Physics, 02.08.2019 00:00 baegel

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?

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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):
$P=\frac{W}{t}$
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.

Answer from: noahrdavila1978

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

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

Answer from: Quest

To use anything transparent

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Imagine a horse pulling a cart. what would happen to the speed of the cart if several bags of cement...