A car can move from rest to 35 m/s in 5.0 seconds. It's acceleration is
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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
Answer:
0.51 m
Explanation:
Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.
Kinetic energy, KE=½kx²
Where k is spring constant and x is the compression of spring
Potential energy, PE=mgh
Where g is acceleration due to gravity, h is height and m is mass
Equating KE=PE
mgh=½kx²
Making x the subject of formula
Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then
Answer:
4.0 m/s
Explanation:
The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.
Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is
where here we have
d = 3.0 m is the horizontal distance covered
vx is the horizontal velocity
t = 1.3 s is the duration of the fall
Solving for vx,
Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by
where
h = 4.0 m is the initial height
vy is the initial vertical velocity
We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy
So now we can find the magnitude of the initial velocity: