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olga55 [171]
3 years ago
7

Most of the energy released by nuclear fission is in the form of: alpha rays, beta rays, or gamma rays?

Physics
1 answer:
Sedbober [7]3 years ago
3 0
I think it is <span>Alpha rays.</span>
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What two sources of friction do you have to overcome when you are walking?
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Gravity is pushing you down on the earth while the friction from the ground is pushing you up. Those are the two frictions you must overcome when walking.






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3 0
4 years ago
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Which state of matter has atoms that are spread out and bouncy?
NikAS [45]
The stage where atoms are spread out and bouncy is the gas stage.

7 0
3 years ago
PHYSICS<br> I need help with number two!!!!
EleoNora [17]

Answer:

mgh₁ + ½mv₁² = mgh₂ + ½mv₂²

Explanation:

Initial total energy = final total energy

PE₁ + KE₁ = PE₂ + KE₂

mgh₁ + ½mv₁² = mgh₂ + ½mv₂²

7 0
3 years ago
The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared
velikii [3]

Answer:

\displaystyle a(5)=-32

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

\displaystyle v(t)=\frac{df}{dt}

And the acceleration is

\displaystyle a(t)=\frac{dv}{dt}

Or equivalently

\displaystyle a(t)=\frac{d^2f}{d^2t}

The given height of a projectile is

f(t)=-16t^2 +238t+3

Let's compute the speed

\displaystyle v(t)=-32t+238

And the acceleration

\displaystyle a(t)=-32

It's a constant value regardless of the time t, thus

\boxed{\displaystyle a(5)=-32}

3 0
3 years ago
A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a
UNO [17]

Answer:

P_{avg} = 6.283*10^{-9} \ W

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

P_{avg} = 6.283*10^{-9} \ W

6 0
4 years ago
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