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Tatiana [17]
3 years ago
11

Can y’all please help

Physics
1 answer:
TiliK225 [7]3 years ago
5 0

Answer:

<em>The cat's speed when it slid off the table was 2.53 m/s</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally at a speed v, from a height h, it describes a curved path ruled by gravity, until it hits the ground.

the range or maximum horizontal distance traveled by the object can be calculated as follows:

\displaystyle d=v\cdot\sqrt{\frac  {2h}{g}}

The situation described in the problem fits into the concept of a horizontal launch when the cat slides off the table and gets to the ground some distance ahead.

It's given the range as d=1.3m and the height h=1.3 m, thus we can find the cat's speed by solving the equation for v:

\displaystyle v=d\cdot\sqrt{\frac  {g}{2h}}

Where g=9.81\ m/s^2

Substituting values:

\displaystyle v=1.3\cdot\sqrt{\frac  {9.81}{2\cdot 1.3}}

\displaystyle v=1.3\cdot\sqrt{\frac  {9.81}{2.6}}

Calculating:

v = 2.53 m/s

The cat's speed when it slid off the table was 2.53 m/s

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   a = 2 lines 1/600

   a = 2/600

    a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm

let's use trigonometry

      tan θ = y / L

as the measured angles are small

      tan θ = sin θ / cos θ sin θ

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we substitute

     a  y/L = λ

     y = λ L / a

for λ = 400 10-9 m

      I = 400 10⁻⁹ 2.9 / 3.33 10⁻³

      i = 346.89 10⁻⁶ m

f

or λ = 700 nm

        y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³

        y_f = 609.609 10⁻⁶ m

the separation of this spectrum

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A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the c
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Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

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the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

B =\mu_o  I(\frac{N}{L} )\\\\B =  \mu_o  I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

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