After rolling off the edge of the cliff and falling ' M ' meters down,
the speed of the boulder is
Square root of ( 19.6 M ) .
If M=111 meters, then the speed is <em>46.64 meters per second</em>.
We have known for roughly 500 years that if there's no air resistance,
the mass of the falling object makes no difference, and all objects fall
with the same acceleration, speed, time to splat, etc.
Answer:
v_f = 0.87 m/s
Explanation:
We are given;
F_avg = -17700 N (negative because it's backward)
m = 117 kg
Δt = 5.50 × 10^(−2) s
v_i = 7.45 m/s
Now, formula for impulse is given by;
I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s
From impulse momentum theory, we know that;
Change in momentum of particle is equal to impulse.
Thus,
Δp = I = m•v_f - m•v_i
Thus,
-973.5= 117(v_f - 7.45)
Thus,
-973.5/117 = (v_f - 7.45)
-8.3205 + 7.45 = v_f
v_f = - 0.87 m/s
We'll take absolute value as;
v_f = 0.87 m/s
Young's double slit experiment(YDSE) can be used for any kind of waves such as electromagnetic waves, sound waves, water waves, gravity waves. YDSE is based on interference. In this experiment, we make two waves interfere in order to obtain bright and dark fringes on the screen(in case of light).
You can carry this out with water, would be great if you try this at pond or water reservoir in order to see perfect ripples.
Answer:

Explanation:
The acceleration of an object is given by Newton's second law:

where
F is the net force applied on the object
m is the mass of the object
For the book in the problem, we have:
is the mass
is the force applied
Substituting into the formula, we find the acceleration:

Answer:

Explanation:
Given data
length=100mm
Diameter=5mm
Thermal conductivity=5 W/m.K
Power=50 W
Temperature=25°C
The temperature of heater surface follows from the rate equation written as:

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

Substitute the given values
![S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B2%5Cpi%20%280.1m%29%7D%7Bln%5B%5Cfrac%7B4%2A0.1m%7D%7B0.005m%7D%20%5D%7D%5C%5C%20S%3D0.143m)
The temperature of heater is then:

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.
