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3 years ago

Your body is receiving backround radiation all the tim. Explain

1 answer:

3 0

The uniform microwave radiation remaining from the Big Bang.

So, your body is always having background radiation and that means space!

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Hello people ~

Pepsi [2]

Answer:

Potential

Explanation:

The most accurate term is Electrostatic potential energy

It is denoted as UC

It's named like this because the force between charges or electrons is called electrostatic force .

5 0

2 years ago

Which statement best describes an isolated system? (1 point)

bonufazy [111]

Answer:

idk

Explanation:

6 0

3 years ago

Read 2 more answers

What does the first law of thermodynamics state about energy?

ruslelena [56]

The first law states that the internal energy change of that system is given by Q − W . Since added heat increases the internal energy of a system, Q is positive when added to the system and negative when removed from the system.

8 0

2 years ago

HELP PLZ

Stells [14]

Answer:

Explanation:

Use the equation

$h(t)=-16t^2+v_0t+h_0$

where h(t) is the height after a certain amount of time goes by, v0t is the initial upwards velocity, and h0 is the initial height of the projectile. For us:

h(t) = 10

v0t = 80

h0 = 3 and filling in:

$10=-16t^2+80t+3$ and get everything on one side to factor:

$0=-16t^2+80t-7$

This factors to

t = .09 sec and 4.9 sec. Let's interpret this.

The time of .09 is when the ball reached 10 feet on the way up, and

the time of 4.9 is when the ball reached 10 feet on the way back down. That's the height we need, 4.9 seconds.

5 0

2 years ago

A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current I. The current in the loop produces a magnetic field at th

rusak2 [61]

Answer:

1.63 A and in clockwise direction

Explanation:

The magnetic field due to the rectangular loop is :

$B=\frac{2 \mu_0 I}{\pi}\left(\frac{\sqrt{L^2+W^2}}{LW}\right)$

Given : W = 4.20 cm

$=4.20 \times 10^{-2} \ m$

L = 9.50 cm

$= 9.50 \times 10^{-2} \ m$

$B = 3.40 \times 10^{-5} \ T$

Rearranging the above equation, we get

$I=\frac{B \pi LW}{2 \mu_0\sqrt{L^2+W^2}}$

$I=\frac{(3.40 \times 10^{-5}) \pi(9.50 \times 10^{-2})(4.20 \times 10^{-2})}{2(4 \pi \times 10^{-7})\sqrt{(9.50 \times 10^{-2})^2+(4.20 \times 10^{-2})^2}}$

I = 1.63 A

So the magnitude of the current in the rectangular loop is 1.63 A.

And the direction of current is clockwise.

4 0

2 years ago