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Goshia [24]
2 years ago
10

A parking lot is going to be 50 m wide and 150 m long. Which dimensions

Physics
1 answer:
Illusion [34]2 years ago
5 0

Answer:

Explanation:

Given:

Length of box = 150 m

Width of box = 50 m

Find:

Dimension used for scale model

Computation:

Always smallest size of a dimension is used for scale model

Small size is easily used for divide so, small size is used for a scale model.

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If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.100
SCORPION-xisa [38]

Answer:

Charge enter a 0.100 mm length of the axon is 8.98\times 10^{-12} C

Explanation:

Electric field E at a point due to a point charge is given by

E=k \frac{q}{r^2}

where k is the constant =9.0 \times 10^9  Nm^2 / C^2

q is the magnitude of point charge and r is the distance from the point charge

Charges entering one meter of axon is 5.\times 10^{11} \times (+e)

Charges entering 0.100 mm of axon is 5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}

substituting the value of +e=1.6\times 10^{-19} C in above equation, we get charge enter a 0.100 mm length of the axon is

q=5.\times 10^{11} \times1.6\times 10^{-19}  \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C

3 0
2 years ago
The nucleus of an atom has all of the following characteristics except that it
Sedaia [141]

Answer:

THE ANSWER IS: contains nearly all of the atom's volume.

Explanation:

3 0
2 years ago
If I0 is the intensity of the unpolarized light incident on the first polarizer, and I1 and I2 denote the intensity of the light
e-lub [12.9K]
E or C 10 hope this helps
7 0
2 years ago
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

5 0
2 years ago
A small crack occurs at the base of a 15.0-m-high dam. The effective area through which water leaves is 2.30 × 10-3 m2. (a) Igno
vova2212 [387]

Answer

given,                                              

height of the dam = 15 m            

effective area of water = 2.3 x 10⁻³ m²

Using energy conservation              

    m g h = \dfrac{1}{2}mv^2

    v= \sqrt{2gh}                  

    v= \sqrt{2\times 9.8 \times 15}

    v= \sqrt{294}              

           v = 17.15 m/s            

 discharge of water

      Q = A V                            

      Q = 2.3 x 10⁻³ x 17.15    

      Q = 0.039 m³/s

3 0
2 years ago
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