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spin [16.1K]
3 years ago
14

A shorter electromagnetic wave is _____. more powerful less powerful hotter colder

Physics
2 answers:
zhannawk [14.2K]3 years ago
5 0
A shorter electromagnetic wave is hotter.
A shorter electromagnetic wave produce heat hotter than ultraviolet rays. Because it produces both gamma rays and ultraviolet rays that makes it hotter that the heat of the sun.
Keith_Richards [23]3 years ago
3 0

the answer is more powerful or A

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1) The smallest part of an element that behaves like the element is the atom.
Mandarinka [93]

Here are the answers: 

1. False - Molecules is the smallest part of an element that behaves like the element. 

2. False - The nucleus contains both protons and neutrons

3. True

4. True

5. A. Nucleus 

6. D. Neutron

7. B. Protons and Neutrons

8. C. Electron

9. C. 6

10. C.6

5 0
3 years ago
What is the charge on 1.0 kg of protons? (e = 1.60 × 10-19 c, mproton = 1.67 × 10-27 kg)?
Andru [333]
First, we need to find the number of protons, which is the total mass divided by the mass of one proton:
N= \frac{m}{m_p}= \frac{1.0 kg}{1.67 \cdot 10^{-27} kg}=6.0 \cdot 10^{26} protons

Then, the total charge is the number of protons times the charge of a  single proton:
Q=Ne = 6.0 \cdot 10^{26}\cdot 1.60 \cdot 10^{-19} C=9.6 \cdot 10^7 C
8 0
3 years ago
Read 2 more answers
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
A student is transmitting sound waves through various materials. Through which metal in the table will the sound waves travel th
MariettaO [177]

Answer:

Aluminum

Explanation:

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5 0
2 years ago
Read 2 more answers
By what factor must the amplitude of a sound wave be decreased in order to decrease the intensity by a factor of 3?
s344n2d4d5 [400]

Answer:

Amplitude is decreased by a factor of \sqrt3 if intensity is decreased by a factor of 3.

Explanation:

Intensity of a sound wave is directly proportional to the square of its amplitude.

Therefore, if intensity is I and amplitude is A, then

I=kA^2, where, k is constant of proportionality.

Now, if intensity of sound wave is decreased by a factor of 3. So,

New intensity is, I_{new}=\frac{I}{3}

I_{new}=kA_{new}^2\\\frac{I}{3}=kA_{new}^2

Plug in kA^2 for I. This gives,

\frac{kA^2}{3}=kA_{new}^2\\A_{new}^2=\frac{A^2}{3}\\A_{new}=\sqrt{\frac{A^2}{3}}=\frac{A}{\sqrt{3}}

Therefore, amplitude is decreased by a factor of \sqrt3.

4 0
3 years ago
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