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frosja888 [35]
3 years ago
10

The electric field that is 0.25m from a small sphere is 450n/c toward the sphere.

Physics
2 answers:
Andrei [34K]3 years ago
8 0
Its b bc its the volume of the sphere 
Hitman42 [59]3 years ago
5 0

PART 1)

A small sphere is considered to be a point charge

So here if electric field is indicating the position of charge itself or it is towards the charge then it must be a negative charge on the sphere

PART 2)

Electric field due to a point charge is given by

E = \frac{KQ}{r^2}

here

k = 9 * 10^9

Q = charge

r = distance of charge from the point where field strength is required

= 0.25 m

E = 450 N/c

now from the above equation we can say

450 = \frac{9*10^9 * Q}{0.25^2}

Q = 3.125 * 10^{-9} C

So magnitude of charge is 3.125 nC on the small sphere

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A spring with spring constant 450 N/m is stretched by 12 cm. What distance is required to double the amount of potential energy
snow_lady [41]

Answer:

<em> The distance required = 16.97 cm</em>

Explanation:

Hook's Law

From Hook's law, the potential energy stored in a stretched spring

E = 1/2ke² ......................... Equation 1

making e the subject of the equation,

e = √(2E/k)........................ Equation 2

Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.

Given: k = 450 N/m, e = 12 cm = 0.12 m.

E = 1/2(450)(0.12)²

E = 225(0.12)²

E = 3.24 J.

When the potential energy is doubled,

I.e E = 2×3.24

E = 6.48 J.

Substituting into equation 2,

e = √(2×6.48/450)

e = √0.0288

e = 0.1697 m

<em>e = 16.97 cm</em>

<em>Thus the distance required = 16.97 cm</em>

6 0
3 years ago
Look at the graph
viva [34]
Here, Carefully look at the graph.
When it is on x=10, it is approximately 10, (slightly less than 10)
Closest value would be 90, so y/x = 90/10 = 9 

So, the density of the graph would be 9 g/cm³

In short, Your Answer would be Option D

Hope this helps!
4 0
3 years ago
Read 2 more answers
What is the force of an object with a mass of 65 kg and an an unknown acceleration?
Alik [6]

We know, F = m.a

F = 65 * a

Where, F = force

a = unknown acceleration

8 0
3 years ago
A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

F, the push applied by the worker

F_f=-\mu mg, the force of friction, with \mu=0.25 being the coefficient of friction, m=30.0 kg being the mass of the crate, and g=9.8 m/s^2. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

where

F = 73.5 N is the force

d = 4.5 m is the displacement

\theta=0^{\circ} is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

W=F_f d cos \theta

where

F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N is the magnitude of the force of friction

d = 4.5 m is the displacement

\theta=180^{\circ} is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

W=F_f d cos \theta

We notice that both gravity and normal force are perpendicular to the displacement: therefore, \theta=90^{circ}, and so

cos \theta=0

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0
3 years ago
The density of an object is dependent upon the object’s mass and ---
kotegsom [21]

Answer:Volume

Explanation:

Density = mass/ Volume

7 0
3 years ago
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