The direction of the magnetic force on the wire is west.
The magnetic force acting on the moving protons acts northward in the horizontal plane. If the thumb is up (current flows vertically up), the wrapped finger will be counterclockwise.
Therefore, the direction of the magnetic field is counterclockwise. Here, the magnetic field is pointing upwards (vertical magnetic field) and the electrons are moving east. Applying Fleming's left-hand rule here, we can see that the direction of force is along the south direction.
As the change in magnetic flux increases upwards, Lenz's law indicates that the induced magnetic field of the induced current must resist and the inside of the loop must be directed downwards. Using the right-hand rule, we can see that a clockwise current is induced.
Learn more about the magnetic fields here: brainly.com/question/7802337
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Distance = speed X time
In this example, the speed of the airplane = 840km. The time (that the question is asking)is how far can it travel in 1 hour.
So just plug in your numbers.
Distance = 840km X 1 hour = 840km/hour or 840km for short.
The height of the tower and the distance from its base form a right angle triangle. Thus,
tan(∅) = height / base
tan(28.81) = height / 1,000 m
height = 549.98 meters
Answer:
A) t = 0.55 s
B) x = 24.8 m
Explanation:
A) We can find the time at which the ball will be in the air using the following equation:
Where:
is the final height= 0
is the initial height= 1.5 m
is the component of the initial speed in the vertical direction = 0 m/s
t: is the time =?
g: is the gravity = 9.81 m/s²
By solving the above equation for t we have:
Hence, the ball will stay 0.55 seconds in the air.
B) We can find the distance traveled by the ball as follows:
Where:
a: is the acceleration in the horizontal direction = 0
is the final position =?
is the initial position = 0
is the component of the initial speed in the horizontal direction = 45 m/s
Therefore, the ball will travel 24.8 meters.
I hope it helps you!
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched
<u>Explanation:</u>
Work, W = 2 J
Initial distance, 
= 30 cm
Final distance, = 42 cm
Force, F = 30 N
Stretched length, x = ?
We know,
W = 1/2 kΔx²
Δx = 42-30 cm = 12 cm = 0.12 m
2 J = 1/2 k X (0.12)²
k = 277.77 N/m
According to Hooke's law,
F = kx
30 N = 277.77 X x
x = 0.108 m
x = 10.8 cm
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.
