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IrinaVladis [17]
3 years ago
14

according to newton's third law, when a horse pulls on a cart, the cart pulls back on the horse with an equal force on the horse

. if in fact the cart pulls back on the horse as hard as the horse pulls forward on the cart, how is it possible for the horse to move the cart?
Physics
1 answer:
Debora [2.8K]3 years ago
8 0
Based on Newton's principle, whenever objects A and B interact with each other, they exert forces upon each other.

When a horse pulls on a cart, t<span>he horse exerts a force only to the cart. But that force applies only to the cart, not to the horse.
 
The cart in turn exerts a force on the horse. But that force applies only to the horse, not the cart also.
</span>
There are two forces resulting from this interaction - a force on the horse and a force on the cart. T<span>he net force on the cart remains as it was --- a positive force in the direction of the horse's movement. Therefore, the cart begins to accelerate and move.</span><span>


</span>
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Answer:

im pretty sure its b

Note that when the acceleration is negative — on the interval [0, 2) — that means that the velocity is decreasing. When the acceleration is positive — on the interval (2, 4] — the velocity is increasing. Speeding up and slowing down.

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A block attached to a spring with unknown spring constant oscilates with a period of 4.2 s. Parts A to D are independent questio
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Answer:

Explanation:

A )

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B )

If the mass is halved , time period becomes 1 / √2 times or .70 times or 3.0 s

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8 0
3 years ago
Read 2 more answers
Two students hear the same sound and their eardrums receive the same power from the sound wave. The sound intensity at the eardr
Margaret [11]

Answer:

a. d₁/d₂ = 1.09 b. 0.054 mW

Explanation:

a. What is the ratio of the diameter of the first student's eardrum to that of the second student?

We know since the power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²

So, I ∝ I/d²

I₁/I₂ = d₂²/d₁² where I₁ = intensity at eardrum of first student, d₁ = diameter of first student's eardrum, I₂ = intensity at eardrum of second student, d₂ = diameter of second student's eardrum.

Given that I₂ = 1.18I₁

I₂/I₁ = 1.18

Since I₁/I₂ = d₂²/d₁²

√(I₁/I₂) = d₂/d₁

d₁/d₂ = √(I₂/I₁)

d₁/d₂ = √1.18

d₁/d₂ = 1.09

So, the ratio of the diameter of the first student's eardrum to that of the second student is 1.09

b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?

We know intensity, I = P/A where P = acoustic power and A = area = πd²/4

Now, P = IA

= I₂A₂

= I₂πd₂²/4

= 1.18I₁πd₂²/4

Given that I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m

So, P = 1.18I₁πd₂²/4

= 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4

= 0.691244π × 10⁻⁴ W/4 =

2.172 × 10⁻⁴ W/4

= 0.543 × 10⁻⁴ W

= 0.0543 × 10⁻³ W

= 0.0543 mW

≅ 0.054 mW

3 0
3 years ago
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mihalych1998 [28]
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8 0
3 years ago
How much work did the movers do (horizontally) pushing a 170-kg crate 10.2 m across a rough floor without acceleration, if the e
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Answer:

Work done, W = 10195.92 Joules

Explanation:

Given that,

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Distance, d = 10.2 m

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Let W is the work done by the mover. It is given by in terms of coefficient of friction as :

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W = 10195.92 Joules

So, the work done by the mover is 10195.92 Joules. Hence, this is the required solution.

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