That which flows on uniformly, of which a day is a logical portion, is called time. True or False

Physics

2 answers:

Sergeu [11.5K]3 years ago

8 0

The anwser is true because it has to be time

mr_godi [17]3 years ago

3 0

The answer to the statement is true because the day is of the logical proportion it has to be time.

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Help on the question (blue letters)

Nataliya [291]

Answer:

The top of the mountian is colder

Explanation:

As air rises, the pressure decreases. It is this lower pressure at higher altitudes that causes the temperature to be colder on top of a mountain than at sea level.

5 0

3 years ago

Two cars, one of mass 1400 kg, and the

Vikki [24]

Suppose that, in the x-y plane, the first car is moving to the right so that its velocity is given by the vector

v₁ = (14 m/s) i

and the second car is moving upward so that its velocity vector is

v₂ = (20 m/s) j

Then the total momentum of two cars before their collision is

m₁v₁ + m₂v₂ = (1400 kg) (14 m/s) i + (2300 kg) (20 m/s) j

= (19,600 i + 46,000 j) kg•m/s

Their momentum after the collision is

(1400 kg + 2300 kg) v = (3700 kg) v

where v is the velocity vector of the wreckage.

By conservation of momentum,

(19,600 i + 46,000 j) kg•m/s = (3700 kg) v

Let a and b be the horizontal and vertical components of v, respectively. Then

19,600 kg•m/s = (3700 kg) a ⇒ a ≈ 5.2973 m/s ≈ 5.3 m/s

46,000 kg•m/s = (3700 kg) b ⇒ b ≈ 12.4324 m/s ≈ 12 m/s

so that the final speed of the wreckage is

||v|| = √(a² + b²) ≈ 13.5139 m/s ≈ 14 m/s

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2 years ago

A test tube is in uniform circular motion in a centrifuge, which is a piece of scientific equipment used to separate solids from

Molodets [167]

Since direction is part of velocity, and the object is moving in a circle, its velocity is constantly changing.

3 0

2 years ago

A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package hits the ground, how high

Lelu [443]

Answer:

The package was released at a height of 1015.296 meters.

Explanation:

The package is dropped at an initial velocity different of zero, decelerated and later accelerated by gravity. Let assume that final height is equal to zero, the final height is given by the following equation of motion:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$