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Radda [10]
3 years ago
11

Which one of the following does not accurately describe the universal gravitational law? Question 10 options: A) Gravitational f

orce is proportional to the square of the distance between the centers of both bodies. B) Every object in the universe that has mass attracts every other object in the universe that has mass. C) Gravitational force is proportional to the product of the two masses. D) The object with the smaller mass does most of the moving because the object with the larger mass has too much inertia to move any noticeable amount.
Physics
2 answers:
just olya [345]3 years ago
5 0

Answer:

Choice C seems to be the right answer.

Explanation:

Kaylis [27]3 years ago
4 0

Answer:

D is right answer

Explanation:

f =  \frac{Gm1m2}{ {r}^{2} }

•Gravitational force directly proportional to product of their masses

•And is inversely proportional to square of distance between them

•Every object in the universe that has mass attracts every other object in the universe that has mass.

THEREFORE OPTION 4 WILL BE THE ANSWER

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A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0
ankoles [38]

Answer:

a.  B = 0.20T

b.  u = 17230.6 J/m³

c.  E = 0.236J

d.  L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

B=\frac{\mu_o n i}{L}               (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current  = 90.0A

You replace the values of the parameters in the equation (1):

B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

E=uV           (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

V=AL

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3

Then, the energy contained in the solenoid is:

E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H

The inductance of the solenoid is 5.84*10^-5 H

3 0
3 years ago
The rectangular coordinates of a point are (5.00, y) and the polar coordinates of
Otrada [13]

Answer:

Explanation:

Polar coordinates formula

Summary. To convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) : x = r × cos( θ ) y = r × sin( θ )

6 0
2 years ago
Consider a simple ideal Rankine cycle with fixed turbine inlet conditions. What is the effect of lowering the condenser pressure
mr Goodwill [35]

Answer:

The effect of lowering the condenser pressure on different parameters is explained below.

Explanation:

The simple ideal Rankine cycle is shown in figure.

Effect of lowering the condenser pressure on

(a). Pump work input :- By lowering the condenser pressure the pump work increased.

(b) Turbine work output :- By lowering the condenser pressure the turbine work increased.

(c). Heat supplied :- Heat supplied increases.

(d). Heat rejected :- The heat rejected may increased  or decreased.

(e). Efficiency :- Cycle  efficiency is increased.

(f). Moisture content at turbine exit :- Moisture content increases.

8 0
2 years ago
Naming covalent compounds <br> P4S5
Andrew [12]

Answer:

what the heck is sakurfa

Explanation:

3 0
2 years ago
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, h
Alex73 [517]

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

5 0
3 years ago
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