Well, one example is that the weight of the rider puts downward force on the motorcycle, which is absorbed by the suspension or shocks or something.
To solve this problem, we know that:
1 Albert = 88 meters
1 A = 88 m
The first thing we have to do is to square both sides of
the equation:
(1 A)^2 = (88 m)^2
1 A^2 = 7,744 m^2
<span>Since it is given that 1 acre = 4,050 m^2, so to reach
that value, 1st let us divide both sides by 7,744:</span>
1 A^2 / 7,744 = 7,744 m^2 / 7,744
(1 / 7,744) A^2 = 1 m^2
Then we multiply both sides by 4,050.
(4050 / 7744) A^2 = 4050 m^2
0.523 A^2 = 4050 m^2
<span>Therefore 1 acre is equivalent to about 0.52 square
alberts.</span>
The only correct statement on the list is choice-A./
A square loop whose sides are long is made of copper wire of radius , given the resistivity of copper is . if the magnetic field perpendicular to the loop changes at a constant rate of I = 14.029 mA.
The basic characteristic of a substance that measures how effectively it resists an electric current is called electrical resistance. A material with low resistance is a material that easily conducts electric current. A Greek letter is often used to indicate resistivity. Electrical resistance is a basic property of a material that measures how strongly it resists an electric current. The SI unit for electrical resistance is the ohmmeter.
We use magnetic field as a tool to describe how the magnetic field is distributed in the space around and inside something of a magnetic nature. A material with low resistance is a material that easily conducts electric current. A Greek letter is often used to indicate resistivity. An ohmmeter is a unit of electrical resistance in the SI system.
Learn more about magnetic field here;
brainly.com/question/24397546
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The complete question is :
A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?
Answer:
B meet A 0.01 km east of flagpole
Explanation:
given data
distance A = 5.7 km west
velocity V1 = 8.9 km/h
distance B = 4.5 km east
velocity V2 = 7 km/h
to find out
How far runners from the flagpole, when paths cross
solution
we know A and B are 5.7 + 4.5 = 10.2 km apart
and we consider here B will run distance x km for meet
so time will be for B is
time B = distance / velocity
time B = x / 7 ...................1
and
for A distance for meet = ( 10.2 - x ) km
so time A = distance / velocity
time A = ( 10.2 - x ) / 8.9 .............2
now equating equation 1 and 2
time A = time B
x / 7 = ( 10.2 - x ) / 8.9
x = 4.490
so distance of B run for meet is 4.490 km
so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km
so B meet A 0.01 km east of flagpole
