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3 years ago

Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speed up

Physics

1 answer:

Amiraneli [1.4K]3 years ago

6 0

Answer: 5 m/s^2

Explanation: You have to use a kinematics equation that Includes what you are given. So in this case you have Initial velocity, final velocity, and time. You’re looking for acceleration so you have everything you need to use an equation, the one without distance.

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Question 14 (2 points)

sveticcg [70]

Answer:

after 2 seconds its velocity is -20 m/s. after 3 seconds its velocity is -30 m/s. after 10 seconds its velocity is -100 m/s.

Explanation:

This is my answer.

7 0

2 years ago

An airplane has a starting velocity of 300m/s. It then accelerates at a rate of 45m/s2 for a time of 10s. What is it's final vel

Olenka [21]

A = (v - u) / t

a = acceleration
v = final velocity
u = initial velocity
t = time

45 = (v - 300) / 10

45 × 10 = v - 300

450 + 300 = v

v = 750 m/s

Hope this helps!

P.S. Let me know if you need an explanation

8 0

2 years ago

What is the average (mean) of these numbers: 13,43, 12, 3,66

algol13

Answer:

27.4

Explanation:

(13+43+12+3+66)/5

137/5

27.4

3 0

3 years ago

Read 2 more answers

What is the resistance at 20°C of a 2.0-meter length of tungsten wire with a cross-sectional area of 7.9 10^-7

Bad White [126]

Answer:

1.4 * 10 ^-1 Ω

Explanation:

Hi,

For this question, we gotta use the formula

R = pL/A

p = The resistivity of your material at 20°C

L = length of the wire

A = cross-sectional area

The resistivity of tungsten is 5.60 * 10^-8 at 20°C

By plugging the values, we get:

R = (5.60 * 10^-8)(2.0)/(7.9*10^-7) = 1.4 * 10 ^-1 Ω

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