Answer is (c), the latent heat of fusion. That is by definition the heat that 1 kg of a substance must absorb to melt in the vicinity of its melting point.
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
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Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
(a)
consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.
m = mass of the tennis ball = 60 g = 0.060 kg
v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s
v = final velocity of the tennis ball after being hit by racket = - 39 m/s
ΔP = change in momentum of the ball
change in momentum of the ball is given as
ΔP = m (v - v₀)
inserting the above values
ΔP = (0.060) (- 39 - 20)
ΔP = - 3.54 kgm/s
hence , magnitude of change in momentum : 3.54 kgm/s
Answer:
ok
Explanation:
the correct answer is C please follow me
