1 Ampere = 1 coulomb/second
1 hour = 3,600 seconds
1 coulomb/second = 3,600 coulomb/hour
10 Amperes x 1 hour = 36,000 coulomb/hour
Answer: mg/Cosθ
Explanation:
Taking horizontal acceleration of wedge as 'a'
FCosΘ = FsinΘ
F = mass(m) × acceleration(a) = ma
For horizontal resolution g = 0
Therefore,
Horizontal = Vertical
maCosΘ = mgSinΘ
aCosΘ = gSinΘ
a = gSinΘ/CosΘ
Recall from trigonometry :
SinΘ/Cosθ = tanΘ
Therefore,
a = gtanΘ
Normal force acing on the wedge:
mgCosΘ + maSinΘ - - - - (y)
Substitute a = gtanΘ into (y)
mgCosΘ + mgtanΘsinΘ
tanΘ = sinΘ/cosΘ
mgCosΘ + mgsinΘ/cosΘsinΘ
mgCosΘ + mgsin^2Θ/cosΘ
Factorizing
mg(Cosθ + sin^2Θ/cosΘ)
Taking the L. C. M
mg[(Cos^2θ + sin^2Θ) /Cosθ]
Recall: Cos^2θ + sin^2Θ = 1
mg[ 1 /Cosθ]
mg/Cosθ
Explanation:
Work done is given by the product of force and displacement.
Case 1,
1. A boy lifts a 2-newton box 0.8 meters.
W = 2 N × 0.8 m = 1.6 J
2. A boy lifts a 5-newton box 0.8 meters.
W = 5 N × 0.8 m = 4 J
3. A boy lifts a 8-newton box 0.2 meters.
W = 8 N × 0.2 m = 1.6 J
4. A boy lifts a 10-newton box 0.2 meters.
W = 10 N × 0.2 m = 2 J
Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.
Answer:
a. E = 
b. 
Explanation:
The important thing to remember is to use Gauss Law. This is a relation that describes the distribution of electric charge to the resultant electric field.
Linear charge density means charge per unit length of material.
Data:
The metal cylinder is hollow.
The unit length is L.
a.The expression will be as follows:
for charge inside the cylinder, where r < R, the expression is:
E =
b. Let's assume that the cylinder is a coaxial cylinder with a radius r > R, then the electrical field strength is given as:
E = 
E = 
This gives:
The solution informs us that there is a surface change taking place on the cylinder. Therefore, there will not be a magnetic field across it.
Answer:
Explanation:
<u>Given:</u>
- Mass,
- Velocity,
where,
are the uncertainties in mass and velocity respectively.
The kinetic energy is given by
The uncertainty in kinetic energy is given as:
