Answer:
proton proton proton proton
Answer:
Convection Currents I think
Explanation:
Hope this helps!!!
and sry if I'm wrong
Answer:
133 kg
Explanation:
Parameters given:
Mass of tackler, m = 133 kg
Initial velocity of tackler, u = 3.4 m/s
Final velocity of tackler and receiver, v = 1.7 m/s
Since momentum is conserved, we apply the principle of conservation of momentum:
Total initial momentum = Total final momentum
mu + MU = (m + M)v
Where U = initial velocity of receiver = 0 m/s
M = mass of receiver
Therefore:
(133 * 3.4) + (M * 0) = (133 + M) * 1.7
452.2 = (133 + M) * 1.7
(133 + M) = 452.2/1.7
133 + M = 266
=> M = 266 - 133
M = 133 kg
The mass of the receiver is 133 kg.
Answer:
a) E = 8628.23 N/C
b) E = 7489.785 N/C
Explanation:
a) Given
R = 5.00 cm = 0.05 m
Q = 3.00 nC = 3*10⁻⁹ C
ε₀ = 8.854*10⁻¹² C²/(N*m²)
r = 4.00 cm = 0.04 m
We can apply the equation
E = Qenc/(ε₀*A) (i)
where
Qenc = (Vr/V)*Q
If Vr = (4/3)*π*r³ and V = (4/3)*π*R³
Vr/V = ((4/3)*π*r³)/((4/3)*π*R³) = r³/R³
then
Qenc = (r³/R³)*Q = ((0.04 m)³/(0.05 m)³)*3*10⁻⁹ C = 1.536*10⁻⁹ C
We get A as follows
A = 4*π*r² = 4*π*(0.04 m)² = 0.02 m²
Using the equation (i)
E = (1.536*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.02 m²)
E = 8628.23 N/C
b) We apply the equation
E = Q/(ε₀*A) (ii)
where
r = 0.06 m
A = 4*π*r² = 4*π*(0.06 m)² = 0.045 m²
Using the equation (ii)
E = (3*10⁻⁹ C)/(8.854*10⁻¹² C²/(N*m²)*0.045 m²)
E = 7489.785 N/C
