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3 years ago

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A 66 kg person is parachuting and experiencing a downward acceleration of 2.1 m/s2. The mass of the parachute is 4.0 kg. (a) Wha

t is the upward force on the open parachute from the air?

1 answer:

UkoKoshka [18]3 years ago

6 0

Answer:

539N

Explanation:

Parameters given:

Mass of person = 66kg

Mass of parachute = 4kg

Downward acceleration = 2.1 m/s²

The total force acting on the system is the sum of the force of gravity and the force due to air acting opposite gravity:

F = mg - F(a)

Where g = acceleration due to gravity

Force is given as:

F = ma

=> ma = mg - F(a)

=> F(a) = mg - ma

F(a) = m(g - a)

m is the mass of the entire sustem

m = 66 + 4 = 70kg

=> F(a) = 70(9.8 - 2.1)

F(a) = 70 * 7.7

F(a) = 539N

The upward force on the open parachute from the air is 539N

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andrew11 [14]

Answer:

a) $F=1.044\times 10^9\ N$

b) $F'=1.044\times 10^9\ N$

c) $F_p=1.0672\times10^{-7}\ N$

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

mass of Mars, $M=6.4\times 10^{23}\ kg$

radius of the Mars, $r=3.4\times 10^{6}\ m$

mass of human, $m=80\ kg$

a)

Gravitation force exerted by the Mars on the human body:

$F=G.\frac{M.m}{r^2}$

where:

$G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2}$ = gravitational constant

$F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}$

$F=1.044\times 10^9\ N$

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

$F'=F$

$F'=1.044\times 10^9\ N$

c)

When a similar person of the same mass is standing at a distance of 4 meters:

$F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}$

$F_p=1.0672\times10^{-7}\ N$

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.

Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.

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2 years ago

Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving

Iteru [2.4K]

Explanation:

Let $m_p$ is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

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r is the radius of path,

$r=\dfrac{mv}{qB}$

Time period is given by :

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi m_p}{qB}$

Frequency of proton is given by :

$f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}$

The wavelength of radiation is given by :

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{2\pi m_pc}{qB}$

So, the wavelength of radiation produced by a proton is $\dfrac{2\pi m_pc}{qB}$ . Hence, this is the required solution.

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3 years ago

A person hits a 45-g golf ball. The ball comes down on a tree root and bounces

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Answer:

Maximum height, h = 10 m

Explanation:

It is given that,

Mass of golf ball, m = 45 g = 0.045 kg

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We need to find the height the ball will rise after the bounce. It is based on the conservation of energy such that,

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Answer:The anwser is B i Promise ok

Explanation:

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2 years ago

Read 2 more answers

Harman [31]

Hey user!

your answer is here..

correct option is A. steel

we know that sounds travel faster in solid as compared to gas and liquids. in gas the molecules are very loosely packed and there is lot of space between so it takes more time to pass sound from each other. and in liquid, the molecules are closer as compared to gas hence it will be little faster and in solid, the molecules are very tightly packed so it will be the fastest. and among these options, steel is the only solid so the speed of sound in steel will be the fastest.

and note that the closer the molecules are to each other ( tightly packed ) makes the bond also tighter and less time to pass sound.

cheers!!

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2 years ago