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Andrews [41]
2 years ago
5

(forces and motion) explain in 3-2 sentences why acceleration is interesting.

Physics
1 answer:
Luden [163]2 years ago
7 0

Answer:

  • Acceleration is a vector and is the rate at which an object changes its velocity. Acceleration is not going fast.
  • If a person is moving fast but not changing speed, he is not accelerating.
  • The person or object has to be going fast sometimes, faster sometimes, or slower sometimes.
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What is the magnitude of the force acting on a spring with a spring constant of 275 N/m that is stretched 14.3 cm?
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Hydrogen atoms are placed in an external magnetic field. The protons can make transitions between states in which the nuclear sp
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Magnetic field = 0.534 T

Explanation:

The solving is on the attach document.

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3 years ago
A system proposed by the U.S. Navy for underwater submarine communication, called ELF (for extremely low frequency), operates wi
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Answer

Wavelength= 30*20^8/30=10^7m

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3 years ago
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A wave is moving at 18 m/s. If its wavelength is 3 meters, what is its frequency?
Vlada [557]

here's the solution,

we know that,

=》

wave \: speed = wavelength \times frequency

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18 = 3 \times f

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f =  \dfrac{18}{3}

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f = 6

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7 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
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