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3 years ago

If the horse lifts a 50 kilogram mass a vertical distance of 3 meters in 2 seconds, what is the horse’s minimum power output?

Physics

1 answer:

yaroslaw [1]3 years ago

7 0

Answer:

735watts

Explanation:

Power = Force*distance/time

Given

Force = mg = 50 * 9.8'

Force = 490N

distance = 3m

time = 2s

Substitute

Power = 490*3/2

Power = 1470/2

Power = 735Watts

Hence the minimum power output is 735watts

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GarryVolchara [31]

Answer:

A.) Equal Forces act in equal times, so the change in momentum for both objects must be equal.

(Hope this helps! Btw, I am the first to answer.)

8 0

3 years ago

An object weigh 40N in air ,weigh 20N when submerged in water,and 30N when submerged in a liquid of unknown liquid density.what

muminat

Answer:

The density is $\rho_u =500 kg /m^3$

Explanation:

From the question we are told that

The weight in air is $W_a = 40 \ N$

The weight in water is $W_w = 20 \ N$

The weight in a unknown liquid is $W_u = 30 \ N$

Now according to Archimedes principle the weight of the object in water is mathematically represented as

$W_w = W_a -m _w g$

Where $m_w$ is he mass of the water displaced

substituting value

$m_w g = 40 -20$

$m_w g = 20 \ N --- (1)$

Now according to Archimedes principle the weight of the object in unknown is mathematically represented as

$W_u = W_a -m _u g$

Where $m_u$ is he mass of the unknown liquid displaced

substituting value

$m_u g = 40 -30$

$m_u g = 10 \ N ---(2)$

dividing equation 2 by equation 1

$\frac{m_ug}{m_wg} = \frac{10}{20}$

$\frac{m_u}{m_w} = \frac{1}{2}$

=> $m_u = 0.5 m_w$

Now since the volume of water and liquid displaced are the same then

$\rho _u = 0.5 \rho_w$

This because

$density = \frac{mass}{volume}$

So if volume is constant

mass = constant * density

Where $\rho_u$ is the density of the liquid

and $\rho_ w$ is the density of water which is a constant with a value $\rho_w = 1000 kg/m^3$

$\rho_u = 1000*0.5$

$\rho_u =500 kg /m^3$

7 0

3 years ago

A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va

makkiz [27]

Answer:

a) In a parallel plate capacitor the capacitance is

$C = \frac{Q}{V} = \epsilon_0\frac{A}{d}$

So, in order to halve the capacitance the plate separation should be twice the original distance.

b) Same question as (a).

c) According to the above equations, the charge stored in a capacitor is directly proportional to the plate separation.

$\frac{Q}{V} = \frac{A}{d}$

In order to halve the charge stored the plate separation should be half the original distance.

d) Same question as (c).

e) The energy stored in a capacitor is

$U = \frac{1}{2}CV^2 = \frac{1}{2}\epsilon_0\frac{A}{d}(Ed)^2 = \frac{1}{2}\epsilon_0AE^2d$

In order to halve the energy stored in the capacitor, the plate separation should be twice the original distance.

f) Same question as (e).

g) The energy density is given by the following equation

$u = \frac{\frac{1}{2}CV^2}{Ad} = \frac{\frac{1}{2}\epsilon_0\frac{A}{d}(E^2d^2)}{Ad} = \frac{1}{2}\epsilon_0E^2.$

As it turns out, the energy density is independent from the geometric factors. This makes sense, because the electric field in a parallel field capacitor is independent from the distance.

3 0

3 years ago

How would you find the horizontal net force for the free body diagram below

tiny-mole [99]

Answer:

Add Ff from Fa

Explanation:

Fnet = sum of all force

horizontal net force = Ff + Fa

7 0

2 years ago

A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta

larisa [96]

Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

$x = 26.16 m$

Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

$y = -75.6 m$

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

Horizontal displacement in 1.03 s

$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

$x = 83.23 m$

Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

$y = -75.6 m$

6 0

3 years ago