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Hunter-Best [27]

2 years ago

6

Find earth's approximate mass from the fact that the moon orbits earth in an average time of 27.3 days at an average distance of

384,000 kilometers. (hint: the moon's mass is only about 180 of earth's.)

1 answer:

Aleks [24]2 years ago

4 0

We can solve the problem by using Kepler's third law, which states:

$\frac{4 \pi^2}{T^2}=\frac{GM}{r^3}$

where T is the period of revolution of the Moon around the Earth, G is the gravitational constant, M the Earth's mass and r the average distance between Earth and Moon.

Using the data of the problem:

$T=27.3 d \cdot 24 \cdot 60 \cdot 60 = 2358720 s=2.36 \cdot 10^6 s$

$r=384000 km=3.84 \cdot 10^8 m$

We can re-arrange the equation and find the Earth's mass:

$M=\frac{4 \pi^2 r^3}{GT^2}=\frac{4 \pi^2 (3.84 \cdot 10^8 m)^3}{(6.67 \cdot 10^{-11})(2.36 \cdot 10^6 s)^2}=6.0 \cdot 10^{24} kg$

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Which statements best describe x-rays? check all that apply. x-rays are electromagnetic waves. x-rays are longitudinal waves. x-

laiz [17]

X ray is one of the electromagnetic waves.

As per Clark Maxwell's electromagnetic theory, all the electromagnetic waves move with the velocity of light i.e c= 3×10^8 m/s

In case of electromagnetic waves,the electric field and magnetic field are perpendicular to each other as well as perpendicular to the direction of propagation.The electromagnetic waves exhibit the property of polarisation. Hence they are transverse in nature.

Hence the best statements about X- ray will be-

1- X -rays are electromagnetic waves

2-X-rays are transverse transverse waves

3- X- rays travel at the speed of light.

5 0

1 year ago

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Sonrisa owns a 300 W television. If the total energy usage for February is 32.4 kWh, how many hours per week does Sonrisa watch

Zielflug [23.3K]

the correct answer is 27 hours per week :) hope this helps

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2 years ago

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A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a const

nataly862011 [7]

a) $-1.54 m/s^2$

b) 803.4 N

Explanation:

a)

At the point C (top position of the loop), the pilot feel weightless, so the normal reaction exerted by the seat is zero:

N = 0

Therefore, the equation of the forces at position C is:

$mg=m\frac{v^2}{r}$

where the term on the left is the weight of the pilot and the term on the right is the centripetal force, and where:

$g=9.8 m/s^2$ is the acceleration due to gravity

$v$ is the velocity of the jet at the top

$r=1200 m$ is the radius of the loop

Solving for v,

$v_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s$

So, this is the velocity of the jet at position C.

The velocity at position A (bottom) is

$v_A=550 km/h =152.8 m/s$

The distance covered by the jet is the length of a semi-circumference of radius r, so

$s=\pi r=\pi(1200)=3770 m$

Since the deceleration of the plane is constant, we can find it by using the following suvat equation:

$v_C^2-v_A^2=2as\\a=\frac{v_C^2-v_A^2}{2s}=\frac{108.4^2-152.8^2}{2(3770)}=-1.54 m/s^2$

b)

The force exerted on the pilot by the seat is equal to the normal force.

At point B (half of the loop), we have:

- The normal force exerted by the seat, N, acting towards the center of the loop

- There are no other forces acting towards the center of the loop, so N must be equal to the centripetal force:

$N=m\frac{v_B^2}{r}$ (1)

where $v_B$ is the velocity at position B.

To find the velocity at position B, we notice that the distance covered by the jet between position A and position B is a quarter of a circle:

$s=\frac{\pi r}{2}=\frac{\pi(1200)}{2}=1885 m$

Since we know the deceleration, we can use the suvat equation to find the velocity at point B:

$v_B^2-v_A^2=2as\\v_B=\sqrt{v_A^2+2as}=\sqrt{152.8^2+2(-1.54)(1885)}=132.4 m/s$

Therefore, we can now use eq.(1) to find the normal force exerted by the seat on the pilot at point B:

$N=(55)\frac{(132.4)^2}{1200}=803.4 N$

6 0

2 years ago

g The current density in the 2.9-mm-diameter wire feeding an incandescent lightbulb is 0.33 MA/m2. Part A What's the current den

Rasek [7]

Answer:

$j_{B} = 917.454\,\frac{mA}{m^{2}}$

Explanation:

The current can be calculated by mutiplying the current density by cross section area. The new current density is obtained by the following relation and considering that same current flows through the new wire:

$j_{A}\cdot A_{A} = j_{B}\cdot A_{B}$

$j_{B} = j_{A}\cdot \frac{A_{A}}{A_{B}}$

$j_{B} = j_{A} \cdot \left(\frac{D_{A}}{D_{B}} \right)^{2}$

$j_{B} = \left(0.33\,\frac{mA}{m^{2}} \right)\cdot \left(\frac{2.9\,mm}{0.055\,mm} \right)^{2}$

$j_{B} = 917.454\,\frac{mA}{m^{2}}$

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1 year ago

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2. A 14.8-kg child sits in a 1.30-kg swing. You pull the swing back, lifting it 52.1 cm vertically, and then let go. Determine t

drek231 [11]

To answer this problem, we have to define the two energies, Potential and Kinetic Energy.

Initially when the child is at rest at 52.1 cm, the potential energy PE is at its highest.

PE = m g h

Where,

m = total mass = 14.8 kg + 1.30 kg = 16.10 kg

g = gravitational acceleration = 9.8 m /s^2

<span> h = height = 52.1 cm = 0.521 m</span>

Computing for PE:

PE = 16.10 * 9.8 * 0.521

PE = 82.29 N

As the swing moves past its lowest point h = 0, then all PE is converted to Kinetic Energy KE. Therefore, at lowest point

KE = 82.29

KE = 0.5 m v^2

0.5 (16.10) v^2 = 82.29

<span>v = 3.20 m/s</span>

6 0

1 year ago