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laiz [17]
4 years ago
6

A simple pendulum consists of a point mass suspended by a weightless, rigid wire in a uniform gravitation field. Which of the fo

llowing statements are true when the system undergoes small oscillations? Check all that apply. a. The period is independent of the length of the wire. b. The period is inversely proportional to the length of the wire. c. The period is independent of the suspended mass. d. The period is proportional to the suspended mass. e. The period is proportional to the square root of the length of the wire. f. The period is inversely proportional to the suspended mass.
Physics
1 answer:
madreJ [45]4 years ago
6 0

Answer:

Option (c) and (e)

Explanation:

In case of a simple pendulum having suspended mass 'M', if the length of the wire is 'L', and the acceleration due to gravity that acts on it is 'g', then the time period, 'T' of the pendulum for one complete oscillation is given by:

T = 2\pi \sqrt{\frac{L}{g}}

From the above equation, we can say that:

  • Time period of the pendulum does not depend on the suspended mass.
  • Time period is in direct proportion to the square root of the  wire length
  • Time period is depends inversely on the acceleration due to gravity.
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The astronaut will take 300 seconds

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, the total momentum of the astronaut+object system must be conserved.

Initially, they are both at rest, so their total momentum is zero:

p=0

After the astronaut throws the object, their total momentum is:

p=MV+mv

where:

M = 80 kg is the mass of the astronaut

V is the final velocity of the astronaut

m = 500 g = 0.5 kg is the mass of the object

v = 8.0 m/s is the velocity of the object

Since momentum is conserved, we can write

0=MV+mv

And solving for V,

V=-\frac{mv}{M}=-\frac{(0.5)(8.0)}{80}=-0.05 m/s

Which means that he starts moving at 0.05 m/s in the direction opposite to the object.

Now the astronaut needs to cover a distance of

d = 15.0 m

And his speed is

v = 0.05 m/s

Therefore, the time taken is

t=\frac{d}{v}=\frac{15.0}{0.05}=300 s

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3 0
4 years ago
Whats the answer for this question
sasho [114]
I don't really know but i think you should ask siri or google
8 0
3 years ago
A stone falls from a ledge and takes 8 seconds to hit the ground. The stone has an original velocity of 0 m/s. How tall is the l
Alex_Xolod [135]
The right answer for the question that is being asked and shown above is that: "D. 1,254.4 meters." A stone falls from a ledge and takes 8 seconds to hit the ground. The stone has an original velocity of 0 m/s. The ledge has the height of <span>D. 1,254.4 meters</span>
7 0
3 years ago
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A sled and rider, gliding over horizontal, frictionless ice at 4.7 m/s , have a combined mass of 80 kg . The sled then slides ov
zvonat [6]
Impulse is the change in momentum.
All we need is the momentum before and after.

Momentum  =  (mass) x (speed)

Before:        =  (80 kg) x (4.7 m/s)  =  376 kg-m/s

After:           =  (80 kg) x (3.4 m/s)  =  272 kg-m/s

Change in momentum  =  (376 - 272) =  <span>104 kg-m/s


</span>Notice that when you work with impulse, you're USUALLY calculating

                           (force) x (time).

The force is in Newtons, and the time is in seconds,
so the impulse is usually in units of

                             Newton-seconds .

So how can I say that impulse and change of momentum are
the same thing ?  Am I trying to pull a fast one on you ?

Remember that    1 Newton is  1 kg-m/s²

So                       (force) x (time) 

                     =  (Newton) x (second) 

                     =  (kg-m/s²) x (second)  =  kg-m/s .

and  Momentum = (mass) x (speed)  =  (kg) x (m/s)

The units of impulse are the same as the units of momentum !

So when you give an object some impulse, you give it exactly
that much momentum.
5 0
4 years ago
2.) Explain why the starting angle doesnt impact the time it takes the pendulum to swing back and forth?​
melomori [17]

The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.

The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)

−gsinθ=lθ¨−gain⁡θ=lθ¨

For small angles, θ≪1,θ≪1, and hence sinθ≈θsin⁡θ≈θ. Hence,

θ¨=−glθθ¨=−glθ

This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos⁡(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.

For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause 

7 0
3 years ago
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