Answer:
9 Brainly hahaha ............huh
Answer:
Options 1 and 5 are correct
Explanation:
Magnetic field lines can never cross, the field is unique at any point in space. Magnetic field lines are continuous, forming closed loops without beginning or end. They go from the north pole to the south pole.
Magnetic field lines form closed loops but do not intersect.
Electric field lines originate at the positive charges and terminate at the negative charges. They move in a straight line and are parallel. Electric field lines neither form closed loops nor intersect.
Since, magnetic field lines form closed loops and move from North to South pole, they come out of north poles outside the magnet and into north poles inside the magnet, they also go into south poles outside the magnet and out of south poles inside the magnet.
Answer: 0.2 hours
Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .
Besides, this battery has a voltage of 12 V
so by using the Ohm law we also know that V=R*I,
Fron this we can obtain:
I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA
then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA in x time in the form:
1hour------- 1800 mA
x hour--------350 mA
time= 350/1800= 0.2 hour
Answer:
Explanation:
Given that
Height = h
Radius = R
From energy conservation
At point B
The minimum speed to complete the the circle
So the kinetic energy at point B
Without falling off at the top (point B)
Answer:
(A) 10052.2 m/s²
(B) 0.00678 seconds
Explanation:
From the question,
(A) Applying
V² = U²+2as..................... Equation 1
Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.
make a the subject of the equation
a = (V²-U²)/2s........................ Equation 2
Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m
Substitute these values into equation 2
a = (68²-0²)/(2×0.230)
a = 10052.2 m/s²
(B) Using,
a = (V-U)/t......................... Equation 3
Where t= time.
make t the subject of the equation
t = (V-U)/a......................... Equation 4
Given: V = 68 m/s, U = 0 m/s, a = 10052.2
Substitute into equation 4
t = (68-0)/10052.2
t = 0.00678 seconds
