While self-gravity pulls the star inward and tries to make it collapse, thermal pressure (heat created by fusion) pushes outward. These two forces cancel each other out in a main sequence star, thus making it stable.
Answer:
2.24 T
Explanation:
From Electromagnetic Field,
F = BILsin∅................ Equation 1
Where F = Force on the wire, B = Field strength, I = current flowing in the conductor, L = length of the conductor, ∅ = The angle the conductor makes with the magnetic field.
Making B the subject of the equation,
B = F/ILsin∅..................... Equation 2
Given: F = 2.15 N, I = 32 A, L = 3.00 cm = 0.03 m, ∅ = 90° ( the wire is perpendicular to the magnetic field)
Substitute into equation 2
B = 2.15/(32×0.03×sin90°)
B = 2.15/0.96
B = 2.24 T.
Hence the Field strength = 2.24 T
Answer:
The answer to your question should be D.
Explanation:
reactants are on the laft side of arrow and products are on right side of arrow
a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J
b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J
Applying Law of Energy conservation :
K 1+U 1
=K 2+U 2
⇒K 1− r 1GmM
=K 2− r 2 GmM
where M=5.0×10 23kg,r1
=> R=3.0×10 6m and m=10kg
(a) If K 1
=5.0×10 7J and r 2
=4.0×10 6 m, then the above equation leads to
K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J
(b) In this case, we require K 2
=0 and r2
=8.0×10 6m, and solve for K 1:K 1
=K 2 +GmM (r 11− r 21)=6.9×10 7 J
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Answer:
The electric field outside the sphere will be 
.
Explanation:
Given that,
Radius of solid sphere = R
Charge = q
According to figure,
Suppose r is the distance between the point P and center of sphere.
If 
be the volume charge density,
Then, the charge will be,
.....(I)
Consider a Gaussian surface of radius r.
We need to calculate the electric field outside the sphere
Using formula of electric field
Put the value from equation (I)
Hence, The electric field outside the sphere will be .
