2. What is the energy that comes from heat?
2 answers:
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Answer:
121.3 cm^3
Explanation:
P1 = Po + 70 m water pressure (at a depth)
P2 = Po (at the surface)
T1 = 4°C = 273 + 4 = 277 K
V1 = 14 cm^3
T2 = 23 °C = 273 + 23 = 300 K
Let the volume of bubble at the surface of the lake is V2.
Density of water, d = 1000 kg/m^3
Po = atmospheric pressure = 10^5 N/m^2
P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2
Use the ideal gas equation
By substituting the values, we get
V2 = 121.3 cm^3
Thus, the volume of bubble at the surface of lake is 121.3 cm^3.
Answer:
T = 0.225 s
Explanation:
The speed of a projectile at the highest point of its motion is the horizontal speed of the projectile. Considering the horizontal motion with negligible air resistance, we can use the following formula:
where,
T = Total time of ball in air = ?
R = Horizontal distance covered = 40 m
= horizontal speed = 9 m/s
Therefore,
<u>T = 0.225 s</u>
Answer:
The pitching speed of your friend is 33.20 m/s
Explanation:
<em>Lets explain how to solve the problem</em>
Your friend throw the ball horizontally that means the vertical initial
component of velocity is zero ().
The ball is thrown from a height 4 meters above the ground.
The height
<u><em>Remember:</em></u> the height is negative value because its below the point of
thrown (initial position)
h = -4 m , and g = -9.8 m/s²(downward)
<em>Substitute these values in the rule above</em>
⇒
⇒ -4 = -4.9t² (multiply both sides by -1)
⇒ 4 = 4.9t² (divide both sides by 4.9)
⇒ 0.81633 = t² (take √ for both sides)
⇒ <em>t = 0.9035</em>
Then the time of the ball to land on the ground is 0.9035 seconds
The range of the ball on the ground is 30 m
The range , where
is the horizontal
component of the initial velocity
R = 30 meters and t = 0.9035
⇒ (divide both sides by 0.9035)
⇒ m/s
<em>The pitching speed of your friend is 33.20 m/s </em>