Answer:
The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.
Explanation:
Given;
mass of block, m = 4 kg
coefficient of kinetic friction, μk = 0.25
angle of inclination, θ = 30°
initial speed of the block, u = 5 m/s
From Newton's second law of motion;
F = ma
a = F/m
Net horizontal force;
∑F = mgsinθ + μkmgcosθ
At the top of the ramp, energy is conserved;
Kinetic energy = potential energy
¹/₂mv² = mgh
¹/₂ v² = gh
¹/₂ x 5² = 9.8h
12.5 = 9.8h
h = 12.5/9.8
h = 1.28 m
Height of the ramp is 1.28 m
Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;
v² = u² + 2ah
v² = 5² + 2 x 7.022 x 1.28
v² = 25 + 17.976
v² = 42.976
v = √42.976
v = 6.56 m/s
Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.
Yes, while he is sitting he is reading something to prepare him to finsh a job. reading instructions can be considered step one in starting a job. ex-before builders are able build, they must read the blueprints in order to know what to do.
Answer:
part A ⇒ u = 1.28 m
part B ⇒v = 0.43 m
Explanation:
for u is the distance to the object from the mirror and v is the distance from the mirror to the image.
Part A:
the mirror equation is given by:
1/f = 1/v + 1/u
but we told that, v = 1/3u:
1/f = 3/v + 1/u = 4/u
1/f = 4/u
f = u/4
u = 4f
= 4×(32×10^-2)
= 1.28 m
Therefore the distance from the mirror to the object is 1.28 m.
part B:
v = 1/3×u = 1/3×(1.28) = 0.43 m
Answer:
The magnitude of the frequency shift (that is the change in frequency) first falls to zero, then increases. This is so because as the projectile moves upward, the position of the sound source is constantly changing and as a result there would be a difference between the frequency of the sound emitted and the sound received by the microphone (that is the frequency shift). A point will be reached in the projectile's motion when the projectile will be momentarily at rest. (At the highest point reached in its flight h).
Explanation:
At this point, the frequency emitted by the source becomes equal to that received by the microphone. This can be seen from the equation 2 for Doppler effect contained in the attachment below.
This moment is short lives and the projectile soon starts a return journey downward under the influence of gravity. The velocity of the projectile is negative ( v < 0 ) and from the equation can be confirmed that the frequency received by the microphone increases as the projectile fall under gravity. This is so because as the projectile falls downward it velocity becomes more and more negative. And from the equivalent in the picture below, the frequency received by the microphone increases
Thank you very much and I hope this was helpful.
Answer:
Explanation:
A )
When empty , H₀ length of barge is inside water .
volume of barge inside water = A x H₀
Weight of displaced water = AH₀ x ρ x g
Buoyant force = weight of displaced water = AH₀ ρg
B)
It should balance the weight of barge
Weight = buoyant force
Weight = AH₀ ρg
mass of barge = weight / g
weight / g = AH₀ ρ
= 550 x .55 x 1000
= 302500 kg