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siniylev [52]
2 years ago
15

What creates a magnetic field?

Physics
1 answer:
kolbaska11 [484]2 years ago
4 0

Answer:

moving electric charges

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A car has a kinetic energy of 1.9 × 10^3 joules. If the velocity of the car is decreased by half, what is its kinetic energy?
VLD [36.1K]
The initial kinetic energy of the car is
E_1 =  \frac{1}{2}mv_1^2 =  1.9 \cdot 10^3 J

Then, the velocity of the car is decreased by half: v_2 =  \frac{v_1}{2}
so, the new kinetic energy is
E_2 =  \frac{1}{2}mv_2 ^2 =  \frac{1}{2} m ( \frac{v_1}{2} )^2= \frac{1}{2}m \frac{v_1^2}{4}= \frac{E_1}{4}
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
E_2 =  \frac{1.9 \cdot 10^3 J}{4}=475 J
5 0
3 years ago
The sum of all forces acting on an object
Zina [86]

Answer:

The net force is the vector sum of all the forces that act upon an object. That is to say, the net force is the sum of all the forces, taking into account the fact that a force is a vector and two forces of equal magnitude and opposite direction will cancel each other out.

Explanation:

put in your own words PLEASE, hope it helps

4 0
3 years ago
7. A 1000 kg car is rolling down the street at 2.5 m/s. How fast would a 2500 kg car have to
babunello [35]

1 m/s

Explanation:

To solve this question we use the following formula:

momentum = mass × velocity

momentum of the first car = 1000 kg × 2.5 m/s

momentum of the second car = 2500 kg × X m/s

To bring the cars at rest the momentum of the first car have to be equal to the momentul of the second car.

momentum of the first car = momentum of the second car

1000 kg × 25 m/s = 2500 kg × X m/s

X (velocity of the second car) = (1000 × 25) / 2500 = 1 m/s

Learn more about:

momentum

brainly.com/question/13378780

#learnwithBrainly

7 0
3 years ago
When all group iia elements lose their valence electrons, the remaining electron configurations are the same as for what family
Orlov [11]
Noble Gases

I hope I helped

ΩΩΩΩΩΩΩΩΩΩ

8 0
3 years ago
Read 2 more answers
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
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