Answer:
Series circuit:
The voltage that is measured across the circuit is different.
The current measured in a series circuit remains the same at all points in the circuit.
Parallel circuit:
The current measured across each resistor varies
The voltage measured across a parallel circuit will remain the same
Explanation:
Series and parallel circuits behave differently when it comes to the circulation of current and the interaction with a potential difference.
In a series circuit, the resistances are connected end to end. As a result, the voltage that is measured across the circuit is different once resistance is encountered. However, the current measured in a series circuit remains the same at all points in the circuit.
A parallel circuit behaves in an exactly opposite manner to the series circuit. In a parallel circuit, the resistances are connected side by side. As a result of this, the current measured across each resistor varies as there are circuit branches through which electric current can flow into. On the other hand, the voltage measured across a parallel circuit will remain the same
The answer is C in this question.
Answer:
Explanation:
A ) When gymnast is motionless , he is in equilibrium
T = mg
= 63 x 9.81
= 618.03 N
B )
When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.
T = mg
= 618.03 N
C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2
Net force on it = T - mg , acting in upward direction
T - mg = m a
T = mg + m a
= m ( g + a )
= 63 ( 9.81 + .6)
= 655.83 N
D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2
Net force acting in downward direction
mg - T = ma
T = m ( g - a )
= 63 x ( 9.81 - .6 )
= 580.23 N
Answer:
option c
Explanation:
The temperature must be constant. Ohms law states that the current running through a conductor is directly proportional to the potential difference across it provided the temperature remains constant
Answer:
2.86×10⁻¹⁸ seconds
Explanation:
Applying,
P = VI................ Equation 1
Where P = Power, V = Voltage, I = Current.
make I the subject of the equation
I = P/V................ Equation 2
From the question,
Given: P = 0.414 W, V = 1.50 V
Substitute into equation 2
I = 0.414/1.50
I = 0.276 A
Also,
Q = It............... Equation 3
Where Q = amount of charge, t = time
make t the subject of the equation
t = Q/I.................. Equation 4
From the question,
4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs
Q = 7.899×10⁻¹⁹ C
Substitute these value into equation 4
t = 7.899×10⁻¹⁹/0.276
t = 2.86×10⁻¹⁸ seconds
