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ch4aika [34]
3 years ago
5

Where does a magnetic field occur in relation to an electrified wire?

Physics
1 answer:
bekas [8.4K]3 years ago
7 0

Answer:

Around the entire length of the wire

Explanation:

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15. A car travelling towards the right has a mass of 1332 kg and has a speed of 25 m/s. A truck is
swat32

Explanation:

Given that,

The mass of a car, m₁ = 1332 kg

The speed of the car, u₁ = 25 m/s (right)

The mass of a truck, m₂ = 3000 kg

The speed of the truck, u₂ = -15 m/s

The total momentum after the crash is given by :

p=m₁u₁ + m₂u₂

Put all the values,

P = 1332(25) + 3000(-15)

= −11700 kg-m/s

So, the total momentum after the crash is equal to 11700 kg-m/s and it is in the left direction.

5 0
2 years ago
(Brainliest) How are radio waves used in cell phone wireless communication technology?
poizon [28]

Radio waves are used in cell phone wireless communication by this: Radio waves are released from the cell phone and travel to satellites in the stratosphere. After they reach the satellite, they are redirected to the recipient of the call/text.

Hope this helps! Can I have brainliest please?

5 0
3 years ago
Match the events related to the formation of the universe with the stages during which they occurred.
igomit [66]

Answer:

does this even make sense

Explanation:

8 0
2 years ago
An athlete completes 1 laps around a track with a radius of 25 meters in 180 seconds. What is the magnitude of the athlete's tan
DanielleElmas [232]

Answer:

0.872<em>m/s</em>

Explanation:

Tangential velocity is given by the formula,

v= 2\pi r/ t

In the question given,

radius= 25meters

time= 180secs

pie= 3.14

number of laps= 1

The magnitude of tangential velocity equals;

\frac{1lap* 2 *3.14 *25m}{180secs}

<em>v </em>= 157<em>m</em>/180<em>secs</em>

Therefore, the magnitude of the tangential velocity

=0.872<em>m/secs</em>

5 0
2 years ago
Pleaseeee Please help, I will love you forever and ever
matrenka [14]

Answer:

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J  = - 63000 J

Cp = 4184 J / kg°C

Formula

                        Q = mCp(T2 - T1)

                         T2 = T1 + Q/mCp    

Substitution

                       T2 = 35 - 63000/(0.5 x 4184)

                        T2 = 35 - 63000/2092

                        T2 = 35 - 30.1

                         T2 = 4.9 °C

6 0
2 years ago
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