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AURORKA [14]
3 years ago
6

Which best describes the runners

Physics
1 answer:
qaws [65]3 years ago
3 0

Answer:

The answer is

Explanation:

A.

Daniela had a 5 meter head start, and Leonard caught up to her at 25 meters.

brainiest would be cool

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What human practices are destroying the rain forest?
marissa [1.9K]
Deforestation, the chopping off of the trees that can take thousands or even millions of years to grow again<span />
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3 years ago
An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate
riadik2000 [5.3K]

Answer: 6.408(10)^{-19} C

Explanation:

This problem can be solved by the following equation:

\Delta K=q V

Where:

\Delta K=7.37(10)^{-17} J is the change in kinetic energy

V=115 V is the electric potential difference

q is the electric charge

Finding q:

q=\frac{\Delta K}{V}

q=\frac{7.37(10)^{-17} J}{115 V}

Finally:

q=6.408(10)^{-19} C

4 0
3 years ago
Which of the following is equivalent to 2.5 meters
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B is the correct answer for sure bro
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3 years ago
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The work-energy principle states that the work done by all the __________ forces acting on an object (or system of objects) caus
CaHeK987 [17]

The work-energy principle states that the work done by all the non-conservative forces acting on an object (or system of objects) causes a change in the total mechanical energy of the object or system.


What is the work-energy principle?
The work-energy principle states that the total work done on a system is equal to the change in kinetic energy of the system. It is given as:

W.D = ΔK.E

       = K.E₁ - K.E₂

where K.E₁ is the initial kinetic energy of the system

            K.E₂ is the final kinetic energy of the system



What is meant by non-conservative forces?

Non-conservative forces as the name suggests are not conserved i.e. these forces cause a loss of mechanical energy from the system. A prime example of non-conservative forces is friction.

The total mechanical energy of the system is the sum of the potential energy and kinetic energy that the system contains. This energy is conserved and follows the work-energy theorem.

Learn more about work and energy here:

<u>brainly.com/question/17290830</u>

#SPJ4

4 0
2 years ago
A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

5 0
3 years ago
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