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Y_Kistochka [10]
2 years ago
5

In part A of the lab we see that the magnetic field of a long straight wire __. a. increases with distance in a linear relations

hip b. increases with distance in a non-linear relationship c. decreases with distance in an inverse (1/r) relationship d. decreases with distance in an inverse-square (1/r2) relationship
Physics
1 answer:
Amanda [17]2 years ago
6 0

Explanation:

a long straight wire __. a. increases with distance in a linear relationship b. increases with distance in a non-linear relationship c. decreases with distance in an inverse (1/r) relationship d. decreases with distance in an inverse-square (1/r2) relationship

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jenyasd209 [6]

The new absloute temperature should be 4t.

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6 0
2 years ago
He diagram shows a person holding a bow and arrow.
olganol [36]

Answer:

B Relesing the string

Explanation:

got it right on edge

5 0
3 years ago
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A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s
kondaur [170]

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

T^2=\frac{4\pi ^2}{GM} a^2

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

M = 3.5M_{sun} = 6.96*10^{30} kg

a=4.2AU = 6.28*10^{11} meters

G = 6.67*10^{-11}m^3kg^{-1}s^{-2}

We put this into Kepler's law and get:

T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.

This when converted to years is 4.6 years.

4 1
3 years ago
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What did the scientists deduce from the fact that the ants eyes of the desert have multiple lenses
Evgesh-ka [11]
That the pupl is smaller than the nulian hope this helped

4 0
3 years ago
You set out to design a car that uses the energy stored in a flywheel consisting of a uniform 101-kg cylinder of radius r that h
Ket [755]
Ok, assuming "mj" in the question is Megajoules MJ) you need a total amount of rotational kinetic energy in the fly wheel at the beginning of the trip that equals
(2.4e6 J/km)x(300 km)=7.2e8 J
The expression for rotational kinetic energy is

E = (1/2)Iω²  

where I is the moment of inertia of the fly wheel and ω is the angular velocity.  
So this comes down to finding the value of I that gives the required energy.  We know the mass is 101kg.  The formula for a solid cylinder's moment of inertia is

 I = (1/2)mR²

We want (1/2)Iω² = 7.2e8 J and we know ω is limited to 470 revs/sec.  However, ω must be in radians per second so multiply it by 2π to get 
ω = 2953.1 rad/s
Now let's use this to solve the energy equation, E = (1/2)Iω²,  for I:
I = 2(7.2e8 J)/(2953.1 rad/s)² = 165.12 kg·m²

Now find the radius R,

 165.12 kg·m² = (1/2)(101)R²,
√(2·165/101) = 1.807m

R = 1.807m
8 0
3 years ago
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