Answer:
30 (kg)
Explanation:
therefore the mass of the ball is 2 so 30 (kg)
Answer:
3.62 V
Explanation:
L = 80 cm = 0.8 m
f = 15 rps
B = 60 m T = 0.060 T
ω = 2 x π x f = 2 x 3.14 x 15 = 94.2 rad/s
v = r ω
here, r be the radius of circular path. Here r = length of rod = L
v = 0.80 x 94.2 = 75.36 m/s
The motional emf is given by
e = B v L = 0.060 x 75.36 x 0.8 = 3.62 V
Answer:
σ₁ =
C/m²
σ₂ =
C/m²
Explanation:
The given data :-
i) The radius of smaller sphere ( r ) = 5 cm.
ii) The radius of larger sphere ( R ) = 12 cm.
iii) The electric field at of larger sphere ( E₁ ) = 358 kV/m. = 358 * 1000 v/m


Q₁ = 572.8
C
Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.
V = constant
∴

=
C
Surface charge density ( σ₁ ) for large sphere.
Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².
σ₁ =
=
=
C/m².
Surface charge density ( σ₂ ) for smaller sphere.
Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² =0.0314 m².
σ₂ =
=
=
C/m²
Answer:
f3 = 102 Hz
Explanation:
To find the frequency of the sound produced by the pipe you use the following formula:

n: number of the harmonic = 3
vs: speed of sound = 340 m/s
L: length of the pipe = 2.5 m
You replace the values of n, L and vs in order to calculate the frequency:

hence, the frequency of the third overtone is 102 Hz
Answer:
v = 66.4 m/s
Explanation:
As we know that plane is moving initially at speed of

now we have




now in Y direction we can use kinematics



since there is no acceleration in x direction so here in x direction velocity remains the same
so we will have


