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natita [175]
3 years ago
11

A wheel of radius 30 cm is rotating at a rate of 2.0 revolutions every 0.080 s. (A) through what angle, in radians, does the whe

el rotate in 1.0 s? (B) what is the linear speed of a point on the wheel's rim? (C) what is the wheel's frequency of rotation?
Physics
1 answer:
ioda3 years ago
7 0

You said 2 revolutions every 0.08 seconds

1 revolution = 2pi radians.

A). The 'unit rate' is    (2 rev) x (2pi / 0.08 sec)  = 50pi radians/sec. =

                                                       157.1 radians per sec (rounded)

B). Radius of the wheel = 30 cm
     Circumference = 2pi R = 60pi cm = 188.5 cm (rounded)

     Rotation speed = 2 revs per 0.08 sec
     Linear speed =    2 x 60pi cm per 0.08 sec

                               (120pi cm) / (0.08 sec) = 47.12 meters per sec

C). Frequency = (revs) per second

                        = (2) / (0.08 sec) = 25 per second .   

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Alona [7]
<span>Venus, Uranus, and Pluto are exceptions</span>
4 0
3 years ago
A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0
ankoles [38]

Answer:

a.  B = 0.20T

b.  u = 17230.6 J/m³

c.  E = 0.236J

d.  L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

B=\frac{\mu_o n i}{L}               (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current  = 90.0A

You replace the values of the parameters in the equation (1):

B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

E=uV           (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

V=AL

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3

Then, the energy contained in the solenoid is:

E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H

The inductance of the solenoid is 5.84*10^-5 H

3 0
3 years ago
An airplane with an airspeed of 120 km/h has a heading of 30 degree east of North in a wind that is blowing toward the east at 6
lesya692 [45]

Answer:

Explanation:

Velocity of plane relative to ground V_pg = ?

Given the velocity in vector form ,

velocity of plane relative to air V_pw = 120 cos30 i + 120sin30j

V_wg = 60 i

V_pg = V_pw +V_wg

= 120 cos30 i + 120sin30j + 60i

= 164 i + 60 j

magnitude

=251 km / h

=

8 0
3 years ago
While an emf source supplies energy to a circuit, that energy is dissipated when the current passes through resistance. In deali
fredd [130]

The question is incomplete, the options are;

RI^2

I^2/R

R/I^2

R/V^2

RV^2

V^2/R

VI

VIR

Select all that apply

Answer:

P=RI^2

P=V^2/R

P=VI

Explanation:

Power is the rate at which energy is changing in a circuit. It is shown by the formulas outlined above from the group of answer choices. Since the current (I), voltage (V), and resistance (R) were mentioned in the question, any of three three formulas could be used to obtain the power drawn by the conductor.

7 0
3 years ago
Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the a
Naily [24]

Answer:

The angular acceleration of the pencil<em> α  = 17 rad·s⁻²</em>

Explanation:

Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:

    τ = I α                              (1)

    W r = I α                          (2)

The weight is that the pencil has is,

   sin 10 = r / (L/2)

   r = L/2(sin(10))

 

The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:

    I = 1/3 M L²

Thus,

   mg(L / 2)sin(10) = (1/3 m L²)(α) 

   α(f) = 3/2(g) / Lsin(10)

   α  = 3/2(9.8) / 0.150sin(10)

  <em> α  = 17 rad·s⁻²</em>

Therefore, the angular acceleration of the pencil<em> </em>is<em> 17 rad·s⁻²</em>

3 0
3 years ago
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