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3 years ago

The pressure at the ice point for a constant-volume gas thermometer is 4.81 x 10^4Pa.

Physics

1 answer:

Artist 52 [7]3 years ago

7 0

Answer:

0 deg C = 4.81E4 pressure at 0 deg

100 deg C = 6.48E4 pressure at steam point

100 deg C - 50 deg C = (6.48 - 4.81) * 10^4 = 1.67E4 Pa

50 deg C = 50 / 100 * 1.67E4 + 4.81E4 = 5.65E4 Pa Just the halfway point between the two given pressures

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Which condition is required for Coulomb's law to hold true?

AleksAgata [21]

The correct answer is:
Point charges must be in a vacuum.

In fact, the usual form for of the Coulomb's law is:
 $F= \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$
where
 $\epsilon_0$ is the permittivity of free space
q1 and q2 are the two charges
q is the separation between the two charges

However, this formula is valid only if the charges are in vacuum. If they are in a material medium, the law is modified as follows:
 $F= \frac{1}{4 \pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{r^2}$
where $\epsilon_r$ is the relative permittivity, which takes into account the dielectric effects of the material.

7 0

3 years ago

"Two uniform identical solid spherical balls each of mass M and radius R" and moment of inertia about its center 2/5 MR2 are rel

adelina 88 [10]

Answer:

he sphere that uses less time is sphere A

Explanation:

Let's start with ball A, for this let's use the kinematics relations

v² = v₀² - 2g (y-y₀)

indicate that the sphere is released therefore its initial velocity is zero and when it reaches the floor its height is zero y = 0

v² = 0 - 2 g (0- y₀)

v = $\sqrt{2g y_o}$

v = $\sqrt{2 \ 9.8\ H}$

v = 4.427 √H

Now let's work the sphere B, in this case it rolls down a ramp, let's use the conservation of energy

starting point. At the highest point, before you start to move

Em₀ = U = m g y

final point. At the bottom of the ramp

Em_f = K = ½ m v² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

Em₀ = Em_f

mg H = ½ m v² + ½ I w²

angular and linear velocity are related

v = w r

w = v / r

the momentorot of inertia indicates that it is worth

I = $\frac{2}{5}$ m r²

we substitute

m g H = ½ m v² + ½ ( $\frac{2}{5}$ m r²) ( $\frac{v}{r}$ )²

gH = $\frac{1}{2}$ v² + $\frac{1}{5}$ v² = $\frac{7}{10}$ v²

v = $\sqrt{\frac{10}{7} \ g H}$

v = $\sqrt{ \frac{10}{7} \ 9.8 \ H}$

v=3.742 √H

Taking the final speeds of the sphere, let's analyze the distance traveled, sphere A falls into the air, so the distance traveled is H. The ball B rolls in a plane, so the distance (L) traveled can be found with trigonometry

sin θ = H / L

L = H /sin θ

we can see that L> H

In summary, ball A arrives with more speed and travels a shorter distance, therefore it must use a shorter time

Consequently the sphere that uses less time is sphere A

5 0

3 years ago

A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

vladimir2022 [97]

Answer:

f = 0.4 Hz

Explanation:

The frequency of rotation of an object in order to achieve required centripetal or radial acceleration, can be found out by using the following formula:

f = (1/2π)√(ac/r)

where,

f = frequency of rotation = ?

ac = radial acceleration = 34.1 m/s²

r = radius = length of beam = 5.55 m

Therefore,

f = (1/2π)√[(34.1 m/s²)/(5.55 m)]

f = 0.4 Hz

6 0

3 years ago

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

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3 years ago

How did putting myths in writing help develop philosophy?

matrenka [14]

Answer: A

Explanation:

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3 years ago