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zavuch27 [327]
3 years ago
14

A balloon filled with helium gas at 20°C occupies 4.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196°C, whil

e the pressure is raised to 5.20 atm. What is the volume of the balloon in the liquid nitrogen?a. 97 Lb. 6.7 Lc. 3.6 Ld. 0.25 L
Physics
1 answer:
mrs_skeptik [129]3 years ago
7 0

Answer:

0.25 L

Explanation:

P_1 = Initial pressure = 1 atm

T_1 = Initial Temperature = 20 °C

V_1 = Initial volume = 4.91 L

P_2 = Final pressure = 5.2 atm

T_2 = Final Temperature = -196 °C

V_2= Final volume

From ideal gas law we have

\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\\Rightarrow P_2=\dfrac{1\times 4.91(273.15-196)}{(20+273.15)\times 5.2}\\\Rightarrow V_2=0.24849\ L\approx 0.25\ L

The pressure experienced by the balloon is 0.25 L

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Equal force is applied to a baseball, a basketball, a tennis ball, and a bowling ball. Which ball will have the GREATEST acceler
mr_godi [17]

according to newton's second law , net force on an object is equal to the product of its mass and its acceleration. the formula is given as

F = ma

where F = force , m = mass  , a  = acceleration

rearranging the equation for acceleration "a" , we get

a = F/m

when force F is kept constant , the acceleration is inversely related to the mass "m".  in other words, greater the mass , smaller will be the acceleration and smaller the mass, greater will be the acceleration.

the masses of baseball, basketball,  tennis ball and bowling ball can in arranged in increasing order as

tennis ball < baseball < basketball < bowling ball.

since acceleration and mass have inverse relation from the formula , the order of the acceleration will also be reverse.

tennis ball > baseball > basketball > bowling ball.

hence the tennis ball will have the greatest acceleration as it has the smallest mass.

4 0
3 years ago
The weight of a person is 500N and his foot imprint area is 0.5m^2.Calculate the total pressure exerted by person when he stands
Ghella [55]

Answer:

Pressure on both feet will be 500N/m^2  

Explanation:

Weight of the person F = 500 N

Area of foot print A=0.5m^2

Area of both the foot A=2\times 0.5=1m^2

We have to find pressure on both the feet

Pressure is equal to ratio of force and area

So pressure P=\frac{F}{A}

P=\frac{500}{1}=500N/m^2

So the pressure on both feet will be 500N/m^2 when person stands on both feet.

7 0
3 years ago
A circular window of 30 cm diameter in a submarine can withstand a maximum force of 5.20 × 105 N. The maximum depth in a lake to
Svetach [21]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

         F = 5.20 \times 10^{5} N

         g = 9.8 m/s

         radius = \frac{diameter}{2}

                    = \frac{30 cm}{2} = 15 cm = 0.15 m   (as 1 m = 100 cm)

Formula to calculate depth is as follows.

        F = \rho \times g \times h \times A

or,      h = \frac{F}{\rho \times g \times A}        

       h = \frac{5.2 \times 10^{5}}{1000 \times 9.8 \times (3.1416 \times (0.15 m^{2})}

           = 751 m

Thus, we can conclude that the maximum depth in a lake to which the submarine can go without damaging the window is closest 750 m.

3 0
3 years ago
Calculate the momentum of a Lion of mass 130-kg and moving at a speed of 22.3 m/s [W]
Sunny_sXe [5.5K]

Answer:

<h2>289.9 kg.m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 130 × 22.3

We have the final answer as

<h3>289.9 kg.m/s</h3>

Hope this helps you

3 0
3 years ago
You are designing a delivery ramp for crates containing exercise equipment. The 1470-N crates will move at 1.8 m/s at the top of
alina1380 [7]

Answer:

The force constant ,I = 2394N/m

Explanation:

Given:

Weight of crate,Wg = 1470N

Theta = 22.0°

Kinetic friction,Fk= Fs(max) = 550N

Total length of ramp =8.0m

If y =0 at the bottom of the ramp

y1 = d Sin theta

y1 = 8 × Sin 22°

y1 = 3.0m

y2 = 0

V1=1.8m/s

V2 = 0

The equation combining the gravitational and elastic potential energy, and the work energy theorem is given by:

K1 + Ugrav1 + Uel1 + W = K2 + Uel1 ..eq1

Where KE is given by: 1/2mv^2

Gravitational potential energy is given by: Ugrav = mgy ...eq2

Elastic potential energy = Uel = 1/2Kx^2 ..eq3

The restoring force constant of a spring compressed by a distance x is given by:

Fx = Kx ..eq4

Workdone by a constant force is given by W = Fscostheta ...eq5

Where s = displacement

Theta = the angle between the force and the displacement

Work,W = Wf = 550 ×8 × cos 180°

W = -4400J

From the weight of the crate.mass is:

Wg/g = m

1470/9.8 = 150kg

K1 = 1/2 ×150×1.8^2 = 243m/s

The crate comes to rest at K2=0

Ugrav1 = 150 × 9.8 × 3 = 4410J

Ugrav2 = 150 × 9.8 x 0 = 0J

Uel1= 0 Spring at equilibrium

Substituting the values of the energies and work

253 + 4410 + 0 - 4400 = 0 + 0 + Uel1

Uel1 = 253J

Substituting into eq3

253 = 1/2 Kx^2 = 1/2 Kx(x)

Kx = 506/x

Since crate remains at rest,we use Newton's 2nd law

Fx = fs + Wsin theta

Fx = 550 + 1470 sin 22

Fx = 1100.7N

Substituting into eq4

Kx = 1100.7

X = 506/1100.7 = 0.46m

Kx = 1100.7

K = 1100.7/0.47

K = 2394N/m

5 0
3 years ago
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