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3 years ago

Please help me again , im sorry I keep asking but please help me please

Physics

1 answer:

choli [55]3 years ago

4 0

He studied a large number of offspring

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What specific space object is shown in the picture below?

PIT_PIT [208]

From context clues, I believe the correct answer is B) Red Dwarf!

3 0

3 years ago

A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 51.6 m/s at an angle of 37.

Andrei [34K]

Answer:

$v_{y}=35.21m/s$

Explanation:

From the exercise we know the cannonball's initial velocity, the angle which its released with respect to the horizontal and its initial height

$v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m$

If we want to know whats the y-component of velocity we need to use the following formula:

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Knowing that $g=-9.8m/s^2$

$v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s$

So, the cannonball's y-component of velocity is $v_{y}=35.21m/s$

6 0

3 years ago

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

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 998

wlad13 [49]

Answer:

1988.05 rad/s^2

Explanation:

The angular speed of the optical disk ω = 998.0 rad/s

the time taken to come to rest t = 0.502 s

The magnitude of the average angular acceleration ∝ = ω/t

∝ = 998.0/0.502 = 1988.05 rad/s^2

7 0

3 years ago

True or false? a physical change is a change of matter from one form to another without a change in chemical properties

Lina20 [59]

The answer is true!!!

6 0

3 years ago