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Marysya12 [62]
3 years ago
7

You need to know the height of a tower, butdarkness obscures the ceiling. You note thata pendulum extending from the ceiling alm

ost touches the floor and that its period is 36 s. What is the height of the tower?
Physics
1 answer:
AysviL [449]3 years ago
8 0

Answer:

The towers height is 322m  

Explanation:

We can assume that a simple pendulum hangs from the top of the tower.  

In this case, the length of the pendulum will be the same as the height of the tower.

We will use the next expression then:

T=(2pi)(\sqrt{L/g})

Where T is the period, L is the pendulum length and g is the gravity (9.81m/s^2)

From the previous expression we will have then that the length will be:

L=(g)*((T/2pi)^2)

Replacing:

L=(9.81m/s^2)*((36s/2pi)^2)

The towers height will be then: 322m

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Which units in the metric system are used to measure distance?
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A block of wood weighs 160 N and has a specific gravity of 0.60. To sink it in fresh water requires an additional downward force
AysviL [449]

Answer:

Your answer is: D) 110N

Hope this helped : )

Explanation:

In order for the wood to sink, it's specific gravity must equal 1 (or greater).

Specific gravity = .6 kg/m^3= Mass/volume

F (Newtons or kg x m/s^2) = Mass*9.8m/s^2.  

Therefore, Mass = 160 N / 9.8 m/s^2= 16.32 kg

Volume = Mass/Specific Gravity = 16.32kg/.6 kg/m^3 = 27.2 m^3

Desired Specific gravity = 1 = (Mass + additional force/9.8m/s^2)/Volume = (16.32 kg +X / 9.8m/s^2)/27.2m^3

Solving for the additional force: X = 106.66...N

The answer is D, 110 N will be required to sink the block of wood.

5 0
3 years ago
An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9000 m/s an
Sergio [31]

Answer:

The distance between earthquake center and the measuring station is 1350 kilometers.

Explanation:

Let the earthquake center be at a distance of 'S' meters from the recording station.

Now from the basic relation of distance, speed and time we know that

Distance=Speed\times Time

For a Primary wave (P wave) let us assume that it appraoches the measuring station after t_{1} minutes

Thus making use of the above relation we have

Distance=V_{p}\times Time\\\\\therefore D=9000\times t_{1}.......(i)

Now since it is given that the secondary wave (S wave) reaches the measuring spot after 2 minutes or 120 seconds thus the time taken by secondary waves to reach recorder equals t_{1}+120 making use of the same relation we get

Distance=V_{s}\times Time\\\\\therefore D=5000\times (t_{1}+120).......(ii)

Solving equation 'i' and 'ii' we get

D=5000\times (\frac{D}{9000}+120)\\\\\therefore \frac{D}{5000}=\frac{D}{9000}+120\\\\\frac{D}{5000}-\frac{D}{9000}=120\\\\\therefore D=\frac{120}{\frac{1}{5000}-\frac{1}{9000}}=1350000meters=1350kilomerers

8 0
3 years ago
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