Ask question

Ask question

All categories

3 years ago

10

The pendulum on a grandfather clock is 0.993 m long and swings to a maximum 4.57° angle. If the bob of the pendulum has mass = 0

.415 kg, how much PE does it have at the top of its swing?

1 answer:

Alla [95]3 years ago

8 0

Answer:

0.0128

Explanation:

You might be interested in

A direct injection (DI) fuel injector system sprays high-pressure fuel, up to _______ psi, into the combustion chamber as the pi

Vika [28.1K]

The answer is not 65. ya moms a hooch

3 0

4 years ago

Read 2 more answers

Minute after minute, hour after hour, day after day, ocean waves continue to splash onto the shore. Explain why the beach is not

MAVERICK [17]

Answer: C. Ocean waves can only bring energy to the shore; the particles of the medium (water) simply oscillate about their fixed position.

Explanation:

The reason why the beach is not completely submerged and the reason why the middle of the ocean has not been depleted of its water supply is due to the fact that water isn't transported by ocean waves.

It should be noted that even a single drop of water cannot be brought by the ocean wave to the shore from the middle of the ocean.

The only thing that the ocean waves can bring to the shore is energy. Hemce, the water particles oscillate in their fixed position which is vital in making sure that the beach isn't piled up with water.

4 0

3 years ago

Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano

Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

<h2>Part a </h2>

Specific energy is given by

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}$

Here

ε is the specific energy
v is the velocity which is given as 2.23 km/s
μ is the gravitational constant whose value is 398600
r is the distance between earth and the meteorite which is 402,000 km

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2$

Value of specific energy is also given as

$\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km$

Orbit formula is given as

$r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0$

Putting values in this equation and solving for e via the quadratic formula gives

$ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69$

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

<h2>Part b</h2>

The radius of trajectory at perigee is given as

$r_p=a(e-1)\\$

Substituting values gives

$r_p=133319 (1.086-1)\\r_p=11465.4 km$

Now for estimation of altitude z above earth is given as

$z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km$

So the altitude at closest approach is 5088 km

<h2>Part c</h2>

radius of perigee is also given as

$r_p=\frac{h^2}{\mu}\frac{1}{1+e}$

Rearranging this equation gives

$h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s$

Now the velocity at perigee is given as

$v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\$

So the velocity at perigee is 8.516 km/s

<h2>Part d</h2>

Turn angle is given as

$\delta =2 sin^{-1} (\frac{1}{e})$

Substituting value in the equation gives

$\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08$

Aiming radius is given as

$\Delta =a \sqrt{e^2-1}$

Substituting value in the equation gives

$\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km$

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0

3 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.23 2.23 times a second. A tack is stuck in the t

elena-14-01-66 [18.8K]

Explanation:

The given data is as follows.

Angular velocity ( $\omega$ ) = 2.23 rps

Distance from the center (R) = 0.379 m

First, we will convert revolutions per second into radian per second as follows.

= 2.23 revolutions per second

= $2.23 \times 2 \times 3.14 rad/s$

= 14.01 rad/s

Now, tangential speed will be calculated as follows.

Tangential speed, v = $R \times \omega$

= 0.379 x 14.01

= 5.31 m/s

Thus, we can conclude that the tack's tangential speed is 5.31 m/s.

8 0

3 years ago

Microwaves and infrared waves are similar because they both have

Orlov [11]

Microwave and Infrared Radiation. Both are forms of electromagnetic radiation. Both are transmitted as waves. Waves are converted to heat.

3 0

3 years ago