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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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What is the difference between a negative feedback system and a positive feedback system?

Anni [7]

Answer:

Explanation:

The negative feedback is responsible for maintaining equilibrium (stability) in a system as it lessens effects, which is contrary to positive feedback which increases reaction and moves a system further away from equilibrium (stability), It does so by amplifying the effects of a product or event and occurs when something needs to happen quickly. e.g

Insulin lowers down blood sugar levels, so when the body detects that it has too much sugar, the pancreas is prompted to release insulin and only stops when balance is achieved; hence, negative feedback.

When there is a tear on the skin, a chemical is released. This chemical causes platelets in the blood to activate, hence they release a chemical which signals more platelets to activate, until the wound is clotted, positive feedback.

7 0

3 years ago

A 3.3 kg ball sits on the ground and is kicked with a FAPP of 36N

jeyben [28]

a) 32.3 N

The force of gravity (also called weight) on an object is given by

W = mg

where

m is the mass of the object

g is the acceleration of gravity

For the ball in the problem,

m = 3.3 kg

g = 9.8 m/s^2

Substituting, we find the force of gravity on the ball:

$W=(3.3)(9.8)=32.3 N$

b) 48.3 N

The force applied

$F_{app} = 36 N$

The ball is kicked with this force, so we can assume that the kick is horizontal.

This means that the applied force and the weight are perpendicular to each other. Therefore, we can find the net force by using Pythagorean's theorem:

$F=\sqrt{W^2+F_{app}^2}$

And substituting

W = 32.3 N

Fapp = 36 N

We find

$F=\sqrt{32.3^2+36^2}=48.3 N$

c) $14.6 m/s^2$

The ball's acceleration can be found by using Newton's second law, which states that

F = ma

where

F is the net force on an object

m is its mass

a is its acceleration

For the ball in this problem,

m = 3.3 kg

F = 48.3 N

Solving the equation for a, we find

$a=\frac{F}{m}=\frac{48.3}{3.3}=14.6 m/s^2$

8 0

4 years ago

(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th

Ghella [55]

Answer:

Assuming that the vertical speed of the ball is 14 m/s we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

Explanation:

a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:

$V_{f}^{2} = V_{0}^{2} - 2gh$

Where:

$V_{f}$ : is the final speed = 14 m/s

$V_{0}$ : is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

$V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s$

b) The maximum height is:

$V_{f}^{2} = V_{0}^{2} - 2gh$

$h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m$

c) The time can be found using the following equation:

$V_{f} = V_{0} - gt$

$t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s$

d) The flight time is given by:

$t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s$

I hope it helps you!

3 0

3 years ago

1. What happens to particle spacing when temperature increase, What we call this process?

enot [183]

Answer:

Particle spacing increases and it's called evaporating

5 0

3 years ago

Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.

Korolek [52]

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

$\int\limits^{10}_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)$

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

4 0

4 years ago