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zalisa [80]
3 years ago
6

If you know that the period of a pendulum is 1.87 seconds, what is the length of that pendulum? (Assume that we are on Earth and

that gravity is 9.81 meters/second².) Select one of the options below as your answer: A. 0.87 centimeters B. 2.1 meters C. 1.6 meters D. 0.87 meters E. 8.3 meters
Physics
2 answers:
shepuryov [24]3 years ago
8 0

Period of an ideal simple pendulum  =  2π √(L / G)

                                          1.87 = 2π √ (L / 9.81)

Divide each side by  2π :      (1.87 / 2π) = √ (L / 9.81)

Square each side:                (1.87 / 2π)²  =  L / 9.81

Multiply each side by  9.81 :      L = (9.81) (1.87 / 2π)²  = <em> 0.869 meter</em>

                                              Choice 'D' is the closest one.


mote1985 [20]3 years ago
8 0

Answer : The correct option is, (D) 0.87 meters

Solution :

Formula used :

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

where,

T = time period of a pendulum = 1.87 seconds

L = length of the pendulum = ?

g = gravity on earth = 9.8m/s^2

Now put all the given values in the above formula, we get the length of the pendulum.

1.87s=2\times \frac{22}{7}\times \sqrt{\frac{L}{9.8m/s^2}}

0.2975=\sqrt{\frac{L}{9.8m/s^2}}

Now squaring on both the sides, we get

L=0.868m=0.87m

Therefore, the length of the pendulum is, 0.87 meters.

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A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine s
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Answer:

11.714 kW

Explanation:

Here is the complete question

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 34.0∘ above the horizontal. The car accelerates uniformly to a speed of 2.25 m/s in 10.5 s and then continues at constant speed. What power must the winch motor provide when the car is moving at constant speed?

Solution

Since the loaded ore car moves along the mine shaft at an angle of θ = 34° to the horizontal, if F is the force exerted on the cable, then the net force on the laoded ore car is F - mgsinθ = ma where  mgsinθ = component of the car's weight along the incline, m = mass of loaded ore car = 950 kg and a = acceleration

F = m(a + gsinθ)  

When the car is moving at constant speed, a = 0

So F = m(a + gsinθ) = F = 950(0 + 9.8sin34) = 5206.1 N

Since it continues at a constant speed of v = 2.25 m/s, the power of the winch motor is P = Fv = 5206.1 N × 2.25 m/s = 11713.7 W = 11.714 kW

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A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio
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Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

F - External force exerted on the sled, measured in newtons.

f - Friction force, measured in newtons.

N - Normal force from the ground on the mass, measured in newtons.

W - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

W = m\cdot g (1)

Where:

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

If we know that m = 50\,kg and g = 9.807\,\frac{m}{s^{2}}, the weight of the sled is:

W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

W = 490.35\,N

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

F_{min,s} = \mu_{s}\cdot W (2)

Where:

\mu_{s} - Static coefficient of friction, dimensionless.

W - Weight of the sled, measured in newtons.

If we know that \mu_{s} = 0.3 and W = 490.35\,N, then the force needed to start the sled moving is:

F_{min,s} = 0.3\cdot (490.35\,N)

F_{min,s} = 147.105\,N

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

F_{min,k} = \mu_{k}\cdot W (3)

Where \mu_{k} is the kinetic coefficient of friction, dimensionless.

If we know that \mu_{k} = 0.1 and W = 490.35\,N, then the force needed to keep the sled moving at a constant velocity is:

F_{min,k} = 0.1\cdot (490.35\,N)

F_{min,k} = 49.035\,N

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

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