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Simora [160]
3 years ago
11

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

force of 130 N to the tire's rim. The mass of the wheel is 1.50 kg and, for the purpose of this problem, assume that all of this mass is concentrated on the outside radius of the wheel. The diameter of the wheel is 55.0 cm. A chain passes over a sprocket that has a diameter of 9.25 cm. In order for the wheel to have an angular acceleration of 5.00 rad/s2, what force, in Newtons, must be applied to the chain
Physics
1 answer:
Sliva [168]3 years ago
4 0

Answer:

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Which one of the following terms is used to describe the bending of waves and subsequent spreading around obstacles or the edges
maks197457 [2]

Answer:

B)Diffraction

Explanation:

The concept of diffraction in the field of physics is defined as the deflection of a wave when it crosses an opening or hits the edge of an opaque element. Diffraction is a phenomenon that involves all waves: electromagnetic, radio, sound, etc., and it is possible to predict their behavior using different mathematical approaches. There is a method of analysis called the Huygens principle, which allows us to understand diffraction as a wavefront that is seen as a series of emitters capable of redirecting the wave when it oscillates and thus promotes propagation. Although the waves produced by the oscillators are spherical, their interference causes a flat wave that moves in the same direction as the initial one.

6 0
3 years ago
Although the evidence is weak, there has been concern in recent years over possible health effects from the magnetic fields gene
Natasha2012 [34]

Answer:

A. B = 6.36 * 10^{-10} T

B. P ≈ 0

Explanation:

In order to calculate the magnetic field strength we have to use the magnetic field strength of a straight wire.

B = \frac{mi* I}{2\pi *d} (eq. I)

B = magnetic field strength at distance d

I = current (A)

mi = represented by the greek letter μ, represents the permeability of the free space, which is: 4 × π 10^(-7) T m/A

d = distance from the wire

By replacing the values in eq I, we have the following:

B = \frac{4\pi  10^{-7} T  m  A^{-1}  200 A}{2\pi *20 m}\\\\B = 6.36 * 10^{-10}  T\\ (eq II)

The earth magnetic field in the surface variates from 25 to 65 microteslas. Thus:

P = Percentage from the wires/percentage of the earth

P = \frac{6.36 * 10^{-10}T}{65* 10^{-3} T}\\ ∵ B ∴

P ≈ 0

5 0
4 years ago
4. A force of 5 N gives a mass m,, an acceleration of 10 m's, and a mass
Art [367]

Answer:

For mass m  1  newton 2nd law

F=m  1  a  1

​5=m  1  ×10

m  1  =  2 1 kg

For mass m  2

​F=m  2  a 2

​5=m  2 ×20

m  2 =  4 1  kg

if tied together  

Total mass =m  1  +m  2  =  1/2 +1/4=3/4kg  

Now

F=M T  Q T

 a  T    =  5/m T =  5×4/3  =  3 /20m/s^2

Explanation:

8 0
3 years ago
What is the current in milliamperes produced by the solar cells of a pocket calculator through which 3.70 C of charge passes in
Kaylis [27]

Answer:

Current flowing in the cell will be equal to 0.1284 mA

Explanation:

We have given charge q = 3.70 C

And time through which charge is flowing = 8 hour

We know that 1 hour = 60 minutes, and 1 minute = 60 sec

So 1 hour = 60×60 = 3600 sec

So 8 hour = 8×3600 = 28800 sec

We know that current is rate time rate of flow of charge

So current i=\frac{q}{t}=\frac{3.70}{28800}=1.284\times 10^{-4}A=0.1284mA

So current flowing in the cell will be equal to 0.1284 mA

3 0
4 years ago
2. A 200.0-kg bear grasping a vertical tree slides down at constant velocity. What is the friction force between the tree and th
Neporo4naja [7]

Answer: 1960 N

Explanation:

The bear is sliding down at constant velocity: this means that its acceleration is zero, so the net force is also zero, according to Newton's second law:

F_{net}=ma=0

There are two forces acting on the bear: its weight W, pulling downward, and the frictional force Ff, pulling upward. Therefore, the net force is given by the difference between the two forces:

F_{net}=W-F_f=0

From the previous equation, we find that the frictional force is equal to the weight of the bear:

F_f=W=mg=(200.0 kg)(9.8 m/s^2)=1960 N


8 0
3 years ago
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