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2 years ago

Why does water require such a higher temperature to raise the pressure than neon?

1 answer:

4 0

Water require such a higher temperature to raise the pressure than neon due to strong bonds between atoms of water.

<h3>Why does water require such a higher temperature to raise the pressure than neon?</h3>

Water require such a higher temperature to raise the pressure than neon because it has strong hydrogen bonds and having high specific capacity as compared to other atoms and molecules.

So we can conclude that water require such a higher temperature to raise the pressure than neon due to strong bonds between atoms of water.

Learn more about temperature here: brainly.com/question/24746268

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A roller coaster weighing 2000 kg is lifted to a height of 28 m within 30 seconds with a kind of elevator.Calculate how many kil

Vladimir [108]

Answer:

Explanation:

The equation for Power is

P = Work/time to do work and the equation for work is

Work = FΔx

We first need to find the amount of work done, then we can find the power it took to do that work.

W = 2000(9.8)(28) so

W = 550,000 N*m

Now we fill that into the power equation:

$P=\frac{550000}{30}$ gives us

P = 18000 Watts. But we need kW, so we divide by 1000 to get

P = 18 kW of power.

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2 years ago

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Adult men have an average height of 69.0 inches with standard deviation of 2.8 inches fins the night of a man with a z-score of

ANTONII [103]

What do you mean I’m confused

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2 years ago

A 0.50-kg red cart is moving rightward with a speed of 40 cm/s when it collides

lukranit [14]

The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.

The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.

The change in momentum of the system of the carts is 0.

<h3>Initial momentum of the carts before collision</h3>

The momentum of the carts before the collision is calculated as follows;

P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s

P(blue) = 1.5 x 0 = 0

<h3>Momentum of the carts after collision</h3>

The momentum of the carts after the collision is calculated as follows;

P(red) = 0.5 x 0.1 = 0.05 kgm/s

P(blue) = 1.5 0.1 = 0.15 kgm/s

<h3>Change in momentum of the carts</h3>

$\Delta P = P_f - P_i$

ΔP = (0.05 + 0.15) - (0.2)

ΔP = 0

Learn more about momentum here: brainly.com/question/7538238

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2 years ago

Can sound travel through a tree

Marizza181 [45]

Answer:

not sure but i think yes

Explanation:

sound waves

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2 years ago

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We want to design a cylindrical vacuum capacitor, with a given radius a for the outer cylindrical shell, that will be able to st

anastassius [24]

Solution :

a). Using Gauss's law :

$E=\frac{Q}{4 \pi \epsilon_0r^2}$ , $$b$ .........(1)

Let $E=E_0,\ r=b$ in equation (1)

Therefore, $Q=4 \pi \epsilon_0b^2E_0$ .............(2)

$V_b-V_a = \int^a_b \vec E. d\vec l$

$=\int^a_b E \ dx$

$=\frac{Q}{4 \pi \epsilon_0} \int^a_b \frac{1}{x^2} \ dx$

$=\frac{Q(a-b)}{4 \pi \epsilon_0 a b}$ ....................(3)

Therefore, $U=\frac{1}{2}Q \Delta V$

$=\frac{1}{2}(4 \pi \epsilon_0 b^2 E_0)\left(\frac{Q(a-b)}{4\pi \epsilon_0 a b}\right)$

$=\frac{4 \pi \epsilon_0}{2a} \ E^2_0 b^3(a-b)$ .............(4)

Now differentiating the equation (4) w.r.t. 'b', we get

$b=\frac{3}{4}a$

Thus the radius for the inner cylinder conductor is $b=\frac{3}{4}a$

b). For the energy storage, substitute the radius in (4), we get

$U = 4 \pi\epsilon_0 \frac{27a^3E^2_0}{512}$

This is the amount of energy stored in the conductor.

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3 years ago