Which element is likely more reactive, and why?

Find the magnitude of the average force ⟨Fx⟩⟨Fx⟩ in the x direction that the particle exerts on the right-hand wall of the conta

hodyreva [135]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

The attached file has a detailed solution of the given problem.

4 0

3 years ago

A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer to the nearest tenth.The

Galina-37 [17]

The speed of the runner is 300 m /38 seconds. You can simplify this answer to be about 7.9 m/s

7 0

3 years ago

If a cart of a roller coaster has a mass of 250kg and is at a height of 14 meters. What is the cart's potential energy?

ahrayia [7]

Answer:

3430000 J

Explanation:

The formula for potential energy is PE=mgh.

M being the mass, g being the force of gravity, and h being the height.

First thing you want to do is convert 250 kg to g (grams).

From there you get 25000g and you have to multiply that by 14m and 9.8m/s^2 (the force of gravity is constant, at least on earth).

5 0

3 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

2 years ago

Read 2 more answers

An object start from rest with a constant acceleration of 8.00 m/s^2 along straight line.

san4es73 [151]

I’ll say c Bc it make more since to find the travel distance

7 0

3 years ago