A 7.3 cm diameter loop of wire is initially oriented so that its plane is perpendicular to a magnetic field of 0.61 T pointing u
1 answer:
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G = 9.81 m/sec^2) g = 9.81
<span>Solving for velocity : </span>
<span> = 2gh </span>
<span>v = </span>
<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>
<span>Solving for velocity : </span>
<span>v = </span>
<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>
Answer:
v_f = 10.85 m/s
Explanation:
We will apply the law of conservation of momentum here:
where,
m₁ = mass of roller skater = 47 kg
m₂ = mass of bag = 6 kg
v_1i = initial speed of roller skater = 12 m/s
v_2i = initial speed of the bag = 0 m/s
v_1f = final speed of the roller skater = ?
v_2f = final speed of the bag = ?
Both the bag and the skater will have same speed at the end because kater is carrying the bag:
v_1f = v_2f = v_f
Therefore, the equation will become:
<u>v_f = 10.85 m/s</u>