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2 years ago

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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Volgvan

Answer:

Explanation:

height of Ellipse $=30 cm$

i.e. $2 a=30$

Width of Ellipse $=22 cm$

i.e. $2 b=22$

Equation of a vertical Ellipse is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

$a=15 ,b =11$

$\frac{x^2}{11^2}+\frac{y^2}{15^2}=1$

at $y=4 cm$

$\frac{x^2}{11^2}+\frac{4^2}{15^2}=1$

$x=\frac{11}{15}\times \sqrt{15^2-4^2}$

$x=10.6 cm$

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3 years ago

Select the correct answer

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Answer:

C or 3

Explanation:

A: no they are called sources of sound. A is incorrect.

B: It does. Many people attest to this. But this is not a property of physics.

C: A media is required is the correct answer.

D: Dogs might. In general we don't.

I would pick C

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3 years ago

A student investigated how the mass of water in an electric kettle affected the time taken for the water to reach boiling point.

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Answer:

1.because of the heat produced by the socat

2. they should have control how they placed the heater

3. because the water is to much

4.because is different from the question

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3 years ago

A cube slides down the surface of a ramp at a constant velocity. What is the magnitude of the

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A . The weight of the cube

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3 years ago

A charged particle A exerts a force of 2.45 μN to the right on charged particle B when the particles are 12.2 mm apart. Particle

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Answer:

$F_2 = 1.10 \mu N$

Explanation:

As we know that the electrostatic force is a based upon inverse square law

so we have

$F = \frac{kq_1q_2}{r^2}$

now since it depends inverse on the square of the distance so we can say

$\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$

now we know that

$r_2 = 18.2 mm$

$r_1 = 12.2 mm$

also we know that

$F_1 = 2.45 \mu N$

now from above equation we have

$F_2 = \frac{r_1^2}{r_2^2} F_1$

$F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)$

$F_2 = 1.10 \mu N$

5 0

4 years ago