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vesna_86 [32]
2 years ago
11

A student made the claim that a 4 gram paintball fired from a paintball gun at 90 m/s could have about the same kinetic energy a

s a 1 gram BB pellet fired from a BB gun at 180 m/s.
Do you agree or disagree with the student's claim? Use evidence and mathematical reasoning to support your response.
Physics
1 answer:
vovikov84 [41]2 years ago
7 0

This question involves the concept of kinetic energy.

The student's claim is "right".

<h3>Kinetic Energy</h3>

The energy possessed by a body, by the virtue of its motion is called kinetic energy. Mathematically it is given by the following formula:

K.E =\frac{1}{2}mv^2

where,

  • K.E = Kinetic energy
  • m = mass
  • v = velocity

Therefore,

For the paintball:

K.E = \frac{1}{2}(4\ g)(90\ m/s)^2

K.E = 16200 J

For the pellet:

K.E = \frac{1}{2}(1\ g)(180\ m/s)^2

K.E = 16200 J

Hence, both paintball and pellet will have same kinetic energy. The student is right.

Learn more about kinetic energy here:

brainly.com/question/12669551

#SPJ1

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ruslelena [56]

The answer for the following problem is mentioned below.

The option for the question is "A" approximately.

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Explanation:

Given:

Spring constant (k) = 240 N/m

amount of the compression (x) = 0.40 m

To calculate:

Elastic potential energy (E)

We know;

<em>According to the formula;</em>

    E = \frac{1}{2} × k × x × x

   <u>E = </u>\frac{1}{2}<u> × k ×(x)²</u>

where;

E represents the elastic potential energy

K represents the spring constant

x represents amount of the compression in the string

So therefore,

Substituting the values in the above formula;

      E = \frac{1}{2} × 240 × (0.40)²

      E =  \frac{1}{2} × 240 × 0.16

      E =  \frac{1}{2} × 38.4

      E = 19.2 J or approximately 20 J

<u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>

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3 years ago
When in use, a lamp is designed to draw
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Answer:

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Suppose that a comet that was seen in 563 A.D. by Chinese astronomers was spotted again in year 1951. Assume the time between ob
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Answer:

a=2.77*10^{13}m

R_a=5.49*10^{13}m

Explanation:

The period of the comet is the time it takes to do a complete orbit:

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writen in seconds:

2514years*\frac{3,154*10^7s}{1year}=7.93 *10^{10}s

Since the eccentricity is greater than 0 but lower than 1 you can know that the trajectory is an ellipse.

Therefore, if the mass of the sun is aprox. 1.99e30 kg, and you assume it to be much larger than the mass of the comet, you can use Kepler's law of periods to calculate the semimajor axis:

T^2=\frac{4\pi^2}{Gm_{sun}}a^3\\ a=\sqrt[3]{\frac{Gm_{sun}T^2}{4\pi^2} } \\a=1.50*10^{6}m

Then, using the law of orbits, you can calculate the greatest distance from the sun, which is called aphelion:

R_a=a(1+e)\\R_a=2.77*10^{13}(1.986)\\R_a=5.49*10^{13}m

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3 years ago
A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float
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Answer:

 t ’= \frac{1450}{0.6499 + 2 v_r},  v_r = 1 m/s       t ’= 547.19 s

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This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

        v_b - v_r = d / t

By the time the boat goes down the river

        v_b + v_r = d '/ t'

let's subtract the equations

       2 v_r = d ’/ t’ - d / t

       d ’/ t’ = 2v_r + d / t

       t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }

In the exercise they tell us

         d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

         d ’= 1.45 km= 1.45 1.³ m

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the speed of river is v_r

      t ’= \frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \  + 2 \ v_r}

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In order to complete the calculation, we must assume a river speed

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The correct answer to your question and how to solve it is
 
The relation between wavelength (λ)and the frequency of electromagnetic oscillation (f) is described by the following expression: λ=c/f, where c–is the speed of light in vacuum = 3*10^8 m/s
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3 0
3 years ago
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