Answer:
The velocity of the gun relative to the ground is 19.66 m/s
Explanation:
Given data,
The mass of the gun, M = 15.0 kg
The mass of the bullet, m = 50 g
The velocity of the train, v = 75 km/h
= 20.83 m/s
The velocity of bullet relative to train, V' = 350 m/s
The velocity of bullet relative to ground, V = 350 + 20
= 370 m/s
According to the law of conservation of momentum,
Mv' + mV' = 0
= -1.17 m/s
Therefore, the velocity of the gun with,
v₀ = V + v'
= 20.83 - 1.17
= 19.66 m/s
Hence, the velocity of the gun relative to the ground is 19.66 m/s
C. <span>north pole of the magnets on the guideway so they repel.</span>
Answer: The 6 kg rock sitting on a 3.2 m cliff.
Explanation:
The potential energy of an object of mass M that is at a height H above the ground us:
U = M*H*g
where g is the gravitational acceleration:
g = 9.8m/s^2
Then:
"An 8 kg rock sitting on a 2.2 m cliff"
M = 8kg
H = 2.2m
U = 8kg*2.2m*9.8 m/s^2 = 172.48 J
"a 6 kg rock sitting on a 3.2 m cliff"
M = 6kg
H = 3.2m
U = 6kg*3.2m*9.8m/s^2 = 188.16 J
You can see that the 6kg rock on a 3.2m cliff has a larger potential energy.
Answer:
496.7 K
Explanation:
The efficiency of a Carnot engine is given by the equation:
where:
is the temperature of the hot reservoir
is the temperature of the cold reservoir
For the engine in the problem, we know that
is the efficiency
is the temperature of the cold reservoir
Solving for , we find:
Answer:
0.545 m
Explanation:
K. E = 26 MeV = 26 x 1.6 x 10^-13 J = 4.16 x 10^-12 J
B = 1.35 T
Let r be the radius of curvature
The formula for the kinetic energy of a cyclotron is given by
m = 1.67 x 10^-27 kg, q = 1.6 x 10^-19 c
r = 0.545 m