Answer:
a
Explanation:
because it has more energy
Answer:
The resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.
Explanation:
We are given 10 voltage power source and we have two Resistors with resistance of 20 ohm and 5ohm.
We need to find the orientation in which these two resistors would be arranged so that the circuit could get a current of 2.5Ampere.
Using ohm's law we have
V = I*R
V= voltage
I= current
R= resistance
10 = 2.5*R
R = 10/2.5 = 4ohm
that means we need a total of 4ohm resistance from these two resistors.
since the net Resistance(4ohm) is lower than the smallest resistance(5ohm) available that means the orientation of the resistors will be in parallel.
R(net) =4ohm
Now the orientation of the resistors are in parallel so the current will be divided.
we know that the current will divide in opposite manner the arm which provides more resistance less current will flow from there and vice versa.
We know that the voltage in parallel remains same
In 20 ohm resistance
again using ohms law
V = i1*R1
10 = i1*20
i1 = 0.5A
in 5ohm resistor
V=i2*R2
10 = I2*5
i2 =2A
and i1+i2 = 0.5+2= 2.5A which means our calculation is correct.
Therefore the resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.
Explanation:
An perfect mass less spring, attached at one end and with a free mass attached at the other end, will have a distinct frequency of oscillation depending on its constant spring and mass. On the other hand, a spring with mass along its length will not have a characteristic frequency of oscillation.
Alternatively, based on its spring constant and mass per length, it will now have a wave Speed. It would be possible to use all wavelengths and frequencies, as long as the component fλ= S, where S is the spring wave size. If that sounds like longitudinal waves, like solid sound waves.
Answer:
∆PE = 749.7 J
At 0.9 m high, PE = 793.8 J
At 1.75 m high, PE = 1543.5 J
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
