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3 years ago

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A force of 3,000 N is applied to a car, resulting in acceleration. In 10 s, the car increases its velocity from 0 m/s to 30 m/s.

What is the mass of the car?

1 answer:

Reil [10]3 years ago

6 0

Answer:

4.5 m/s

Explanation:

I am not sure if it correct.

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A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

Semenov [28]

Answer:

$f_n=3.75N$

Explanation:

From the question we are told that:

Frictional force $F=0.150N$

Coefficient of kinetic friction $\mu=0.04$

Generally the equation for Normal for is mathematically given by

$f_n=\frac{F}{\mu}$

Therefore

$f_n=\frac{0.150}{0.04}$

$f_n=3.75N$

5 0

3 years ago

Which would BEST describes what occurs when a ball is thrown against a wall? A) The ball will not bounce off the wall. B) The ba

oee [108]

Answer:

D) The ball exerts a force on the wall and the wall exerts a force back.

Explanation:

Newton's third law of motion states that:

"When an object A exerts a force on another object B, then object B exerts an equal and opposite force on object A"

In this problem, we can identify (for instance) object A with tha ball and object B with the wall. Therefore, if we apply Newton's third law, we get:

The ball (object A) exerts a force on the wall (object B), therefore the wall (object B) exerts an equal and opposite force on the ball (object A). So, option D is the correct one.

3 0

3 years ago

Read 2 more answers

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago

A piano mover raises a 1000-n piano at a constant speed using a very light rope in a frictionless pulley system, as shown in the

Rina8888 [55]

Here As we can see the figure that the end of the rope is pulled by some force F

Now as we can see that Piano is connected by a pulley which is passing over the pulley so effectively net force on the piano upwards will be 2F as it is connected by 2 ropes by the pulley

Now for constant velocity of the piano we will say

$2F - mg = ma$

since velocity is constant so acceleration must be ZERO

so here we have

$2F - mg = 0$

as we know here that

mg = 1000 N

so we will have

$2F - 1000 = 0$

$F = 500 N$

so here force must be 500 N

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3 years ago

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Fenómeno químico en virtud del cual se transforma un cuerpo o compuesto por la accion de un oxidante

uranmaximum [27]

Chemical phenomenon by virtue of which a body or compound is transformed by the action of an oxidant

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3 years ago