To provide a greater certainty that the observed results are not by chance.
Answer:
If the ship speed is doubled, then the power developed is 8 times the initial value.
Explanation:
ship power is roughly proportional to the cube of the speed, so
P ∝ v³
If the speed is doubled, then the power developed becomes
P ∝ (2)³ = 8 times
Therefore, if the ship speed is doubled, then the power developed is 8 times the initial value.
Answer:
Explanation:
We have given work is done on the gas is 1200 J
So work done will be ( as work is done on the gas )
It is given that internal energy is increases by exactly 700 J
So
From thermodynamic equation
So
Here negative sign indicates that heat flow out of the gas
If heat was negative then heal was flowing in the gas
Because the two paths are perpendicular, therefore the
target proton's new path must be at 30 degrees from the original
direction.
Using the law of conservation of momentum about the original direction:
m (400 m/s) = m (v1) cos(60) + m (v2) cos(30)
Cancelling m since the two protons have similar mass.
(v1)cos(60) + (v2)cos(30) = 500 m/s ---> 1
Now by using the law conservation of momentum perpendicular to the original
direction:
m (0 m/s) = m (v1) sin(60) – m (v2) sin(30)
Which simplifies to:
(v1)sin(60) - (v2)sin(30) = 0 m/s
v2 = v1 * sin(60) / sin(30) = v1 * sqrt(3) ---> 2
Plugging equation 2 to equation 1:
(v1) (1/2) + (v1 * sqrt(3)) sqrt(3)/2 = 500 m/s
(1/2) (v1) + (3/2) (v1) = 500 m/s
2 (v1) = 500 m/s
v1 = 250 m/s
Thus, from equation 2:
v2 = v1*sqrt(3) = (250 m/s) sqrt(3) = 433.01 m/s
So,
A. The target proton's speed is about 433 m/s
B. The projectile proton's speed is 250 m/s
Answer:
437.5Kjoules
Explanation:
K.E=half multiply by mass multiply by square of velocity
=437.5Kjoules