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andrew11 [14]
3 years ago
5

Which of these components is present in this circuit schematic?

Physics
2 answers:
Alchen [17]3 years ago
7 0

Answer:

the answer is d for plato users

Explanation:

Svetlanka [38]3 years ago
4 0
Its resistor :}...............

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20 cm long 10 cm wide and 5 cm thick as a mass of 500 g determine the greatest pressure that can be exerted by block on the flat
uysha [10]

100000 Pascal

Explanation:

pressure= force/area

Max pressure= force/min area

so f=5

min area= 5×10^-5

5÷5*10^-5 = 100000pascal

8 0
2 years ago
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
2 years ago
Fat Albert (the TV show character) runs up the stairs on Monday. On Tuesday, he walks up the same set of stairs. Which day did h
lys-0071 [83]

Answer:

Tuesday bc instead of running he/she was walking bc he/she might not have as much energy

Explanation:

8 0
2 years ago
A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels
Crank

Answer:

v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }

Explanation:

The average velocity is total displacement divided by time:

v_{avg} =\dfrac{D_{tot}}{t}

And in the case of vertical v_{avg}

v_{avg}=\dfrac{y_{tot}}{t}

where y_{tot} is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height H with initial velocity v_0 is given by:

y=H+v_0t-\dfrac{1}{2} gt^2

The time it takes for the rock to reach maximum height is when y'(t)=0, and it is

t=\frac{v_0}{g}

The vertical distance it would have traveled in that time is

y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2

y_{max}=\dfrac{2gH+v_0^2}{2g}

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

y_{down}=\dfrac{2gH+v_0^2}{2g}+H

Therefore, the total displacement throughout the rock's journey is

y_{tot}=y_{max}+y_{down}

y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H

\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

t=\dfrac{v_0}{g},

and it should take the rock the same amount of time to return to the roof, and it takes another t_0 to go from the roof of the building to the ground; therefore,

t_{tot}=2\dfrac{v_0}{g}+t_0

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

Therefore the total time is

t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH}  }{g}}

Now the average velocity is

v_{avg}=\dfrac{y_{tot}}{t}

v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }

\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }

5 0
3 years ago
A speeding Thunderbird left skid marks on the road that were 76.7 m long when it came to a stop. If the acceleration of the car
FrozenT [24]

Answer: 39.2 m/s

Explanation:

You can use the kinematic equation:

v_f^2=v_i^2+2a*d

We know the final velocity because it says it came to a stop. So now all we gotta do is plug in.

0^2=v_i^2+2(-10)(76.7)\\v_i^2=1,534\\v_i=\sqrt{1534} \\=39.166 m/s

4 0
3 years ago
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