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Oksi-84 [34.3K]
3 years ago
15

Anyone play Among US?

Physics
2 answers:
dusya [7]3 years ago
7 0

Answer:

sure, but I dunno if your still here

ipn [44]3 years ago
5 0

Answer:

Yes

Explanation:

Its a good game.

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What is the measure of how difficult it is for current to flow in a circuit
LenaWriter [7]
The resistance of a circuit is a measure of the energy
that a current loses while flowing through the circuit. 

Ease or difficulty is not a helpful concept in the case of
an electrical circuit.
3 0
4 years ago
The energy per unit volume in an electromagnetic wave is: ________
Marianna [84]

Answer:

c.

Explanation:

Electromagnetic waves are waves that are generated when an electric field and a magnetic field vibrate together. They are usually created whenever an electric field collides with a magnetic field.

In more generic terms for an electric field, the energy per unit volume is given by:

Energy density = \dfrac{1}{2}\varepsilon _oE^2

For magnetic field:

Energy density =\dfrac{\beta^2}{2 \mu _o}

∴

For the electromagnetic wave (u):

Total energy density is:

u = \dfrac{1}{2}\varepsilon_oE^2 + \dfrac{\beta ^2}{2 \mu_o}

Due to the fact that the energy related with both fields is equivalent:

Then:

E = cB

6 0
3 years ago
How does an atom of bromine-79 become a bromide ion with a -1 charge?
Naily [24]

with a -1 iron charge it becomes an electrode

6 0
3 years ago
Read 2 more answers
An ambulance is currently traveling at 25 m/s, and is accelerating with a constant acceleration of 5 m/s^2. The ambulance is att
bazaltina [42]

Answer:

PLEASE MARK ME BRAINLIEST!!!

Explanation:

Given initial velocity of ambulance

v0=18m/s , acceleration  a=5m/s2  

To find distance,  x=?  

First we need to calculate the time for which it acquires 30\,m/s. For that use equation

v=v0+at  

30=18+5×t  

⇒t=30−185=125seconds  

Distance travelled by the ambulance

x=v0t+12at2  

x=(18×125)+12×5×(125)2  

x=43.2+2.5×5.76=43.2+14.4  

x=57.6m  

Therefore the ambulance has to travel 57.6 m to match the velocity of car.

5 0
2 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
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