r^2 = (G*2*m1*4*m2)/r^2 = 8*(G*m1*m2)/r^2 = 8*F_g
Answer:
When the ball hits the ground, the velocity will be -34 m/s.
Explanation:
The height and velocity of the ball at any time can be calculated using the following equations:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height of the ball at time "t".
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity. (-9.8 m/s² considering the upward direction as positive).
v = velocity at time "t".
If we place the origin of the frame of reference on the ground, when the ball hits the ground its height will be 0. Then using the equation of height, we can calculate the time it takes the ball to reach the ground:
y = y0 + v0 · t + 1/2 · g · t²
0 = 60 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²
0 = 60 m - 4.9 m/s² · t²
-60 m / -4.9 m/s² = t²
t = 3.5 s
Now, with this time, we can calculate the velocity of the ball when it reaches the ground:
v = v0 + g · t
v = 0 m/s - 9.8 m/s² · 3.5 s
v = -34 m/s
When the ball hits the ground, the velocity will be -34 m/s.