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hoa [83]
3 years ago
15

A uniform diameter rod is weighted at end and is floating in a liquid. Determine if the liquid is (a) lighter than water, (b) wa

ter, or (c) heavier than water. Justify your answer.
Physics
1 answer:
Yakvenalex [24]3 years ago
4 0

(C) Heavier than water

<u>Explanation:</u>

Given

Diameter of the rod is uniform

The road is floating in a liquid.

Let the area of cross-section of the rod = A

Let the weight density of water = γw

Let the weight density of liquid = γl

Let the density of water = ρw

g is the gravitational force

So, weight of rod = density of the rod X volume

Wrod = ρw X g X ( A X 2L) + 2 ρwg (A X L)

Wrod = 4ρwgAL

Wrod = 4γwAL

We can see that the weight of the rod is more than the density of water. So, in water, the rod will sink.

If the rod is floating on the liquid then the liquid is heavier than the water.

Therefore, The liquid is heavier than the water.

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When a car slows down suddenly, passengers in the car tend to move toward the front of the car. What is this due to?
Neko [114]
Hey there Kendrell!

Yes, this is very true, when the car slows down, our bodies will tend to lean forward a little bit, and this is actually due to the "motion of inertia".

Inertia allows for this to happen, this is why in this case, we have this case.

Hope this helps.
~Jurgen


4 0
3 years ago
Use the motion map to answer the question.
Lana71 [14]

Answer:

The object starts away from the origin and then moves toward the origin at a constant velocity. Next, it stops for one second. Finally, it moves away from the origin at a greater constant velocity.

3 0
3 years ago
A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist
Paha777 [63]

Answer:

Part a)

h' = \frac{10}{14} h

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

h' = \frac{10}{14} h

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

\frac{1}{2}mv^2 = mgh'

now we know that

I = \frac{2}{5}mr^2

\omega = \frac{v}{r}

so we have

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh

\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh

\frac{7}{10}mv^2 = mgh

\frac{1}{2}mv^2 = \frac{10}{14}mgh

so the height on the smooth side is given as

h' = \frac{10}{14} h

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

h' = \frac{10}{14} h

so on smooth it will go to lower height

6 0
3 years ago
Read 2 more answers
A wave with a speed of 9 m/s and a frequency of 0.5 Hz has a λ of what?
patriot [66]

Wave speed = (wavelength) x (frequency)

Wavelength = (wave speed) / (frequency)

Wavelength = (9 m/s) / (0.5 Hz)

<em>Wavelength = 18 m</em>

6 0
2 years ago
What is the mass of an object that has a weight of 110N ?
fgiga [73]
  • Weight (W) = 110 N
  • Acceleration due to gravity (g) = 9.8 m/s^2
  • Let the mass of the object be m.
  • By using the formula, W = mg, we get,
  • 110 N = 9.8 m/s^2 × m
  • or, m = 110 N ÷ 9.8 m/s^2
  • or, m = 11.2 Kg

<u>Answer:</u>

<em><u>The </u></em><em><u>mass </u></em><em><u>of </u></em><em><u>the </u></em><em><u>object </u></em><em><u>is </u></em><em><u>1</u></em><em><u>1</u></em><em><u>.</u></em><em><u>2</u></em><em><u> </u></em><em><u>Kg.</u></em>

Hope you could get an idea from here.

Doubt clarification - use comment section.

3 0
2 years ago
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