"If air in a pump is squeezed more, then the air gets hotter because energy is added to it" is a good hypothesis that could lead to new experimentation.
<u>Option: C</u>
<u>Explanation:</u>
If we use a pump to inflate a basketball, we initially pull the handle to draw air to fill the sphere in. As we move it down we apply a great deal of force to pump in air through the pin's tiny hole because of this resistance force in the air we find the tube warmed.
A needle of ball pump is a metal tube in which air, from an inflating pump to a sports ball, moves through it. In continuous-flow operation, pumps are often used and built to produce comparatively little pressure towards a free-flowing environment with limited back pressure. Such pumps have a fixed configuration and work freely along their power curve as circumstances change.
Answer:
24445.85 J/s
Explanation:
Area, A = 300 m^2
T = 33° C = 33 + 273 = 306 k
To = 18° C = 18 + 273 = 291 k
emissivity, e = 0.9
Use the Stefan's Boltzman law
Where, e be the energy radiated per unit time, σ be the Stefan's constant, e be the emissivity, T be the temperature of the body and To be the absolute temperature of surroundings.
The value of Stefan's constant, σ = 5.67 x 10^-8 W/m^2k^4
By substituting the values
E = 24445.85 J/s
Answer:
The time taken for the race is 17.20 s.
Explanation:
It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.
Calculate the final speed of the sprinter.
The expression for the equation of the motion is as follows;
Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.
Put u= 0, s=30 m and .
Calculate time taken to cover 30 m distance.
The expression for the equation of motion is as follows;
Put u= 0, s=30 m and .
t=6.45 s
Calculate the time taken to complete his race.
T= t+t'
Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.
Put s= 30 m, and s'= 100 m.
T= 17.20 s
Therefore, the time taken for the race is 17.20 s.
I believe the correct gravity on the moon is 1/6 of Earth.
Take note there is a difference between 1 6 and 1/6.
HOWEVER, we should realize that the trick here is that the
question asks about the MASS of the astronaut and not his weight. Mass is an
inherent property of an object, it is unaffected by external factors such as
gravity. What will change as the astronaut moves from Earth to the moon is his
weight, which has the formula: weight = mass times gravity.
<span>Therefore if he has a mass of 50 kg on Earth, then he will
also have a mass of 50 kg on moon.</span>