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Mademuasel [1]
2 years ago
13

A bullet (m=20g) shot with a speed of 800 m/s hits an oak tree and sticks 4cm inside it.

Physics
1 answer:
Tcecarenko [31]2 years ago
6 0

Hi there!

We can begin by converting cm to m:

4cm = 0.04 m

Use the following equation:

vf² = vi² + 2ad

The final velocity is 0 m/s, so:

0 = vi² + 2ad

vi² = -2(a)d

Plug in given values to solve for the acceleration:

800²/2(0.04) = -a

a = -8,000,000 m/s²

Find the force using the equation:

F = ma

F = (0.02)(-8,000,000) = -160,000 N

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Answer and Explanation:

In optics, a CoC(Circle of Confusion) is defined the minimum cross section of a paraxial bundle of rays made by a lens which is sphero-cylindrical type and can be viewed as an optical spot, which do not converge perfectly at the focus  while a point source is being imaged due to spherical aberration.

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When it comes to how ray lines are drawn, what makes the convex lens and concave mirror similar to each other?
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Answer:

Convex lens and convex mirrors are similar that

1. They have the same image characteristics at various object positions

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3. Both their ray lines converge to a particular focal point

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3 years ago
Calculate the frequency of visible light having a wavelength of 410 nm
maria [59]
Given: Wavelength λ = 410 nm  convert to Meters m = 4.10 x 10⁻⁷ m

Speed of light c = 3 x 10⁸ m/s

Required: Frequency  f = ?

Formula: c = λf 

               f = c/λ

               f = 3 x 10⁸ m/s/4.10 x 10⁻⁷ m

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3 years ago
A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet
lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

  • <em>Current flowing in the wire, I = 4.00 mA</em>
  • <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
  • <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
  • <em>Length of wire, L = 2.00 m</em>
  • <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2

The final area of the copper wire;

A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2

The initial drift velocity of the electrons is calculated as;

v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

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Natalka [10]
Low coefficient of friction

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3. swimming (water & skin)
4. rowing a boat (water and boat)

6 0
3 years ago
Read 2 more answers
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