Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.
The question can be solved, using Newton's second law of motion.
Note: Momentum of the cannon = momentum of the cannonball.
<h3>
Formula:</h3>
- MV = mv................. Equation 1
<h3>Where:</h3>
- M = mass of the cannon
- m = mass of the cannonball
- V = velocity of the cannon
- v = velocity of the cannonball
Make v the subject of the equation.
- v = MV/m................ Equation 2
From the question,
<h3>Given: </h3>
- M = 500 kg
- V = 3 m/s
- m = 10 kg.
Substitute these values into equation 2.
- v = (500×3)/10
- v = 150 m/s.
Hence, The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.
Learn more about Newton's second law here: brainly.com/question/25545050
Answer:
Output power of the charger is 9.85 watts.
Explanation:
It is given that,
Output current of the USB charger, I = 1.97 A
The standard USB output voltage, V = 5 V
We need to find the output power of the charger. It can be determined using the following formula as :
P = V × I

P = 9.85 watts
The output power of the charger is 9.85 watts. Hence, this is the required solution.
Answer:
Negative intrapleural pressure is the correct answer
Explanation:
Intrapleural pressure is more subatmospheric in the uppermost part of the thorax than in the lowermost parts in the standing horse.
Air moves from a region of higher pressure to one of lower pressure. Therefore, for air to be moved into or out of the lungs, a pressure difference between the atmosphere and the alveoli must be established. If there is no pressure difference, no airflow will occur.
Under normal circumstances, inspiration is accomplished by causing alveolar pressure to fall below atmospheric pressure. When the mechanics of breathing are being discussed, atmospheric pressure is conventionally referred to as 0 cm H2O, so lowering alveolar pressure below atmospheric pressure is known as negative-pressure breathing.
Answer:
If you throw a pebble into a pond, ripples
spread out from where it went in. These
ripples are waves travelling through the
water. The waves move with a transverse
motion.
Explanation: