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3 years ago

the sputnik 1 satellite orbited earth (mass=5.98 x 10^24 kg) in a circle of radius 6.96 x 10^6 m. what was its orbital velocity?

Physics

2 answers:

user100 [1]3 years ago

8 0

Answer:

v = 4.79 10³ m / s

Explanation:

For this exercise we will use the universal law of gravitation and Newton's second law

F = G M m / r²

where G is the gravitational constant, m the mass of the satellite, M the mass of the Earth, r the distance from the center of the planet to the satellite

F = m a

since the satellite has a circular path, the acceleration is centripetal

a = v² / r

we substitute

G M m / r² = m v² / r

G M / r = v²

We calculate

v² = 6.67 10⁻¹¹ 5.98 10²⁴ / 6.96 10⁶

v = √ (22.94 10⁶)

v = 4.79 10³ m / s

galben [10]3 years ago

7 0

Answer:

7570 m/s

Explanation:

take the square root of: 6.67E-11× 5.98E24/6.96E6

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stealth61 [152]

Answer :

(a) The acceleration of the car is, $-2m/s^2$

(b) The distance covered by the car is, 150 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

The acceleration of the car is, $-2m/s^2$

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

By solving the term, we get:

$s=150m$

The distance covered by the car is, 150 m

8 0

3 years ago

It says find the slope for each line I'm stuck on number one can you help me

Allushta [10]

$\text{slope}=\frac{y_2-y_1}{x_2-x_1}$

(-2, -1)(3, 1)

Therefore,

$\text{slope}=\frac{1-(-1)}{3-(-2)}=\frac{1+1}{3+2}=\frac{2}{5}$

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1 year ago

What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$

Substituting:

$\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}$

Calculating:

$\displaystyle \Delta x=\sqrt{3.0175}$

$\boxed{\displaystyle \Delta x=1.74\ m}$

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3 years ago

What is the quantity of motion ??

horsena [70]

Answer:

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Explanation:

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docker41 [41]

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3 years ago

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