<span>Since the wheel start from rest. angular acceleration,
θ=1/2αt²
14=1/2α x 8.7²
α= 0.3699 rad/s²
moment of inertia of loop= mr²= 4.1x0.37=1.517 kgm²
torque=T= lα
T= 0.5611Nm= 0.56Nm to significant figure
Disc
moment of inertia of disc= 1/2mr²
Required torque value= 0.28Nm
So,
I= 1/2X 4.1X 0.37²= 0.280 Kgm²
T= Iα = 0.280 X 0.3699= 0.10 to two significant figure</span>
Answer:
density of water is much greater than density of air
Explanation:
when body is in fluid it carries weight which is always equal to upthrust
upthrust is given as
upthrust= density of fluid ×strength of gravitational field g × volume of body in fluid
as volume of ship and aeroplane is same and strength of gravitational field or acceleration due to gravity is same hence upthrust depens upon density of fluid .
density of air is about 1.2 kg/m^3 and density of water is 1000kg/m^3 hence ship carries more load
Answer:
b ac power source
Explanation:
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Answer:
the maximum voltage induced in the coil is 2.574 × 10⁻⁵ V
Explanation:
Given the data in the question;
Number of turns N = 10
major axis Ma = 13 cm = 0.13 m
a = 0.13/2 = 0.065 m
Minor axis Mi = 6 cm = 0.06 m
b = 0.06/2 = 0.03 m
we know that; 1 RPM = 0.10472 rad/s
rate of rotation R = 73rpm = 7.64 rad/s
Magnetic field = 55 uT
we know that, Area of ellipse = π × a × b
we substitute
A = π × 0.065 m × 0.03 m
A = 0.006126 m²
so
Maximum Voltage = N × Area × Magnetic field × rate of reaction
we substitute
Maximum Voltage = 10 × 0.006126 × ( 55 × 10⁻⁶ ) × 7.64
Maximum Voltage = 2.574 × 10⁻⁵ V
Therefore, the maximum voltage induced in the coil is 2.574 × 10⁻⁵ V
Answer:
Speed =0.283m/ s
Direction = 47.86°
Explanation:
Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane
MU1 =MU2cos38 + MV2cos y ...x plane
0 = MU2sin38 - MV2sin y .....y plane
Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2
Substitute into equation above
.46 = .34cos38 + V2cos y ...equ1
.34sin38 = V2sin y...equ2
.19=V2cos Y...x
.21=V2sin Y ...y
From x
V2 =0.19/cost
Sub V2 into y
0.21 = 0.19(Sin y/cos y)
1.1052 = tan y
y = 47.86°
Sub Y in to x plane equ
.19 = V2 cos 47.86°
V2=0.283m/s