Answer:
The magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
Explanation:
Given;
radius of the wire, r = 0.45 m
current on the loop, I = 2.4 A
angle of inclination, θ = 36⁰
torque on the coil, τ = 1.5 N.m
The torque on the coil is given by;
τ = NIBAsinθ
where;
B is the magnetic field
Area of the loop is given by;
A = πr² = π(0.45)² = 0.636 m
τ = NIBAsinθ
1.5 = (1 x 2.4 x 0.636 x sin36)B
1.5 = 0.8972B
B = 1.5 / 0.8972
B = 1.67 T
Therefore, the magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
Answer
given.
Mass of big fish = 15 Kg
speed of big fish = 1.10 m/s
mass of the small fish = 4.50 Kg
speed of the fish after eating small fish =?
a) using conservation of momentum
m₁v₁ + m₂v₂ = (m₁+m₂) V
15 x 1.10 + 4.50 x 0 = (15 + 4.5)V
16.5 = 19.5 V
V = 0.846 m/s
b) Kinetic energy before collision
KE₁ = 9.075 J
Kinetic energy after collision
KE₂ = 6.98 J
Change in KE = 6.98 - 9.075 = -2.096 J
hence,
mechanical energy was dissipated during this meal = -2.096 J
The speed at the sound barrier is 343 m/s
Answer:
the extension recorded by the student would be smaller than the actual extension of the spring
Answer:
t = 1.16 s.
Explanation:
Given,
speed of conveyor belt, v = 3.2 m/s
coefficient of friction,f = 0.28
Using newton second law
f = ma
and we also know that frictional force
f = μ N
f = μ m g
equating both the force equation
a = μ g
a = 0.28 x 9.81
a = 2.75 m/s²
Using Kinematic equation
v = u + at
3.2 = 0 + 2.75 x t
t = 1.16 s.
Time taken by the box to move without slipping is 1.16 s.
