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andreev551 [17]
3 years ago
7

Spacecraft have been sent to Mars in recent years. Mars is smaller than Earth and has correspondingly weaker surface gravity. On

Mars, the free-fall acceleration is only 3.8m/s2. What is the orbital period of a spacecraft in a low orbit near the surface of Mars?

Physics
1 answer:
Lady bird [3.3K]3 years ago
5 0

Answer:

5900sec = 98mins

Explanation:

The solution is in the attached file below

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3 years ago
015 10.0 points
Salsk061 [2.6K]

Answer:

20.996 m

Explanation:

Given:

Initial velocity, u=0 \textrm{ m/s}

Final velocity, v=12.4062 \textrm{ m/s}

Total time taken, t_{Total} = 7.13 s.

∴ Acceleration is given as,

a=\frac{v-u}{t_{Total}}=\frac{12.4062-0}{7.13}=1.74 m/s²

Now, using Newton's equation of motion, we find the displacement.

Displacement is given as:

s=ut+\frac{1}{2} at^{2}

Plug in 0 for u, 4.91257 for t and 1.74 for a. Solve for s.

This gives,

s=0+\frac{1}{2} \times 1.74 \times (4.91257)^{2}=20.996 \textrm{ m}

Therefore, the train's displacement in the first 4.91257 s of motion is 20.996 m.

7 0
3 years ago
A shopping cart with two nice kids, rolls off a horizontal roof ledge that is 50
Alisiya [41]

Answer:

9.39 m/s

Explanation:

Using the y-direction, we can solve for the time t it takes for the cart to reach the ground.

Assume the up direction is positive and the down direction is negative.

  • v₀ = 0 m/s
  • a = -9.8 m/s²
  • Δy = -50 m
  • t = ?

Find the constant acceleration equation that contains these four variables.

  • Δy = v₀t + 1/2at²  

Substitute known values into this equation.

  • -50 = (0)t + 1/2(-9.8)t²

Multiply and simplify.

  • -50 = -4.9t²

Divide both sides of the equation by -4.9.

  • 10.20408163 = t²

Square root both sides of the equation.

  • t = 3.194382825

Now we can use this time t and solve for v₀ in the x-direction. Time is most often our link between vertical and horizontal components of projectile motion.  

List out known variables in the x-direction.

  • v₀ = ?
  • t = 3.194382825 s
  • a = 0 m/s²
  • Δx = 30 m

Find the constant acceleration equation that contains these four variables.

  • Δx = v₀t + 1/2at²  

Substitute known values into the equation.

  • 30 = (v₀ · 3.194382825) + 1/2(0)(3.194382825)²

Multiply and simplify.

  • 30 = v₀ · 3.194382825

Divide both sides of the equation by 3.194382825.

  • v₀ = 9.391485505

The cart was rolling at a velocity of 9.39 m/s (initial velocity) when it left the ledge.

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