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4 years ago

If a horse does 4000j of work over 20m how much force does the horse use

1 answer:

3 0

Hello! My name is Zalgo and I shall be helping you out on this mystical day~! The answer that you are looking for is... "In physics, a force is said to do work if, when acting, there is a displacement of the point of application in the direction of the force. It is calculated by the formula W = f x d. Therefore, we calculate the problem above as follows:

W = f x d

4000 = f x 20

f = 200 N"

I hope this answer helps! :P

"Stay Brainly and stay proud!" - Zalgo

(By the way, can you give me Brainliest? I'd greatly appreciate it, Thank you! X3)

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Answer:

RT = 17 ohms

Explanation:

For two parallel resistances in a circuit the combined resistance is given by:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}\\\\\frac{1}{R_T}=\frac{R_2+R_1}{R_1R_2}\\\\R_T=\frac{R_1R_2}{R_1+R_2}$

R1 = 25 ohms

R2 = 50 ohms

You replace the values of R1 and R2 in the formula for RT:

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4 years ago

A makeshift swing is constructed by making a loop in one end of a rope and tying the other end to a tree limb. A child is sittin

sveticcg [70]

Answer:

a) 264.74 N

b) 91.15 N

c) 20.12°

Explanation:

Given:

Angle between the rope and the vertical, θ = 19°

Tension in the rope, T = 280 N

For the system to be in equilibrium,

The net force in vertical as well as in horizontal direction, should be zero

Therefore,

a) For Vertical direction

Weight of the child = vertical component of the tension

W = T cosθ ..............(1)

W = 280 cos19° = 264.74 N ............(a)

b) For horizontal

force on the child, F = T sinθ .............(2)

F = 280 sin19° = 91.15 N

c) Now, on dividing (1) and (2), we have

W/F = Tcosθ/Tsinθ

tanθ = F/W

now, for F = 97 N

tanθ = 97/264.74 = 0.3663 (W from (a))

θ = tan⁻¹(0.3663)

θ = 20.12°

3 0

3 years ago

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To give solution to the exercise we must use the concepts of Torque, Vector magnitude and vector direction of the forces.

For the given problem we have to

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$\tau = \frac{dL}{dt} = 8\hat{i}-8.9\hat{j}$

The net torque acting on the particle is

$\tau_{net} = \sqrt{T_i^2+T_j^2}$

$\tau_{net} = \sqrt{(8)^2+(-8.9)^2}$

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PART B) The direction of the torque is given by,

$tan\theta = \frac{y}{x}$

$\theta = tan^{-1}\frac{y}{x}$

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Therefore the torque direction is 48.04° below the x axis.

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4 years ago

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svetoff [14.1K]

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3 years ago