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3 years ago

The air temperature in a 28 in.3 container with a free sliding piston is initially measured at 45 °

1 answer:

8 0

To solve this we assume that the gas inside is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

V2 = T2 x V1 / T1

V2 = 659.7 x 28 / 504.7

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Particles q1 = +8.0 UC, 92 = +3.5 uc, and

Olin [163]

Answer:

F_total = 29.4 N, directed to the right of particle 2

Explanation:

We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.

Let's use Coulomb's law to calculate each force

F = $k \frac{q_1q_2}{r_{12}^2}$

particles 1 and 2

q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m

F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²

F₁₂ = 2.59 10¹ N

Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.

particles 2 and 3

q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m

we calculate

F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²

F₂₃ = 3.5 N

as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2

Now we add the forces as vectors

F_total = ∑ F = F₁₂ + F₂₃

F_total = 25.2 +3.5

F_total = 29.4 N

directed to the right of particle 2

5 0

3 years ago

Read 2 more answers

A projectile is launched vertically from the surface of the Moon with an initial speed of 1360 m/s. At what altitude is the proj

LenaWriter [7]

Answer:

485520 m

Explanation:

$v_{o}$ = initial velocity of the projectile = 1360 m/s

$v_{f}$ = final velocity of the projectile = $\left ( \frac{2}{5} \right )v_{_{o}}$ = $\left ( \frac{2}{5} \right )(1360)$ = 544 m/s

a = acceleraton due to gravity on moon = - 1.6 m/s²

h = Altitude of the projectile

Using the kinematics equation

$v_{f}^{2} = v_{o}^{2} + 2 a h$

Inserting the values

$544^{2} = 1360^{2} + 2 (-1.6) h$

h = 485520 m

3 0

4 years ago

What type of forces accelerate masses?

leva [86]

Answer:

i would say that the answer would be B

5 0

3 years ago

Can you answer fast plz

tatyana61 [14]

Answer:

wavelength frequency

energy

frequency

Explanation:

7 0

3 years ago

How to find a planet’s gravitational field strength using its radius?

grin007 [14]

The gravitational field strength is approximately equal to 10 N.

<u>Explanation:</u>

Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.

So the gravitational field strength will be equal to the gravitational force acting on the object.

The formula for gravitational field strength is

$g = \frac{F}{m}$

Here g is the gravitational field strength, m is the mass of the object placed on the surface and F is the gravitational force acting on the object.

Since, the mass of any object placed on the surface of earth will be negligible compared to the mass of Earth, so the mass of the object is considered as 1 kg.

Then the g = F

And $F =\frac{GMm}{r^{2} }$

Here G is the gravitational constant, M is the mass of Earth and m is the mass of the object placed on the surface, while r is the radius of the Earth.

$g = F = \frac{6 \times 10^{24} \times 6.67 \times 10^{-11} \times 1}{(6.6 \times 10^{6}) ^{2} }$

$g = 0.977 \times 10^1= 9.77\ N$

So, the gravitational field strength is approximately equal to 10 N.

5 0

3 years ago