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navik [9.2K]
3 years ago
11

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

Physics
1 answer:
mash [69]3 years ago
6 0

a) The acceleration of gravity is 4.96 m/s^2

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

g=\frac{GM}{(R+h)^2}

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2

b)

The critical velocity for a satellite orbiting around a planet is given by

v=\sqrt{\frac{GM}{R+h}}

where we have again:

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s

Converting into km/h,

v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

T=\frac{2\pi (R+h)}{v}

where

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

v = 6668 m/s

Substituting,

T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s

d)

One day consists of:

t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

n=\frac{t}{T}=\frac{86400}{8452}=10.2

e)

The escape velocity for an object in the gravitational field of a planet is given by

v=\sqrt{\frac{2GM}{R+h}}

where here we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

Substituting, we find

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s

f)

We can apply again the formula to find the escape velocity for the rocket:

v=\sqrt{\frac{2GM}{R+h}}

Where this time we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=0, because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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cluponka [151]

The distance traveled by the sprinter in meters is determined as 1.88 m.

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The acceleration of the sprinter is the rate of change of velocity of the sprinter with time.

The acceleration of the sprinter is calculated as follows;

Apply Newton's second law of motion as follows;

F = ma

a = F/m

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a = 693 N / 64 kg

a = 10.83 m/s²

<h3>Distance traveled by the sprinter</h3>

The distance traveled by the sprinter is calculated as follows;

s = ut + ¹/₂at²

where;

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s = ¹/₂at²

where;

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s = (0.5)(10.83)(0.59²)

s = 1.88 m

Thus, the distance traveled by the sprinter in meters is determined as 1.88 m.

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1 year ago
The bonds of oxygen molecules are broken by sunlight. The minimum energy required to break the oxygen-oxygen bond is 495 kJ/mol.
makvit [3.9K]

Answer:

The wavelength of sunlight that can cause this bond breakage is 242 nm

Explanation:

The minimum energy of the sunlight that'll break Oxygen-oxygen bond must match 495 KJ/mol

But 1 mole of any molecule contains 6.02 × 10²³ molecules/mol

Each molecule of Oxygen will require (495 × 10³)/(6.02 × 10²³) = 8.22 × 10⁻¹⁹ J

E = hf

v = fλ

f = v/λ

f = frequency of the sunlight

λ = wavelength of the sunlight

v = speed of light = 3.0 × 10⁸ m/s

E = hv/λ

λ = hv/E

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

λ = (6.63 × 10⁻³⁴)(3 × 10⁸)/(8.22 × 10⁻¹⁹)

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5 0
3 years ago
17. A 25 kg block is initially at rest on a rough, horizontal surface. A horizontal force of 75 N is required to set
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Answer:

0.30581

0.24464

Explanation:

\mu_s = Coefficient of static friction

\mu_k = Coefficient of kinetic friction

F_f = 75 N

F_k = 60 N

Normal force

F_n=mg\\\Rightarrow F_n=25\times 9.81\\\Rightarrow F_n=245.25\ N

Frictional force

F_f=\mu_sF_n\\\Rightarrow \mu_s=\frac{F_f}{F_n}\\\Rightarrow \mu_s=\frac{75}{245.25}\\\Rightarrow \mu_s=0.30581

The coefficient of static friction is 0.30581

Kinetic force

F_k=\mu_kF_n\\\Rightarrow \mu_k=\frac{F_k}{F_n}\\\Rightarrow \mu_s=\frac{60}{245.25}\\\Rightarrow \mu_s=0.24464

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The gravitational force between Mars and the Sun is 1.65\cdot 10^{21} N

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where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

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r is the separation between them

In this problem, we have:

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r=229\cdot 10^6 km = 229\cdot 10^9 m is the average distance Mars-Sun

Substituting into the equation, we find the gravitational force:

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So, the closest answer is

1.65\cdot 10^{21} N

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