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AfilCa [17]
3 years ago
12

SI unit for electrical current

Physics
1 answer:
Mashcka [7]3 years ago
5 0
I believe it is called an ampere.
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The answer is D plasma
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In a transfer of energy, there is no change in the energy's, what?
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I think it is there is no change in the energy's sunlight

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Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts
Nata [24]

(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.

<h3>Net force on the third child</h3>

Apply Newton's second law of motion;

∑F = ma

where;

  • ∑F is net force
  • m is mass of the third child
  • a is acceleration of the third child

∑F = 96 N - 75 N - 12 N = 9 N

Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;

  • the wagon
  • the children outside the wagon

<h3>Free body diagram</h3>

           →                 →              Ф                         ←

         1st child      friction       wagon                2nd child

<h3>Acceleration of the  child and wagon system</h3>

a = ∑F/m

a = 9 N / 27 kg

a = 0.33 m/s²

<h3>When the frictional force is 21 N</h3>

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

a = 0 m/s²

Learn more about net force here: brainly.com/question/14361879

#SPJ1

7 0
2 years ago
An electromagnetic wave has a frequency of 4.0 x 10^18 Hz. What is the wavelength of the wave?
LenaWriter [7]

Answer:

7.5 × 10^-11 m

Explanation:

Hope this helps !

3 0
3 years ago
Read 2 more answers
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
2 years ago
Read 2 more answers
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