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2 years ago

1 HELP

Physics

1 answer:

11Alexandr11 [23.1K]2 years ago

6 0

Answer:

The force applied to the spring is 35,780 N

Explanation:

Given;

extension of the spring, x = 200 mm = 0.2 m

energy stored in the spring, E = 3578 J

The force applied to the spring is calculated as;

E = ¹/₂Fx

where;

F is the applied force

F = 2E/x

F = ( 2 x 3578) / 0.2

F = 35,780 N

Therefore, the force applied to the spring is 35,780 N

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A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m

Damm [24]

Answer

given,

Tension of string is F

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consider the centrifugal force acting on the stone

= $\dfrac{mv^2}{r}$

now centrifugal force is balanced by tension

T = $\dfrac{mv^2}{r}$

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3 years ago

1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.

harkovskaia [24]

Answer:

(a) The net force is 80.394 N

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(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, $F_y = Fsin \theta = 250sin45 = 176.78 \ N$

Applied force in x-direction, $F_x = Fcos \theta = 250cos45 = 176.78 \ N$

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

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The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

$F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N$

Apply Newton's second law of motion;

F = ma

$a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2$

(b) the velocity of the crate after 5.0 s

$F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s$

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3 years ago

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galina1969 [7]

Answer:

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