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EastWind [94]
2 years ago
11

Calculate the acceleration of a train travelling from rest to 24 m/s in 12 seconds.

Physics
1 answer:
Vilka [71]2 years ago
3 0
Answer: A 2m/s^2

Steps: Formula for acceleration. (Velocity Final - Initial Velocity) / Time

(24 - 0) / 12 = 2
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Which process is used to make a semiconductor into a transistor?
Klio2033 [76]

Answer:

B

Explanation:

Adding a dopant is correct on edge.

6 0
3 years ago
Read 2 more answers
During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot ca
Dmitriy789 [7]

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

a=\frac{v^2}{r}

r is the radius of the loop

on substituting the respective values, we get

8\times9.81=\frac{340^2}{r}

or

r = 1472.98 m

5 0
4 years ago
The question is in the picture
Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

v_0-gt = 0.

where v_0 is the initial velocity.

Solving for t we get

t = \dfrac{v_0}{g}

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time t_0 since it had reached the highest point, the ball has traveled downwards and the velocity v_f it has gained is

v_f = gt_0,

and we are told that this is twice the initial velocity v_0; therefore,

v_f = 2v_0  = gt_0

which gives

t_0 = \dfrac{2v_0}{g}.

Thus, the total time taken to reach velocity 2v_0 is

t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}

t_{tot} = \dfrac{3v_0}{g}.

This t_{tot}, we are told, is 36 seconds; therefore,

36= \dfrac{3v_0}{g},

and solving for v_0 we get:

v_0 = \dfrac{36g}{3}

v_0 = \dfrac{36s(10m/s^2)}{3}

\boxed{v_0 = 120m/s}

which from the options given is choice e.

7 0
3 years ago
What is the force acting on a 10kg object that accelerates from 5 m/s to 20 m/s in 5s?
vitfil [10]

Answer:

Option C

Explanation:

v= u + at

20 = 5 + a(5)

15= a(5)

a= 3 m/s²

Force = mass × acceleration

= 10 × 3

= 30 N

3 0
3 years ago
Can anybody help me solve this problem? Thank you so much!
ser-zykov [4K]
Please ignore my comment -- mass is not needed, here is how to solve it. pls do the math

at bottom box has only kinetic energy
ke = (1/2)mv^2
v = initial velocity
moving up until rest work done = Fs
F = kinetic fiction force = uN = umg x cos(a)
s = distance travel = h/sin(a)
h = height at top
a = slope angle
u = kinetic fiction
work = Fs = umgh x cot(a)
ke = work (use all ke to do work)
(1/2)mv^2 = umgh x cot(a)
u = (1/2)v^2 x tan (a) / gh
4 0
3 years ago
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