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3 years ago

12

Which of the following did you include in your response? Check all that apply. Absolute strength measures maximum force. Dynamic

strength measures repeated motions Elastic strength measures exertion of force rapidly Strength endurance measures fatigue resistance

2 answers:

alexandr1967 [171]3 years ago

4 0

Answer:

d

Explanation:

Alexeev081 [22]3 years ago

3 0

Answer:

any answer would be correct. its a choose your own answer question lol

Explanation:

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Students are asked to design an experiment about Newton’s 2nd Law. One student decides to roll a marble down a ramp into a pile

sveta [45]

Answer:

A

Explanation:

Trust me I just took it !

8 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

3 years ago

Read 2 more answers

Please help I have no idea how to do this

rewona [7]

Answer:

So an object with mass is attracted to another object with mass, and the gravitational force is directly proportional to the masses of the two objects, and inversely proportional to the <em>square</em> of the distance between the two objects.

If distance were to increase, than the gravitational force would decrease. If mass were to increase, so would the gravitational force.

Explanation:

5 0

3 years ago

Read 2 more answers

A hiker throws a ball at an angle of 21.0° above the horizontal from a hill 21.0 m high. The hiker’s height is 1.750 m. The magn

nydimaria [60]

Answer:

38.8 m

Explanation:

h = -(21 + 1.75) = - 22.75 m

g = - 9.8 m /s^2

Ux = 14.004 m/s

Uy = + 5.376 m/s

Let the ball hits the ground in time t and at a distance d from the base of hill.

Use second equation of motion

h = Uyt + 0.5 at^2

- 22.75 = 5.376 t - 0.5 x 9.8 t^2

4.9 t^2 - 5.376 t - 22.75 = 0

$t=\frac{5.376\pm \sqrt{5.376^{2}+4\times 4.9\times 22.75}}{9.8}$

By solving

t = 2.77 second

So, horizontal distance

d = Ux t

d = 14.004 x 2.77 = 38.8 m

6 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge −8.00 ✕ 10⁻⁷ C at x = 6.00 cm, and particle 2 of charge +8.00 ✕ 10⁻⁷

victus00 [196]

Answer:

0 N/C

Explanation:

Parameters given:

$q_1 = -8.00 * 10^{-7} C$

$x_1 = 6.00 cm$

$q_2 = +8.00 * 10^{-7} C$

$x_2 = 21 cm$

The distance between $q_1$ and $q_2$ is

$21 - 6 = 15cm$

Electric field is given as

$E = \frac{kq}{r^2}$

r = 15/2 = 7.5cm = 0.075m

The electric field at their midpoint due to $q_1$ is:

$E = \frac{9 * 10^9 * -8.0 * 10^{-7}}{0.075^2}$

$E_1$ = $-1.28 * 10^6 N/C$

The electric field at the midpoint due to $q_2$ is:

$E = \frac{9 * 10^9 * 8.0 * 10^{-7}}{0.075^2}$

$E_2$ = $1.28 * 10^6 N/C$

The net electric field will be:

E = $E_1$ + $E_2$

E = $-1.28 * 10^6$ + $1.28 * 10^6$

E = 0 N/C

7 0

4 years ago