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Vsevolod [243]
3 years ago
12

A ball is dropped from the top of a tall building. As the ball falls, the upward force of air rsistance becomes equal to the dow

nward pull of gravity. When these two forces become equal in magnitude the ball will: a) flatten due to the forces b) fall at a constant speed c) continue to speed up d) slow to a stop
Physics
1 answer:
iren [92.7K]3 years ago
5 0
Fall at a constant speed

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The volume flow rate of the water supplied by a well is 2.0×10−4m3/s.The well is 40.0 m deep. (a) What is the power output of th
MaRussiya [10]

Answer:

a). P=78.4W

b). P=392kPa

c.) It must be at the bottom

Explanation:

Given:

Volume flow V_f=2.0x10^{-4}m^3/s

Well depp h=40.m

a.

The power output of the pum

W=F*d

F=m*g

m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg

W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2

W=78.4J

P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W

b.

The pressure of difference the pum

ΔP=p*g*y'

ΔP=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa

P=392kPa

c.

It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way  

4 0
3 years ago
A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di
siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

  • The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\

part a)

Electric Field strength at point P: a = 0.75 m

E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

6 0
3 years ago
In Ampere's law, ∮????⃗∙????????⃗=????0???? the direction of the integration around the path:
kupik [55]

Answer:

C) must be such as to follow the magnetic field lines.

Explanation:

Ampere's circuital law helps us to calculate magnetic field due to a current carrying conductor. Magnetic field due to a current forms closed loop around the current . If a  net current of value I creates a magnetic field B around it , the line integral of magnetic field around a closed path becomes equal to μ₀ times the net current . It is Ampere's circuital law . There may be more than one current passing through the area enclosed by closed curve . In that case we will take net current by adding or subtracting them according to their direction.

It is expressed as follows

∫ B.dl = μ₀ I . Here integration is carried over closed path . It may not be circular in shape. The limit of this integration must follow magnetic field lines.

the term ∫ B.dl is called line integral of magnetic field.

3 0
3 years ago
The orbital period of an object is 2 x 10^7 and its total radius is 4 x 10^4 m.
SVETLANKA909090 [29]

Answer:

12566

Explanation:

Use formula: vt=2pir/t

7 0
3 years ago
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A student has paper sitting at rest on their desk. Which of the following statements best describes this situation?
ValentinkaMS [17]

C. The forces acting on the paper are balanced .

-There are several forces acting on your paper, but they balance each other.

8 0
3 years ago
Read 2 more answers
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