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3 years ago

A 50N girl pushes a 10,000 N car with force of 200N. What is the force the car pushes back at the girl? *

Physics

1 answer:

miss Akunina [59]3 years ago

8 0

Answer:

200N

Explanation:

EVERY FORCE IS OPPOSED BY AN EQUAL FORCE, REGARDLESS OF THE WEIGHT OF THE OBJECTS APPLYING THE FORCE.

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The primary coil of an ideal transformer has 100 turns and its secondary coil has 400 turns. if the ac voltage applied to the pr

Nana76 [90]

The formula used in calculations relating to transformers is:

<span>Secondary voltage (Vs)/ Primary voItage (VP) = Secondary turns (nS)/ Primary turns (nP)</span>

Substituting the given values to find for Vs,

Vs / 120 V = 400 turns / 100 turns

6 0

4 years ago

What does the force of friction between two objects in context depend on?

xenn [34]

The amount of friction depends on the force pushing the surfaces together. If this force increases, the hills and valleys of the surfaces can come into closer contact. The close contact increases the friction between the surfaces.

7 0

3 years ago

A particle is moving with (SHM) of period 8.0s and amplitude5.0m

nadezda [96]

Answer:

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$

Explanation:

Given the description of period and amplitude, the SHM could be described by:

$f(x)=5\,sin(\frac{\pi}{4}x)$

and its angular velocity can be calculated doing the derivative:

$f(x)=5\, \,sin(\frac{\pi}{4}x)\\f'(x)=5\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

And therefore, the tangential velocity is calculated by multiplying this expression times the radius of the movement (3 m):

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$ and is given in m/s.

Then the maximum speed is obtained when the cosine function becomes "1", and that gives:

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

The acceleration is found from the derivative of the velocity expression, and therefore given by:

$acceleraton(x)=-15\,\frac{\pi^2}{16}\,sin(\frac{\pi}{4}x)$

and the maximum of the function will be obtained when the sine expression becomes "-1", which will render:

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$

6 0

3 years ago

A 0.23-f capacitor is desired. What area must the plates have if they are to be separated by a 3.8-mm air gap?

tia_tia [17]

The area of the plates must have is(A)= 9.91×10⁷ m²

<h3 /><h3>How can we calculate the value of a area of a capacitor?</h3>

To calculate the the value of a area of the plates of a capacitor, we are using the formula,

C= $\frac{\epsilon_0 A}{d}$

Or, A= $\frac{C\times d}{\epsilon_0}$

Here we are given,

C= The desired capacitance of a capacitor.

= 0.23F

d=distance of separation between the plates.

=3.8mm= 0.0038m.

$\epsilon_0$ = permittivity of the vacuum.

=8.854×10⁻¹²F/m

We have to calculate the area of the plates must have = A m².

Now we put the known values in the above equation, we can get

A= $\frac{C\times d}{\epsilon_0}$

Or, A= $\frac{0.23\times 0.0038}{8.854\times 10^{-12}}$

Or, A= 9.91×10⁷ m²

From the above calculation, we can conclude that the area of the plates must have is(A)= 9.91×10⁷m²

Learn more about Capacitor:

brainly.com/question/13578522

#SPJ4

5 0

2 years ago

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.12 m whose potential is 230

slamgirl [31]

Answer:

(a) q=3.07 nC

(b) σ=17 nC/m²

Explanation:

Given data

Radius r=0.12m

Potential V=230 V

To find

(a) Charge q

(b) Charge density σ

Solution

For Part (a)

As we know that potential is:

$V_{potential}=k_{constat}\frac{q_{charge}}{r_{radius}} \\\\q_{charge}=\frac{V_{potential*r_{radius}}}{k_{constant}}$

Substitute the given values

$q_{charge}=\frac{(0.12m)(230V)}{9*10^{9} }\\ q_{charge}=3.07*10^{-9}C\\or\\q_{charge}=3.07nC$

For Part (b)

The charge density is given by:

σ=q/(4πr²)

Substitute the given values and value of q to find charge density

$=\frac{3.07*10^{-9}C}{4\pi (0.12m)^2}\\ =1.69*10^{-8}C/m^2\\or\\=17nC/m^2$

σ=17 nC/m²

3 0

3 years ago