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A force F is exerted at a distance D from the axis of rotation of a wrench and is exerted at an angle θ to the wrench (θ is the

Katyanochek1 [597]

Answer:

Torque = R X F = R * F sin theta

-f * 2R will exert an equal opposite torque

-f * 2 = F sin theta

f = -F sin theta / 2

4 0

3 years ago

From the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The init

Talja [164]

Answer:

$\vec{a} = -(9.8~{\rm m/s^2})\^y$

Explanation:

Regardless of the initial velocity of the pebble, the acceleration of the pebble is equal to the gravitational acceleration which is equal to 9.8 m/s2 towards downwards direction.

This can be shown by Newton's Second Law. According to the law, the net force applied on an object is equal to mass times acceleration of that object.

During the downward motion, the only force acting on the pebble is the gravitational force, hence its acceleration is equal to gravitational acceleration.

8 0

4 years ago

A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the a

finlep [7]

Solution :

Let $m_1=m_2=4$ kg

$u_1 = 5$ m/s

Let $v_1$ and $v_2$ are the speeds of the disk $m_1$ and $m_2$ after the collision.

So applying conservation of momentum in the y-direction,

$0=m_1 .v_1_y -m_2 .v_2_y$

$v_1_y = v_2_y$

$v_1 . \sin 60=v_2. \sin 30$

$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$

$v_2=1.732 \times v_1$

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

$m_1.u_1=m_1.v_1_x+m_2.v_2_x$

$u_1=v_1_x+v_2_x$

$5=v_1. \cos 60 + v_2 . \cos 30$

$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$

$v_1 = 2.50$ m/s

So, $v_2 = 1.732 \times 2.5$

= 4.33 m/s

Therefore, speed of the disk 2 after collision is 4.33 m/s

5 0

3 years ago

What term best describes the function of

Alja [10]

Lol idkddddddddddddddddd

4 0

4 years ago

Read 2 more answers

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

3 years ago