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Answer:
Explanation:
1 ) Magnetic field due to a circular coil carrying current
= μ₀I / 2r
I is current , r is radius of the wire , μ₀ = 4π x 10⁻⁷
= 4π x 10⁻⁷ x 15 / (2 x 3.5 x 10⁻²)
= 26.9 x 10⁻⁵ T
2 )
Negative z direction .
The direction of magnetic field due to a circular coil having current is known
with the help of screw rule or right hand thumb rule.
3 )
If we decrease the radius the magnetic field will:__increase _____.
A. Increase.
Magnetic field due to a circular coil carrying current
B = μ₀I / 2 r
Here r is radius of the coil . If radius decreases magnetic field increases.
So magnetic field will increase.
Answer:
2.83 x 10^4m/s
Explanation:
First, let us calculate the time taken by the object to hit the surface of the earth.
H = 4.1 x 10^7m
g = 9.8m/s2
t = √(2H/g)
t = √((2x 4.1 x 10^7) /9.8)
t = 2892.64secs
Now, we can find the velocity with which the object strikes the earth as follows:
V = gt
V = 9.8 x 2892.64
V = 2.83 x 10^4m/s
Answer:
A. 2.8 m/s
Explanation:
Suppose that at the height of 0 m, the path of the pendulum is lowest.
If we use law of conservation of energy, the pendulum will have zero kinetic energy or K.E when it is at highest point, because K.E happens during movement of object and at the highest point all the energy will be P.E
P.E= mgh
Similarly, when the pendulum reaches at the lowest point, the height becomes zero and the P.E also becomes zero. Now all the energy will be K.E
K.E= 1/2 m v^2
In question, we are asked about the speed as the pendulum it reaches the lowest point of its path. Like we mentioned P.E will be zero at lowest point because of zero height. And also we will use law of conservation of energy because no energy has been lost from system.
K.E= P.E
1/2 m v^2 = mgh
Taking sq.root at both sides
v= Under root 2 gh
v=Under root 2x 9.8 m/s x0.4 m
v=Under root 7.84
v=2.8 m/sec
Hope it helps!
