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3 years ago

What are some examples of uniform motion

Physics

1 answer:

Flauer [41]3 years ago

7 0

Some examples are:

-Uniform translation

-Steady circular motion

-Steady Rotation

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An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 1760 1760 × 103 seconds (a

Katen [24]

Answer:

$a_c=6.471\frac{m}{s^{2}}$

Explanation:

We can calculate the magnitude of the tangential velocity of the moon, using the equation:

$v=\frac{2\pi R}{\tau}$

(This works, since velocity is defined as distance over time; in this case the distance is 2πR and the time is the period τ)

Next, from the equation of the centripetal acceleration we have:

$a_c=\frac{v^{2} }{R}\\\\a_c=\frac{4\pi^{2}R}{\tau^{2} }$

Be careful, the radius of the orbit R is equal to the distance from the center of the moon to the center of the planet. So we have to sum the distance from the center of the moon to the surface of the planet and the radius of the planet to obtain R:

$R=285.0*10^{6} m+3.50*10^{6} m=288.5*10^{6}m$

Finally, plugging the given values into the centripetal acceleration formula, we have:

$a_c=\frac{4(3.141)^{2}(288.5*10^{6}m)}{(1760*10^{3}s)^{2}} =6.471\frac{m}{s^{2}}$

In words, the moon's radial acceleration a_c is 6.471m/s².

3 0

3 years ago

The rectangular coils in a 345-turn generator are 12 cm by 12 cm. Part A

Likurg_2 [28]

Answer:

118.166 volt

Explanation:

We have given number of turns N =345

Sides of the rectangular coils is 12 cm =0.12 m

So area A =0.12×0.12=0.0144 $m^2$

Magnetic field B =0.45 T

Angular speed =505 rpm

Speed in rad/sec $\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 505}{60}=52.8566rad/sec$

The emf is given by $E=NAB\omega SIN\omega t$

For maximum emf sinwt =1

So $E=345\times 0.0144\times 0.45\times 52.8566=118.166volt$

4 0

3 years ago

If you wanted to brighten a room, which of the following actions would you most likely take? increase the medium density in the

seropon [69]

Increase the photons in the room.

3 0

3 years ago

Read 2 more answers

A gas bottle contains 0.650 mol of gas at 730 mm Hg pressure. If the final pressure is 1.15 atm, how many moles of gas were adde

AlekseyPX

Answer:

0.779 mol

Explanation:

Since the gas is in a bottle, the volume of the gas is constant. Assuming the temperature remains constant as well, then the gas pressure is proportional to the number of moles:

$p \propto n$

so we can write

$\frac{p_1}{n_1}=\frac{p_2}{n_2}$

where

p1 = 730 mm Hg = 0.96 atm is the initial pressure

n1 = 0.650 mol is the initial number of moles

p2 = 1.15 atm is the final pressure

n2 is the final number of moles

Solving for n2,

$n_2 = n_1 \frac{p_2}{p_1}=(0.650 mol)\frac{1.15 atm}{0.96 atm}=0.779 mol$

6 0

4 years ago

Read 2 more answers

A car travels at 12 m/s. The car then accelerates at 3 m/s2 until it reaches a speed of 18m/s.

Galina-37 [17]

Answer:

С. 30 m

Explanation:

Given the following data;

Initial velocity, U = 12m/s

Final velocity, V = 18m/s

Acceleration, a = 3m/s²

To find the distance, we would use the third equation of motion;

V ² = U² + 2aS

Substituting into the equation, we have

18² = 12² + 2*3*S

324 = 144 + 6S

6S = 324 - 144

6S = 180

S = 180/6

Distance, S = 30 meters.

3 0

3 years ago