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coldgirl [10]
3 years ago
12

A 62 kg skydiver moving at terminal speed falls 50 m in 1 s. What power is the skydiver expending on the air?

Physics
1 answer:
IgorLugansk [536]3 years ago
3 0

Answer:

P = 30380 W

Explanation:

given,

mass of the skydiver, m = 62 Kg

distance, s = 50 m

time, t = 1 s

we know,

Power= \dfrac{work\ done}{time}

Power= \dfrac{F.s}{t}

F = m g

Power= \dfrac{m g.s}{t}

inserting all the values

P = \dfrac{62\times 9.8\times 50}{1}

P = 30380 W

hence, Power the skydiver expending on the air is 30380 W

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

\rho (r) \  \alpha  \  \delta (r -R)

\rho (r) = k \  \delta (r -R) \ \  at \ \  (r = R)

\rho (r) = 0\ \ since \ r< R  \ \ or  \ \ r>R---- (1)

To find the constant k, we  examine the total charge Q which is:

Q = \int \rho (r) \ dV = \int \sigma \times dA

Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2

∴

\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta  \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2

\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

Thus;

k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

k * \int ^{R}_{0}  \delta (r -R)  r^2dr = \sigma \times  R^2

k * R^2= \sigma \times  R^2

k  =   R^2 --- (2)

Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

\mathbf{\rho (r) =0 \ \  at   \ \  rR}

To verify the units:

\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \  sin \theta  \ dr \ d\theta \  d \phi  \\ \\  Q = \int ^{2 \pi}_{0} \  d \phi  \int ^{\pi}_{0} \ sin \theta  \int ^R_{0} \rho (r) r^2 \ dr

Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  \int ^R_0  * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  *R^2    since  ( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )

\mathbf{Q = 4 \pi R^2  \sigma  }

6 0
3 years ago
a mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. if the mass of the object is 0.20kg and the spr
Reil [10]

Answer:

4.06 Hz

Explanation:

For simple harmonic motion, frequency is given by

f=\frac {1}{2\pi}\times \sqrt{\frac {k}{m}} where k is spring constant and m is the mass of the object.

Substituting 0.2 Kg for mass and 130 N/m for k then

f=\frac {1}{2\pi}\times \sqrt{\frac {130}{0.2}}=4.057670803\\f\approx 4.06 Hz

5 0
3 years ago
The periodic table arranges elements by increasing _________.
Olegator [25]

Answer:

Atomic number

Explanation:

Hope it helps you in your learning process.

4 0
3 years ago
You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.
dem82 [27]
  • Angle (θ) = 60°
  • Force (F) = 20 N
  • Distance (s) = 200 m
  • Therefore, work done
  • = Fs Cos θ
  • = (20 × 200 × Cos 60°) J
  • = (20 × 200 × 1/2) J
  • = (20 × 100) J
  • = 2000 J

<u>Answer</u><u>:</u>

<u>2</u><u>0</u><u>0</u><u>0</u><u> </u><u>J</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0
2 years ago
Read 2 more answers
Which statement explains why astigmatism causes objects to appear blurry?
Elis [28]
'C' is a simple explanation of astigmatism.
4 0
3 years ago
Read 2 more answers
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