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2 years ago

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

1 answer:

astraxan [27]2 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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The majority of earthquakes worldwide occur at all but one location. That is A)at tectonic plate boundaries.B)where sediments de

DENIUS [597]

Answer:b

Explanation:

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2 years ago

3.25 kcal is the same amount of energy as

PIT_PIT [208]

<span>1 cal = 4,185 J
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8 0

2 years ago

Read 2 more answers

A golf ball reaches a height of 150 m before it stops rising and starts to fall to the ground. What is the golf balls speed (rou

artcher [175]

Answer:

v = 54 m/s

Explanation:

Given,

The maximum height of the flight of golf ball, h = 150 m

The velocity at height h, u = 0

The velocity of the golf ball right before it hits the ground, v = ?

Using the III equations of motion

<em> v² = u² + 2gh</em>

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= 2940

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4 0

2 years ago

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sp2606 [1]

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3 years ago

Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length

SVEN [57.7K]

Answer with Explanation:

We are given that

Length of wire 1= $L_1$

Length of wire 2= $L_2$

Resistivity of copper wire= $\rho_1=1.7\times 10^{-5}\Omega-m$

Resistivity of aluminum wire= $\rho_2=2.82\times 10^{-5}\Omega-m$

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

Resistance = $\frac{\rho l}{A}$

When diameter of wires are same then their cross section area are also same .

$l=\frac{RA}{\rho}$

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity is small then the length of wire will be large.

$\rho_1$

Therefore, $L_1>L_2$

Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

$\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66$

$L_1=1.66L_2$

7 0

2 years ago