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4 years ago

What causes air resistance when it comes to parachutes? Use ideas about particles in your answer.

Physics

1 answer:

Kobotan [32]4 years ago

6 0

Answer:

When a parachute opens, it is a second force that works against gravity. This is air resistance. Air collects under the fabric parachute, pushing it up as gravity pulls the heavy object attached to it down. This pushing slows the fall of the object by resisting the air under the parachute.

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2. A test model for an experimental gasoline engine does 45 J of work in one cycle

loris [4]

Answer: im just trying to get some points so i can answer my question.

sorry

Explanation:

7 0

3 years ago

A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 2.5 m away. He then wants to take a p

Temka [501]

Answer:

0.004 m away from the film

Explanation:

u = Object distance

v = Image distance

f = Focal length = 50 mm

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m$

The image distance is 0.051 m

When u = 50 cm

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m$

The image distance is 0.055 m

The lens has moved 0.055-0.051 = 0.004 m away from the film

3 0

3 years ago

Determine the magnitude of the resultant force acting on a 5 −kg particle at the instant t=2 s, if the particle is moving along

Alex787 [66]

Answer:

F = 296.7N

Explanation:

The x and y component of the position vector are given by:

$x(t) =rcos\phi, y(t) = rsin\phi\\ \frac{dx}{dt} =\frac{dr}{dt} cos\phi - rsin\phi\frac{d\phi}{dt} , \frac{dy}{dt}=\frac{dr}{dt}sin\phi + rcos\phi\frac{d\phi}{dt}\\ \frac{d^2x}{dt^2} = \frac{d^2r}{dt^2} cos\phi - \frac{dr}{dt} sin\phi\frac{d\phi}{dt} -\frac{dr}{dt} sin\phi\frac{d\phi}{dt} - r(cos\phi\frac{d^2\phi}{dt^2} + sin\phi\frac{d^2\phi}{dt^2})$

$\frac{d^2y}{dt^2} = \frac{d^2r}{dt^2} sin\phi + \frac{dr}{dt} cos\phi\frac{d\phi}{dt}+\frac{dr}{dt}cos\phi\frac{d\phi}{dt}+r(-sin\phi\frac{d^2\phi}{dt^2} +cos\phi\frac{d^2\phi}{dt^2})$

At t = 2s:

$\phi(2) = 1.5t^2-6t= -6\\ \frac{d\phi}{dt}(2) = 3t-6=0\\ \frac{d^2\phi}{dt^2}=3\\r(2)=2t+10=14\\ \frac{dr}{dt}=2\\\frac{d^2r}{dt^2} = 0$

Plugging in:

$\frac{d^2x}{dt^2}=-42(cos(-6) + sin(-6))=-52\frac{m}{s^2} \\\frac{d^2y}{dt^2} = 42(cos(-6)-sin(-6))=28.6\frac{m}{s^2}$

The resulting force F is:

$F = m\sqrt{(\frac{d^2x}{dt^2})^2 + (\frac{d^2y}{dt^2})^2}=296.7N$

5 0

3 years ago

Iron oxide reacts with aluminum to give aluminum oxide and iron. What kind of chemical reaction is this?

dexar [7]

a displacement reaction

5 0

3 years ago

A pilot, whose mass is 84.0 kg, makes a loop-the-loop in a fast jet. Assume that the jet maintains a constant speed of 345 m/s a

arsen [322]

Answer:

The apparent weight is 5 times greater than the original weight at the bottom.

Explanation:

Given:

Mass of the pilot, m = 84 kg

Velocity of the jet, v = 345 m/s

Radius of the loop, R = 3.033 km = 3.033 * 10^3 m

We have to find the apparent weight that the pilot feels.

Let the apparent weight be "N" .

Apparent weight :

It is based on where is the position of the pilot in the loop-the-loop.
The apparent weight is the highest at the bottom of the loop-the-loop.
Because the weight acts down and the normal force acts towards the center of the circle.

From the FBD shown we can say that :

apparent weight (N)

⇒ $N=mg+\frac{mv^2}{R}$