<span>"prevent the engine from over speeding the armature"
hopes this help :) :D :)</span>
Answer:
The ball will be at 700 m above the ground.
Explanation:
We can use the following kinematic equation
.
where y(t) represent the height from the ground. For our problem, the initial height will be:
.
The initial velocity:
,
take into consideration the minus sign, that appears cause the ball its thrown down. The same minus appears for the acceleration:
So, the equation for our problem its:
.
Taking t=6 s:
.
.
.
.
.
So this its the height of the ball 6 seconds after being thrown.
Answer:
B. counterclockwise
Explanation:
We can solve the problem by using the right-hand rule:
- put your thumb finger of the right hand in the same direction of the current in the wire (upward)
- wrap the other fingers around the thumb
- the direction of the other fingers will give the direction of the magnetic field lines
By doing these steps, we see that the other fingers form concentric circles in a counterclockwise direction (seen from above), so this is the direction of the magnetic field lines.
Answer:
0 to 145 degrees
Explanation:
The normal range of flexion and extension is from 0 to 145 degrees.
Given Information:
Magnetic field = B = 1×10⁻³ T
Frequency = f = 72.5 Hz
Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m
Required Information:
Maximum Emf = ?
Answer:
Maximum Emf = 20.66×10⁻¹² volts
Explanation:
The maximum emf generated around the perimeter of a cell in a field is given by
Emf = BAωcos(ωt)
Where A is the area, B is the magnetic field and ω is frequency in rad/sec
For maximum emf cos(ωt) = 1
Emf = BAω
Area is given by
A = πr²
A = π(d/2)²
A = π(7.60×10⁻⁶/2)²
A = 45.36×10⁻¹² m²
We know that,
ω = 2πf
ω = 2π(72.5)
ω = 455.53 rad/sec
Finally, the emf is,
Emf = BAω
Emf = 1×10⁻³*45.36×10⁻¹²*455.53
Emf = 20.66×10⁻¹² volts
Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts
