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ycow [4]
3 years ago
11

As the water molecules heat up they become more dense and float to the top. True or False

Physics
1 answer:
miss Akunina [59]3 years ago
8 0

Answer: False but read why

Heated water molecules expands ( become less dense) but they do float to the top.

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A cannon fires a cannonball 500.0 m downrange when set at a 45.0o angle. At what velocity does the cannonball leave the cannon?
kobusy [5.1K]

Answer:

v = 70 m/s

Explanation:

Range on the cannon ball is given as

d = 500.0 m

here the angle of the projection of the ball is given as 45 degree

now we know that if the velocity of the ball is "v" then its two components will be given as

v_x = vcos45

v_y = vsin45

so here time of flight of the motion is given as

T = \frac{2v_y}{g}

T = \frac{2vsin45}{g}

also the range is given as

R = v_x T

R = (vcos45)(\frac{2vsin45}{g})

now plug in all data in this equation

500.0 = \frac{v^2(2sin45cos45)}{g}

v = 70 m/s

4 0
3 years ago
This solid layer of the earth is made of mostly iron and nickel.
Studentka2010 [4]

Answer

Explanation:

Yes, it's true that the solid layer of the earth is known as the most dense part as it is made up of the heavy metals like iron and nickel. Inner part is the hotter part due to the high pressure and temperature. It has the temperature of about 5,200°C and the pressure of 3.6 million atm but still the iron and nickel are present there in the solid form as they withstand such high temperature and pressure values.

5 0
3 years ago
Someone please hellpp!!! Will mark brainliest
____ [38]

<em>Answer:</em>

<em>r=x+y</em>

<em>sorry if its not correct you can delete if you want.</em>

6 0
3 years ago
. Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copp
MA_775_DIABLO [31]

Answer:

a) Yes

b) No

Explanation:

In the first case, part a, yes we can say for certainty that cylinderical symmetry holds. Why so? You may ask. This is because from the question, we are told that the length of the rod is 300 cm. And this said length is longer than the distance to the point from the center of the rod, which is 5 cm.

In the second half of the question, I beg to disagree that cylindrical symmetry holds. Again, you may ask why, this is because the length of the rod in this case, is having the same order of magnitude as the distance to the center of the rod. Thus, it is not symmetrical.

6 0
2 years ago
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
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