a pole vaulter at the top of his vault is 6.15 m in the air . if he has a gravitational potential energy of 4o42 j , what is his
2 answers:
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The capacitance is defined as the maximum charge stored in a capacitor, Q, divided by the voltage applied, V:
The capacitor is initially charged with the battery of 108 V, so the the initial charge on the capacitor can be found by re-arranging the previous formula:
The capacitor is initially charged with the battery of 108 V, so the the initial charge on the capacitor can be found by re-arranging the previous formula:
Answer:
(A) 0.279N at angle 38.02°
(B) 0.701N
(C) 14.19°
Explanation:
(A) The net force on q3 is given as:
F = Fxi + Fyj
Fx is the x component of the force
Fy is the y component of the force
Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0
Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx
First let us find y and angle x from the diagram.
Using Pythagoras theorem,
y² = 0.3² + 0.4²
y² = 0.25
y = 0.5m
Using SOHCAHTOA to find x,
sinx = 0.4/0.5
x = 53.13°
Electrostatic force, F is given as:
F = kqQ/r²
Where k = Coulumbs constant
F(1,3) = (k*q1*q3) / r²
F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)
F(1, 3) = 0.288N
F(2,3) = (k*q2*q3) / r²
F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)
F(2, 3) = 0.45N
Therefore,
Fx = -0.288cos36.87 + 0.45
Fx = 0.22N
Fy = 0.288cos53.13
Fy = 0.172N
=> F = 0.22i + 0.172j
The magnitude of the force will be
F(mag) = √(0.22² + 0.172²)
F(mag) = 0.279N
The direction of the force makes will be
tanθ = Fy/Fx
tanθ = 0.172/0.22 = 0.781
θ = 38.02° to the x axis.
(B) q2 = - 2.0 * 10^(-6)
This implies that:
F(2,3) = (k*q2*q3) / r²
F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)
F(2, 3) = -0.45N
Therefore,
Fx = -0.288cos36.87 - 0.45
Fx = -0.68N
Fy = 0.172N
=> F = - 0.68i + 0.172j
The magnitude of the force will be
F(mag) = √((-0.68)² + 0.172²)
F(mag) = 0.701N
(C) The direction of the force makes will be
tanθ = 0.172/0.68
θ = 14.19° to the x axis
The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g
<u>Explanation:</u>
Given,
Temperature, T = 0°C
Initial mass, Mi = 62kg
Speed, s = 5.48m/s
Distance, x = 26.8m
Friction is present.
Mass of ice melted = ?
We know,
The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice
and
Therefore,
KE = 930.94 Joules
Ice melting lateral heat is 334 kJ/kg = 334000 J/kg.
Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.
Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g