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hjlf
3 years ago
8

Physics help please

Physics
1 answer:
olchik [2.2K]3 years ago
8 0

Answer:

i think the answer is 0.001m³

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A a healthy diet and exercise have an impact on the overall health of
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Answer:

This would be answer choice A

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3. The figure below shows the motion of a car. It starts from the origin, O travels 8m
Nuetrik [128]

Answer:

i. -4m

ii. 20m

Explanation:

The car travels 8m to the east, then travels 12m to the west which is the opposite of the east. Going west, the car travels 8m back to the origin point and then another 4m due west to make 12m. The displacement from the origin point is -4 (the negative sign shows the direction because displacement is a vector quantity)

Total distance = 8m going east + 8m back to origin + 4m west = 20m

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What is the magnitude of fs on an object lying on a flat surface without moving, on
Degger [83]

The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

  • magnitude of force on the object, F = 10 N
  • angle between the object and the horizontal flat surface = 0⁰

Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N

Therefore, the magnitude of the force acting on the object is 10 N.

Learn more here: brainly.com/question/19887955

7 0
2 years ago
The planet Krypton has a mass of 7.6 × 1023 kg and radius of 1.7 × 106 m. What is the acceleration of an object in free fall nea
olga55 [171]

Answer:

17.55 m/s²

Explanation:

Parameters given:

Mass of Krypton, M = 7.6 * 10^23 kg

Radius, R = 1.7 * 10^6 m

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Acceleration due to gravity of planet of mass M is given as:

g = GM/R²

Since the object is close to the surface of Krypton, we can say that the distance from the Centre of Krypton is the radius of the planet Krypton.

Therefore,

g = (6.6726 * 10^(-11) * 7.6 * 10^23)/(1.7 * 10^6)²

g = 17.55 m/s²

5 0
3 years ago
If the room radius is 4.5 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out, what is the mini
kondaur [170]
<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration. The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N. a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r. The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec. So a_N = 114 m/sec^2. g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>
3 0
3 years ago
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