Answer:
Doing work' is a way of transferring energy from one object to another, energy is transferred when a force moves through a distance.
Explanation: So more energy, more work done bc u transferred more energy to move the object and doing the work. and if you only use a little of energy, the work done also only a little.
39.2 J
Explanation:
Step 1:
To find the potential energy the following formula is used.
Potential Energy = m × g × h
Where,
m = Mass
g = Acceleration due to gravity
h = Height
Step 2:
Here m = 4 kg, g = 9.8 m/s², h = 1 m
Potential Energy = ( 4 × 9.8 × 1)
= 39.2 J
Answer:
B) 4500 Pa
Explanation:
As pressure is force per unit area,
P = F/A
It stands to reason that the smallest pressure for a given force is when it is shared by the largest area.
The possible areas are
0.30(0.40) = 0.12 m²
0.30(0.50) = 0.15 m²
0.40(0.50) = 0.20 m²
The pressure when the face with the largest area (0.20 m²) is down is
P = 900 / 0.20 = 4500 N/m² or 4500 Pa
the other possible pressures would be
900/0.15 = 6000 Pa
900/0.12 = 7500 Pa
which are both larger than our solution.
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
At the frequency of 5 MHz, the period of the oscillations is 1/5meg. That's a period of 1/5 microsecond.
There are 5 full cycles in one full microsecond, and there are 2.5 full cycles in a 0.5 us pulse.
You'll have to decide for yourself how damped a pulse of 2.5 cycles is, because the parameters of the definition are corrupted in the question.
