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2 years ago

How much force is needed to stretch a spring 1. 2 m if the spring constant is 8. 5 N/m? 7. 1 N 7. 3 N 9. 7 N 10. 2 N.

1 answer:

6 0

The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N). Force is needed to stretch spring is 10.2 N.

<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 =?

K is the spring constant= 8.5 N/m

x is the length by which spring got stretched = 1.2m

$\rm F_S=Kx \\\\ \rm F_S=8.5 \times 1.2 \\\\ \rm F_S=10.2 N$

Hence the force is needed to stretch the spring is 10.2 N.

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

brainly.com/question/4291098

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3 years ago

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(11%) Problem 5: A uniform stationary ladder of length L = 2.7 m and mass M = 11 kg leans against a smooth vertical wall, while

jeka94

Answer:

970.2 N

Explanation:

We are given that

Length of ladder=2.7 m

Mass,M=11 kg

Coefficient of friction= $\mu=0.45$

$\theta=51^{\circ}$

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Distance from base=d

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3 years ago

The force between a pair of charges is 900 newtons. The distance between the charges is 0.01 meters. If one of the charges is 2e

Maksim231197 [3]

Answer:

$\fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}$

Explanation:

Given:

Force between pair of charges= 900 newtons

The distance between the charges = 0.01 meters

Strength of Charge first q1 = 2e-10 Coulomb

To find:

Strength of Charge second q2 = ____ Coulomb?

Solution:

We know that,

Force between two charges separate by distance r is given by the equation,

$|F| = K_e \frac{q1 \cdot \: q2}{ {r}^{2} } \\ 900 =K_e \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times {10}^{ - 4} = K_e {(2e - 10)\cdot \: q2} \\ q2 = \frac{9 \times {10}^{ - 2} }{(2e - 10) K_e} \\ \\ \fbox{We \: know \: that \: e = 2.71 } \\ substituting \: the \: value \: \\ q2 = \frac{9 \times {10}^{ - 2} }{(2 \times 2.71 - 10)K_e} \\ q2 = \frac{0.09}{ - 4.58 K_e} \\ q2 = \frac{-0.0196}{K_e}\: coulomb$

$\fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}$

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2 years ago

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A. true

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3 years ago

A steam power plant with a power output of 200 MW consumes coal at a rate of 700 tons/h. If the heating value of the coal is 40,

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Answer:

η = 2.57%

Explanation:

Given:

Output power of the steam power plant, P(out) = 200 MW

Consumption of coal, m = 700 tons/h = (700 × 10³)/3600 kg/s= 194.44 kg/s

Heating capacity of the coal, Cv = 40000 kJ/kg

now,

Input power, P(in) = mCv

P(in) = 194.44 kg/s × 40000 = 7777.77 MW

thus,

efficiency, η = P(out)/P(in) = (200 MW) / (7777.77 MW) = 0.02571

η = 2.57%

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3 years ago