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Ipatiy [6.2K]
3 years ago
14

Please help answer question​

Physics
1 answer:
nika2105 [10]3 years ago
4 0

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, v_t=60\ m/s

Area of cross section, A=0.33\ m^2

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

\dfrac{1}{2}\rho CAv_t^2=mg

Where

\rho is the density of air = 1.225 kg/m³

C is drag coefficient

So,

C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

So, the drag coefficient is 1.01.

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We are given with:
Sensitivity: 100uJ/cm2
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Source Optical Power: 50uW

Required: exposure time

Solution:
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3 0
3 years ago
Astronauts who traveled to the moon were able to float slightly as they walked along its surface. Why were the astronauts able t
Maslowich

Answer:

the weight of astronauts on earth is greater than his weight on d moon

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3 years ago
Question 4 A car of mass 820 kg has a maximum power of 30 kW and moves against a constant resistance of motion to 910 N. Calcula
wolverine [178]

Answer:

P = W / t = m g s / t = m g v       where work by auto = m g s

30 kw = 30000 watts = 30000 J / s

Work wasted = F v       as shown above relating work and power

Work done against incline = m g s sin 8.75  and power against incline

= m g v sin 8.75 = 1222 v Joules / sec

power in moving auto = power available - power lost to friction

power in moving auto = 30000 - resistance = 30000 - 910 v

1222 v = 30000 - 910 v

v = 30000 / 2132 = 14 m/s

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7 0
3 years ago
PLEASE PLEASE HELP!!!Answer the following questions
adoni [48]

Answer:

u have to stop

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6 0
3 years ago
A box of mass 72 kg is at rest on a horizontal frictionless surface. A constant horizontal force F then acts on the box and acce
zvonat [6]

Answer:

162\ N

Explanation:

We\ are\ given\ that,\\Mass\ of\ the\ box=72\ kg\\Distance/Displacement\ travelled\ by\ the\ box\ during\ the\ application\ of\\ Force=13\ m\\Time\ taken\ for\ it\ to\ displace=3.4 \seconds\\Now,\\As\ we\ know\ that,\\Force=Mass*Acceleration\\In\ order\ to\ know\ force\ we\ need\ to\ know\ the\ acceleration.\ Lets\ find\\ that\ out.\\We\ would\ use\ Newton's\ Third\ Equation\ Of\ Motion\ and\ Newton's\\ First\ Equation\ Of\ Motion:\\2as=v^2-u^2\\v=u+at

Hence,\\First\ lets\ consider\ Newtons\ First\ Equation\ Of\ Motion:\\v=u+at\\v-u=at\\Hence\ now\ lets\ move\ on\ to\ Newton's\ Third\ Law\ Of\ Motion:\\2as=v^2-u^2\\2as=(v+u)(v-u)\\Substituting\ (v-u)=at,\\2as=at(v+u) \\Hence,\ as\ the\ body\ moves\ from\ rest,\ u=0\\So,\\2as=at*v\\Cancelling\ a\ at\ both\ the\ sides\ we\ get,\\2s=vt\\Hence,\\Lets\ plug\ in\ the\ values\ of\ Displacement\ and\ Time,\ Shall\ we?\\Hence,\\2*13=3.4v\\26=3.4v\\\frac{26}{3.4}=v\\v \approx 7.647\ m/s

Now,\\As\ we\ know\ that,\\a=\frac{v-u}{t} [Equation\ for\ acceleration]\\Hence,\\a=\frac{v}{t} [As\ u=0]\\Hence,\\Acceleration=\frac{7.647}{3.4} \\Acceleration \approx 2.25\ m/s^2\\

Hence,\\Now,\\As\ by\ using\ expression\ for\ Force,\\Force= Mass*Acceleration\\Here,\\Force\ exerted\ on\ the\ box=72*2.25= 162 N

3 0
3 years ago
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