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3 years ago

A train travels 2km in 100 seconds. find the velocity of the train

Physics

1 answer:

AVprozaik [17]3 years ago

3 0

Answer:

20m/s = 72km/h

Explanation:

change 2km=2000m

The speed of the train is

v.t=S <=> v=S/t=2000/100=20 m/s

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What is the average speed of an object with a total distance traveled of 45 meters in 50 seconds

Anettt [7]

Answer: 0.9m/s

Explanation:

Use equation for speed:

V=S/t

S-distance; S=45m

t-time; t=50 s

V=S/t

V=45m/50s

V=0.9 m/s

7 0

4 years ago

What objects do balanced forces act on?

leva [86]

Answer: Stationary or constant velocity

Explanation:

Objects with balanced forces acting on them experience no change in motion, or no acceleration. So these objects could either be stationary at rest or have a constant velocity. These include a hanging object, a floating object, an object on a table that doesn't move, and a car moving at a constant 10 mph

4 0

3 years ago

To understand the formula representing a traveling electromagnetic wave.

Mumz [18]

Answer:

Explanation:

Time period is the reciprocal of frequency

T = 1/F

F = 1/T

but angular frequency w = 2πF

F = w/2π

The detailed steps is as shown in the attached file

8 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

When an object like a tree is illuminated by the sun, and you are looking toward the tree, light rays leave the object _________

Dima020 [189]

Objects absorb and reflect light differently depending on their physical characteristics, such as their shape or composition. Thanks to the reflection we can see the objects. Reflection can be defined as the change of direction of a wave, which, when in contact with the separation surface between two changing means, returns to the point where it originated. When the light illuminates the object, such as the tree, the rays of light will disperse in all directions allowing observation.

The correct answer is A. From every point on the surface of the tree, and in every direction

6 0

4 years ago