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3 years ago

13

On a hot summer day several swimmers decide to jump into a river. They step off the bridge and strike the water at 1.5s later. I

gnore air resistance. What is the height of the bridge?
I did it but I don’t know if it’s right

1 answer:

xxTIMURxx [149]3 years ago

6 0

That is right because base*length is how you get the height

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A vertically hung spring has a spring constant of 150. newtons per meter. A

stiks02 [169]

What fraction of stipend triangle is a shaded triangle? What fraction of the spotted triangle is a shaded triangle? Use >,

5 0

3 years ago

Convert 3402kgm/s to 20000Newtons

oee [108]

The 3,402 has units of kg-m/s. That's momentum. The 20,000 has units of Newtons. That's force. Momentum and force are different physical things, and you can't convert them from one to the other.

The best I can do for you is something like this:

Let's say you have a moving object with 3,402 kg-m/s of momentum, and you want to STOP it completely. You want to stand in front of it and push back on it, hard enough and for long enough to CHANGE its momentum from 3,402 kg-m/s to zero.

Also ... there's a limit to how hard you can push. The most force you can exert is 20,000 Newtons.

The amount you'll change its momentum is called the <u><em>impulse</em></u> you give it. The quantity of impulse is (force) x (length of time you push on it).

So you need to keep pushing it back for (T seconds) long enough so that

(20,000 Newtons of force) x (T seconds) = 3,402 kg-m/s of momentum .

Divide each side of that equation by (20,000 Newtons). Then it says:

(T seconds) = (3,402 kg-m/s) / (20,000 Newtons)

<em>T = 0.1701 second</em>

And that's how you provide just enough impulse to stop the flying object ... push on it with 20,000 Newtons of force for exactly 0.1701 second, and it loses all its momentum and falls out of the air onto the ground at your feet.

This story is the closest I can come to anything that looks like "convert"ing momentum into force.

3 0

3 years ago

7. 13 a turbine receives steam at 6 mpa, 600°c with an exit pressure of 600 kpa. Assume the turbine is adiabatic and neglect kin

AnnyKZ [126]

The work done by the turbine will be 708.2 kJ/kg. The work done by the turbine is the difference of the enthalpy at inlet and exit.

<h3 /><h3>What is temperature?</h3>

Temperature directs the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles, more the temperature.

If the given turbine is assumed to be reversible;

$\rm P_I$ (Initial pressure)=60 mpa = 60 bar

$\rm T_i$ (Initial temperature)=600° C

$\rm P_e$ (Exit pressure)=600 kpa=6 bar

The heat balance equation is;

$\rm q-W_t=h_e-h_i\\\\ q=0\\\\\ W_t=h_i-h_e$

The change in the entropy is;

$\rm S_2-S_1=\frac{\delta q}{dt}$

The work done by the turbine is;

$\rm W_t = h_i-h_e\\\\ W_t =3658.4 -29.5019\\\\ W_t =708.2 \ kJ/kg$

Hence,the work done by the turbine will be 708.2 kJ/kg.

To learn more about the temperature, refer to the link;

brainly.com/question/7510619

#SPJ4

6 0

2 years ago

The base of a pyramid covers an area of 13.0 acres (1 acre = 43,560 ft2) and has a height of 481 ft. If the volume of a pyramid

Allisa [31]

V = 1/3 Bh v = 1/3 (13 ac)(43560ft^2/ac)(481ft) v = 90793560 ft^3 * 0.3048m/ft * 0.3048m/ft * 0.3048m/ft = 2570987m^3

3 0

3 years ago

Read 2 more answers

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce

jasenka [17]

Answer:

$\lambda= 506.25 nm$

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

$y=\frac{m \lambda D}{a}$

Where:

$y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1$

Solving for λ:

$\lambda=\frac{y*a}{mD}$

Replacing the data provided by the problem:

$\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm$

7 0

3 years ago