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SIZIF [17.4K]
3 years ago
8

The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if i

t spins at 1200 rev/min? (b) What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac? (c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of g .
Physics
1 answer:
Sergio [31]3 years ago
7 0

Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius=r=\frac{d}{2}=\frac{2.3}{2}=1.15 m

a.We have to find the angular velocity in radians per second if it spins=1200 rev/min

Frequency=\frac{1200}{60}=20 Hz

1 minute=60 seconds

Angular velocity=\omega=2\pi f

Angular velocity=2\times \frac{22}{7}\times 20=125.7 rad/s

b.We have to find the linear speed of its tip at this  angular velocity if the plane is stationary on the tarmac.

v=r\omega=1.15\times 125.7=144.56 m/s

c.Centripetal acceleration=\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2

Centripetal acceleration==\frac{18170.56\times g}{9.81}=1852.25 g m/s^2

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Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t
padilas [110]

Answer:

\omega=32.14\ rad/s

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

v=\omega\times A

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

\omega=\dfrac{v}{A}

\omega=\dfrac{2.08\ m/s}{0.0647\ m}

\omega=32.14\ rad/s

So, the angular frequency of the spring is 32.14 rad/s.

4 0
3 years ago
Can I put an “I” for an argumentative essay?
masha68 [24]

Answer:

There is no place for first person in formal academic argumentative essay. However, you can use first person only if it has been asked to provide your personal opinion or reflection about the topic.

Explanation:

4 0
3 years ago
In an experiment, a large number of electrons are fired at a sample of neutral hydrogen atoms and observations are made of how t
kakasveta [241]

Answer:

N = 1036 times

Explanation:

The radial probability density of the hydrogen ground state is given by:

p(r) = \frac{4r^{2} }{a_{0} ^{3} } e^{\frac{-2r}{a_{0} } }

p(\frac{a_{0} }{2} ) = \frac{4(\frac{a_{0} }{2} )^{2} }{a_{0} ^{3} } e^{\frac{-2(\frac{a_{0} }{2} )}{a_{0} } }

p(2a_{0} ) = \frac{4(2a_{0}) ^{2} }{a_{0} ^{3} } e^{\frac{-4a_{0} }{a_{0} } }

N = 1300\frac{p(2a_{0}) }{p(\frac{a_{0} }{2} )}

N = 1300\frac{(2a_{0}) ^{2}e^{\frac{-4a_{0} }{a_{0} } }  }{(\frac{a_{0} }{2} )^{2} e^{\frac{-a_{0} }{a_{0} } }}

N = 1300(16) e^{-3}

N = 1035.57

N = 1036 times

6 0
3 years ago
If potential energy at point A is 12 joules, what is the potential energy at D?
Irina-Kira [14]

Answer:

In D: 3J

Explanation:

Potential energy: Ep=mgh where m is the mass, h altitude.

In point A: h=20cm=0.2m

Epa=12=0.2×mg. Thus mg=12/0.2=60N

For point D: hd=5cm=0.05m

Epd=mg×0.05=60×0.05=3J

5 0
2 years ago
A 150 g copper bowl contains 210 g of water, both at 24.0°C. A very hot 430 g copper cylinder is dropped into the water, causing
Dahasolnce [82]

Answer:

A. 15969.22 cal

B. 1052,22 cal

C. 528,87 °C

Explanation:

To solve this kind of question, a proper method is to work from the data that you have towards the data that you need. Also, it is recommended to analyze related equations as they could give us clues on how to find the missing information or the information that the problem is asking us.

Let us start with Question A. It is important to remember that energy transfers with the environment are being neglected; this means that all the energy that the cylinder lose is picked up by the water and the copper bowl. To find the amount of energy transferred to the water, we first find the amount of energy necessary to raise the water’s temperature to 100°C and then we find the amount of energy necessary to evaporate the 17.1 g of water indicated by the question. This would be:

Q = m_water * CP_water *∆T =210g *1 cal/(g K) * (100°C-24°C) = 15960 cal

Q_evap = m_wat * L = 17,1 g * 539 cal/kg* (1 kg)/(1000 g) =9.2169 cal

Therefore, the total energy that was transferred to the water is the sum of these components, that would be Q_tot = 15960 cal + 9.2159 cal = 15969.22 cal.  Let´s also remember that a temperature difference in K is equal to a temperature difference in ° C

To solve Question B, we use the same method. We must find the amount of energy necessary to raise the temperature from its initial temperature to the one stated by the problem to be the equilibrium temperature of the system (100°C):

Q= m_copper *CP_copper *∆T = 150g * 0.0923 cal/(g K) * (100°C-24°C) = 1052,22 cal

If we add the components we just found in questions A and B, we can find the amount of energy than the Copper cylinder lost, this would be: Q_tot = 15969.22 cal + 1052.22 cal = 17021.44 cal.

The question C asks us to find the initial temperature of the cylinder and Q_tot will help us to find it.

We know that Q_tot is the energy lost by the cylinder and we also know that Q_tot = m_cylinder * CP_copper * ∆T. Therefore, what we need to do  is clear the last term of the equation and find the initial temperature.

Q_tot = m_cylinder *CP_copper *∆T → T_fin-T_initial = Q_tot/(m_cylinder*CP_copper ) = (-17021.44 cal)/(430g*0.0923 cal/(g K))

→ T_initial = 100°C + (-17021.44 cal)/(430g * 0.0923 cal/(g K)) = 528,87 °C

If we convert the 100°C to K before we do the calculation, the result would be the same one, You would only need to add 273,15 to the final result to check it out.  

Hope everything was clear. If you have any further question, I'll be happy to help :D

5 0
3 years ago
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