PHYSICS COLLEGE

A weather balloon is designed to expand to a maximum radius of 16.2 m when in flight at its working altitude, where the air pressure is 0.0282 atm and the temperature is â65âC. If the balloon is filled at 0.873 atm and 21âC, what is its radius at lift-off?

Answers

Answer 1

Answer:

Answer:

5.78971 m

Explanation:

$P_1$ = Initial pressure = 0.873 atm

$P_2$ = Final pressure = 0.0282 atm

$V_1$ = Initial volume

$V_2$ = Final volume

$r_1$ = Initial radius = 16.2 m

$r_2$ = Final radius

Volume is given by

$(4)/(3)\pi r^3$

From the ideal gas law we have the relation

$(P_1V_1)/(T_1)=(P_2V_2)/(T_2)\n\Rightarrow (0.873* (4)/(3)\pi r_1^3)/(294.15)=(0.0282(4)/(3)\pi r_2^3)/(208.15)\n\Rightarrow (0.873r_1^3)/(294.15)=(0.0282* 16.2^3)/(208.15)\n\Rightarrow r_1=(0.0282* 16.2^3* 294.15)/(208.15* 0.873)\n\Rightarrow r_1=5.78971\ m$

The radius of balloon at lift off is 5.78971 m

Answer 2

Answer:

Final answer:

To find the radius of the weather balloon at lift-off, the ideal gas law can be used. Using the equation P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off, the radius at lift-off can be calculated to be approximately 4.99 m.

Explanation:

To find the radius of the weather balloon at lift-off, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

In this case, we know that the number of moles is constant, as the balloon is filled with the same amount of helium at lift-off and in flight. Therefore, we can write the equation as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off.

Plugging in the given values, we have (0.873 atm)(V1) = (0.0282 atm)(16.2 m)^3. Solving for V1, we find that the volume at lift-off is approximately 110.9 m^3. The radius can then be calculated using the formula for the volume of a sphere: V = (4/3) * π * r^3, where r is the radius.

Therefore, the radius at lift-off is approximately 4.99 m.

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If the diameter of the black marble is 3.0cm, and bye using the formula for volume, what is a good approximation if it’s volume? Record to the ones place

Answers

Complete question is;

If the diameter of the black marble is 3.0 cm, and by using the formula for volume, what is a good approximation of its volume?

Answer:

14 cm³

Explanation:

We will assume that this black marble has the shape of a sphere from online sources.

Now, volume of a sphere is given by;

V = (4/3)πr³

We are given diameter = 3 cm

We know that radius = diameter/2

Thus; radius = 3/2 = 1.5 cm

So, volume = (4/3)π(1.5)³

Volume ≈ 14.14 cm³

A good approximation of its volume = 14 cm³

A mass of 0.14 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.28 m)cos[(8 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass .

Answers

Answer:

The amplitude of oscillation for the oscillating mass is 0.28 m.

Explanation:

Given that,

Mass = 0.14 kg

Equation of simple harmonic motion

$x(t)=(0.28\ m)\cos[(8\ rad/s)t]$ ....(I)

We need to calculate the amplitude

Using general equation of simple harmonic equation

$y=A\omega \cos\omega t$

Compare the equation (I) from general equation

The amplitude is 0.28 m.

Hence, The amplitude of oscillation for the oscillating mass is 0.28 m.

Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off

Answers

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.

$\to (\$60)/((30 \ days)/(24\ hours)) = \$0.08 / kwh.$

Thus, $(\$0.08)/(\$0.12) = 0.694 \ kW * 0.694 \ kW * 1000 = 694 \ W.$

The electricity used is continuously 694W over 30 days.

If just resistor loads (no reagents) were assumed,

$\to I = (P)/(V)= (694\ W)/(120\ V) = 5.78\ A$

Energy usage reduction percentage = $((60\ W)/(694\ W) * 100\%)$

This bulb accounts for $8.64\%$ of the energy used, hence it saves when you switch it off.

The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom rises from pin A to B, at angle phi from horizontal. A cable goes up from the load to B and then left to a pulley. Determine the greatest weight of the crate that can be supported without causing the cable to fail if ϕ=30∘. Neglect the size of the winch.

Answers

Explanation:

Let us assume that forces acting at point B are as follows.

