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blsea [12.9K]
3 years ago
5

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

Physics
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

6 0
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.

4 0
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
7 0
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
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