Answer:
1.045 m from 120 kg
Explanation:
m1 = 120 kg
m2 = 420 kg
m = 51 kg
d = 3 m
Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.
By use of the gravitational force
Force on m due to m1 is equal to the force on m due to m2.
3 - y = 1.87 y
3 = 2.87 y
y = 1.045 m
Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.
Answer:
A Shunt is a passive element, usually resistive, that is used to bypass current around another element, like a meter, that is not able to handle the full current flow. ... A Multiplier is an active element that amplifies a voltage or current to enable a less sensitive device or circuit to make use of it.
Answer:
Explanation:
Given that,
Two resistor has resistance in the ratio 2:3
Then,
R1 : R2 = 2:3
R1 / R2 =⅔
3 •R1 = 2• R2
Let R2 = R
Then,
R1 = ⅔R2 = 2/3 R
So, if the resistor are connected in series
Let know the current that will flow in the circuit
Series connection will have a equivalent resistance of
Req = R1 + R2
Req = R + ⅔ R = 5/3 R
Req = 5R / 3
Let a voltage V be connect across then, the current that flows can be calculated using ohms law
V = iR
I = V/Req
I = V / (5R /3)
I = 3V / 5R
This the current that flows in the two resistors since the same current flows in series connection
Now, using ohms law again to calculated voltage in each resistor
V= iR
For R1 = ⅔R
V1 =i•R1
V1 = 3V / 5R × 2R / 3
V1 = 3V × 2R / 5R × 3
V1 = 2V / 5
For R2 = R
V2 = i•R2
V2 = 3V / 5R × R
V2 = 3V × R / 5R
V2 = 3V / 5
Then,
Ratio of voltage 1 to voltage 2
V1 : V2 = V1 / V2 = 2V / 5 ÷ 3V / 5
V1 : V2 = 2V / 5 × 5 / 3V.
V1 : V2 =2 / 3
V1:V2 = 2:3
The ratio of their voltages is also 2:3
- Mass=1167kg
- Initial velocity=u=10m/s
- Acceleration=a=4m/s^2
- Work done=105J=W
- Final velocity=v=?
- Force=F
- Distance=d
Apply Newton's second law
Now
Now
According to third equation of kinematics
Answer:
C) Both the charge on the plates of the capacitor and its capacitance would change.
Explanation:
The capacitance of parallel plate capacitor without dielectric material is given as;
A parallel plate capacitor with a dielectric between its plates has a capacitance given by;
where;
C is the capacitance
K is the dielectric constant
ε₀ is permittivity of free space
A is the area of the plates
d is the distance of separation of the two plates
- first point to note, is that the capacitance increases when dielectric material is inserted by a factor 'k'
Again, Q = CV (without dielectric material)
(with dielectric material)
- second point to note, is that charge stored in the plates increases due to presence of dielectric material.
Finally, we can conclude that both the charge on the plates of the capacitor and its capacitance would change.
