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Over [174]
2 years ago
15

Which of the following is not an electromagnetic wave

Physics
1 answer:
SVETLANKA909090 [29]2 years ago
6 0

Answer:

ocean wave

Explanation:

ocean waves are not a wavelength but produced from the pull of the moons gravitational pull

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The atomic bomb dropped on Hiroshima converted about 7.00x10-4kg of mass to energy. How much energy did that bomb produce?
strojnjashka [21]

Answer:

\sf \: given  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \\  \bf \: mass \:  \:  \: m \:  = 7.00 \times  {10}^{ - 4}  \: kg \\  \\  \bf \: E=mc^2 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\ \\   =  > E=7.00 \times  {10}^{ - 4}  \times  ({3 \times  {10}^{8} })^{2}  \\  \\  =  > \green{ \boxed{ E = 6.3 \times  {10}^{13}  \: J}}

6 0
2 years ago
Difference between acceleration and decceleration​
mario62 [17]

acceleration is considered to describe an increase or positive change of speed or velocity But deceleation is considered to describe a decrease or negative change of speed or velocity

8 0
3 years ago
Find a numerical value for ρearth, the average density of the earth in kilograms per cubic meter. use 6378km for the radius of t
Andre45 [30]

Answer:

5501 kg/m^3

Explanation:

The value of g at the Earth's surface is

g=\frac{GM}{R^2}=9.70 m/s^2

where G is the gravitational constant

M is the Earth's mass

R=6378km = 6.378 \cdot 10^6 m is the Earth's radius

Solving the formula for M, we find the value of the Earth's mass:

M=\frac{gR^2}{G}=\frac{(9.81 m/s^2)(6.378\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=5.98\cdot 10^{24}kg

The Earth's volume is (approximating the Earth to a perfect sphere)

V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6.378\cdot 10^6 m)^3=1.087\cdot 10^{21} m^3

So, the average density of the Earth is

\rho = \frac{M}{V}=\frac{5.98\cdot 10^{24} kg}{1.087\cdot 10^{21} m^3}=5501 kg/m^3

4 0
3 years ago
Read 2 more answers
A 0.22-caliber handgun fires a 28-g bullet at a velocity of 765 m/s. Calculate the de Broglie wavelength of the bullet.
ser-zykov [4K]

Answer:

de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m

Explanation:

Given;

mass of bullet, m = 28 g = 0.028 kg

velocity of the bullet, v = 765 m/s

de Broglie wavelength of the bullet is given by;

λ = h / mv

where;

λ is de Broglie wavelength of the bullet

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

λ = h / mv

λ = (6.626 x 10⁻³⁴ ) / (0.028 x 765)

λ = 3.093 x 10⁻³⁵ m

Therefore, de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m

8 0
2 years ago
A 15.0 kg cart is moving with a velocity of 7.50 m/s down a level hallway. A constant force of 10.0 N acts on the cart, and its
mezya [45]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below are the answers:

a. <span>DE=421.875-76.8=-345.075 joules (negitive sign means that the energy is transfering out of your system ie slowing down) 

b. </span><span>work=DE (work done is change in energy)=-345.075 joules 

c. </span><span>W=f*d </span>

<span>d=w/f </span>

<span>d=345.075/10 </span>

<span>d=34.5075 meters</span>
3 0
3 years ago
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