a. 0.5 T
- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position
- The period T is the time the system takes to complete one oscillation
During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.
So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion
and solving for t we find
b. 1.25T
Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that
- the mass takes a time of 1 T to cover a distance of 4A
we can set the following proportion:
And by solving for t, we find
Explanation:
Mass of the astronaut, m₁ = 170 kg
Speed of astronaut, v₁ = 2.25 m/s
mass of space capsule, m₂ = 2600 kg
Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :
initial momentum = final momentum
Since, initial momentum is zero. So,
So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.
Answer:
14.7 m/s.
Explanation:
From the question given above, the following data were obtained:
Time (t) = 1.5 s
Acceleration due to gravity (g) = 9.8 m/s².
Height = 11.025 m
Final velocity (v) = 0 m/s
Initial velocity (u) =?
We, can obtain the initial velocity of the penny as follow:
H = ½(v + u) t
11.025 = ½ (0 + u) × 1.5
11.025 = ½ × u × 1.5
11.025 = u × 0.75
Divide both side by 0.75
u = 11.025/0.75
u = 14.7 m/s
Therefore, the penny was travelling at 14.7 m/s before hitting the ground.
Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
The work done by the three students is 3,000 J.
The energy transferred in the process is 3,000 J.
<h3>What is work done?</h3>
- Work done is the product of force and distance moved by the object.
W = Fd
The work done by the three students is calculated as follows;
W = 300 x 10
W = 3,000 J
<h3>What is energy transfer?</h3>
- This is means by which energy is converted from one form to another.
The energy transferred in the process is determined by work energy theorem.
E = W
E = 3,000 J
Learn more about work-energy theorem here: brainly.com/question/22236101
