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3 years ago

PULLEYS UNIT: HELP!

Physics

1 answer:

Alex787 [66]3 years ago

4 0

Well A goes clock wise, B goes counter clock wise, and C goes clock wise.

Hope this helps!

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Jaiden is writing a report about the structure of the atom. In her report, she says that the atom has three main parts and two s

USPshnik [31]

No because an atom consists of two main parts and three subatomic particles - protons, neutrons, electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - e.g. Hydrogen has 1 proton and 1 electron. Neutrons do not affect the overall charge of the atom, and only increase the atomic mass.

7 0

3 years ago

Read 2 more answers

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

3 years ago

What is the net force on a car if the force of friction is 15 N and the forward force due to the engine is 20 N?

anygoal [31]

D. 35n forwards....................

8 0

3 years ago

Certain neutron stars (extremely dense stars) are believed to be rotating at about 1000 rev/s. If such a star has a radius of 14

Sever21 [200]

Answer:

minimum mass of the neutron star = 1.624 × 10^30 kg

Explanation:

For a material to remain on the surface of a rapidly rotating neuron star, the magnitude oĺf the gravitational acceleration on the material must be equal to the magnitude of the centripetal acceleration of the rotating neuron star.

This can be represented by the explanations in the attached document.

minimum mass of the neutron star = 1.624 × 10^30 kg

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3 years ago

describe the time of day that an early explorer might have planned to enter a harbor and when he might have planned to leave

inessss [21]

He would try to enter as the tide is rising, and leave as the tide is falling. Those things happen at all different times of day during a month.

4 0

3 years ago