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pochemuha
3 years ago
6

The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper e

nd. Each arm supports a seat suspended from a cable 5.00 m long, the upper end of the cable being fastened to the arm at a point 3.00 m from the central shaft. Does the angle depend on the weight of the passenger for a given rate of revolution?
Physics
2 answers:
Mashcka [7]3 years ago
8 0

Answer:

Explanation:

When the central shaft rotates , the seat along with passenger also rotates . Their rotation requires a centripetal force of mw²R where m is mass of the passenger and w is the angular velocity and R is radius of the circle in which the passenger rotates.

This force is provided by a component of   T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,

T cos θ will provide the centripetal force . So

Tcosθ = mw²R

Tsinθ component will balance the weight .

Tsinθ = mg

Dividing the two equation

Tanθ = \frac{g}{\omega^2R}

Hence for a given w , θ depends upon g or weight .

lesya692 [45]3 years ago
5 0

Answer:

In the expression the mass is cancelled out and angle is independent of the weight of the person

Explanation:

Let the swing is rotating with constant angular speed

now due to rotation of swing the string will make some angle with the vertical

now we have

Tcos\theta = mg

T sin\theta = m\omega^2 r

now we will have

tan\theta = \frac{\omega^2 r}{g}

here we know that

r = R + L sin\theta

where

R = 3.00 m

L = 5.00 m

Now as we can see the above expression the mass is cancelled out and angle is independent of the weight of the person

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3 years ago
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At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar
swat32

Answer:

C=1,25\cdot 10^{5} kJ/^{\circ}C

Explanation:

First of all let's define the specific molar heat capacity.

C = \frac{-Q}{n\cdot \Delta T} (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system  

Now, we can find n with the molar mass (M) the mass of the compound (m).

n=\frac{m}{M}=6.95\cdot 10^{-3} moles      

Using (1) we have:

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6 0
3 years ago
Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o
Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.  

\lambda_{proton} = 6.05x10^{-14}m < \lambda_{electron} = 1.10x10^{-10}m

Part B:

The proton has a smaller wavelength than the electron.

\lambda_{proton} = 1.29x10^{-13}m < \lambda_{electron} = 5.525x10^{-12}m

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

\lambda = \frac{h}{p} (1)

Where h is the Planck's constant and p is the momentum.

\lambda = \frac{h}{mv} (2)

Part A

Case for the electron:

\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}

But J = Kg.m^{2}/s^{2}

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}

\lambda = 1.10x10^{-10}m

Case for the proton:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}

\lambda = 6.05x10^{-14}m

Hence, the proton has a smaller wavelength than the electron.  

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

KE = \frac{1}{2}mv^{2}

2KE = mv^{2}

v^{2} = \frac{2KE}{m}

v = \sqrt{\frac{2KE}{m}}  (3)

Case for the electron:

v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}

but 1J = kg \cdot m^{2}/s^{2}

v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}

v = 1.316x10^{8}m/s

Then, equation 2 can be used:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}    

\lambda = 5.525x10^{-12}m

Case for the proton :

v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}

But 1J = kg \cdot m^{2}/s^{2}

v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}

v = 3.07x10^{6}m/s

Then, equation 2 can be used:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}

\lambda = 1.29x10^{-13}m    

Hence, the proton has a smaller wavelength than the electron.

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3 years ago
An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/
Katen [24]
<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}    (1)

Where:

y  is the instrument's final position  

y_{o}=0  is the instrument's initial position

V_{o}=15m/s is the instrument's initial velocity

t is the time the parabolic movement lasts

g=2.5\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of planet X.

As we know y_{o}=0  and y=0 when the object hits the ground, equation (1) is rewritten as:

0=V_{o}.t-\frac{1}{2}g.t^{2}    (2)

Finding t:

0=t(V_{o}-\frac{1}{2}g.t^{2})   (3)

t=\frac{2V_{o}}{g}   (4)

t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}   (5)

Finally:

t=12s

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3 years ago
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