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Reptile [31]
3 years ago
15

When a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm.

Physics
1 answer:
Readme [11.4K]3 years ago
8 0
F=K*X,
F=M*a 

M*a=K*X

2.5*9.81=K*0.0276

24.525=K*0.0276

24.525/0.0276=K

K= 888.6 N/m ---- force constant 

assuming 2.5 refers to the new extension, just divide F/ 0.025
to get

981N/m 


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The graph in the accompanying figure (Figure 1) shows the magnitude of the force exerted by a given spring as a function of the
serious [3.7K]

The elastic potential energy stored in the stretched spring is 1 J.

<h3>What is Hooke's law?</h3>

Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.

Given that;

Force on the spring = 350 Newton

Distance stretched = 7 centimeters or 0.07 m

Hence;

F = ke

k = F/e = 350 Newton/0.07 m = 5000 N/m

Work done in stretching a spring = 1/2ke^2

= 0.5 × 5000 × (2 × 10^-2)^2 =1 J

Learn more about elastic potential energy: brainly.com/question/156316

4 0
2 years ago
An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2

Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

v_f=\sqrt{2(9.8)(3.1)}

v_f = 7.79m/s

Now from Newton's second law we know that

F = ma

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

F = m\frac{dv}{dt}

Replacing we have that,

F = (89)\frac{7.79}{0.5}

F = 1386.62N

Therefore the force that the water exert on the man is 1386.62

3 0
3 years ago
Steroid use in females can cause masculine characteristics, such as lower voice and increased body hair." True False
s344n2d4d5 [400]

Answer:

True

Explanation:

Side affects can range from

Problems with periods to Loss of breasts

7 0
3 years ago
Anyone please help me with this question ASAP
Katyanochek1 [597]

Answer:

increases

Explanation:

it would have to work harder to get to two points together

7 0
3 years ago
Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W
jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

Where;

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24}kg is the mass of the Earth

r  is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}

V_{X}=18224.783 m/s (9)

<u>For Satellite Y:</u>

V_{Y}=\sqrt{\frac{GM}{r_{Y}}} (10)

V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}

V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0
3 years ago
Read 2 more answers
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