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arsen [322]
3 years ago
9

3.

Physics
1 answer:
Flura [38]3 years ago
8 0

Answer:

about 4 km

Explanation:

15 minutes is a quarter of an hour, so you divide 16km by 4 to get your answer

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Chang sees the following formula on a website about Newton’s second law.
yarga [219]
The correct answer is the correct answer is A
3 0
3 years ago
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Electricity The power P, in watts, that a circular solar cell produces and radius of the cell in centimeters are related by the
gavmur [86]

Answer:

9.05 W

Explanation:

The given formula is r=\sqrt {\frac {P}{0.02\pi}} where r is in centimeters and P is in Watts

Making Power, P the subject from the above formula

P=0.02\pi r^{2}

Substituting r with 12 cm then

P=0.02\pi 12^{2}=
9.047786842  W\approx 9.05 W

5 0
3 years ago
A hypothesis is
weqwewe [10]

Answer:

D a conclusion

Explanation:

The definition of hypothesis is a supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation

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3 years ago
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A uniformly charged, straight filament 5.10 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder
Ratling [72]

Answer:

70509.8039216 N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

q = Charge = 2.00 µC

l = Length of filament = 5.1 m

r = Radius of cylinder = 10 cm

\lambda=\dfrac{q}{l}

Electric field is given by

E=\dfrac{2k\lambda}{r}\\\Rightarrow E=\dfrac{2\times 8.99\times 10^9\times \dfrac{2\times 10^{-6}}{5.1}}{10\times 10^{-2}}\\\Rightarrow E=70509.8039216\ N/C

The electric field at the surface of the cylinder is 70509.8039216 N/C

7 0
3 years ago
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A person pushes a 15.7-kg shopping cart at a constant velocity for a distance of 25.9 m on a flat horizontal surface. She pushes
valentinak56 [21]

Answer:

a)    F = 35.7 N, b)   W = 846.7 J, c)   W = - 846.9 J, d) W=0

Explanation:

a) For this exercise let's use Newton's second law, let's set a reference frame with the x-axis horizontally

let's break down the pushing force.

        cos (-23,7) = Fₓ / F

        sin (-237) = F_y / F

        Fₓ = F cos 23.7 = F 0.916

        F_y = F sin (-23.7) = - F 0.402

         

Y axis  

       N- W - F_y = 0

       N = W + F 0.402

X axis

       Fₓ - fr = 0

       F 0.916 = fr

       F = fr / 0.916

       F = 32.7 / 0.916

       F = 35.7 N

It is asked to calculate several jobs

b) the work of the pushing force

       W = fx x

       W = 35.7 cos 23.7 25.9

       W = 846.7 J

c) friction force work

        W = F x cos tea

friction force opposes movement

        W = - fr x

         W = - 32.7 25.9

         W = - 846.9 J

d) The work of the force would gravitate, as the displacement and the force of gravity are at 90º, the work is zero

          W = 0

6 0
3 years ago
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