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2 years ago

A weight of 200 n is hung from a spring with a spring constant of 2500 n/m and lowered slowly. How much will the spring stretch?

2 answers:

8 0

The length by which the spring got stretched will be 0.08 m. The force is directly propotional to the distance by which the spring stretched.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =200

K is the spring constant= 2500 N/m

x is the length by which spring got stretched =?

The stretch of the spring is found as;

$\rm F=kx \\\\ x = \frac{F}{k} \\\\\ x= \frac{200}{2500} \\\\ x=0.08 \ m$

Hence the length by which the spring got stretched will be 0.08 m.

To learn more about the spring force refer to the link;

brainly.com/question/4291098

#SPJ4

leva [86]2 years ago

3 0

8n/m this is because 200×2500 all over 100 . 0's will cancile 0's living 200×2500×1 ,25 will go into 200 8 times ×1 =8n/m

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A soccer ball with mass 0.450 kg is initially moving with speed 2.20 m/s. A soccer player kicks the ball, exerting a constant fo

Alinara [238K]

Answer:

0.187 m

Explanation:

We'll begin by calculating the acceleration of the ball. This can be obtained as follow:

Mass (m) = 0.450 Kg

Force (F) = 38 N

Acceleration (a) =?

F = m × a

38 = 0.450 × a

Divide both side by 0.450

a = 38 / 0.450

a = 84.44 m/s²

Finally, we shall determine the distance. This can be obtained as follow:

Initial velocity (u) = 2.20 m/s.

Final velocity (v) = 6 m/s

Acceleration (a) = 84.44 m/s²

Distance (s) =?

v² = u² + 2as

6² = 2.2² + (2 × 84.44 × s)

36 = 4.4 + 168.88s

Collect like terms

36 – 4.84 = 168.88s

31.52 = 168.88s

Divide both side by 168.88

s = 31.52 / 168.88

s = 0.187 m

Thus, the distance is 0.187 m

6 0

3 years ago

With detailed explaniation

belka [17]

Ø=37°
Initial velocity=u=20m/s
g=10m/s²

$\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}$

$\\ \rm\Rrightarrow H_{max}=20sin^237$

$\\ \rm\Rrightarrow H_{max}=7.2m$

$\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}$

$\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}$

$\\ \rm\Rrightarrow R=40sin74$

$\\ \rm\Rrightarrow R=38.5m$

$\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}$

$\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}$

$\\ \rm\Rrightarrow T=4sin37$

$\\ \rm\Rrightarrow T=2.4s$

Now

$\\ \rm\Rrightarrow v=u-gt$

$\\ \rm\Rrightarrow v=20-10(2.4)$

$\\ \rm\Rrightarrow v=20-24$

$\\ \rm\Rrightarrow v=-4m/s$

4 0

2 years ago

An elevator is being lowered at steadily decreasing speed by a steel cable attached to an electric motor. There is no air resist

shtirl [24]

Answer:

F = W + ma a> 0

Explanation:

For this exercise let's use Newton's second law

we assume the upward direction as positive

F - W = m a

F = W + ma

F = m (g + a)

In this case they indicate that the speed is less and less as it goes down, therefore the acceleration must be opposite to the speed, that is, the acceleration is upwards, consequently it is positive

We can see that since a> 0 the force F must have greater than the weight of the elevator

6 0

2 years ago

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If 0.250 L of a gas weighs 0.308 g under normal conditions of pressure and temperature, what is its molecular weight?

Phantasy [73]

Answer:

=28

Explanation:

.25/22=0,11 mol

0.308/0,11=28

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2 years ago

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Dafna11 [192]

Answer:

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3 years ago