Ask question

Ask question

All categories

3 years ago

8

How long does it take a wheel that is rotating at 33.3 rpm to speed up to 78.0 rpm if it has an angular acceleration of 2.15 rad

/ s 2?

1 answer:

Rzqust [24]3 years ago

7 0

initial angular speed is given by 33.3 rpm

$w_0 = 2\pi \frac{33.3}{60}$

$w_0 = 3.49 rad/s$

final angular speed is given by 78 rpm

$w_f = 2\pi \frac{78}{60}$

$w_f = 8.17 rad/s$

now by using kinematics we will have

$w_f = w_0 + \alpha * t$

$8.17 = 3.49 + 2.15 * t$

$t = 2.17 seconds$

You might be interested in

Alien A lifts a 500-newton child from the floor to a height of 0.40 meters in 2 seconds

vivado [14]

Strong alien you got there good luck bud you never asked a question

4 0

3 years ago

In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys

nydimaria [60]

Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J

Solution :

(1) The equation used is,

$\Delta U=q+w\\\\U_{final}-U_{initial}=q+w$

where,

$U_{final}$ = final internal energy

$U_{initial}$ = initial internal energy

q = heat energy

w = work done

(2) The known variables are, q, w and $U_{initial}$

initial internal energy = $U_{initial}$ = 2000 J

heat energy = q = 1000 J

work done = w = 500 J

(3) Now plug the numbers into the equation, we get

$U_{final}-(2000J)=(1000J)+(500J)$

(4) By solving the terms, we get

$U_{final}-(2000J)=(1000J)+(500J)$

$U_{final}-(2000J)=1500J$

$U_{final}=2000J+1500J$

$U_{final}=3500J$

(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J

5 0

3 years ago

A diagram of a closed circuit with a power source on the left labeled 120 V. There are 3 resistors in parallel, separate paths,

Zina [86]

1) The equivalent resistance is $2.73\Omega$

2) The voltage in the circuit is 120 V

3) The total current in the circuit is 44.0 A

Explanation:

1)

Resistors are said to be in parallel when they have the same potential difference at their terminals.

The formula to calculate the equivalent resistance for three resistors in parallel is:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

where $R_1, R_2, R_3$ are the resistances of the three resistors.

In this problem, the resistance of each resistor is:

$R_1 = 5.0 \Omega$

$R_2 = 10.0 \Omega$

$R_3 = 15.0 \Omega$

Substituting,

$\frac{1}{R_T}=\frac{1}{5}+\frac{1}{10}+\frac{1}{15}=\frac{11}{30}$

So the equivalent resistance is

$R_T = \frac{30}{11}=2.73 \Omega$

2)

In this problem we are told that the three resistors are connected in parallel. This means that their terminals are connected to the same point of the circuit: therefore, this means that they also have the same potential difference across them.

Therefore, the voltage in each branch containing each resistor is the same, and it is equal to the voltage of the battery, which is

$V=120 V$

So, the voltage in the circuit is 120 V.

3)

The total current in the circuit can be found by using Ohm's law:

$V=RI$

where

V is the voltage

R is the equivalent resistance of the circuit

I is the current

In this circuit, we have:

V = 120 V

$R=2.73\Omega$

Re-arranging the equation for I, we find the current:

$I=\frac{V}{R}=\frac{120}{2.73}=44.0 A$

Learn more about current and potential difference:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and th

gizmo_the_mogwai [7]

Answer:

λ = 5.2 x 10⁻⁷ m = 520 nm

Explanation:

From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:

Δx = λL/d

where,

Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m

L = Distance between slits and screen = 3.1 m

d = Separation between slits = 0.0005 m

λ = wavelength of light = ?

Therefore,

0.00322 m = λ(3.1 m)/(0.0005 m)

λ = (0.00322 m)(0.0005 m)/(3.1 m)

<u>λ = 5.2 x 10⁻⁷ m = 520 nm</u>

5 0

3 years ago

A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t

Jobisdone [24]

Answer:

(a) E=0 : 0 cm from the center of the sphere

(b) E= 227.8*10³ N/C : 10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C : 40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C : 59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

$E=\frac{K*Q}{r^{2} }$ : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

$E=k*\frac{Q}{a^{3} } *r$ :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q: Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at 0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

$E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1$

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }$

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere

r>a , We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }$

E= 411.84 * 10³ N/C

4 0

3 years ago