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Irina18 [472]
1 year ago
12

A horse does 860 j of work in 420 seconds while pulling a wagon. what is the power output of the horse? round your answer to the

nearest whole number.
Physics
2 answers:
kicyunya [14]1 year ago
6 0

the correct answer is 2w

GenaCL600 [577]1 year ago
6 0

By definition, the power is given by:

P = \frac{W}{t}

Where,

W: work done

t: time spent doing work

Therefore, substituting values we have:

P = \frac{860}{420}\\P = 2.05 watts

Rounding the nearest whole we have:

P = 2 watts

Answer:

The power output of the horse is:

P = 2 watts

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A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi
tekilochka [14]
This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg. 
We know that the normal component of Fg is just Fg, which is equal to as 1110N. 
From the geometry, the normal component of Fp can be calculated: 
Fpn = Fp * cos(θp) 
= 1016.31 N * cos(53) 
= 611.63 N 

The total normal force Fn then is: 
Fn = Fg + Fpn 
= 1110 + 611.63
= 1721.63 N</span>

 

> Find the friction force on the crate

<span>We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely,  so we can say that the horizontal component is zero: 

Fh = Fph + Fgh 
= (Fp * sin(θp)) + 0 
= 1016.31 N * sin(53) 
= 811.66 N</span>

 

> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

 

μs = Fh / Fn

= 811.66 N / 1721.63 N

<span>= 0.47</span>

8 0
2 years ago
Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur
d1i1m1o1n [39]

Answer:

The speed of bullet and wooden bock coupled together, V = 22.22 m/s

Explanation:

Given that,

Mass of the bullet, m = 0.04 Kg

Mass of the wooden block, M = 0.5 Kg

The initial velocity of the bullet, u = 300 m/s

The initial velocity of the wooden block, U = 0 m/s

The final velocity of the bullet and wooden bock coupled together, V = 0 m/s

According to the conservation of linear momentum, the total momentum of the body after impact is equal to the total momentum before impact.

Therefore,

                              mV + MV = mu + MU

                               V(m+M) = mu

                                 V = mu/(m+M)

Substituting the values in the above equation,

                                V = 0.04 Kg x 300 m/s  / (0.04 Kg+ 0.5 Kg)

                                    = 22.22 m/s

Hence, the speed of bullet and wooden bock coupled together, V = 22.22 m/s

8 0
1 year ago
The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls
kifflom [539]
Impulse = Integral of F(t) dt from 0.012s to 0.062 s

Given that you do not know the function F(t) you have to make an approximation.

The integral is the area under the curve.

The problem suggest you to approximate the area to a triangle.

In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s

The height is the peak force: 35 N.

Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s

Answer> 0.875 N*s
6 0
2 years ago
Consider a ball rolling around in a circular path on the inner surface of a cone. the weight of the ball is shown by the vector
zimovet [89]

Answer:

option 2 is righttttt.................

5 0
1 year ago
Can someone help with my physics homework? please
Murrr4er [49]

Answer:

a) 19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) 24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

Explanation:

a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:

W = F\cdot \Delta s (1)

Where:

F - Force applied by the cart, measured in newtons.

\Delta s - Distance travelled by the car, measured in meters.

W - Work done on the cart, measured in joules.

If we know that F = 888\,N and \Delta s = 22\,m, then the work done on the cart is:

W =(888\,N)\cdot (22\,m)

W = 19536\,J

19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) If we know that F = 1111\,N and \Delta s = 22\,m , then the work on the cart is:

W = (1111\,N)\cdot (22\,m)

W = 24442\,N

24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

3 0
1 year ago
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