A weight lifter does 700J of work on a weight that he lifts in 3.1s. What is the power with

Which of the following is an inclined plane wrapped around a rod?

zlopas [31]

Alright well the Answer to your question is A). Screw

Hope this helps have a nice day : )

If u want i can explain why

8 0

3 years ago

Read 2 more answers

Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin

andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

$W=Fd$

Substitute the value in the above equation.

$\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}$

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

$\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}$

The expression for the normal reaction force is given as:

$N=mg\cos \theta$

The expression for the mass of the box is given as:

$\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}$

Substitute the value in the above equation.

$\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}$

Thus, the mass of the box is 12.9 kg.

7 0

1 year ago

The gravitational field at the Moon in N/kg due to the Earth is approximately (G = 6.67 × 10-11 N m2/kg2, the mass of the Earth

rosijanka [135]

Answer:

F = 2.69 10⁻³ m [ N]

Explanation:

This exercise asks to calculate the gravitational field of the Earth on the lunar surface, let's use the universal gravitation law

F = G m M / r²

where m is the mass of the body, M the mass of the Earth and r the distance between the Earth and the Moon

F = (G M / r²) m

F = (6.67 10⁻¹¹ 5.98 10²⁴ / (3.85 10⁸)² ) m

F = 2.69 10⁻³ m [ N]

This force is directed from the Moon towards the Earth, therefore it reduces the weight of the body

8 0

3 years ago

A metal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction.

kherson [118]

Answer:

a) k = 120 N / m

, b) f = 0.851 Hz

, c) v = 1,069 m / s

, d) x = 0

, e) a = 5.71 m / s²

, f) x = 0.200 m

, g) Em = 2.4 J

, h) v = -1.01 m / s

Explanation:

a) Hooke's law is

F = k x

k = F / x

k = 24.0 / 0.200

k = 120 N / m

b) the angular velocity of the simple harmonic movement is

w = √ k / m

w = √ (120 / 4.2)

w = 5,345 rad / s

Angular velocity and frequency are related.

w = 2π f

f = w / 2π

f = 5.345 / 2π

f = 0.851 Hz

c) the equation that describes the movement is

x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

x = 0.2 cos wt

Speed is

v = dx / dt

v = -A w sin wt

The speed is maximum for sin wt = ±1

v = A w

v = 0.200 5.345

v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

x = A cos wt = 0

x = 0

e) the acceleration is

a = d²x / dt² = dv / dt

a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

a = A w²

a = 0.2 5.345

a = 5.71 m / s²

f) the position for this acceleration is

x = A cos wt

x = A

x = 0.200 m

g) Mechanical energy is

Em = ½ k A²

Em = ½ 120 0.2²

Em = 2.4 J

h) the position is

x = 1/3 A

Let's calculate the time to reach this point

x = A cos wt

1/3 A = A cos 5.345t

t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

a = -1.91 m / s²

5 0

3 years ago

Josie walks 15 meters north down the freshman hallway. She then walks 25 meters east. Lastly, she walks 15 meters south. What is

hichkok12 [17]

Answer:

the answer is yours a calculated

3 0

3 years ago