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Mnenie [13.5K]
4 years ago
3

A vertical piston–cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 5 kg and a diameter of 12 cm

. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will double the pressure of the gas inside the cylinder.
Physics
1 answer:
Hunter-Best [27]4 years ago
3 0

Answer:

A. 95.6 kPa.

B. 115.36 kg.

Explanation:

A.

P1 = Patm + F1/A

Therefore,

Patm = P1 - (m*g)/(D^2 * π/4)

= 100 - (5 * 9.81)/(0.12^2 * π/4) x 10^-3 kPa

= 100 - 4.336

= 95.6 kPa.

B.

Δm = m2 - m1

= (D^2 * π/(4 * g)) * (P2 - Patm) - m1

= (0.12^2 * π/(4 * 9.81)) * (2*100 - 95.6) x 10^3 - 5

= 115.36 kg.

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How many photons will be required to raise the temperature of 1.8 g of water by 2.5 k ?'?
tatyana61 [14]
Missing part in the text of the problem: 
"<span>Water is exposed to infrared radiation of wavelength 3.0×10^−6 m"</span>

First we can calculate the amount of energy needed to raise the temperature of the water, which is given by
Q=m C_s \Delta T
where
m=1.8 g is the mass of the water
C_s = 4.18 J/(g K) is the specific heat capacity of the water
\Delta T=2.5 K is the increase in temperature.

Substituting the data, we find
Q=(1.8 g)(4.18 J/(gK))(2.5 K)=18.8 J=E

We know that each photon carries an energy of
E_1 = hf
where h is the Planck constant and f the frequency of the photon. Using the wavelength, we can find the photon frequency:
\lambda =  \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{3 \cdot 10^{-6} m}=1 \cdot 10^{14}Hz

So, the energy of a single photon of this frequency is
E_1 = hf =(6.6 \cdot 10^{-34} J)(1 \cdot 10^{14} Hz)=6.6 \cdot 10^{-20} J

and the number of photons needed is the total energy needed divided by the energy of a single photon:
N= \frac{E}{E_1}= \frac{18.8 J}{6.6 \cdot 10^{-20} J} =2.84 \cdot 10^{20} photons
4 0
3 years ago
A force of 400-N pushes on a 25-kg box horizontally. The box accelerates at 9 m/s? Find the coefficient of kinetic friction betw
umka2103 [35]

Answer:

<h3>0.69</h3>

Explanation:

Using the Newtons law of motion;

\sum Fx = ma_x\\Fm - Ff = ma_x

Fm is the moving force = 400N

Ff is the frictional force = μR

μ is the coefficient of kinetic friction

R is the reaction = mg

m is the mass

a is the acceleration

The equation becomes;

Fm - \mu R = ma_x\\Fm - \mu mg = ma_x\\400- \mu (25)(9.8) = 25(9)\\400 - 254.8 \mu = 225\\- 254.8 \mu = 225 - 400\\- 254.8 \mu = -175\\ \mu = \frac{-175}{- 254.8} \\\mu = 0.69

Hence the coefficient of kinetic friction between the box and floor is 0.69

7 0
3 years ago
What does the term magnetic environment refer to?
Leokris [45]

Answer:

D is the correct answer! Hope this helps you!!!!

Explanation:

7 0
4 years ago
A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from i
murzikaleks [220]

Answer:

f" = 40779.61 Hz

Explanation:

From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;

from the Doppler effect equation, we can calculate the initial observed frequency as:

f' = f(1 - (v_o/v))

We are given;

f = 46.2 kHz = 46200 Hz

v_o = 21.8 m/s

v is speed of sound = 343 m/s

Thus;

f' = 46200(1 - (21/343))

f' = 43371.4285 Hz

In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;

Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;

f" = f'/(1 + (v_o/v))

f" = 43371.4285/(1 + (21.8/343))

f" = 40779.61 Hz

4 0
3 years ago
You collect some data on horse racing along a straight track. You are able to fit the motion of the horse to a function during t
Digiron [165]

Answer:

The equation is missing in the question. The equation is $10 m  + 5(m/s^2)t^2+3(m/s^3)t^3$

a). $v=10 t +9t^2$ , the horse will not turn.

b). a(t) = 10 + 18t

Explanation:

Given :

$x(t)=10 m  + 5(m/s^2)t^2+3(m/s^3)t^3$

∴ At t =0, x = 10 m

a). Velocity as a function of time

$v = \frac{dx}{dt} $

  = $10 t +9t^2$

Turning velocity must be zero.

v(t) = 0

$10 t +9t^2=0$

$\therefore t = 0 \text{ or}\ t =-\frac{10}{9}$

Taking the positive value of time.

The horse will not turn.

b). Acceleration as a function of time.

$a(t)=\frac{dv}{dt}$

     = 10 + 18t

∴ a(t) = 10 + 18t

6 0
4 years ago
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