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1 year ago

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Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250

rpm in 5.00 s. (a) Calculate the angular acceleration in rad/s2. (b) If she now slams on the brakes, causing an angular acceleration of –87.3rad/s2, how long does it take the wheel to stop

1 answer:

shtirl [24]1 year ago

3 0

Answer: a) angular acceleration, a = 5.24rad/s^2

b) time taken for the wheel to stop, ∆t = 0.30s

Explanation:

All shown in the attachment.

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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

A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x fro

MArishka [77]

Answer:

Resistance = 3.35* $10^{-4}$ Ω

Explanation:

Since resistance R = ρ $\frac{L}{A}$

whereas $\rho(x) = a + bx^2$

resistivity is given for two ends. At the left end resistivity is $2.25* 10^{-8}$ whereas x at the left end will be 0 as distance is zero. Thus

$2.25*10^{-8} = a + b(0)^2\\ 2.25*10^{-8} = a + 0 \\2.25*10^{-8} = a$

At the right end x will be equal to the length of the rod, so $x = 1.50\\8.50*10^{-8} = (2.25*10^{-8}) + ( b* (1.50)^2 )\\8.50*10^{-8} - (2.25*10^{-8}) = b*2.25\\\frac{6.25*10^{-8}}{2.25} = b\\b = 2.77 *10^{-8}$

Thus resistance will be R = ρ $\frac{L}{A}$

where A = π $r^2$

so,

$R = \frac{8.50*10^{-8} * 1.50}{3.14*(1.10*10^{-2})^2} \\R=3.35 * 10 ^{-4}$

6 0

1 year ago

A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri

MaRussiya [10]

Answer:

$dz=19217687.07\ m$

Explanation:

Given:

initial gauge pressure in the container, $P_0=2.02\times 10^{5}\ Pa$

atmospheric pressure at sea level, $P_a=1.01\times 10^5\ Pa$

initial volume, $V_0=4.4\times 10^{-4}\ m^3$

maximum pressure difference bearable by the container, $dP_{max}=2.26\times 10^{5}\ Pa$

density of the air, $\rho_a=1.2\ kg.m^{-3}$

density of sea water, $\rho_s=1.2\ kg.m^{-3}$

<u>The relation between the change in pressure with height is given as:</u>

$\frac{dP_{max}}{dz} =\rho_a.g_n$

where:

dz = height in the atmosphere

$g_n$ = standard value of gravity

<em>Now putting the respective values:</em>

$\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8$

$dz=19217.687\ km$

$dz=19217687.07\ m$

Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)

7 0

1 year ago

Suppose you are myopic (nearsighted). You can clearly focus on objects that are as far away as 52.5 cm away. You can clearly foc

Lilit [14]

Answer:

Explanation:

Image of distant object will be made at far point or at 52.5 so

object distance u = infinity

image distance v = - 52.5 cm

focal length required = f

Lens formula

1 / v - 1 / u = 1 / f

1 / - 52.5 - 0 = 1 / f

f = -52.5 cm

= -.525 m

Power P = 1 / f = - 1 / .525

= - 1.90

now , for eye with glass we shall find new near point .

v = ?

u = - 17.2 cm

f = - 52.5 cm

1 / v - 1 / u = 1 / f

1 / v + 1 / 17.2 = - 1 / 52.5

1 / v = - 1 / 17.2 - 1 / 52.5

= - .05813 - .019

= - .07713

u = - 12.96 cm

so new near point will be 12.96 cm

5 0

1 year ago

Which trailer has more downward pressure where it attaches to the car?

VARVARA [1.3K]

The one that is loaded worst. The overall weight is not important; tongue weight is a matter of loading. Our 12,000 lb snow cat trailer, which has stops to position the cat properly, has under 100 lbs tongue weight. Excessive tongue weight is a Bad Thing because it reduces weight on the towing vehicle's front wheels, leading to instability.

4 0

1 year ago