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Kruka [31]
4 years ago
11

Rt a - the forces acting on an oil droplet watch the animation. identify the types of forces acting on an oil droplet while the

voltage is turned on and the droplet is either falling or rising.
Physics
1 answer:
topjm [15]4 years ago
3 0

When the voltage is turned on and the droplet either rises or falls, the type of forces which are acting on the droplet are:

 1) Gravitational force, 

2) Electrical repulsion

<span>When the drop rises, it means that the upward electrical force is acting on it and a gravitational force is acting in downward direction. But in this case, the upward electrical force will be greater than the downward gravitational force.</span>

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What does addition of two vectors give you?
mash [69]

Answer:

Explanation:

To add or subtract two vectors, add or subtract the corresponding components. Let →u=⟨u1,u2⟩ and →v=⟨v1,v2⟩ be two vectors. The sum of two or more vectors is called the resultant. The resultant of two vectors can be found using either the parallelogram method or the triangle method .

4 0
4 years ago
We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,
larisa [96]

Answer:

The final charges of each sphere are:   q_A = 3/8 Q , q_B = 3/8 Q ,               q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

                q_A = Q / 2

                q_B = Q / 2

Now sphere A touches sphere C, ending with half the charge

                q_A = ½ (Q / 2) = ¼ Q

                q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

                  q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

                  q_A = 3/8 Q

                  q_B = 3/8 Q

The final charges of each sphere are:

                q_A = 3/8 Q

                q_B = 3/8 Q

                q_C = 3/4 Q

7 0
3 years ago
An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t
Ede4ka [16]

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0

4 0
4 years ago
Which of the following is the same in all frames of reference?
Ksju [112]
The correct option is B.
The length of an object, the mass of an object and the rate of time passage for an object can change depending on the situation which the object is subject to. For instance in space, the mass and the velocity of an object usually change. But, the value of the speed of light in the space is the same for all observers regardless of the motion of an object, that is, the speed of light is a constant.<span />
4 0
4 years ago
A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w
Scilla [17]

Answer:

v_f = 20 m/s

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2

now for pure rolling condition we will have

v = R\omega

also we have

I = mR^2

now we will have

KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}

KE = mv^2

now by work energy theorem we can say

W = KE_f - KE_i

842 J = mv_f^2 - mv_i^2

842 = 3(v_f^2) - 3\times 11^2

now solve for final speed

v_f = 20 m/s

3 0
3 years ago
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