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Leya [2.2K]
3 years ago
5

A car initially traveling at 15 m/s speeds up at a constant rate of 2.0 m/s2 for 3 seconds. The velocity of the car at the end o

f the 3 second interval is _________ m/s.
Physics
1 answer:
timofeeve [1]3 years ago
8 0

Answer:

Vf = 21 m/s

Explanation:

Data:

  • Initial Velocity (Vo) = 15 m/s
  • Acceleration (a) = 2.0 m/s²
  • Time (t) = 3 s
  • Final Velocity (Vf) = ?

Use formula:

  • Vf = Vo + a * t

Replace:

  • Vf = 15 m/s + 2.0 m/s² * 3s

Multiply the acceleration with time:

  • Vf = 15 m/s + 6 m/s

Solve the sum:

  • Vf = 21 m/s

The velocity of the car at the end of the 3 second interval is <u>21 meters per second.</u>

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Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

Answer:

E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

Then the electric field at the point of interest is estimated as:

E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}

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