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svetlana [45]
3 years ago
7

you measured the torque caused by the gyroscope's weight by lifting up the end of the gyroscope (at 7.5 inches). If you measured

at a different distance, say you attached the spring scale at 5 inches from the pivot point, would the spring scale read a different value? Would the torque be a different value? Does this explain why you do not need to measure the spring scale at the center of mass location, where the moment of gravitational force is located? (provide an argument as well as your measurement report.)
Physics
1 answer:
DIA [1.3K]3 years ago
3 0

Answer:

i) No, the spring scale does not read a different value

ii) The torque will read a different value, it will reduce

iii) The spring scale does not need to be measured at the center of mass location.

Explanation:

The torque caused by the gyroscope can be given by the relation,

\tau = r × f

\tau = rf cos \theta

The torque measured by the gyroscope varies directly with the distance, r.

A decrease in the distance r will also cause a decrease in the value of the torque measured. When the distance, r is reduced from 7.5 inches to 5 inches, the torque caused by the gyroscope's weight also reduces.

The weight of the gyroscope remains constant despite the reduction in the distance because the weight  of the gyroscope is not a function of the distance from the gyroscope. Therefore, the spring scale will not read a different value.

Yes, the spring scale does not need to be measured from the center of mass location because the weight does not depend on the location of measurement. The reading of the sprig scale remains constant.

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3. Ultraviolet can kill these. <br><br>8words and 2nd letter starts with an A​
Karo-lina-s [1.5K]

Answer:

Bacteria

Explanation:

UV kills bacteria and its 8 letters

7 0
3 years ago
An 85-kg man plans to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that h
Lelu [443]

Answer:

The greatest acceleration the man can give the airplane is 0.0059 m/s².

Explanation:

Given that,

Mass of man = 85 kg

Mass of airplane = 109000 kg

Distance = 9.08

Coefficient of static friction = 0.77

We need to calculate the greatest friction force

Using formula of friction

F=\mu mg

Where, m = mass of man

g = acceleration due to gravity

Put the value into the formula

F = 0.77\times85\times9.8

F= 641.41\ N

We need to calculate the acceleration

Using formula of newton's second law

F = ma

a=\dfrac{F}{m}

Put the value into the formula

a=\dfrac{ 641.41}{109000}

a=0.0059\ m/s^2

Hence, The greatest acceleration the man can give the airplane is 0.0059 m/s².

3 0
3 years ago
A 2100 kg car starts from rest and accelerates at a rate of 2.6 m/s2 for 4.0 s. Assume that the force acting to accelerate the c
Reika [66]
When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.

a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s

Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ
8 0
3 years ago
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor
const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  \tau = 34.3 \  N\cdot m

Explanation:

From the question we are told that

   The mass of the steel ball is  m  =  3.0 \  kg

    The length of arm is  l =  70 \ cm  = 0.7 \  m

    The mass of the arm is m_a  = 4.0 \  kg

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       r = \frac{l}{2}

=>    r = \frac{ 0.7}{2}  

=>    r = 0.35 \ m  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      \tau =  m_a * g * r  + m * g *  L

=>    \tau =  4 * 9.8 * 0.35 + 3 * 9.8 *  0.70

=>    \tau = 34.3 \  N\cdot m

5 0
2 years ago
A active-satellite, one that sends out electromagnetic waves, can be the energy down toward the ocean surface and determine how
Anna71 [15]

Answer: This is called backscatter which refers to the ability of big waves to reflect the energy in order to give back the signal .

Explanation:

What is meant by backscatter?

Backscatter is the process where by the waves or signal is reflected back to the original direction and get scattered in all directions.

Backscatter allows us to receive signal and be able to view all the channels that are connected through the satellite.

4 0
3 years ago
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