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Bingel [31]
3 years ago
7

An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom o

f the ramp slows the object until it comes to a stop, how much work must the factory worker have done
Physics
1 answer:
ruslelena [56]3 years ago
8 0

Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

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The power in an electrical circuit is given by the equation P= RR, where /is
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Answer:D

Explanation:Electric power=I*I*R

=12*12*100

=14400watts

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3 years ago
A roller coaster cart of mass m = 223 kg starts stationary at point A, where h1 = 26.8 m and a while later is at B, were h2 = 14
Tresset [83]

Answer:

vB = 15.4 m/s

Explanation:

Principle of conservation of energy:

Because there is no friction the mechanical energy is conserve

ΔE = 0

ΔE : mechanical energy change (J)

K : Kinetic energy (J)

U: Potential energy (J)

K = (1/2)mv²

U = m*g*h

Where :

m: mass (kg)

v : speed (m/s)

h : hight (m)

Ef - Ei = 0

(K+U)final - (K+U)initial =0

(K+U)final = (K+U)initial

((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:

((1/2)vB² + g*hB = (1/2 )vA²+ g*hA

(1/2) (vB)² + (9.8)*(14.7) =  0 + (9.8)(26.8 )

(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)

(vB)² = (2)(9.8)(26.8 - 14.7)

(vB)² = 237.16

v_{B} = \sqrt{237.16}

vB = 15.4 m/s : speed of the cart at B

4 0
3 years ago
How is the surface area of an average person is 2m^2 in the chapter pressure
11Alexandr11 [23.1K]

Answer:

We have to show surface area =2m^2,with few conditions that is by considering Force =200000\ N and Pressure =100000\ Pa to be respectively.

Explanation:

The atmospheric pressure is =10^{5}\ Pa on Earth's surface.

The magnitude of the force exerted on a person by the atmosphere is =2\times 10^{5}\N (or)\ 200kN\ (or)\ 20\ ton.

Now to calculate surface area we can find it from pressure=\frac{force}{area} and re-arranging it to.

area=\frac{force}{pressure}

So plugging the values,

Surface area =\frac{20000}{10000}=2\ m^{2}

Hence from the above calculations we can say that surface area is 2m^2.

So the surface area of an average person can be said to have 2m^2, using the concept of pressure and force.

3 0
3 years ago
Read 2 more answers
A woman climbs up a ladder in 1.37 s at 2.20 m/s. How tall is the ladder?
ArbitrLikvidat [17]

Answer:

The ladder is 3.014 m tall.

Explanation:

To solve this problem, we must use the following formula:

v = x/t

where v represents the woman’s velocity, x represents the distance she climbed (the height of the ladder), and t represents the time it took her to move this distance

If we plug in the values we are given for the problem, we get:

v = x/t

2.20 = x/1.37

To solve this equation for x (the height of the ladder), we must multiply both sides by 1.37. If we do this, we get:

x = (2.20 * 1.37)

x = 3.014 m

Therefore, the ladder is 3.014 m tall.

Hope this helps!

6 0
3 years ago
Read 2 more answers
A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang
olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: s_c = \omega_ct = 0.233t

For the horse: s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

s_c = s_h

0.233t = 1.57 + 0.0068t^2

0.0068t^2 - 0.233t + 1.57 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}

t= \frac{0.233\pm0.11}{0.0136}

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

6 0
3 years ago
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