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Helga [31]
3 years ago
15

A skydiver of mass m jumps from a hot air balloon and falls a distance d before reaching a terminal velocity of magnitude v . As

sume that the magnitude of the acceleration due to gravity is g .
a. What is the work (Wd) done on the skydiver, over the distance , by the drag force of the air?
b. Find the power (P d) supplied by the drag force after the skydiver has reached terminal velocity v.
Physics
1 answer:
stiks02 [169]3 years ago
5 0

Answer:

Explanation:

work done by gravitational force during fall of distance d = mgd

work done by drag = w ( let )

mgd - w = 1/2 m v²

w = m g d - 1/2 m v²

so work done by drag force

= m g d - 1/2 m v²

b ) when terminal velocity is reached , drag force = mg

power supplied by drag force = force x velocity

= mg x v

P_d = mgv

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Please help, thank you!
VARVARA [1.3K]

gurlll this need way more points

8 0
3 years ago
A 50.0-turn circular coil of radius 5.00 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0
Blababa [14]

Answer:

The maximum torque in the coil is 4.9\times 10^{-3}\ N-m.

Explanation:

Given that,

Number of turns in the circular coil, N = 50

Radius of coil, r = 5 cm

Magnetic field, B = 0.5 T

Current in coil, I = 25 mA

We need to find the magnitude of the maximum possible torque exerted on the coil. The magnetic torque is given by :

\tau=NIAB\ \sin\theta

For maximum torque, \theta=90^{\circ}

\tau=NIAB\\\\\tau=50\times 25\times 10^{-3}\times \pi (0.05)^2\times 0.5\\\\\tau=4.9\times 10^{-3}\ N-m

So, the maximum torque in the coil is 4.9\times 10^{-3}\ N-m.

4 0
3 years ago
Identify the energy transformation is image below
-BARSIC- [3]

Answer:

chemical to mechanical

hope it helps have a nice day

3 0
2 years ago
An airplane has a speed of 135 mi/h in still air. It is flying straight north so that it is always directly above a north-south
vlabodo [156]

Answer:

The answer is below

Explanation:

Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.

The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph

vₐ = vₙ + vₐₙ

vₐ = vₐₙ

Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.

The direction of the wind θ is:

sin(θ / 2) = vₙ / 2vₐ

sin(θ / 2) = 70/ (2*135)

sin(θ / 2) = 0.2593

θ / 2 = sin⁻¹(0.2593) = 15

θ = 30⁰

2α = 180° - 30°

2α = 150°

α = 75°

a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.

8 0
3 years ago
A temperature of 200°F is equivalent to approximately
aev [14]
<span>93.3°C
A temperature in Fahrenheit (°F) can be converted to Celsius (°C), using the formula
[°C] = ([°F] − 32) ×  5⁄9. Here we have to convert a temperature of 200°F in to Celsius. Thus Subtract 32 from Fahrenheit and multiply by 5 then divide by 9 . That is (200°F - 32) × 5/9=168 × 5/9
                                          =840/9
                                          =93.333333333°C
                                          = 93.3°C</span>
7 0
2 years ago
Read 2 more answers
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