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3 years ago

A satellite is in orbit 3.110106 m from the center of Earth. The mass of Earth is 5.98011024 kg Calculate the orbital

Physics

1 answer:

TEA [102]3 years ago

3 0

Answer:

9194.4278 seconds

Explanation:

Orbital period is given by :

T² = (4π²r³) /GM

G = Gravitational constant = 6.67 * 10^-11

M = 5.9801 * 10^24 kg

r = (3.110 * 10^6 m + 6378000 m) =3110000 + 6378000 = 9488000 m

T² = (4π² * 9488000^3) / (6.67* 10^-11 * 5.9801*10^24)

T² = 3.37197 * 10^22 / 398872670000000

T² = 84537504.161415

T = sqrt(84537504.161415)

T = 9194.4278 seconds

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A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

8 0

3 years ago

50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the bui

soldi70 [24.7K]

Answer:

98.13m

Explanation:

Complete question

Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building

CHECK THE ATTACHMENT

From the figure, using trigonometry

Tan(θ ) = opposite/adjacent

Where Angle (θ )= 63°

Opposite= X = height of the building

Adjacent= 50 m

Then substitute the values we have

Tan(63)= X/50

1.9626= X/50

X= 1.9626 × 50

X= 98.13m

Hence, the height of the building is 98.13m

8 0

2 years ago

An electric field of 280000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -

nordsb [41]

<h2>The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.</h2>

Explanation:

Electric field is the ratio of force and charge.

Electric field, E = 280000 N/C

Charge, q = -7.9 C

We have

$E=\frac{F}{q}\\\\280000=\frac{F}{7.9}\\\\F=280000\times 7.9\\\\F=2.21\times 10^6N$

The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.

4 0

3 years ago

A parked car begins to roll down a hill, what can you conclude from that observation?

Fynjy0 [20]

Answer:

its The rolling friction is greater than the force of the car’s weight against the hill.

and A force was required to start the car rolling.

Explanation:

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2 years ago

Which of the following shows a balanced nuclear reaction?

miskamm [114]

Answer:

No. 2

Explanation:

6 0

2 years ago

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