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3 years ago

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uppose a wheel is initially rotating at 10.0 rad/s while undergoing constant angular acceleration reaching a speed of 30.0 rad/s

after 36.0 seconds have elapsed. How long after the initial time has the wheel undergone half of the angular displacement that it will have gone through during the entire 36.0-second interval

1 answer:

Paul [167]3 years ago

8 0

Answer:

t = 22.2 s

Explanation:

angular distance covered in the 36.0 s is

θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians

720/2 = 360 radians

α = Δω/t = (30 - 10)/36 = 5/9 rad/s²

θ = ω₀t + ½αt²

360 = 10.0t + ½(5/9)t²

0 = (5/18)t² + 10.0t - 360

0 = t² + 36t - 1296

t = (-36 ±√(36² - 4(1)(-1296))) / 2

t = (-36 ±√(6480)) / 2

t = -18 ±√1620

we ignore the negative time result as it occurs before we care.

t = -18 + √1620 = 22.249223... s

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(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you ca

Colt1911 [192]

a) 52.5 m

b) 16.0 m/s

Explanation:

a)

The motion of a car slowed down by friction is a uniformly accelerated motion, so we can use the following suvat equation:

$v^2-u^2=2as$

where

v = 0 is the final velocity (the car comes to a stop)

u = 28.7 m/s is the initial velocity of the car

a is the acceleration

s is the stopping distance

For a car acted upon the force of friction, the acceleration is given by the ratio between the force of friction and the mass of the car, so:

$a=\frac{-\mu mg}{m}=-\mu g$

where:

$\mu=0.80$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting and solving for s, we find:

$s=\frac{v^2-u^2}{-2\mu g}=\frac{0-(28.7)^2}{-2(0.80)(9.8)}=52.5 m$

b)

In this case, the car is moving on a wet road. Therefore, the coefficient of kinetic friction is

$\mu=0.25$

Here we want the stopping distance of the car to remain the same as part a), so

$s=52.5 m$

We can use again the same suvat equation:

$v^2-u^2=2as$

And since the final velocity is zero

u = 0

We can find the initial velocity of the car:

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3 years ago

An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

s2008m [1.1K]

Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, $Q_{L}= mL_{i}$

$Q_{L}= 0.555\times334\times10^{3}= 185370 J$

<u>Now the time required for supply of 185370 J:</u>

$t=\frac{Q_{L}}{q}$

$t=\frac{185370}{820}$

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

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What can be caused by nuclear explosions and can easily penetrate into the human body causing severe damage?

dedylja [7]

Answer:

Nuclear explosions produce air-blast effects similar to those produced by conventional explosives. The shock wave can directly injure humans by rupturing eardrums or lungs or by hurling people at high speed, but most casualties occur because of collapsing structures and flying debris.

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murzikaleks [220]

The answer would be a nail rusting, but really it could be any of these.

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4 years ago

A 17.3 eV electron has a 0.295 nm wavelength. If such electrons are passed through a double slit and have their first maximum at

4vir4ik [10]

Answer:

0.541 nm

Explanation:

The condition for maxima is,

$dsin\theta=m\lambda$

Here, m=0,1,2,.....

And d is the slit separation, m is the order of maxima, $\lambda$ is the wavelength.

Given that, the 17.3 eV electron posses a wavelength of

$\lambda=0.295 nm\\\\\lambda=0.295\times 10^{-9}m$

And the order of maxima is $m=1$ .

And the angle at which first order maxima occur is, $\theta=33^{\circ}$ .

Put these values in maxima condition while solving for d.

$d=\frac{1\times 0.295\times 10^{-9}m}{sin33^{\circ}} \\d=\frac{0.295\times 10^{-9}m}{0.545} \\d=0.541\times 10^{-9}m}\\d=0.541 nm$

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3 years ago