Answer:
= -32.53 m / s
this velocity is directed downwards
Explanation:
This is a free fall exercise, let's use the expression
= v_{oy}^{2} + 2 g (y -yo)
where we are assuming that there is friction with the air, as the body falls its initial velocity is zero
v_{oy} = √ 2g (y - y₀)
let's calculate
v_{y} = √ (2 9.8 (0-54.0))
= -32.53 m / s
this velocity is directed downwards
Answer:
t = 2.58*10^-6 s
Explanation:
For a nonconducting sphere you have that the value of the electric field, depends of the region:
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
R: radius of the sphere = 10.0/2 = 5.0cm=0.005m
In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:
Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:
with this values of a you can use the following formula:
hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s
Answer:
The momentum would be doubled
Explanation:
The magnitude of the momentum of the freight train is given by:
where
m is the mass of the train
v is its speed
In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:
therefore, the new momentum is
so, the momentum has also doubled.
Answer:
The maximum induced emf in the rotating coil = 29.66V
The induced emf in the rotating coil when (t = 1.00 s) = 26.66V
The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s
Explanation:
Lets state the parameters we are being given right from the question:
Number of rectangular coil, (N) = 44
Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m
Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m
Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T
Angular Speed of Coil, (ω) = 64 rad/s
(a)
To calculate the induced emf in the rotating cell,we can use the formula:
emf = NBAωsin(ωt)
For maximum induced emf, the value of sin(ωt) will be 1
= NBAω ; if (A = l × w) , we have:
= NB(l × w)ω
subsitituting the parameters into the above equation; we have:
= 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64
= 29.66V
(b)
At t = 1s, the induced emf is calculated as:
emf = NBAωsin(ωt)
substituting the parameters into the equation, we have:
emf = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)
=26.66V
(c)
To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.
i.e =
since = 29.66 and N = 44; we have:
29.66 =
=
= 0.674 Wb/s