Answer:
a) 
b) 
Explanation:
From the question we are told that:
Wire Length 
Resistance 
Force 
Power 
a)
Generally the equation for Power is mathematically given by

Therefore



b)
Generally the equation for Magnetic Field is mathematically given by



Answer:
z = 0.8 (approx)
Explanation:
given,
Amplitude of 1 GHz incident wave in air = 20 V/m
Water has,
μr = 1
at 1 GHz, r = 80 and σ = 1 S/m.
depth of water when amplitude is down to 1 μV/m
Intrinsic impedance of air = 120 π Ω
Intrinsic impedance of water = 
Using equation to solve the problem

E(z) is the amplitude under water at z depth
E_o is the amplitude of wave on the surface of water
z is the depth under water



now ,


taking ln both side
21.07 x z = 16.81
z = 0.797
z = 0.8 (approx)
It would be 1.5 meters im sure form that distance to me is that nswe
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
Learn more about Kinetic energy on:
brainly.com/question/12337396
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Answer:
1000 cm.
Explanation:
To obtain the estimated tree height :
(Height of rod / length of rod shadow) = (height of tree / length of tree shadow)
Substituting values into the formula :
(150cm / 120 cm) = (height of tree / 800 cm)
Using cross multiplication :
Height of tree * 120 = 150 * 800
Height of tree = (150 * 800) / 120
Height of tree = 120,000 / 120
Height of tree = 1000
Hence, estimate height of tree = 1000 cm