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Arlecino [84]
3 years ago
7

When a star goes from the main sequence to the red giant phase, it becomes larger rather than smaller. This is true because grav

ity is getting temporarily overpowered by
A) Antigravity
B) Electron Degeneracy
C) Energy of Fusion
D) Energy of Fission
Physics
1 answer:
GuDViN [60]3 years ago
4 0
I believe this is electron degeneracy, because the star is essentially having too many reactions too fast and collapses in on itself eventually.
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A 22-turn circular coil of wire has diameter 1.02 m. It is placed with its axis along the direction of the Earth's magnetic fiel
ollegr [7]

Answer:

ξ = 0.00845020162 V or 8.4 mV

Explanation:

Magnetic flux measures the total magnetic field that passes through a known area. Magnetic flux describe the effect of magnetic field in a given area. Mathematically,

magnetic flux (Ф) = BA cos ∅

where

A = test area

B = magnetic field

before the flip

Ф = Bπr²N

N = number of turn

magnitude of induced emf = N |ΔФ/Δt|

ξ  = 2Nπr²B/dt

ξ  = 2 × 22 × π × (1.02/2)² × 0.000047/0.2

ξ = 44 × π × 0.51² × 0.000047/0.2

ξ = 44 × π × 0.2601  × 0.000047/0.2

ξ = 0.0005378868  × 3.142/0.2

ξ = 0.00169004032/0.2

ξ = 0.00845020162 V or 8.4 mV

8 0
3 years ago
During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t
leonid [27]

The instant it was dropped, the ball had zero speed.

After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered <em>4.9 meters</em>.

ANYTHING you drop does that, if air resistance doesn't hold it back.

7 0
3 years ago
Read 2 more answers
A satellite with a mass of 2,000 ㎏ is inserted into an orbit that is twice the Earth' s radius. W hat is the force of gravity on
dolphi86 [110]

Answer:

option (b) 4900 N

Explanation:

m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R

F = G Me x m / (R + h)^2

F = G Me x m / 2R^2

F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2

F = 4900 N

4 0
3 years ago
How much work is done when a 214 newton force pushes a sleeping cow 37m across a field.
nata0808 [166]
Hello!

Answer: 
7918 J

Explanation:

We are assuming that the floor (field) is completely horizontal since there's no information about that in the statement. 

We are going to use the following formula:

W= F . Cos \alpha . D

Where:

F=214 N
\alpha =0º
D= 37m

Then, by substituting we have:

W=214N . Cos (0).37m= 7918 N.m=7918 J

8 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
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