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dusya [7]
3 years ago
10

A 0.250 kg car rest rolls down a frictionless incline from a starting height of 0.75 m what is the final velocity of the car? SH

OW ALL WORK.
Also what difference would it make if it is not frictionless? Please and thank you!
Physics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

v = 3.83 m/s

Explanation:

Given that,

The mass of a car, m = 0.250

It rolls down a frictionless incline from a starting height of 0.75 m

We need to find the final velocity of the car. Let v be the velocity of the car. It can be calculated using the third equation of motion i.e.

v^2-u^2=2as

Here,

u = 0 and a = g

v=\sqrt{2gs} \\\\v=\sqrt{2\times 9.8\times 0.75} \\\\v=3.83\ m/s

So, the final velocity of the car is 3.83 m/s.

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P1V1=P2V2
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7 0
3 years ago
A small glass bead has been charged to +20 nC. A small metal ball bearing 1.0 cm above the bead feels a 0.018 N downward electri
Alla [95]

Answer:

q=1\times10^{-8}C

Explanation:

Let the charge on the ball bearing is q.

charge on glass bead, Q = 20 nC = 20 x 10^-9 C

Force between them, F = 0.018 N

Distance between them, d = 1 cm = 0.01 m

By use of Coulomb's law in electrostatics

F=\frac{KQq}{d^{2}}

By substituting the values

0.018=\frac{9\times10^{9}\times20\times10^{-9}q}{0.01^{2}}

q=1\times10^{-8}C

Thus, the charge on the ball bearing is q=1\times10^{-8}C

7 0
2 years ago
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

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A <u>Law</u> is an affirmation (something established) based on repeated long-term observation of a phenomenon that has been studied and verified.  

That is: A law is present in all known theories and therefore is considered universal. In addition, a law can not be refuted, nor changed, because its precepts have been proven through various studies.  

<u>Unlike theory</u>, which is the set of rules and principles that describe and explain a particular phenomenon and <u>is subject to changes as new evidence emerges that gives meaning to it.  </u>

Then, based on what is explained above, the law of universal gravitation is a statement that exists because it was rigorously tested and verified, therefore it can not be refuted.

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the light from andromeda galaxy takes about 2.6 million years to reach earth. which of these statements is correct about the and
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A. it is <span>located at a distance of 2.6 million light years from earth</span>
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