Answer:
The correct answer is "6666.67 N".
Explanation:
The given values are:
Mass,
m = 0.100
Relative speed,
v = 4.00 x 10³
time,
t = 6.00 x 10⁻⁸
As we know,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.
The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)
Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
Answer:
10.23m/s^2
Explanation:
GIven data
mass of elevator = 2125 kg
Force= 21,750 N
Required
The maximum acceleration upward
F= ma
a= F/m
a=21,750/2125
a= 10.23m/s^2
Hence the acceleration is 10.23m/s^2
Potential energy due to gravity = Ep = mgh [symbols have their usual meaning ]
Evidently, HALVING the mass will make Ep , HALF its previous value. So, It will be halved.
Answer:

Explanation:
Given that


Emissivity of surfaces(∈) = 1
We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies

So now by putting the values



So rate of heat transfer per unit area
