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1 year ago

Why must the test charge used to determine the electric field be very, very small?.

1 answer:

8 0

The test charge used to determine the electric field is very, very small so that its presence does not affect the electrical field because of the supply price. The electrical rate that produces the electric area is known as a source charge.

An electric-powered area is a physical area that surrounds electrically charged debris and exerts pressure on all differently charged particles inside the area, both attracting or repelling them. It also refers back to the physical field for a machine of charged particles.

Electric-powered discipline is described as the electrical force in keeping with unit fee. The path of the sector is taken to be the route of the force it might exert on a fine check price. the electric discipline is radially outward from a nice rate and radially in toward a bad point rate.

An electric discipline is a vicinity of space around an electrically charged particle or object in which an electric charge might sense force. An electric field is a vector amount and may be visualized as arrows going in the direction of or away from fees.

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Using information about natural laws, explain why some car crashes produce minor injuries and others produce catastrophic injuri

Nataly_w [17]

Don't they have to pay insurance company if something happens to there car.

5 0

3 years ago

Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c

grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg

Force applied by the motor, $F_A=1000\ N$

The static and dynamic friction coefficient is, $\mu=0.5$

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

$F_A-\mu mg=ma$

$\dfrac{F_A-\mu mg}{m}=a$

$a=\dfrac{1000-0.5(1000)(9.81)}{1000}$

$a=-3.905\ m/s^2$

So, the acceleration of the car is $-3.905\ m/s^2$ . Hence, this is the required solution.

6 0

2 years ago

A stunt driver drives a car horizontally off the edge of a cliff at 3.8m/s and reaches the water below 2.5s later.

andreyandreev [35.5K]

A. The cliff was 30.7 m high
B. I also got 9.5 as the horizontal distance

Here is my work, I find making charts like this one to find knowns and unknowns can be helpful

4 0

3 years ago

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

Mila [183]

Answer:

11.4 m/s

Explanation:

The expression for the Centripetal acceleration is :

$a=\frac{v^2}{R}$

Where, a is the accleration

v is the velocity around circumference of circle

R is radius of circle

In the given question,

a = g = Acceleration due to gravity as the car is at top = $9.81\ m/s^2$

v = ?

R = 13.2 m

So,

$9.81=\frac{v^2}{13.2}$

$v^2=9.81\times {13.2}$

8 0

3 years ago

A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi

OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn = normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

The sled required almost 10 s to travel down the slope.

8 0

3 years ago