Answer:
As the temperature increases, the kinetic energy of the particles increases.
Explanation:
When the temperature of the substance increases, the velocity increases which makes the movement of the particles to speed up. This causes the particles to increase. Therefore, as the temperature increases, the kinetic energy of the particles also increases.
Answer: distance d = 4.73e10m
Explanation: Suppose the charge on the black hole is 5740 C which is a positive charge.
Using electric potential V formula:
V = kq / d
Where K = 9.05×10^9Nm^2/C
And e = 1.6×10^-19C
But you don't need to substitute it.
1090 V = 8.99e9N·m²/C² * 5740C /d
Make d the subject of formula
d = 4.73e10 m
Answer:
i dont know
Explanation:
i dont know since you didn't provide something to base off of
Answer:
(a) Vf = 128 ft/s
(b) K.E = 122.8 Btu
Explanation:
(a)
In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = 32.2 ft/s²
h = height = 253 ft
Vf = Final Velocity = ?
Vi = Initial Velocity = 10 ft/s
Therefore,
(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²
16293.2 ft²/s² + 100 ft²/s² = Vf²
Vf = √(16393.2 ft²/s²)
<u>Vf = 128 ft/s</u>
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(b)
The kinetic energy of the object before it hits the surface of earth is given by:
K.E = (0.5)(m)(Vf)²
where,
m = mass of object = 375 lb
K.E = Kinetic energy of object before it strikes the surface of earth = ?
Therefore,
K.E = (0.5)(375 lb)(128 ft/s)²
K.E = 3073725 lb.ft²/s²
Now, converting this to Btu:
K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)
<u>K.E = 122.8 Btu</u>
