I believe it she should use the first aid kit next
Answer:
The maximum value of the induced magnetic field is
.
Explanation:
Given that,
Radius of plate = 30 mm
Separation = 5.0 mm
Frequency = 60 Hz
Suppose the maximum potential difference is 100 V and r= 130 mm.
We need to calculate the angular frequency
Using formula of angular frequency

Put the value into the formula


When r>R, the magnetic field is inversely proportional to the r.
We need to calculate the maximum value of the induced magnetic field that occurs at r = R
Using formula of magnetic filed

Where, R = radius of plate
d = plate separation
V = voltage
Put the value into the formula


Hence, The maximum value of the induced magnetic field is
.
90+66=156
156/2=78
Reply:78kilometers in 2 hours.
Answer:
A cosmic year is 365.25 days, some times called a side real year and is just the time it takes for us to go round the sun once.
A light year is the distance light travels in a year. Now light travels at about 186,262 miles a Second! Which is not slow by any ones book.
An experiment was conducted just after Christmas a few years ago. Two girls were selected from the audience and went into two phone boxes a few feet apart. They could only hear each other via the phones. The phone call went to a ground station about 200 miles away, then up to a geostationary coms satellite, back to a ground station 1/3 of the way around the world, then repeated, with a third satellite before being sent from another ground station back to London and the other phone box. We the audience could hear both sides of the conversation from both boxes. And could hear the delay between sending and receiving. So even at the speed of light, there was about 1.5 seconds of delay. So because distances in space are so vast that saying a star is x millions of miles away causes problems, you run out of zero’s! So our nearest other star is about 4.5 light years away. Our sun (our nearest start) is about 8 light minuets away. Varies slightly as our orbit is not 100% cirular.
I HOPE THIS IS HELPFUL.
Given:
u(initial velocity)=0
a=5.54m/s^2
v(final velocity)=2 m/s
v=u +at
Where v is the final velocity.
u is the initial velocity
a is the acceleration.
t is the time
2=0+5.54t
t=2/5.54
t=0.36 sec