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3 years ago

What can we conclude from the attractive nature of the force between a positively charged rod and an object?

Physics

1 answer:

Paraphin [41]3 years ago

6 0

Answer:

E; The object is negatively charged

Explanation:

Here, we want to state the conclusion that can be drawn from a positively charged rod being attracted to an object.

Generally as we know, oppositely charged materials attract while the ones with same charges repel each other.

Thus, in this case, for the rod to attract the object, there must have been an opposite charge of negativity on the object

So we conclude that the reason why the rod attracted the object was because of the presence of opposing charges on both of them. And since the rod has taken the positive charge, it is only correct to state that the object is negatively charged

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A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the p

frez [133]

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

$a(t)=(5-10t)\ m/s^2$

(a) Since, $a=\dfrac{dv}{dt}$

v = velocity

$dv=a.dt$

$v=\int(a.dt)$

$v=\int(5-10t)(dt)$

$v=5t-\dfrac{10t^2}{2}=5t-5t^2$

(b) $v=\dfrac{dx}{dt}$

x = position

$x=\int v.dt$

$x=\int (5t-5t^2)dt$

$x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3$

(c) Velocity function is given by :

$v=5t-5t^2$

$5t-5t^2=0$

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

7 0

3 years ago

What is the potential energy of a 2 kg ball 15 m in the air?

oee [108]

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 2 × 10 × 15

We have the final answer as

Hope this helps you

7 0

3 years ago

The gravitational potential energy of a particle of mass m moving under the influence of a fixed mass M is given by - , where G

djverab [1.8K]

-GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.

Given

A particle of mass m moving under the influence of a fixed mass's M, gravitational potential energy of formula -GMm/r, where r is the separation between the masses and G is the gravitational constant of the universe.

As the Gravity Potential energy of particle = -GMm/r

Total energy of particle = Kinetic energy + Potential Energy

As we know that

Kinetic energy = 1/2mv²

Also, v is equals to square root of GM/r

v = √GM/r

Put the value of v in the formula of kinetic energy

We get,

Kinetic Energy = GMm/2r

Total Energy = GMm/2r + (-GMm/r)

= GMm/2r - GMm/r

= -GMm/2r

Hence, -GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.

Learn more about Gravitational Potential Energy here brainly.com/question/15896499

#SPJ4

3 0

1 year ago

Can anyone answer this fast pls

nekit [7.7K]

I believe the answer would be 4.5. because it wouldnt be c or d. and 2 seems too small.

3 0

3 years ago

At which point on this electric field will a test charge show the maximum strength?

Juli2301 [7.4K]

Answer:

Explanation:

The figure shows the electric field produced by a spherical charge distribution - this is a radial field, whose strength decreases as the inverse of the square of the distance from the centre of the charge:

$E\propto \frac{1}{r^2}$

More precisely, the strength of the field at a distance r from the centre of the sphere is

$E=k\frac{Q}{r^2}$

where k is the Coulomb's constant and Q is the charge on the sphere.

From the equation, we see that the field strength decreases as we move away from the sphere: therefore, the strength is maximum for the point closest to the sphere, which is point A.

This can also be seen from the density of field lines: in fact, the closer the field lines, the stronger the field. Point A is the point where the lines have highest density, therefore it is also the point where the field is strongest.

8 0

4 years ago