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zaharov [31]
3 years ago
14

You are standing on a moving bus, facing forward, and you suddenly fall forward. You can imply form this that the bus’s

Physics
1 answer:
aleksley [76]3 years ago
8 0
I think has stopped maybe
You might be interested in
What is the difference in KE between a 52.5 kg person running 3.50 m/s and a 0.0200 kg bullet flying 450 m/s?
rusak2 [61]

Answer:

Ek = 1705.28 [J]

Explanation:

In order to solve this problem, we must remember that kinetic energy can be calculated by means of the following equation.

E_{k}=\frac{1}{2} *m*v^{2}

where:

m = mass [kg]

v = velocity [m/s]

Ek = kinetic energy [J] (Units of Joules)

<u>For the person running</u>

<u />E_{k} =\frac{1}{2}*52.2*(3.5)^{2} \\ E_{k} =319.72[J]<u />

<u />

<u>For the bullet</u>

<u />E_{k} =\frac{1}{2} *m*v^{2}<u />

<u />E_{k} =\frac{1}{2} *0.02*(450)^{2} \\E_{k}=2025 [J]<u />

<u />

The difference in Kinetic energy is equal to:

Ek = 2025 - 319.72

Ek = 1705.28 [J]

8 0
2 years ago
PLEASE HELP ASAP!!
kirza4 [7]

Answer:

A. Distance over which the force is applied

Explanation:

As we know that in pulley system the mass of the car is balanced by the tension in the string

so here we will have

T = r \times F

so here in order to decrease the force needed to lift the car we have to increase Distance over which the force is applied

So here if we increase the distance over which force is applied then it will reduce the effort applied by us in this pulley system as the torque will be more if the distance is more.

7 0
3 years ago
An Olympic track runner starts from rest and has an acceleration of 2.4 m/s2 for 3.6 s, then has zero acceleration for the remai
rjkz [21]

Answer:

The runner's speed at the following times would remain 8.64 m/s.

Explanation:

Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.

In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.

First equation of motion:        Vf = Vi + a×t

Vf stands for final velocity

Vi stands for initial velocity

a stands for acceleration

t stands for time

In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.

The acceleration (a) is given as 2.4 m/s²

The time (t) is given as 3.6 s

Now put the values of Vi, a and t in first equation of motion

                       Vf = Vi + a×t

                       Vf = 0 + 2.4×3.6

                       Vf = 2.4×3.6

                       Vf = 8.64 m/s

So,the runner's speed at the following times would remain 8.64 m/s.

5 0
3 years ago
You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through
Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 =  \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

6 0
3 years ago
A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i
Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

v^2=u^2+2as

0^2=14.5^2+2\times a\times 18.25

a=-5.76m/sec^2

We know that acceleration is given by

a=\mu g

5.76=\mu\times  9.81

\mu =0.587

So coefficient of friction will be 0.587

6 0
3 years ago
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