A satellite with a mass of 5.60 E5 kg is orbiting the Earth in a circular path at 7.037E 6 meters from

Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a

Viktor [21]

Answer:

Power, P = 924.15 watts

Explanation:

Given that,

Length of the ramp, l = 12 m

Mass of the person, m = 55.8 kg

Angle between the inclined plane and the horizontal, $\theta=25^{\circ}$

Time, t = 3 s

Let h is the height of the hill from the horizontal,

$h=l\ sin\theta$

$h=12\times \ sin(25)$

h = 5.07 m

Let P is the power output necessary for a person to run up long hill side as :

$P=\dfrac{E}{t}$

$P=\dfrac{mgh}{t}$

$P=\dfrac{55.8\times 9.8\times 5.07}{3}$

P = 924.15 watts

So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.

3 0

2 years ago

An object is observed for a time interval of 20 seconds. From time 6.7 s the object experiences a Force of 106 N that lasts unti

shusha [124]

Answer:

1654 kg m/s

Explanation:

The impulse experienced by an object is equal to the product between the force exerted on the object and the time during which the force lasts:

$I=F\Delta t$

where:

I is the impulse

F is the force exerted on the object

$\Delta t$ is the time during which the force is applied

For the object in this problem, we have

$F=106 N$ (force applied)

$\Delta t= 15.6 s$ (time interval)

Therefore, the impulse experienced by the object is:

$I=(106)(15.6)=1654 kg m/s$

3 0

3 years ago

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$

The locations are +5.9 cm or -5.9 cm

4 0

2 years ago

1. Identify the significance of the following terms to the subject of the Civil War.

DanielleElmas [232]

What following terms I don’t see any please make your question clear

4 0

3 years ago

Latent heat of fusion definition and explanation

pogonyaev

Answer:

The enthalpy of fusion of a substance, also known as heat of fusion, is the change in its enthalpy resulting from providing energy, typically heat, to a specific quantity of the substance to change its state from a solid to a liquid, at constant pressure.

5 0

3 years ago