Answer:
Part 1)

Part 2)

Part 3)

Part 4)
Since torque on right side is more so here it will turn and slip over it
Explanation:
As we know that the block A is placed at distance
d = 50 cm from the hinge at 70 cm mark
So torque due to weight of A is given as

the block B is placed at distance
d = 30 cm from the hinge at 70 cm mark
So torque due to weight of B is given as

Now torque due to weight of the scale is given as


now torque on left side of scale is given as


Torque on right Side is given as

Since torque on right side is more so here it will turn and slip over it
Answer:
Momentum Packet Answer KEY - Science Online
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Answer:
(a) 1.16 s
(b)0.861 Hz
Explanation:
(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.
From the question,
If 1550 cycles is completed in (30×60) seconds,
1 cycle is completed in x seconds
x = 30×60/1550
x = 1.16 s
Hence the period is 1.16 seconds.
(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).
Mathematically, Frequency is given as
F = 1/T ........................... Equation 1
Where F = frequency, T = period.
Given: T = 1.16 s.
Substitute into equation 1
F = 1/1.16
F = 0.862 Hz
Hence thee frequency = 0.862 Hz
Answer:
15.75 m
Explanation:
First, let's look at the top brick by itself. In order for it not to tip over the bottom brick, its center of gravity must be right at the edge of the bottom brick. So the edge of the top brick must be 10.5 m from the edge of the bottom brick.
Now let's look at both bricks as a combined mass. We know the total length of this combined brick is 10.5 m + 21 m = 31.5 m. And we know that for it to not tip over the edge of the surface, its center of gravity must be at the edge. So the edge of the combined brick must be 31.5 m / 2 = 15.75 m from the edge of the surface.
Answer:
3.38 m/s
Explanation:
Mass of child = m₁ = 25
Initial speed of child = u₁ = 5 m/s
Initial speed of cart = u₂ = 0 m/s
Mass of cart = m₂ = 12 kg
Velocity of cart with child on top = v
This is a case of perfectly inelastic collision

Velocity of cart with child on top is 3.38 m/s