Your roommate is working on his bicycle and has the bike upside down. He spins the 60 cm

1 answer:

5 0

Diameter = 60 cm, Radius = 60/2 = 30 cm = 30/100 = 0.3 m.

The pebble in the tread goes by 3 times every second.

This is the same as 3 times per second.

Recall the unit of frequency is Hertz or per second, s⁻¹

So 3 times per second, Frequency, f = 3s⁻¹ or 3 Hertz

For angular motion:

Angular speed, ω = 2πf

   = 2*π*3

   = 6π   rad/s

Linear speed, v = ωr = 6π * 0.3 = 1.8π m/s

Linear acceleration, a = v² / r

   a = 1.8π * 1.8π / 0.3 = 10.8π² m/s²

Angular acceleration α = a/r = 10.8π² / 0.3 = 36π² rad/s²

Angular speed = 6π rad/s ≈ 18.840 rad/s

The linear speed of the pebble = 1.8π m/s ≈ 5.655 m/s

The angular acceleration = 36π² rad/s² ≈ 355.306 rad/s²

The linear acceleration of the pebble = 10.8π² m/s ≈ 106.592 m/s²

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Each of the two coils has 320 turns. The average radius of the coil is 6 cm. The distance from the center of one coil to the ele

Temka [501]

Answer:

Magnetic field, $B=2.39\times 10^{-3}\ T$

Explanation:

It is given that,

Number of turns, N = 320

Radius of the coil, r = 6 cm = 0.06 m

The distance from the center of one coil to the electron beam is 3 cm, x = 3 cm = 0.03 m

Current flowing through the coils, I = 0.5 A

We need to find the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils. The magnetic field midway between the coils is given by :

$B=\dfrac{\mu_oINr^2}{(x^2+r^2)^{3/2}}$

$B=\dfrac{4\pi \times 10^{-7}\times 0.5\times 320\times (0.06)^2}{(0.03^2+0.06^2)^{3/2}}$

B = 0.00239 T

$B=2.39\times 10^{-3}\ T$

So, the magnitude of the magnetic field at a location on the axis of the coils, midway between the coils is $2.39\times 10^{-3}\ T$ . Hence, this is the required solution.

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3 years ago

Why are objects that fall near earth’s surface rarely in free fall?

Sloan [31]

Answer:

Because of the presence of air resistance

Explanation:

When an object is in free fall, ideally there is only one force acting on it:

- The force of gravity, W = mg, that pushes the object downward (m= mass of the object, g = acceleration of gravity)

However, this is true only in absence of air (so, in a vacuum). When air is present, it exerts a frictional force on the object (called air resistance) with upward direction (opposite to the motion of free fall) and whose magnitude is proportional to the speed of the object.

Therefore, it turns out that as the object falls, its speed increases, and therefore the air resistance acting against it increases too; as a result, the at some point the air resistance becomes equal (in magnitude) to the force of gravity: when this happens, the net acceleration of the object becomes zero, and so the speed of the object does not increase anymore. This speed reached by the object is called terminal velocity.

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3 years ago

A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

Ne4ueva [31]

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N