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dsp73
3 years ago
11

If a horse does 4000 J of work over 20 m, how much force does the horse use?

Physics
1 answer:
astraxan [27]3 years ago
5 0
In physics, a force is said to do work<span> 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 <----- FIRST OPTION</span>
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How much work is done when a 214 newton force pushes a sleeping cow 37m across a field.
nata0808 [166]
Hello!

Answer: 
7918 J

Explanation:

We are assuming that the floor (field) is completely horizontal since there's no information about that in the statement. 

We are going to use the following formula:

W= F . Cos \alpha . D

Where:

F=214 N
\alpha =0º
D= 37m

Then, by substituting we have:

W=214N . Cos (0).37m= 7918 N.m=7918 J

8 0
3 years ago
Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges becom
jeka57 [31]

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

       

we substitute

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3 0
3 years ago
A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro
svetoff [14.1K]

Answer:

The equation of motion is x(t)=-\frac{1}{3} cos4\sqrt{6t}

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is 32ft/s^2

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet x=\frac{4}{12} =\frac{1}{3}feet

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

W=kx ⇒ k=\frac{W}{x}

The spring constant , k=\frac{24}{1/3}

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    m\frac{d^2x}{dt} +kx=0

    \frac{3}{4} \frac{d^2x}{dt} +72x=0

  \frac{d^2x}{dt} +96x=0

Auxiliary equation is, m^2+96=0

                                 m=\sqrt{-96}

                               =\frac{+}{} i4\sqrt{6}

Thus , the solution is x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}

                                 x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2  4\sqrt{6} cos4\sqrt{6t}

The mass is released from the rest x'(0) = 0

                    =-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2 4\sqrt{6} cos4\sqrt{6(0)} =0

                                                    c_2 4\sqrt{6} =0

                                     c_2=0

Therefore , x(t)=c_1 cos 4\sqrt{6t}

Since , the mass is released from the rest from 4 inches

                    x(0)= -4 inches

c_1 cos 4\sqrt{6(0)} =-\frac{4}{12} feet

   c_1=-\frac{1}{3} feet

Therefore , the equation of motion is  -\frac{1}{3} cos4\sqrt{6t}

7 0
2 years ago
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