Answer:
B = 0.546 T, F = 2.59 10⁻¹² N
Explanation:
The magnetic force is
F = q v x B
We can calculate the magnitude of the force and find the direction by the right hand rule
F = q v B sin θ
Let's use Newton's second law
F = m a
Acceleration is centripetal
a = v² / r
We substitute
q v B sin θ = m v² / r
The angle between the field and the radius of the circle is 90º so sin 90 = 1
q B = m v / r
B = m v / q r
Let's calculate ’
B = 1.67 10⁻²⁷ 2.97 10⁷ / (1.60 10⁻¹⁹ 0.568)
B = 0.546 T
The foce is
F = q v B
F = 1.60 10⁻¹⁹ 2.97 10⁷ 0.546
F = 2.59 10⁻¹² N
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tray for dissection is the smoother side.
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a worm, </span><span>lay the worm on your </span>dissecting
tray<span> <span>with its dorsal side facing up.</span></span>
Answer:
i. 15.6 m/s
ii. I = 1.44 KNs
Explanation:
The impulse, I, on a body is the product of force applied on it and the time it acts.
i.e I = F x t
Impulse is sometimes expressed as the change in momentum of a body. It is measured in Ns.
i. mass, m, of the player = 92 kg
initial velocity of the player, u = 9.4 m/s
final velocity of the player, v = 6.2 m/s
Since he bounces back on hitting the pole, then the sign of initial and final velocities are of opposite sign.
So that,
change in velocity of the player = final velocity - initial velocity
= 6.2 - (-9.4)
= 6.2 + 9.4
= 15.6 m/s
change in velocity of the player is 15.6 m/s
ii. Impulse, I = m(v - u)
= 92 x 15.6
= 1435.2
Impulse on the player is 1.44 KNs.
Answer: 0.8 m
Explanation:
Velocity of throw = 4m/s
Maximum Height attained(h) =?
Downward acceleration experienced = 10m/s^2
Using the relation:
v^2 = u^2 + 2aS
v = final Velocity = 0 (at maximum height)
u = Initial Velocity = 4
a = g downward acceleration = - 10
0 = 4^2 + 2(-10)(S)
0 = 16 - 20S
20S = 16
S = 16 / 20
S = 0.8m
Maximum Height attained = 0.8m
Answer:
Explanation:
The magnetic moment of the coil M = current x area of loop
= I x L x L
M = L² I .
Magnetic field = B
Torque = M x B
= MB sinθ where θ is angle between M and B . Direction of M is perpendicular to face of loop
Torque = MB sin90
= MB
= L² I B .
