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pentagon [3]
2 years ago
10

When does the particle have a displacement of +1 meter from its initial position (at t = 0)?

Physics
1 answer:
anzhelika [568]2 years ago
8 0

000000000000000000000000

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A .2kg Basketball is pitched with a velocity or 40 m/s and then bed and into the picture with a velocity of 60 m/s. What is the
Tamiku [17]

Answer:

40kgm

Explanation:

∆p = m(v - u)

= 2(60 - 40)

= 2 × 20

= 40kgm/s

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At which location would a bowling ball have the greatest weight?
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When you are rolling it
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A certain water wave has a wavelength of 25 m and a frequency of 4.0 Hz. What is its velocity?
babymother [125]
Wave speed = frequency * wavelength
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8 0
3 years ago
What is the difference between a hurricane and a typhoon
Lera25 [3.4K]

Answer:

The answer is below

Explanation:

Well, a hurricane is a storm with violent winds (more less with a force of 12 in the scale that measures winds) and it happens especially in the Caribbean sea. While a typhoon is also a strong storm but it happens especially in the Indian region and the western pacific oceans.

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3 0
2 years ago
Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha
Westkost [7]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, A_p = 180 cm^2

- The charge on each plate, q = 17 * 10^-^6 C

- Permittivity of free space, e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}

- The radius for the flux region, r = 3.3 cm

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

                             A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ = \frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

                         E+ = E- = \frac{sigma}{2*e_o} \\\\

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

                        E_n_e_t = (E+)  + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C}  \\

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C

5 0
2 years ago
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