Assuming this coin is on earth and that it wasn’t dropped forcefully:
Use the formula d = 1/2at^2. Rewriting using a=g and solving for height h gets us h = 1/2(9.8)t^2.
In this case that would get that the change in height h is 0.5(9.8)(0.3^2) = 0.441 m.
Answer:
I should be active for 15 hours to meet the physical activity requirement.
Explanation:
Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.
So, t = t₀/√(1 - β²)
t = 9/√(1 - (v/c)²)
= 9/√(1 - (0.8c/c)²)
= 9/√(1 - (0.8)²)
= 9/√(1 - (0.64)
= 9/√0.36
= 9/0.6
= 15 hr
So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.
Answer:
increase.
Explanation:
According to the newton’s second law of motion force is expressed as product of mass and acceleration.
F = m a
If the force acting is constant, then.
m∝ 
That is if the mass of object increases the acceleration decreases and vice versa. The above equation is used when the force acting on the body is constant.
As the thrust force from the rocket engine is constant throughout there will be a variation in the mass or acceleration.
Thus, it won't stay the same.
As the weight of the car is maximum at the start because of the fuel present in the rocket engine and minimum at the end as the fuel burns throughout the journey of the car. Weight will be minimum at the end and hence acceleration is maximum at the end.
Thus, it won't decrease.
As the acceleration is going from minimum at the start to maximum at the end, therefore it is continuously increases throughout its journey.
Thus, it will increase.
Answer:
I = 2 kgm^2
Explanation:
In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:
(1)
I: moment of inertia of the door
α: angular acceleration of the door = 2.00 rad/s^2
τ: torque exerted on the door
You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:
(2)
F: force = 5.00 N
d: distance to the hinges = 0.800 m
You replace the equation (2) into the equation (1), and you solve for α:

Finally, you replace the values of all parameters in the previous equation for I:

The moment of inertia of the door around the hinges is 2 kgm^2