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kondaur [170]
3 years ago
5

Ben franklin showed that 1 teaspoon of oil would cover about 0.50 acre of still water. if you know that 1.0 x 104m2 = 2.47 acres

, and that there are 5.0 cm3 in a teaspoon, what is the thickness (in cm) of a later of oil?
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

To solve for the thickness of the latter of oil, we can simply divide the volume by the total surface area:

Thickness = Volume / Surface Area

We are both given the volume and the surface area, all we have to do now is to convert the units in like terms so that we can cancel those out.

Surface Area = 0.50 acres (1.0 x 10^4 m^2 / 2.47 acres)

Surface Area = 2,024.29 m^2

Further converting this into cm:

Surface Area = 2,024.29 m^2 (100 cm / m)^2

Surface Area = 20,242,914.98 cm^2

 

Therefore the thickness is:

Thickness = 5.0 cm^3 / 20,242,914.98 cm^2

Thickness = 2.47 x 10^-7 cm

or

<span>Thickness = 2.47 x 10^-9 m = 2.47 nm = 24.7 A</span>

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3 years ago
A transmission diffraction grating with 420 lines/mm is used to study the light intensity of di event orders (n). A screen is lo
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Answer:

Explanation:

Diffraction grating is used to form interference pattern of dark and bright band.

Distance between adjacent slits (a ) = 1 / 420 mm

= 2.38 x 10⁻³ mm

2.38 x 10⁻⁶ m

wave length of red light

= 680 x 10⁻⁹ m

For bright red band

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= n λD / a ,         n = 0,1,2,3 etc

D = distance of screen

putting n = 1 , 2 and 3 , we can get three locations of bright red band.

x₁ = λD / a

=  680 x 10⁻⁹ x 2.8 / 2.38 x 10⁻⁶

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3 years ago
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A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis
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Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

Y= VoT+0.5gt^{2}

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Vo = Initial speed =0

T = time

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y = height=60m

solving for time

Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

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c)

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t    

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d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S

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It is 25 I’m sure of it
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