Answer:
Part A : E =
ε₀ Q₁/R₁² Volt/meter
Part B : V =
ε₀ Q₁/R₁ Volt
Explanation:
Given that,
Charge distributed on the sphere is Q₁
The radius of sphere is R
₁
The electric potential at infinity is 0
<em>Part A</em>
The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.
If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by
F =
ε₀ Q₁/R₁²
Then the electric field at that point is
E = F/1
E =
ε₀ Q₁/R₁² Volt/meter
Part B
The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.
Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation
V =
ε₀ Q₁/R₁ Volt
Answer:
Show attached picture
Explanation:
Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call
its internal resistance) and R indicates the resistance of the light bulb.
We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:
(1)
Both the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:

Using Ohm's law,
, we can rewrite the previous equation as:

where
is the current in the meter
is the current in the bulb
Using (1), this equation becomes

so, the current in the meter is 1000 times less than through the bulb.
:<span> </span><span>30.50 km/h = 30.50^3 m / 3600s = 8.47 m/s
At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg)
So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road ..
Fn = mg - mv²/R
Fn = m(g - v²/R) .. .. 1800kg (9.80 - 8.47²/20.20) .. .. .. ►Fn = 11 247 N (upwards)
(b)
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero.
ie .. mg = mv²/R .. .. v² = Rg .. .. 20.20m x 9.80 = 198.0(m/s)²
►v = √198 = 14.0 m/s</span>
Answer:
160N
Explanation:
Moments must be conserved - so.


Answer:
I = 2 kgm^2
Explanation:
In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:
(1)
I: moment of inertia of the door
α: angular acceleration of the door = 2.00 rad/s^2
τ: torque exerted on the door
You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:
(2)
F: force = 5.00 N
d: distance to the hinges = 0.800 m
You replace the equation (2) into the equation (1), and you solve for α:

Finally, you replace the values of all parameters in the previous equation for I:

The moment of inertia of the door around the hinges is 2 kgm^2