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gregori [183]
3 years ago
8

A skydiver of mass 87 kg falls straight down. If air resistance exerts 355 N of force on him as he falls, what is the net force

on the skydiver?
Physics
2 answers:
alukav5142 [94]3 years ago
6 0

Answer:

497.6 N

Explanation:

From the question,

The net force on the skydiver = weight of the skydiver- the resistive force of air

F' = W-F...................... Equation 1

Where W = weight of the skydiver, F = resistive force of air.

But,

W = mg................ Equation 2

Where m = mass of the skydiver, g = acceleration due to gravity.

Substitute equation 2 into equation 1

F' = mg-F............ Equation 3

Given: m = 87 kg, F = 355 N, g = 9.8 m/s²

Substitute these values into equation 3

F' = 87(9.8)-355

F' = 852.6-355

F' = 487.6 N

SpyIntel [72]3 years ago
4 0

Answer:

498 N down

Explanation:

You might be interested in
How do I convert 498.82 cg to mg
mezya [45]
<span><u><em>Answer:</em></u>
498.82 cg is equivalent to 4988.2 mg

<u><em>Explanation:</em></u>
cg stands form centigrams
mg stands for milligrams

From the standards of conversion, to convert from centi to milli, we multiply the amount ny 10

<u>This means that:</u>
1 centigram = 10 milligram

To convert 498.82 cg to mg, all we have to do is <u>cross multiplication</u> as follows:
1 cg ..................> 10 mg
498.82 cg .........> ?? mg

498.82 cg = </span>\frac{498.82*10}{1}<span> = 4988.2 mg

Hope this helps :)</span>
3 0
3 years ago
Read 2 more answers
a 2kg object is moving horizontally with a speed of 4 m/s.how much net force is required to keep the object moving at this speed
Elza [17]
If the object is moving in a straight line with constant speed,
that's a description of " acceleration = zero ".

Zero acceleration means zero net force on the object.

NO net force is 'required' to keep an object moving in a straight line
at constant speed.  In fact, if there IS any net force on the object,
then either its speed or its direction MUST change ... there's no way
to avoid it.

None of this depends on the object's mass, or on the speed or direction
of its motion.
8 0
3 years ago
A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un
Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within  1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

7 0
3 years ago
A 2 kg ball is moving 3 m/s when it starts rolling up a hill.
AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]

Replacing the values on the equation we have:

Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\

Now we have:

h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

6 0
3 years ago
The moon's surface gravity is one-sixth that of the earth. Calculate the weight on the moon of an object that has a mass of 24 k
ad-work [718]
When we say "<span>The moon's surface gravity is one-sixth that of the earth.",
we mean that the acceleration of gravity on the Moon's surface is 1/6 of
the acceleration of gravity on the Earth's surface.

The acceleration of gravity is (9.8 m/s</span>²) on the Earth's surface, so
<span>it would be (9.8/6 m/s</span>²) on the Moon's surface.
<span>
The weight of any object, right now, is

(object's mass) </span>· (acceleration of gravity where the object is located now) .
<span>
If the object's mass is 24 kg and the object is on the Moon right now,
then its weight is 

(24 kg) </span>· (9.8/6 m/s²)

= (24 · 9.8 / 6) kg-m/s²

= 39.2 Newtons
7 0
3 years ago
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