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tester [92]
3 years ago
9

imagine a horse pulling a cart. What would happen to the speed of the cart if several bags of cement were adding to the cart?

Physics
2 answers:
iogann1982 [59]3 years ago
7 0
Assuming the power delivered by the horse does not change, the speed of the cart will decrease.
In fact, the power delivered by the horse is the work done by the horse (W) per unit time (t):
P=\frac{W}{t}
<span>If several bags are added to the cart, the horse must do more work to transport them. Therefore, W in the fraction increases. But if the power P of the horse is constant, then it means that the time t must increase as well. So, the horse will take more time to transport the car, and this means that the speed of the cart has decreased.</span>
boyakko [2]3 years ago
4 0

Answer:

Horse will become slower as we add more number of bags on the trolley

Explanation:

If a horse is pulling the cart then here horse must have to apply sufficient force on the cart so that the applied force is just counterbalancing the friction force on cart.

Now here we know that friction force is given as

F_f = \mu mg

where

m = total mass of the cart

Now if we assume here that horse will is giving constant power to the cart

so we will have

P = F.v

Power is given as product of applied force by horse on cart and its velocity

now as we know that if we put more bags on the cart then it will increase the weight of cart.

Due to increase in weight of cart there will be more friction force on it so horse must have to apply more force to counterbalance the friction force.

Now in order to maintain constant power velocity of horse must have decrease.

So horse will become slower as we add more number of bags on the trolley

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Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3
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Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

J = - D \frac{\Delta c}{\Delta x}

Where \frac{\Delta c}{\Delta x} is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

D= D_0 e^{\frac{E_d}{RT}}

Where D_0 is the proportionality constant,

R is the gas constant, T the temperature and E_d is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}

Rearrange the equation to get the value of temperature,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

We have all the values in our equation.

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\Delta x = 9.7*10^{-3}m

E_d = 82000J

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Substituting,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

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T=1118.07K=844\°C

4 0
4 years ago
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