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Whitepunk [10]
3 years ago
12

Which has the greater acceleration, a person going from 0 m/s to 10 m/s in 10 seconds or an ant going from 0 m/s to 0.25 m/s in

2.0 seconds?​
Physics
1 answer:
monitta3 years ago
4 0

<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>1</u>

  • Initial velocity=u=0m/s
  • Final velocity=v=10m/s
  • Time=10s=t

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{10-0}{10}

\\ \sf\longmapsto Acceleration=\dfrac{10}{10}

\\ \sf\longmapsto Acceleration=1m/s^2

<u>P</u><u>e</u><u>r</u><u>s</u><u>o</u><u>n</u><u>-</u><u>2</u>

  • initial velocity=0m/s=u
  • Final velocity=v=0.25m/s
  • Time=t=2s

\\ \sf\longmapsto Acceleration=\dfrac{0.25-0}{2}

\\ \sf\longmapsto Acceleration=\dfrac{0.25}{2}

\\ \sf\longmapsto Acceleration=0.125m/s^2

Person-1 is accelerating faster.

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Answer:

true

Explanation:

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3 years ago
Which product of nuclear decay has mass but no charge?
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Neutrons have a zero charge but consist of mass.
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Suppose you have a coffee mug with a circular cross section and vertical sides (uniform radius). What is its inside radius if it
Vlad [161]

Answer:

r = 4.62 cm

Explanation:

Mass of a coffee, m = 370 g

Height of the mug, h = 5.5 cm

We need to find the inside radius of the mug. Density of an object is equal to the mass per unit volume. It can be given by :

d=\dfrac{m}{V}\\\\d=\dfrac{m}{\pi r^2 h}\\\\r=\sqrt{\dfrac{m}{\pi dh}}\\\\r=\sqrt{\dfrac{370\ g}{\pi \times 1\ g/cm^3\times 5.5\ cm}}\\\\r=4.62\ cm

So, the inside radius is 4.62 cm.

8 0
3 years ago
Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge th
statuscvo [17]

Answer:

 P /K = 1,997 10⁻³⁶  s⁻¹

Explanation:

For this exercise let's start by finding the radiation emitted from the accelerator

       \frac{dE}{dt} = \frac{q^{2} a^{2} }{6\pi  \epsilon_{o} c^{2}    }

the radius of the orbit is the radius of the accelerator a = r = 0.530 m

let's calculate

       \frac{dE}{dt} = [(1.6 10⁻¹⁹)² 0.530²] / [6π 8.85 10⁻¹² (3 108)³]

      P= \frac{dE}{dt}= 1.597 10⁻⁵⁴ W

Now let's reduce the kinetic energy to SI units

       K = 5.0 10⁶ eV (1.6 10⁻¹⁹ J / 1 eV) = 8.0 10⁻¹⁹ J

the fraction of energy emitted is

      P / K = 1.597 10⁻⁵⁴ / 8.0 10⁻¹⁹

      P /K = 1,997 10⁻³⁶  s⁻¹

3 0
3 years ago
How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s
poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

W = \Delta KE

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

Here,

m = mass

v_{f,i} = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2

W_1 = 150(m) J

Second case) When the particle goes from 20m/s to 30m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2

W_1 = 250(m) J

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0
4 years ago
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