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Olegator [25]
3 years ago
12

A car travelled a distance of 348 km in 6.0 hours. What was it’s speed in (a) km/h and (b) m/sec?

Physics
1 answer:
zavuch27 [327]3 years ago
3 0

Answer:

[(a=58)(b=16)] is the homogeneous mixture

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Which variables are involved in understanding Kepler's third law of motion? (1 point)
Zielflug [23.3K]

The variables which are involved in understanding Kepler's third law of

motion are

  • Orbital velocity
  • Distance to sun
<h3 /><h3>What is Kepler's third law of motion?</h3>

Kepler's third law of motion states that the the square of the orbital period of

a planet is proportional to the cube of the semi-major axis of its orbit. He

also inferred that the greater the distance, the slower the orbital velocity.

This thereby makes option D the most appropriate option as it contains the

orbital velocity and distance to sun variables.

Read more about Kepler's third law of motion here brainly.com/question/777046

7 0
2 years ago
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You must record your daily workouts on a
larisa [96]

I do a yoga mat that could be different to your answer

5 0
3 years ago
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A grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. The angular speed is then increased to 12
PSYCHO15rus [73]

Answer: 6.36

Explanation:

Given

Radius of grindstone, r = 4 m

Initial angular speed of grindstone, w(i) = 8 rad/s

Final angular speed of the grindstone, w(f) = 12 rad/s

Time used, t = 4 s

Angular acceleration of the grinder,

α = Δw / t

α = w(f) - w(i) / t

α = (12 - 8) / 4

α = 4/4 = 1 rad/s²

Number of complete revolution in 4s =

Δθ = w(i).t + 1/2.α.t²

Δθ = 8 * 4 + 1/2 * 1 * 4²

Δθ = 32 + 1/2 * 16

Δθ = 32 + 8

Δθ = 40 rad/s

40 rad/s = 40/2π rpm = 6.36 rpm

Therefore, the grindstone does 6.36 revolutions during the 4 s interval

6 0
3 years ago
Can the object be in motion if the net force acting on it is zero? explain.
artcher [175]
When object travels with uniform velocity, no force acts on it. hence , yes.
3 0
3 years ago
A 100 N force is applied to a 50 kg crate resting on a level floor. The coefficient of kinetic friction is 0.15. What is the acc
Nookie1986 [14]

The acceleration of the crate after it begins to move is 0.5 m/s²

We'll begin by calculating the the frictional force

Mass (m) = 50 Kg

Coefficient of kinetic friction (μ) = 0.15

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) = mg = 50 × 10 = 500 N

<h3>Frictional force (Fբ) =?</h3>

Fբ = μN

Fբ = 0.15 × 500

<h3>Fբ = 75 N</h3>

  • Next, we shall determine the net force acting on the crate

Frictional force (Fբ) = 75 N

Force (F) = 100 N

<h3>Net force (Fₙ) =?</h3>

Fₙ = F – Fբ

Fₙ = 100 – 75

<h3>Fₙ = 25 N</h3>

  • Finally, we shall determine the acceleration of the crate

Mass (m) = 50 Kg

Net force (Fₙ) = 25 N

<h3>Acceleration (a) =?</h3>

a = Fₙ / m

a = 25 / 50

<h3>a = 0.5 m/s²</h3>

Therefore, the acceleration of the crate is 0.5 m/s²

Learn more on friction: brainly.com/question/364384

8 0
2 years ago
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