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Mashutka [201]
4 years ago
11

A scientist suspends a spring from a force sensor. She then pulls on the spring causing the spring to stretch 1.0 cm. She record

s the amount of force applied to the spring. She then stretches the spring 2.0 cm and measures the new force. She repeats the process several times, each time increasing the length the spring is stretched.
Physics
1 answer:
Nadusha1986 [10]4 years ago
8 0
This situation describes the Hooke's Law which states that "When an elastic object - such as a spring - is stretched, the increased length is called its extension. The extension of an elastic object is directly proportional to the force applied to it". The formula is <span>F = k × e , F for the force, k for spring constant expressed in N/m, e for extension in m. This equation works for as long the spring is not stretch too much because once it exceeded its limit, the spring will not return to its original length the moment the load is removed.</span>
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vredina [299]

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6 0
3 years ago
If the mass of a wheel is increased by a factor of 7 and the radius is increased by a factor of 15, by what factor is the moment
inn [45]

Answer:

The moment of inertia will be increased by a factor of 1575

Explanation:

Let the moment of inertia of a wheel be given as;

I = \frac{1}{2} mr^2\\\\

where;

I is the moment of inertia of the wheel

m is mass of the wheel

r is radius of the wheel

when the mass is increased by a factor of 7 and the radius is increased by a factor of 15

I_1 = \frac{1}{2}mr^2\\ \\I_2 = \frac{1}{2}(7m)(15r)^2\\\\I_2 =  \frac{1}{2}(7m)(225r^2)\\\\I_2 = 7\times 225[\frac{1}{2}(mr^2)]\\\\I_2 = 1575[\frac{1}{2}(mr^2)]\\\\Recall, \frac{1}{2}(mr^2) = I_1 \\\\I_2 = 1575 I_1

Thus, the moment of inertia will be increased by a factor of 1575

3 0
3 years ago
True or false, not all substances are soluble in water.
lions [1.4K]

Answer:

A.true

Explanation:

7 0
3 years ago
Read 2 more answers
A 2.2 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determi
iogann1982 [59]

Answer:

a) k = 2851.2\,\frac{N}{m}, b) v = 0.54\,\frac{m}{s}

Explanation:

a) According to the Principle of Energy Conservation, the kinetic energy of the box is transformed into elastic potential energy.

\frac{1}{2}\cdot (2.2\,kg)\cdot (1.8\,\frac{m}{s} )^{2} = \frac{1}{2}\cdot k \cdot (0.05\,m)^{2}

The spring constant is:

k = 2851.2\,\frac{N}{m}

b) The initial speed needed to compress the spring by 1.5 centimeters is:

\frac{1}{2}\cdot (2.2\,kg)\cdot v^{2} = \frac{1}{2}\cdot (2851.2\,\frac{N}{m} )\cdot (0.015\,m)^{2}

v = 0.54\,\frac{m}{s}

5 0
4 years ago
Two equal forces are applied to a door. The first force is applied at the midpoint of the door;
fredd [130]

Answer:

B

Explanation:

Torque exerted on an object depends upon three things.

1. Moment arm

2. Magnitude of applied force

3. Direction of force

T=r x F

T= r * F sinФ

Let T1 is force due to force applied at mid point of door.

let d is the width of door, so

T1= \frac{d}{2}* F * sin(90°)                  (∴ moment arm =d)

T1=\frac{d* F}{2}  __________(2)       (∴ sin 90°=1)

Let T2 is the torque due to force applied at the doorknob.

T2=  d* F * sin(90)                                (∴ moment arm =d)

T2=d*F __________(2)                      (∴ sin 90°=1)

From equation (1) and equation (2)

2T1=T2

or

T1= T2/2

As T2 is greater so option B is the correct answer.

5 0
4 years ago
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