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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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DIA [1.3K]

Answer:

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Explanation:

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ΣF = ma

we need to find 'a' first, using the definition of 'a' we get equation:

a = (Vf-Vo)/Δt

a = (30m/s)/10s

a = 3 m/s^2

now substitute into top equation

ΣF = ma

Fengine = (1200kg)(3m/s^2)

Fengine = 3600N

5 0
2 years ago
Read 2 more answers
Q 1 . How many significant figures are in the following measurement? 0.0009(1 point)
Crazy boy [7]

Here we have some questions about experimental errors.

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0.0009

The number of significant figures is the number of known digits that are not the leading zeros.

Here we can see four leading zeros, and a single-digit different than zero, which is a 9.

Then we have only one significant figure, the 9.

Q2) Here we will use the measure that is the less exact, as the error of that measure may be larger than the smaller significant figures of the other measures.

Then:

31.2 lb + 38.02lb + 45 lb

The worst measure is 45lb, so the smallest significant figure that we should use is the first one at the left of the decimal point, then we need to round the other two measures to the next whole number, we will get:

31 lb + 38 lb + 45 lb = 114lbs

Q3) We know that the measure is 11.5 seconds and the uncertainty of 1.7%, then the uncertainty will be the 1.7% of the above measure:

(1.7%/100%)*11.5 s = 0.1955 s

Notice that our measure has one significant figure after the decimal point, so we need to round the error to the same significant figure.

0.1955 s ≈ 0.2s

Then the measure is:

11.5 s ± 0.20 s

Q4) We have the measure:

312.0 mph ± 3.9 mph.

The percent uncertainty will be the quotient between the error and the measure times 100%, or:

(3.9 mph/312.0 mph)*100%  = 1.25%

This is a percent error, we do not need to round this.

If you want to learn more, you can read:

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Which is a freshwater region characterized as a large standing body of water? A. lake B. river C. ocean D. stream
Alexus [3.1K]
This is A.) lake. A river is a small amount of water that isn't always fresh water. A stream is too small. And an ocean is made of salt water.
4 0
3 years ago
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Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.320 mm wide. The diffraction pattern is observed
DanielleElmas [232]

Answer:

W = 10.28\ mm

Explanation:

Given,

Red light wavelength = 633 nm

width of slit = 0.320 mm

distance,d = 2.60 m

Condition of first maximum

a sin \theta_1 = m\lambda

\theta_1 =sin^{-1}(\dfrac{m\lambda}{a})

m = 1

\theta_1 =sin^{-1}(\dfrac{633\times 10^{-9}}{0.32\times 10^{-3}})

\theta_1 = 0.1133^\circ

Width of the first minima

y_1 = L tan \theta_1

y_1 = 2.60\times tan( 0.11331)

y_1 = 5.14 \ mm

Now, width of the central region

W = 2 y_1

W = 2\times 5.14

W = 10.28\ mm

8 0
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a student pushes a 0.500 kg trolley along a frictionless surface and accelerates it from rest to 4m/s. how much kinetic energy d
aalyn [17]
  • mass=m=0.5kg
  • velocity=4m/s

\\ \sf\longmapsto KE=\dfrac{1}{2}mv^2

\\ \sf\longmapsto KE=\dfrac{1}{2}(0.5)(4)^2

\\ \sf\longmapsto KE=0.25(16)

\\ \sf\longmapsto KE=4J

7 0
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