A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? Wha
1 answer:
Answer:
1) Hence, the period is 0.33 s.
2) The amplitude is 10 cm.
Explanation:
1) The period is given by:
Where:
f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz
Hence, the period is 0.33 s.
2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:
Therefore, the amplitude is 10 cm.
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Answer:
R = 9.85 ohm , r = 0.85 ohm
Explanation:
Let the two resistances by r and R.
when they are connected in series:
V = 12 V
i = 1.12 A
The equivalent resistance when they are connected in series is
Rs = r + R
So, By using Ohm's law
V = i Rs
Rs = V / i = 12 / 1.12 = 10.7 ohm
R + r = 10.7 ohm .... (1)
When they are connected in parallel:
V = 12 V
i = 9.39 A
The equivalent resistance when they are connected in parallel
So, By using Ohm's law
V = i Rp
Rp = V / i = 12 / 9.39 = 1.28 ohm
.... (2)
by substituting the value of R + r from equation (1) in equation (2), we get
r R = 8.36 ..... (3)
..... (4)
By solvng equation (1) and (4), we get
R = 9.85 ohm , r = 0.85 ohm
The amount of work done by two boys who apply 200 N of force in an unsuccessful attempt to move a stalled car is 0.
Answer: Option B
<u>Explanation: </u>
Work done is the measure of work done by someone to push an object from its present position. We can also define work done as the amount of forces needed to move an object from its present position to another position. So the amount of work done is directly proportionate to the product of forces acting on the object and the displacement of the object.
So in this present case, as the two boys have done an unsuccessful attempts to push a stalled car so that means the displacement of the car is zero as there is no change in the position of the car. But they have applied a force of 200 N each. So the amount of work done will be
Thus, the amount of work done by two boys will be zero due to their unsuccessful attempt to move a stalled car.
Refer to the diagram shown below.
Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.
The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m
The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²
The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s
Answer: 1.375 s
Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.
The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m
The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²
The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s
Answer: 1.375 s