Answer:581.87 K
Explanation:
Given
Sphere is melted to form a square
Let the radius of sphere be r and square has a side a
Therefore

Surface area of sphere 
Surface area of cube 
Total emmisive remains same
Thus 







Answer:
After 1 sec = 4.9 m
After 2 sec = 19.6 m
After 3 sec = 44.1 m
After 4 sec = 78.4 m
After 5 sec = 122.5 m
Explanation:
After 1 sec:
<em>u=0m/s t=1 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(1) + (1/2)(9.8)(1²) = 4.9m
After 2 sec:
<em>u=0m/s t=2 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(2) + (1/2)(9.8)(2²) = 19.6m
After 3 sec:
<em>u=0m/s t=3 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(3) + (1/2)(9.8)(3²) = 44.1m
After 4 sec:
<em>u=0m/s t=4 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(4) + (1/2)(9.8)(4²) = 78.4m
After 5 sec:
<em>u=0m/s t=5 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(5) + (1/2)(9.8)(5²) = 122.5m
Answer:
B. NET force: 2 resultant motion: left
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C. Net force: 3 Resultant motion: Left
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D. Net Force: 7 Resultant motion: right
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E. Net Force:0 resultant motion: NO MOTION
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F. NET Force: 3 resultant motion: Down
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G. NET FORCE: 10 resultant motion: up
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H. Net force: 3 Resultant motion: left
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I. Net force: 50 Resultant motion: right
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J. NET FORCE: 75 Resultant motion: down
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K. Net force :200 Resultant motion: Right
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L. Net force: 0 resultant motion:No motion
Explanation:
Given Information:
Length of wire = 132 cm = 1.32 m
Magnetic field = B = 1 T
Current = 2.2 A
Required Information:
(a) Torque = τ = ?
(b) Number of turns = N = ?
Answer:
(a) Torque = 0.305 N.m
(b) Number of turns = 1
Explanation:
(a) The current carrying circular loop of wire will experience a torque given by
τ = NIABsin(θ) eq. 1
Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.
We know that area of circular loop is given by
A = πr²
where radius can be written as
r = L/2πN
So the area becomes
A = π(L/2πN)²
A = πL²/4π²N²
A = L²/4πN²
Substitute A into eq. 1
τ = NI(L²/4πN²)Bsin(θ)
τ = IL²Bsin(θ)/4πN
The maximum toque occurs when θ is 90°
τ = IL²Bsin(90)/4πN
τ = IL²B/4πN
torque will be maximum for N = 1
τ = (2.2*1.32²*1)/4π*1
τ = 0.305 N.m
(b) The required number of turns for maximum torque is
N = IL²B/4πτ
N = 2.2*1.32²*1)/4π*0.305
N = 1 turn