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3 years ago

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a 0.05-kg starts from rest at a height of 0.95m. assuming no friction, what is the kinetic energy of the car when it reaches the

bottom of the hill? (Assume g = 9.81 m/s)

1 answer:

Alex73 [517]3 years ago

3 0

Fhubugukhkbjhvlfipy0iy ft

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A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.

Arturiano [62]

Answer:

a) $\alpha=0.5117\ rad.s^{-2}$

b) $\theta=8.4\ rad$

Explanation:

Given:

initial rotational speed of phonograph, $\omega_i=0\ rad.s^{-1}$

final rotational speed of phonograph, $N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}$

time taken for the acceleration, $t=5.73\ s$

a)

Now angular acceleration:

$\alpha=\frac{\omega_f-\omega_i}{t}$

$\alpha=\frac{2.932-0}{5.73}$

$\alpha=0.5117\ rad.s^{-2}$

b)

Using eq. of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2$

$\theta=8.4\ rad$

5 0

3 years ago

In reaching her destination, a backpacker walks with an average velocity of 1.41 m/s, due west. This average velocity results, b

V125BC [204]

The total average velocity $v=+1.41 m/s$ (I assume west as positive direction) is given by the total displacement, S, divided by the total time taken, t:
$v= \frac{S}{t}= \frac{S_1+S_2}{t_1+t_2}$
where:
-The total displacement S is the algebraic sum of the displacement in the first part of the motion ( $S_1=6.30 km=6300 m$ , due west) and of the displacement in the second part of the motion ( $S_2$ , due east).
-The total time taken t is the time taken for the first part of the motion, $t_1$ , and the time taken for the second part of the motion, $t_2$ . $t_1$ can be found by using the average velocity and the displacement of the first part:
$t_1= \frac{S_1}{v_1}= \frac{6300 m}{2.49 m/s}=2530 s$

$t_2$ , instead, can be written as $\frac{S_2}{v_2}$ , where $v_2=-0.630 m/s$ is the average velocity of the second part of the motion (with a negative sign, since it is due east).

Therefore, we can rewrite the initial equation as:
$v=1.41 = \frac{6300+S_2}{2530- \frac{S_2}{0.630} }$
And by solving it, we find the displacement in the second part of the motion (i.e. how far did the backpacker move east):
$S_2=-844 m=-0.844 km$

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3 years ago

A 0.14 kilogram baseball is throw in a straight line at a velocity of 30 m/sec.what is the momentum of the baseball

Masja [62]

4.2 kg m/s hope this helps

4 0

3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 27 ft/s. Its height

Ahat [919]

Answer:

$V_{3.01}=-93.2m/s$

$V_{3.005}=-93.1m/s$

$V_{3.002}=-93.04m/s$

$V_{3.001}=-93.02m/s$

$V_{3}=-93m/s$

Explanation:

To calculate average velocity we need the position for both instants t0 and t1.

Now we will proceed to calculate all the positions we need:

$Y_{3}=-99m/s$

$Y_{3.01}=-99.932m/s$

$Y_{3.005}=-99.4655m/s$

$Y_{3.002}=-99.18608m/s$

$Y_{3.001}=-99.09302m/s$

Replacing these values into the formula for average velocity:

$V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s$

$V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s$

$V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s$

$V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s$

To know the actual velocity, we derive the position and we get:

$V=27-40t = -93m/s$

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4 years ago

please help me asap!!! please i will mark brailiest and give lots of points thank you so much and may God be with you all!!!

Anna35 [415]

Answer: 9) when water is put inside a pot and put on a stove, the fire heats the pot, heat is transferred through the pot by conduction, the next step here in heat transfer by convection which happens in the water in the pot, when heat is transferred from the pot to the water the water molecules gain kinetic energy and moves up while the heavier molecules of the water comes down to get heated, this continues until heat is transferred throughout the water by convection, radiation occur when the quantity of the water is getting reduced.

10) same principle applies to that of the stove, but the difference is that the heat comes from sun and not from fire, the ice-cream is heated by conduction and heat is spread throughout the ice cream by convection , and radiation comes from the sun to the ice cream

Explanation:

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3 years ago