The amount of power required to move an object can be increased without changing the
1 answer:
Answer:
Yes it is possible to increase the power with out changing the amount of work.
Explanation:
The power is defined by the amount of power divided by the time. This time is the one needed to do the work. We can understand this issue by analyzing an example with numeric values.
Work = 500 [J]
Time = 5 [s]
Power will be:
Now if we change the time to 2 seconds:
As we can see, the power was increased without the need to change the work.
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Complete question:
At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.
[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]
Answer:
The time it will take the particle to pass through point (P) again is 1.639 ns.
Explanation:
F = qvB
Also;
solving this two equations together;
where;
m is the mass of electron = 9.11 x 10⁻³¹ kg
q is the charge of electron = 1.602 x 10⁻¹⁹ C
B is the strength of the magnetic field = 3.47 x 10⁻³ T
substitute these values and solve for t
Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.
Answer: beta particles
Explanation:
Given mass = 14.0 g
Molar mass = 137 g/mol
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number of particles.
1 mole of cesium contains atoms =
0.102 moles of cesium contains atoms =
The relation of atoms with time for radioactivbe decay is:
Where =atoms left undecayed
= initial atoms
t = time taken for decay = 3 minutes
= half life = 30.0 years =
minutes
The fraction that decays :
Amount of particles that decay is =
Thus beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.
Answer:
v = 0.41 m/s
Explanation:
- In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
- At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
- So, we can write the following general equation, taking the initial and final values of the energies:
- Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
- ⇒ Kf = 1/2*m*vf² (2)
- The change in the potential energy, can be written as follows:
where k = force constant = 815 N/m
xf = final displacement of the block = 0.01 m (taking as x=0 the position
for the spring at equilibrium)
x₀ = initial displacement of the block = 0.03 m
- Regarding the work done by the force of friction, it can be written as follows:
where μk = coefficient of kinettic friction, Fn = normal force, and Δx =
horizontal displacement.
- Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
- Fn = Fg= m*g (5)
- Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:
- Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get: