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3 years ago

The moon is 250,000 miles away. How many feet is it from earth

Physics

2 answers:

beks73 [17]3 years ago

4 0

(250,000 miles) x (5,280 feet/mile) = 1,320,000,000 feet

( 1.32 x 10⁹ )

Fynjy0 [20]3 years ago

3 0

The conversion between miles and feet is:

$1 mi = 5280 ft$

So we can set up a proportion to find how many feet corresponds to 250000 miles:

$1 mi: 5280 ft = 250000 mi :x$

and solving it, we find

$x=\frac{5280 ft \cdot 250000 mi}{1 mi} =1.32 \cdot 10^9 ft$

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She sights two sailboats going due east from the tower. The angles of depression to the two boats are 42o and 29o. If the observ

Reika [66]

Answer:

The boats are 934.65 feet apart

Explanation:

Given:

The angles of depression to the two boats are 42 degrees and 29 degrees

Height of the observation deck i = 1,353 feet

To Find:

How far apart are the boats (y )= ?

Solution:

Step 1 : Finding the value of x(Refer the figure attached)

We can use the tangent ratio to find the x value

$tan(42^{\circ}) = \frac{1353}{x}$

$x = \frac{1353}{tan(42^{\circ}) }$

x = 590.47 feet

Step 2 : Finding the value of z (Refer the figure attached)

$tan(29^{\circ}) = \frac{1353}{z }$

$z = \frac{1353}{tan(29^{\circ})}$

z = 1525.12 feet

Step 3 : Finding the value of y (Refer the figure attached)

y = z -x

y = 1525.12 - 590.47

y = 934.65 feet

Thus the two boats are 934.65 feet apart

3 0

3 years ago

Does anyone know what #2 and #3 is

EleoNora [17]

Answer:

He can figure out if they heat to the same temp with the thermometer.

Explanation:

They have different boiling points and heat up at different rates so they would have different temps at the same time

6 0

2 years ago

What does it mean if an object is in motion?

Alina [70]

Explanation:

basically it's moving I think

6 0

3 years ago

A girl standing on a bridge throws a stone vertically downward with an initial velocity of 15.0 m/s into the river below. If the

hodyreva [135]

Vi = 15 m/s
t = 2 s
a = 9.8 m/s^2
y = ?

The kinematic equation that has all of our variables is d = Vi*t + 0.5*a*t^2
y = 15*2 + 0.5*9.8*2^2 = 49.6 m

6 0

3 years ago

A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i

timurjin [86]

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

$I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2$

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

$I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2$

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

$I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2$

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

$I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}$

$\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}$

$\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s$

4)Kinetic energy before:

$E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J$

Kinetic energy after:

$E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J$

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) $a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2$

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

3 0

3 years ago