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Cloud [144]
2 years ago
13

4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity

.]
Physics
1 answer:
Slav-nsk [51]2 years ago
8 0

Answer:

w_f= 22.41rad/s

Explanation:

First, we know that:

a = 4 rad/s^2

S = 10 rev = 62.83 rad

Now we know that:

w_f^2-w_i^2=2aS

where w_f is the final angular velocity, w_i the initial angular velocity, a is the angular aceleration and S the radians.

Replacing, we get:

w_f^2-(0)^2=2(4)(62.83)

Finally, solving for w_f:

w_f= 22.41rad/s

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The following times are given using metric prefixes on the base SI unit of time: the second. Rewrite them in scientific notation
Leno4ka [110]

Answer: a=9.8*10^-10s

b=9.8*10^-13s

c=1.7*10^-8s

d=5.57*10^-4s

Explanation:

a) given 980 ps

Expected answer is 980 * 10^-12

Therefore, 980ps = 9.8*10^-10s

b) given 980 fs

Expected answer is 980 * 10^-15

Therefore, 980fs = 9.8*10^-13s

c) given 17 ns

Expected answer is 17 * 10^-9

Therefore, 17ns = 1.7*10^-8s

d) given 577 μs

Expected answer is 577 * 10^-6

Therefore, 577μs = 5.57*10^-4s

a=9.8*10^-10s

b=9.8*10^-13s

c=1.7*10^-8s

d=5.57*10^-4s

8 0
2 years ago
Read 2 more answers
Before leaving the house in the morning, you plop some stew in your slow cooker and turn it on Low. The slow cooker has a 160 Oh
guajiro [1.7K]

Answer:

Total charge flow through the cooker is 21600 C

Explanation:

As we know that the current flow through the cooker is given by Ohm's law

here it is given as

V = i R

i = \frac{V}{R}

i = \frac{120}{160}

i = \frac{3}{4} A

now the charge flow through it is given as

Q = i t

total time is t = 8 hours

Q = \frac{3}{4}(8 \times 60 \times 60)

Q = 21600 C

7 0
2 years ago
To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V
aleksley [76]

Answer:

The maximum energy stored in the combination is 0.0466Joules

Explanation:

The question is incomplete. Here is the complete question.

Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Energy stored in a capacitor is expressed as E = 1/2CtV² where

Ct is the total effective capacitance

V is the supply voltage

Since the capacitors are connected in series.

1/Ct = 1/C1+1/C2+1/C3

Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF

1/Ct = 1/11.7 + 1/21.0 + 1/28.8

1/Ct = 0.0855+0.0476+0.0347

1/Ct = 0.1678

Ct = 1/0.1678

Ct = 5.96μF

Ct = 5.96×10^-6F

Since V = 125V

E = 1/2(5.96×10^-6)(125)²

E = 0.0466Joules

8 0
2 years ago
pitot tube on an airplane flying at a standard sea level reads 1.07 x 105 N/m2. What is the velocity of the airplane?
Allushta [10]

Answer:

V_infinty=98.772 m/s

Explanation:

complete question is:

The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?

<u>solution:</u>

<u>given:</u>

<em>p_o=1.07*10^5 N/m^2</em>

<em>ρ_infinity=1.23 kg/m^2</em>

<em>p_infinity=1.01*10^5 N/m^2</em>

p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2

V_infinty^2=9756.097

V_infinty=98.772 m/s

8 0
2 years ago
A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov
vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}

Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.  

if the -5% is applied to both resistors the Voltage is still 5V because the quotient  has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V

Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V

R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140

R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260

so.

V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}

4 0
2 years ago
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