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djverab [1.8K]
3 years ago
13

How do i do this question?just number 5

Physics
1 answer:
larisa86 [58]3 years ago
3 0

5 a)

Start by arranging the materials by the sonic speed and then their physical state:

  • 4600 \; \text{m} \cdot \text{s}^{-1}- Copper (solid)
  • 4500 \; \text{m} \cdot \text{s}^{-1}- Glass (solid)
  • 4000 \; \text{m} \cdot \text{s}^{-1}- Wood (solid)
  • 1500 \; \text{m} \cdot \text{s}^{-1}- Sea Water (liquid)
  • 1200 \; \text{m} \cdot \text{s}^{-1}- Acetone (liquid)
  • 1100 \; \text{m} \cdot \text{s}^{-1}- Alcohol (liquid)
  • 972 \; \text{m} \cdot \text{s}^{-1}- Helium (gas)
  • 267 \; \text{m} \cdot \text{s}^{-1}- Carbon dioxide (gas)

What trend do you identify from these data? Here's what I've got:

\text{Sonic Speed}: \text{solid} > \text{liquid} > \text{gas}

5 b)

The way microscopic particles are arranged in a substance helps distinguish between different physical states:

  • Particles in a solid are held tightly in place with small separation in between; it's hard for particles in a solid to move past one another; solids therefore have shapes that persists over time.
  • Particles in a gas are highly mobile- they keep moving AT ALL TIMES. There are large separations between individual particles and therefore gases tend to show no definite shape or volume.
  • The arrangement of particles in a liquid is located somewhere in between that of solids and gases. The exact configuration is dependent on the nature of the liquid- for example, molecules in maple syrup are held way closer to each other than those in distilled water are.

Sound travels as a longitudinal wave. As a sound wave passes through a medium, individual particles become excited and gain energy; as they run into others they transfer their energy to the next particle; the sound wave thus propagate across the medium. With a lower average distance between individual particles this action can proceed at a greater rate in average solids than in average liquids, and in average liquids than in average gases. Hence the trend.

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The quickest braking by acceleration is 8.8 m/s set on a car driven at 32m/s to rest. How long did it take the car to brake?
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The area of the pond is approximately equal to the area of a circle with radius 297m. Find the mass of the ice. Answer in kilogr
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Answer:

<em>mass of the ice is 254980463.8T kg</em>

<em>where T is the value of the thickness omitted in the question.</em>

Explanation:

The ice on Walden Pond is .......... thick. The area of the pond is approximately equal to the area of a circle with radius 297 m. Find the mass of the ice.  Answer in kg.

<em>The value of the thickness of the ice T is omitted, but I will show the solution, and the real answer can be gotten by multiplying the final calculated answer here by the thickness of the ice omitted.</em>

Given the radius of the equivalent circle of the ice = 297 m'

the area of the ice can be gotten from area A = \pi r^{2} = 3.142*297^{2} = 277152.678 m^2

recall that the density of ice p ≅ 920 kg/m^3

also,

density of ice p = (mass of ice, m) ÷ (volume of ice, v)

i.e p = m/v

and,

m = pv

substituting the value of the density of water p into the equation, we have,

mass of the ice, m = 920v ....... equ 1

The volume of the ice above will be = (area of the ice, A) x (thickness of the ice, T)

i.e v = AT

substituting the value of area A into the equation, we have

v =  277152.678T  ......equ 2

substitute value of v into equ 1

mass of the ice, m = 920 x (277152.678T)

mass of the ice, m = 254980463.8T kg

where T is the thickness of the ice

NB: To get the mass, multiply this answer with the thickness T given in the question.

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3 years ago
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti
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Use Newton's second law to determine the acceleration being applied to the sled. There are three forces at work on the sled (its weight, the force normal to the ground, and friction) but two of them cancel, leaving friction as the only effective force. This vector is pointed in the opposite direction of the sled's movement, so if we take the direction of its movement to be the positive axis, we would find the acceleration due to the friction to be

\vec F_G+\vec F_N+\vec F_F=m\vec a\iff-12\,\mathrm N=(20\,\mathrm{kg})a\implies a=-0.6\,\dfrac{\rm m}{\mathrm s^2}

Now we use the formula

{v_f}^2-{v_i}^2=2a(x_f-x_i)

to find the distance it travels. The sled comes to a rest, so v_f=0, and let's take the starting position x_i=0 to be the origin. Then the distance traveled x_f-x_i=x_f is

-\left(4.5\,\dfrac{\rm m}{\rm s}\right)^2=2\left(-0.6\,\dfrac{\rm m}{\mathrm s^2}\right)x_f\implies x_f\approx17\,\mathrm m

6 0
3 years ago
7. A scientist studying a squid observes that the squid at rest
Rzqust [24]

The average force on the squid during the ejection of 0.60 kg of water at a velocity of 15.0 m/s in 0.15 seconds is 60 N.              

We can calculate the average force with the average acceleration as follows:

F = m\overline{a}   (1)

Where:

  • m: is the mass of water = 0.60 kg
  • \overline{a}: is the average acceleration

The <em>average acceleration</em> is given by the change of velocity in an interval of time

\overline{a} = \frac{v_{f} - v_{i}}{t_{f} - t_{i}}   (2)

Where:

  • v_{i}: is the initial velocity = 0 (the squid is at rest)
  • v_{f}: is the final velocity = 15.0 m/s
  • t_{i}: is the initial time = 0  
  • t_{f}: is the final time = 0.15 s

Now we can find the <em>average force</em> after entering equation (2) into (1)

F = m(\frac{v_{f} - v_{i}}{t_{f} - t_{i}}) = 0.60 kg(\frac{15.0 m/s - 0}{0.15 s}) = 60 N  

Therefore, the average force on the squid during the propulsion is 60 N.

Find more about average force here:

  • brainly.com/question/20902034
  • brainly.com/question/12916904

I hope it helps you!

3 0
3 years ago
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