All categories

3 years ago

What is an object doing at the moment its instantaneous acceleration is zero?

Physics

1 answer:

Oliga [24]3 years ago

3 0

Answer: Option A

Explanation:

Acceleration is defined as the rate of change of the vellocity.

Then, if the acceleration is equal zero, the object has a constant velocity or a velocity equal to zero (also constant, of course)

Then, te correct option would be option A; resting, because an object resting has instantaneuos acceleration of zero.

You might be interested in

7. What is the velocity of an object with a distance of 90m south and a time of 5s?

IrinaK [193]

Answer:

Explanation:

v= s/t

V =90m/5s

V = 8m/s

4 0

3 years ago

An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec

lana66690 [7]

Answer:

91.87 m/s

Explanation:

Given:

x = initial distance of the electron from the proton = 6 cm = 0.06 m

y = initial distance of the electron from the proton = 3 cm = 0.03 m
u = initial velocity of the electron = 0 m/s

Assume:

m = mass of an electron = $9.1\times 10^{-31}\ kg$
v = final velocity of the electron
e = magnitude of charge on an electron = $1.6\times 10^{-19}\ C$

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

$\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\$

$\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\$

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0

3 years ago

A paper pinwheel is spinning in the wind. Which statement is correct about the forces responsible for the rotation?

VikaD [51]

Answer:

Only the perpendicular component of gravity is responsible for the rotation because wind points toward the pivot.

Explanation:

3 0

4 years ago

A positively charged metal sphere, A, is held close to but not touching and identical uncharged sphere, sphere B. Sphere A is no

Yuri [45]

Answer:

The sphere C carries no net charge.

Explanation:

When brougth close to the charged sphere A, as charges can move freely in a conductor, a charge equal and opposite to the one on the sphere A, appears on the sphere B surface facing to the sphere A.
As sphere B must remain neutral (due to the principle of conservation of charge) an equal charge, but of opposite sign, goes to the surface also, on the opposite part of the sphere.
If sphere A is removed, a charge movement happens in the sphere B, in such a way, that no net charge remains on the surface.
If in such state, if the sphere B (assumed again uncharged completely, without any local charges on the surface), is touched by an initially uncharged sphere C, due to the conservation of charge principle, no net charge can be built on sphere C.

3 0

3 years ago

A man pulls a 10 kg box at constant speed across the floor with a rope. The tension in the rope is 120 N at an angle of 30°.

Murrr4er [49]

Answer:

98n

Explanation:

5 0

3 years ago