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ivann1987 [24]
3 years ago
10

A fan blade rotates with angular velocity given by ωz(t)= γ − β t2, where γ = 4.90 rad/s and β = 0.750 rad/s3 . part a calculate

the angular acceleration as a function of time.
Physics
2 answers:
Klio2033 [76]3 years ago
8 0
The angular velocity as a function of time is given by
\omega (t)=\gamma-\beta t^2
where \gamma=4.90 rad/s and \beta=0.750 rad/s^3. The angular acceleration as a function of time is equal to the derivative of the angular velocity. If we calculate the derivative of w(t), we find:
\alpha(t)= \frac{d\omega}{dt} =-2\beta t
<span>and this is the angular acceleration of the fan blade.</span>
Umnica [9.8K]3 years ago
7 0

Answer:

Angular acceleration, α = -2βt

Explanation:

Angular velocity of fan is \omega_{z(t)}=\gamma -\beta t^2

\gamma=4.90\ rad/s

\beta=0.750\ rad/s^3

Angular acceleration is given by :

\alpha=\dfrac{d\omega}{dt}

\alpha=\dfrac{d(\gamma -\beta t^2)}{dt}

\alpha=-2\beta t

Hence, the above equation is the angular acceleration as a function of time.   

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Las condiciones iniciales de un gas son 3000 cm3
slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

PV=nRT

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C

hence, T'=92.70°C

8 0
3 years ago
A hockey puck with a mass of 0.175 kg slides over the ice. The puck initially slides with a speed of 5.25 m/s, but it comes to a
Neko [114]

Answer:

1.70 J

Explanation:

The heat dissipated is the difference in the kinetic energies.

This is given by

E = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

v_i and v_f are the initial and final velocities.

With <em>m</em> = 0.175 kg,

E = \frac{1}{2}\times0.175(2.85^2 - 5.25^2) = -1.701\text{ J}

The negative sign appears because energy is lost.

7 0
3 years ago
Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co
Illusion [34]

Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

         k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

          q = \sqrt{ \frac{G}{k} }    m

we substitute

           q = \sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }   m

            q = \sqrt{0.7419 \ 10^{-20}}   m

           q = 0.861 10⁻¹⁰ m

           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

             

this is a very small charge value so it should be easy to create in each one

6 0
2 years ago
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Bad, it's bring more hatred to the world
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Svet_ta [14]
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