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2 years ago

I need help with this physics question

Physics

2 answers:

faltersainse [42]2 years ago

5 0

Answer:

It would have to be B. i hope dis helps :) hruu??

Explanation:

OlgaM077 [116]2 years ago

4 0

Answer:

its B hope you have a good Day

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For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo

Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

$\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}$

so that its period is $T=3.644\,\frac{\mathrm s}{\mathrm{rev}}$ (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration $\Delta t$ . Denote by $\theta$ the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

$\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ$

$\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ$

$\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ$

We can then compute the magnitude of the velocity vector differences $\Delta\vec v$ for each time interval by using the law of cosines:

$|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta$

$\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

and in turn we find the magnitude of the average acceleration vectors to be

$\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

5 0

3 years ago

Which of the following best explains how the Ptolemy’s (geocentric) and Copernicus’s (heliocentric) models of the solar system d

TiliK225 [7]

Answer:

The correct answer is c

Explanation:

In these two different models of movement of the planets

Ptolemy raises the Earth as the center of the solar system

In the Copernicus system, it poses the Sun as the center of the solar system.

Copernicu's system was accepted for giving a simpler and more complete explanation of the problem

The correct answer is c

7 0

4 years ago

A thief is trying to escape from a parking garage after completing a robbery, and the thief's car is speeding (v = 13 m/s) towar

rodikova [14]

<u>Answer</u>

Yes, the car reaches the door before the gate closes.

<u>Explanation</u>

The time taken by the car to reach at the door.

Time = distance / time

= 22/13

= 1.6923 seconds

Time taken by the door to close up to the height of the car.

Distance the door has to move to prevent the car from escaping = 9.1.4 = 7.6 m

From newton's 2nd law of motion;

s = ut + 1/2 gt²

7.6 = 0.6t + 1/2 × 10t²

7.6 = 0.6t + 5t²

50t² + 6t - 76 = 0

Solving this quadrilatic equation,

t = 28.537 seconds

Answer: Yes, the car reaches the door before the gate closes.

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3 years ago

A 2 kg block moves along x axis . if acceleration as function of its

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3 years ago

A person rolls a barrel up an inclined plane with a length of 6m and a height of 3m/ How much force is needed to roll up the bar

snow_lady [41]

Answer:

Length (l) = 6 m

height (h) = 3 m

Load(L) = 500 N

Effort (E) = ?

we know the principal that

E * l = L * h

6 E = 500 * 3

6E = 1500

E = 250

therefore 250 N work is done on the barrel.

8 0

4 years ago