We simply asked to name three uses for mercury.
The most common and well-known use of mercury is the production of thermometers. It's property to stay liquid at room temperature makes it ideal for a temperature indicator. However, the use of mercury is thermometers has been phased out due to health hazards.
It is also used to form an amalgam which is the result of its combination with silver or gold. Mercury has been used to mine gold and silver. This application has also been phased out.
Today's use of mercury includes mercury-vapor lamps which are the bright lamps used in high-ways.
Answer:
F = 8.6 10⁻¹² N
Explanation:
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Em₀ = U = q ΔV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v²
Em₀ = Emf
e ΔV = ½ m v²
v =√ 2 e ΔV / m
v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)
v = √(1.8075 10¹⁶)
v = 1,344 10⁸ m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Emo = U = q DV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v2
Emo = Emf
.e DV = ½ m v2
.v = RA 2 e DV / m
.v = RA (2 1.6 10-19 51400 / 9.1 10-31)
.v = RA (1.8075 10 16)
.v = 1,344 108 m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10-19 1,344 108 0.4
F = 8.6 10-12 N
You haven't said what 'high' resistance or 'low' current means, so there's way not enough info to nail the statement as true or false. The most precise answer is "certainly could be but not necessarily". Anyway, the current in the circuit depends on BOTH the resistance AND the voltage. So without knowing the voltage too, you can't say anything about the current.
Answer:
Explanation:
ACCORDING TO NEWTONS SECOND LAW;
F = mass * acceleration
F = m(v-u/t)
m is the mass = 0.15kg
v is the final velocity = 11m/s
u is the initial velocity = 0m/s
t is the time = 0.015
Substitute;
F = 0.15(11-0)/0.015
F = 0.15(11)/0.015
F = 1.65/0.015
F = 110N
Hence the net force is 110N
Here is the full question:
The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:
The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.
Answer:
a) 0.85 m
b) 0.98 m
c) 0.76 m
Explanation:
Given that: the radius of gyration
So, moment of rotational inertia (I) of a cylinder about it axis = 
k = 0.8455 m
k ≅ 0.85 m
For the spherical shell of radius
(I) = 
k = 0.9797 m
k ≅ 0.98 m
For the solid sphere of radius
(I) = 
k = 0.7560
k ≅ 0.76 m
