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Black_prince [1.1K]
2 years ago
12

In downtown Chicago, the east-west blocks are 400 ft long while the north-south blocks are 280 ft long. Because of the many one-

way streets, it can be challenging to get around. Veronica starts at the corner of Dearborn and Ohio Streets. She drives four blocks north to Superior, two blocks east to Wabash, then a block south to get to her destination at Wabash and Huron.
Required:
What is the straight-line distance from her starting point?
Physics
1 answer:
pickupchik [31]2 years ago
8 0

Answer:

The answer is "1160 ft".

Explanation:

Using the Pythagoras theorem:

\to a= 280 \times 3= 840\\\\\to b= 400 \times 2= 800\\\\

\bold{\to y^2= a^2+b^2}\\\\

         =840^2 +800^2\\\\= 705600+640000\\\\=1345600\\\\=1160 \ ft\\

  \to y=1160 \ ft

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An astronaut drops a rock from the top of a crater on the moon. When the rock is halfway down to the bottom of the crater, its s
Alexxx [7]

Answer: vf1/vf2= 1/ sqrt(2)

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if the rock travels H to the bottom we can calculate velocity:

vo=0m/s (drops the rock)  , yo=0

vf*vf= vo*vo+2g(y-yo)

when the rock is halfway  y = H/2 so:

vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)

when the rock reach the bottom y=H so:

vf2*vf2=2*g*H so vf2 = sqrt(2gH)

so vf1/vf2= 1/ sqrt(2)

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8 0
3 years ago
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A container is filled with water to a depth of 26.2 cm. On top of the water floats a 16.1 cm thick layer of oil with a density of
Solnce55 [7]

Answer:

1.34 x 10^3 Pa

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g = 9.81 m/s^2

height of oil column = 16.1 cm = 0.161 m

Pressure on the surface of water = height of oil column x density of oil x g

                                                      = 0.161 x 0.85 x 10^3 x 9.81 = 1.34 x 10^3 Pa

Thus, the pressure on the surface of water is 1.34 x 10^3 Pa.

4 0
3 years ago
A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)
bezimeni [28]

Answer:

205 V

V_{R} = 2.05 V

Explanation:

L = Inductance in Henries, (H)  = 0.500 H

resistor is of 93 Ω so R = 93 Ω

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V_{L} = - IwLsin(wt)

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

 = 11.0 V / 500 rad/s (0.500 H)

 = 11.0 / 250

I = 0.044 A

Now

V_{R} = IR

    = (0.044 A) (93 Ω)

V_{R} = 4.092 V

Deriving formula for voltage across the resistor

The derivative of sin is cos

V_{R} = V_{R} cos (wt)

Putting V_{R} = 4.092 V and w = 500 rad/s

V_{R} = V_{R} cos (wt)

    = (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V_{R} = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

    =  (4.092 V) (cos (500 rads/s)(0.00209))

    = (4.092 V) (cos(1.045))

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V_{R} = 2.05 V

8 0
2 years ago
Is Radioactive decay a way to find the relative age of a rock?<br><br> True<br> False
galina1969 [7]
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5 0
3 years ago
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Schach [20]

Answer:

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From the question we are told that

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=>   E  A  =  \frac{q}{\epsilon_o }

Given that  the electric field is pointing downward  , the equation become

    - E  A  =  \frac{q}{\epsilon_o }

Here   q is the excess charge on the surface of the earth

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Where r is the radius of the earth which has a value r = 6.3781*10^6 m

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So

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Here \epsilon_o s the permitivity of free space with value

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substituting values

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     q  =  -461532.5 \ C

6 0
3 years ago
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