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allsm [11]
3 years ago
15

A car travels 45 km due north and 70 km west. What is the car's displacement? 6 points 24.7 km northeast 83.2 km northwest 76.5

km northwest 115 km north
Physics
1 answer:
HACTEHA [7]3 years ago
8 0

Answer:

3150

Explanation:

if if you were two times 45 times 70 it would give you that answer

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Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth
Semmy [17]

Answer:

\lambda=6.83\times 10^{-5}\ m

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m

So, the wavelength of the infrared radiation is 6.83\times 10^{-5}\ m.

5 0
3 years ago
Three forces of magnitude 10N, 5N and 4N act on an object in the directions North, West and East respectively. Find the magnitud
Gekata [30.6K]

Answer:

19N to the south

Explanation:

F =10N + 5N + 4N

8 0
3 years ago
What is the correct order of the structures of the motor neuron?
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Dendrites, the cell body, axon, terminal branches of the axon
7 0
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Write<br>any<br>to increase<br>magnitude of current in dynamo​
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4 0
3 years ago
The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan
ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

I = \frac{P}{A}

The area of a sphere is given by

A = 4\pi r^2

So replacing we have to

I = \frac{P}{4\pi r^2}

Since the question tells us to find the proportion when

r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}

So considering the two intensities we have to

I_1 = \frac{P_1}{4\pi r_1^2}

I_2 = \frac{P_2}{4\pi r_2^2}

The ratio between the two intensities would be

\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}

The power does not change therefore it remains constant, which allows summarizing the expression to

\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2

Re-arrange to find I_2

I_2 = I_1 (\frac{r_1}{r_2})^2

I_2 = 100*(\frac{1}{5})^2

I_2 = 4W/m^2

Therefore the intensity at five times this distance from the source is 4W/m^2

3 0
3 years ago
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