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3 years ago

In your own words, explain newtons law of universal gravitation

1 answer:

3 0

His law states that every particle is attracted to every other particle in the universe using a force that is proportional to the product of their masses <span />

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The potential difference between two points is 100 V. If a particle with a charge of 2 C is transported from one of these points

Aleonysh [2.5K]

Answer: 200 J

Explanation: In order to explain this we have consider that the work done in a electric field is given by:

Work= Q*ΔV=2*100=200J

4 0

3 years ago

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Brainlist nd 20 points

nadezda [96]

Answer:

I'm not sure but I think it's 35-39

4 0

3 years ago

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To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(

DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number = 2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2 m

and when x = x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is 0.0549 m

5 0

3 years ago

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

3 years ago

A monatomic gas is adiabatically compressed to 0.250 of its initial volume. Do each of the following quantities change?

Len [333]

Answer:

Given that

V2/V1= 0.25

And we know that in adiabatic process

TV^န-1= constant

T1/T2=( V1 /V2)^ န-1

So = ( 1/0.25)^ 0.66= 2.5

Also PV^န= constant

So P1/P2= (V2/V1)^န

= (1/0.25)^1.66 = 9.98

A. RMS speed is

Vrms= √ 3RT/M

But this is also

Vrms 2/Vrms1= (√T2/T1)

Vrms2=√2.5= 1.6vrms1

Lambda=V/4π√2πr²N

Lambda 2/lambda 1= V2/V1 = 0.25

So the mean free path can be inferred to be 0.25 times the first mean free path

C. Using

Eth= 3/2KT

So Eth2/Eth1= T2/T1

Eth2= 2.5Eth1

Using CV= 3/2R

Cvf= Cvi

So molar specific heat constant does not change

5 0

3 years ago

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