D). located out side the nucleus
Answer:
217.43298 m/s
Explanation:
= Mass of bullet = 19 g
= Mass of bob = 1.3 kg
L = Length of pendulum = 2.3 m
= Angle of deflection = 60°
u = Velocity of bullet
Combined velocity of bullet and bob is given by

As the momentum is conserved

The speed of the bullet is 217.43298 m/s
Answer:
the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1
Explanation:
Given the data in the question;
Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.
We know that from Newton's Second Law;
Force = mass × Acceleration
F = ma
Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.
Now,
Mass
× Acceleration
= Mass
× Acceleration
so
Mass
/ Mass
= Acceleration
/ Acceleration
given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,
Mass
/ Mass
= 1 / 1.26
Mass
/ Mass
= 0.7937 or [ 0.7937 : 1 ]
Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]
To solve this problem, let us recall that the formula for
gases assuming ideal behaviour is given as:
rms = sqrt (3 R T / M)
where
R = gas constant = 8.314 Pa m^3 / mol K
T = temperature
M = molar mass
Now we get the ratios of rms of Argon (1) to hydrogen (2):
rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)
or
rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))
rms1 / rms2 = sqrt (T1 M2 / T2 M1)
Since T1 = 4 T2
rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)
rms1 / rms2 = sqrt (4 M2 / M1)
and M2 = 2 while M1 = 40
rms1 / rms2 = sqrt (4 * 2 / 40)
rms1 / rms2 = 0.447
Therefore the ratio of rms is:
<span>rms_Argon / rms_Hydrogen = 0.45</span>
Answer:

Explanation:
To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:
Δx=vit+(1/2)at²

As we have find the time taken now we need to find the final velocity vf from below equation as

So the depth of lake is given by:
first we need to find total time as
t=3.0-1.01 =1.99 s
