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MaRussiya [10]
1 year ago
11

an object is acted on by a drag force with a magnitude that is proportional to the speed. the object accelerates downward at 3.0

0 m/s2 when it is falling with a speed 20.0 m/s. what is the terminal speed of the object as it is falling?
Physics
1 answer:
Nutka1998 [239]1 year ago
7 0

The terminal speed of the object falling down is 66.67 m/s.

The terminal speed acquired by the body when,

Weight of the body = Drag force of the body

It is given,

Drag force is directly proportional to the speed,

So,

F = CV

Where F is drag force,

V is the speed,

C is the constant,

So, it can be written as C = F/V.

The weight of the body = mg

The weight of the body = 10m

M is the mass and g is the acceleration due to gravity,

The drag force when the speed is 20m/s.

Drag force = ma

a is the acceleration during the drag force which is given to be 3m/s²,

Drag force = 3m

Now we can write,

F₁/V₁ = F₂/V₂

F₁ is the drag force at 20m/s speed.

F₂ is the weight of the body and V₂ is the terminal speed,

Now, it can be written,

3m/20 = 10m/V₂

V₂ = 66.67 m/s.

So, the terminal speed is 66.67m/s.

To know more about terminal speed, visit,

brainly.com/question/14605362

#SPJ4

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loris [4]

Answer:

The angle for the forward Mach line is 19.47°

The angle for the rearward Mach line is 5.21°

Explanation:

From table A-1 (Modern Compressible Flow: with historical perspective):

\frac{P_{o} }{P_{1} } =36.73 (M₁ = 3)

If Po₁ = Po₂

\frac{P_{o2} }{P_{2} } =\frac{P_{o1} }{P_{1} } *\frac{P_{1} }{P_{2} } =36.73*\frac{1}{4} =91.825

Table A-1:

\frac{P_{o2} }{P_{2} } =91.825,M_{2} =3.63

Table A-5:

v₁ = 49.76°

μ₁ = 19.47°

v₂ = 60.55°

μ₂ = 16°

θ = 60.55 - 49.76 = 10.79°

The angle for the forward Mach line is:

μ₁ = 19.47°

The angle for the rearward Mach line is:

θr = μ₂ - θ = 16 - 10.79 = 5.21°

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3 years ago
Yes or no
Tju [1.3M]

Answer:

1. No, it is a colloid

2. Yes

Explanation:

8 0
2 years ago
Acceleration is generally defined as the time rate of change of velocity. When can it be defined as the time rate of change of s
Lena [83]

Answer:

When the velocity doesn't change its direction

Explanation:

Since velocity vector has 2 components: direction and magnitude, and speed is the velocity's magnitude. So if the velocity doesn't change its direction, we essentially use its magnitude, aka speed, to calculate the rate of change for acceleration.

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If an astronaut has a mass of 16 Kg on Earth, what would be his mass on the moon and on the space station
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Answer:

The astronaut's mass is 16 kg.

Explanation:

Mass can be defined as a measure of the amount of matter an object or a body comprises of. The standard unit of measurement of the mass of an object or a body is kilograms.

Irrespective of the location of an object or a body at a given moment in time, the mass (amount of matter that they're made up of) is constant. This ultimately implies that, whether you're in the moon, space, earth or any other place, your mass remains the same (constant).

Therefore, if an astronaut has a mass of 16 Kg on Earth, his mass on the moon and on the space station would remain the same, as his original mass of 16 Kg because mass is indestructible.

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3 years ago
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Based on the formula for calculating impulse, the impulse of the speed bump to the car is 2500 Ns.

<h3>What is the impulse of the speed bump?</h3>

  • Impulse = change in momentum
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Change in momentum = mu - mv

where u is initial velocity

v is final velocity

Impulse = 500 * 20  - 500 * 15

Impulse = 2500 Ns

Therefore, the impulse of the speed bump to the car is 2500 Ns.

Learn more about impulse at: brainly.com/question/297527

5 0
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