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3 years ago

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Gamma rays and X–rays are similar because they both have

1 answer:

VARVARA [1.3K]3 years ago

5 0

<span>High photon energy, high frequency, sub-optical wavelength</span>

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1. An electron travels 4.82 meters in 0.00360 seconds. What is its average speed?

vredina [299]

Answer:

speed =distance /time

speed =4.82/0.00360

speed =1338.8m/s

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3 years ago

The relative highness or lowness of a sound is called ______. Multiple Choice pitch timbre dynamics octave

Assoli18 [71]

Pitch is the answer…….

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2 years ago

You are working at a company that manufactures electri- cal wire. Gold is the most ductile of all metals: it can be stretched in

Alona [7]

Explanation:

We know that the relation between volume and density is as follows.

Volume = $\frac{\text{mass}}{\text{density}}$

So, V = $\frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}$

= $5.181 \times 10^{-8} m^{3}$

Now, we will calculate the area as follows.

Area = $\frac{\text{volume}}{\text{length}}$

= $\frac{5.181 \times 10^{-8} m^{3}}{2.4 \times 10^{3}}$

= $2.15 \times 10^{-11} m^{2}$

Formula to calculate the resistance is as follows.

R = $\rho \frac{l}{A}$

= $\frac{2.44 \times 10^{-8} \times 2400}{}2.15 \times 10^{-11}}$

= $2.71 \times 10^{6} ohm$

Thus, we can conclude that the resistance of given wire is $2.71 \times 10^{6} ohm$ .

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3 years ago

If a magnet repels a metal object what can you say about the pole of the magnet and the pole of the object?

sleet_krkn [62]

Poles are the same D is the right answer

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3 years ago

Read 2 more answers

The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,

kondaur [170]

Answer:

$a= -2\ m/s^2$

$a=-12.5\ m/s^2$

x=2.17 m

x=8.4 m

Explanation:

Given that

$v=A+Bt^{-1}$

$v=2+0.25t^{-1}$

To find acceleration :

we know that

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=0-0.5t^{-2}$

$a=-0.5t^{-2}$

Acceleration at t= 2 s

$a=-0.5\times 2^{-2}$

$a= -2\ m/s^2$

Acceleration at t= 5 s

$a=-0.5\times 5^{-2}$

$a=-12.5\ m/s^2$

We know that

$v=\dfrac{dx}{dt}$

$dx=\left(2+\dfrac{1}{4t}\right)dt$

Position at t= 2 s:

$\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{2}$

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

$\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{5}$

x=8+0.25 ln5

x=8.4 m

4 0

3 years ago