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2 years ago

When he drove home his car broke down it wasnt due to a problem with the wheels it was the?

Physics

2 answers:

Goshia [24]2 years ago

3 0

It would have to be the engine

Nikitich [7]2 years ago

3 0

Engine. As that is what powers the car.

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A spherically spreading EM wave comes from an 1800-W source. At a distance of 5.0 m, what is the intensity, and what is the rms

Aleonysh [2.5K]

Explanation:

It is given that,

Power of EM waves, P = 1800 W

We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.

Intensity,

$I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2$

The formula that is used to find the rms value of the electric field is as follows :

$I=\epsilon_o cE^2_{rms}$

c is speed of light and $\epsilon_o$ is permittivity of free space

So,

$E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m$

Hence, this is the required solution.

4 0

2 years ago

A robot standing on a cliff shoots a ball upwards with an initial speed of 30 m/s. What is the height of the cliff given that th

valina [46]

Answer:

C 80 m

Explanation:

Given:

v₀ = 30 m/s

a = -10 m/s²

t = 8 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²

Δy = -80 m

The ball lands 80 m below where it started. So the height of the cliff is 80 m.

5 0

2 years ago

I need help on these 3 questions plz 20 points

Dmitry_Shevchenko [17]

Answer: the first one is speed

the second one is acceleration

the last is velocity

Explanation:

i got you

4 0

2 years ago

Weather stays the same all the time

VikaD [51]

False. Weather has periods of change, imagine a rainy day and a sunny day, it is a change in weather

7 0

2 years ago

A bugle can be thought of as an open-end pipe. If a bugle were straightened out, it would be 2.65 m long. a. If the speed of sou

vitfil [10]

Answer

given,

Length of pipe, L = 2.65 m

speed of sound, v = 343 m/s

Pipe is open end Pipe

a) Lowest frequency

condition, for open end pipe

$\lambda_1 = 2 L$

$\lambda_1 = 2\times 2.65$

$\lambda_1= 5.30 m$

For lowest frequency

$f_1 = n\dfrac{v}{\lambda_1}$

n = 1

$f_1 = \dfrac{343}{5.30}$

f₁ = 64.72 Hz

b) For second frequency in bugle

n = 2

$f_2 =2\times \dfrac{343}{5.30}$

f₂ = 129.43 Hz

for n = 3

$f_3 =3\times \dfrac{343}{5.30}$

f₃ = 194.16 Hz

5 0

3 years ago