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10 months ago

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A load of 50N attached to a spring hanging vertically stretches the spring is 5cm . The spring is now placed horizontally on a t

able and stretched 11cm . What force is required to stretch the spring?

1 answer:

jasenka [17]10 months ago

5 0

When the spring is now placed horizontally on a table and stretched 11cm then the force required to stretch the spring would be 110 N.

<h3>What is the spring constant?</h3>

The spring constant is used to define the stiffness of the spring, the greater the value of the spring constant stiffer the spring and it is more difficult to stretch the spring.

The mathematical relation for calculating the spring constant is as follows

F = - Kx

where F represents the Force

x represents the amount of stretch in the spring

K represents the value of the spring constant

For the given problem that a load of 50N attached to a spring hanging vertically stretches the spring is 5cm

F= 50 N and x = 5 cm

K = F/x

=50/5 N/ cm

=10 N / cm

The spring constant for the spring is 10 N/ cm

Now the spring is placed horizontally on a table and stretched 11cm

by using the same formula we used earlier to calculate the spring constant

K = F/x

As we calculated the value of the spring constant (K) as 10 N / cm

by substituting K=10 N / cm and x =11 cm

10 =F/11

F = 110 N

Thus, a force of 110 N force is required to stretch the spring when it is placed horizontally

Learn more about spring constant from here

brainly.com/question/14670501

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A weightlifter does 450 J of work on a barbell in 3 s. How much power is the weightlifter generating?

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The pupil of an eagle’s eye has a diameter of 6.0 mm. Two field mice are separated by 0.010 m. From a distance of 185 m, the eag

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time is 5.973826 sec

Explanation:

Given data

diameter D = 6.0 mm 6× $10^{-3}$ m

separated d = 0.010 m

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speed s = 16 m/s

wavelength = 550 nm = 550 $10^{-9}$ m

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How much time passes

solution

we know that for resolution we use Rayleigh's Criterion i.e

θ = 1.22 wavelength / diameter = separated / distance 1

we calculate distance 1 by put value wavelength, diameter and separated

distance 1 = diameter × separated / 1.22 wavelength

distance 1 = 6× $10^{-3}$ × 0.010 / 1.22 × 550 × $10^{-9}$

distance 1 = 89.418778

so time will be i.e = distance (dis) - distance 1 / speed

time = ( 185 - 89.418778) / 16

time = 5.973826 sec

time is 5.973826 sec

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3 years ago

How long will it take for a spacecraft traveling at

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16 2/3 hr

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2 years ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

leonid [27]

Answer:

$KE_A=33\ J$

$KE_B=99\ J$

Explanation:

Given:

Let mass of the particle B be, $m_B=m$

then the mass of particle A, $m_A=3m$

Energy stored in the compressed spring, $E=132\ J$

Now when the compression of the particles with the spring is released, the spring potential energy must get converted into the kinetic energy of the particles and their momentum must be conserved.

Kinetic energy:

$\frac{1}{2}m_A.v_A^2+\frac{1}{2}m_B.v_B^2=132$

$3m.v_A^2+m.v_B^2=264$ .............................(1)

<u>Using the conservation of linear momentum:</u>

$m_A.v_A+m_B.v_B=0$

$3m.v_A+m.v_B=0$ .............................(2)

Put the value of $v_A$ from eq. (2) into eq. (1)

$3m\times (\frac{-v_B}{3})^2+m.v_B^2=264$

$v_B^2=\frac{198}{m}$ ...........................(3)

<u>Now the kinetic energy of particle B:</u>

$KE_B=\frac{1}{2}\times m_B\times v_B^2$

$KE_B=\frac{1}{2}\times m\times \frac{198}{m}$

$KE_B=99\ J$

Put the value of $v_B^2$ form eq. (3) into eq. (1):

$v_A^2=\frac{22}{m}$

<u>Now the kinetic energy of particle A:</u>

<u /> $KE_A=\frac{1}{2}m_A.v_A^2$ <u />

<u /> $KE_B=\frac{1}{2}\times 3m\times \frac{22}{m}$ <u />

$KE_A=33\ J$

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