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Natali [406]
3 years ago
12

When you whirl a small rubber stopper on a cord in a vertical circle, you find a critical speed at the top for which the tension

in the cord is zero. At this speed. the force of gravity on the object 1S itselfisufficient to supply the necessary centripetal force. How slowly can you swing a 15-g stopper like this so that It will just follow a circle with a radius of m?​
Physics
1 answer:
Sauron [17]3 years ago
6 0

Answer:

  √(9.8/m) radians per second

Explanation:

The centripetal acceleration required for a radius of m is ...

  a = mω²

so, the angular speed required is ...

  ω = √(a/m) =  √(g/m) . . . . . where g is the acceleration due to gravity

  ω = √(9.8/m) . . . radians per second

__

The speed is independent of the mass. For a 1 m radius, the speed is about 29.9 RPM.

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What happens when the data in an investigation does not support the origanal hypothesis
alexgriva [62]

Answer: The scientist gives up and starts an investigation on a new topic.

Explanation:

The data is altered so that it supports the original hypothesized. The data is then altered so that it supports the original hypothesis.

5 0
3 years ago
Which of the follow represent a chemical changes
mixer [17]

Answer:

C) steel turning to rust in salt air

Explanation:

The missing options are:

A) ice melting to form liquid water

B) water boiling to form steam

C) steel turning to rust in salt air

D) sugar dissolving into hot coffee

In a chemical change the atoms of the reacting compounds are reordered forming new compounds. In a chemical change, new compounds appear, but in a physical change not.

Then, change of states like ice melting and water boiling are not chemical changes.

During steel rust, components of steel, like iron, are oxidized, that is, reacts with oxygen forming oxides.

The dissolution of sugar into hot coffee is a physical change in which sugar molecules get further apart in the coffee,  but they don't change.

5 0
4 years ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
3 years ago
An electromagnetic wave has a wavelength of 1.3m what is the frequency of wave
larisa86 [58]
V = f * wavelength
as we know electromagnetic wave has speed equal to light, so

3 * 10^8 = f * 1.3
f = 2.3 * 10^8 hertz
f = 230 mega hertz
6 0
3 years ago
Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.
Burka [1]
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time  t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is  7/3 .              v1 = 7/3 m/s .

At time  t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
                                     v2 = zero .

At time  t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is  -3/5 .            v3 = -0.6 m/s .              
7 0
3 years ago
Read 2 more answers
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