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Alex Ar [27]
3 years ago
14

What do the four terrestrial planets and the four gas giant planets have in common?

Physics
1 answer:
vodomira [7]3 years ago
4 0
They all have the same aphelion distances
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A 5kg box slides across the floor with an initial velocity of 5m/s. If the coefficient of kinetic friction between the box and t
4vir4ik [10]

The net force on the block perpendicular to the floor is

∑ F[perp] = F[normal] - mg = 0

so that

F[normal] = (5 kg) g = 49 N

Then

F[friction] = 0.1 F[normal] = 4.9 N

so that the net force parallel to the floor is

∑ F[para] = -4.9 N = (5 kg) a

Solve for the acceleration a :

a = (-4.9 N) / (5 kg) = -0.98 m/s²

Starting with an initial velocity of 5 m/s, the box comes to a stop after time t such that

0 = 5 m/s - (0.98 m/s²) t

⇒   t ≈ 5.1 s

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2 years ago
Which feature of the sun is a section that is cooler than its surroundings?
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Sunspot. Sunspots are temporary cooler areas on the Sun's surface caused by changes in its magnetic field.
7 0
3 years ago
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Answer question quick
Anettt [7]

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Yes , insulation has no role

5 0
3 years ago
In which scenario will the two objects have the greatest gravitational force
Brilliant_brown [7]

Answer: I think the answer C

Explanation:

7 0
2 years ago
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A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to
Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

p_{bot}=p_{atm}=1.013\cdot 10^5 Pa

So, the net pressure acting on the hatch is

p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa

which acts from above.

The area of the hatch is given by:

A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N

8 0
3 years ago
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