$\sum F_(x) = 0$

$T + F_(AB) Sin 60$ = 0 ...... (1)

$\sum F_(y)$ = 0

$F_(AB) Cos 60 + W$ = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

$\sigma_(allow) = (T)/(A)$

T = $(20 * 1000) (\pi)/(4) * (0.5)^(2)$

= 3925 kip

From equation (1), $F_(AB)Sin (60^(o))$ = -3925

$F_(AB) * -0.304 8$ = -3925

$F_(AB)$ = 12877.29 kip

From equation (2), -12877.29 (Cos 60) + W = 0

$-12877.29 kip * (1)/(2) + W$ = 0

W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

To determine the greatest weight of the crate that can be supported without causing the cable to fail, calculate the normal stress on the cable using σ = F/A, where σ is the normal stress, F is the force on the cable, and A is the cross-sectional area of the cable. Then, compare the calculated normal stress to the allowable normal stress. Consider the angle phi in this calculation by using the equation F = W / sin(ϕ), where F is the force on the cable, W is the weight of the crate, and ϕ is the angle with respect to the horizontal.

To determine the greatest weight of the crate that can be supported without causing the cable to fail, we need to calculate the normal stress on the cable and compare it to the allowable normal stress. The normal stress can be calculated using the formula σ = F/A, where σ is the normal stress, F is the force on the cable, and A is the cross-sectional area of the cable. In this case, the force on the cable is equal to the weight of the crate, and the cross-sectional area of the cable can be calculated using the formula A = π*(d/2)^2, where d is the diameter of the cable.

Given that the diameter of the cable is 0.5 in and the allowable normal stress is 21 ksi, we can substitute these values into the equations and solve for the force on the cable:

Calculate the cross-sectional area of the cable: A = π*(0.5/2)^2 = π*(0.25)^2 = 0.1963 in^2

Plug the cross-sectional area and the allowable normal stress into the formula for normal stress: σallow = F/A → 21 ksi = F/0.1963 in^2

Solve for the force on the cable: F = 21 ksi * 0.1963 in^2 = 4.1183 ksi*in^2

Therefore, the greatest weight of the crate that can be supported without causing the cable to fail is equal to the force on the cable, which is 4.1183 ksi*in^2. However, it's important to note that we also need to consider the angle phi (ϕ) in this calculation. Since the cable goes up from the load to point B and then left to a pulley, the weight of the crate will create a vertical component and a horizontal component. To determine the weight of the crate that corresponds to the calculated force on the cable, we need to consider the trigonometric relationship between the force and the weight at an angle. In this case, the angle is 30 degrees, so we can use the equation F = W / sin(ϕ), where F is the force on the cable, W is the weight of the crate, and ϕ is the angle with respect to the horizontal.

Given that ϕ = 30 degrees and F = 4.1183 ksi*in^2, we can substitute these values into the equation and solve for the weight of the crate:

Plug the values into the equation: 4.1183 ksi*in^2 = W / sin(30)

Solve for the weight of the crate: W = 4.1183 ksi*in^2 * sin(30)

Therefore, the greatest weight of the crate that can be supported without causing the cable to fail and at an angle of 30 degrees is equal to the force on the cable, which is 4.1183 ksi*in^2, multiplied by the sine of 30 degrees.

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Consider a vertical elevator whose cabin has a total mass of 800 kg when fully loaded and 150 kg when empty. The weight of the elevator cabin is partially balanced by a 400-kg counterweight that is connected to the top of the cabin by cables that pass through a pulley located on top of the elevator well. Neglecting the weight of the cables and assuming the guide rails and the pulleys to be frictionless, determine (a) the power required while the fully loaded cabin is rising at a constant speed of 1.2 m/s and (b) the power required while the empty cabin is descending at a constant speed of 1.2 m/s. What would your answer be to (a) if no counterweight were used? What would your answer be to (b) if a friction force of 800 N has developed between the cabin and the guide rails?

Answers

Answer:

Part a)

$P = 4.71 * 10^3 Watt$

Part b)

$P = 2.94 * 10^3 W$

Part c)

$P = 9.4 * 10^3 W$

Part d)

$P = 3.9 * 10^3 W$

Explanation:

Part a)

When cabin is fully loaded and it is carried upwards at constant speed

then we will have

net tension force in the rope = mg

$T = (800)(9.81)$

$T = 7848 N$

now it is partially counterbalanced by 400 kg weight

so net extra force required

$F = 7848 - (400 * 9.81)$

$F = 3924 N$

now power required is given as

$P = Fv$

$P = 3924 (1.2)$

$P = 4.71 * 10^3 Watt$

Part b)

When empty cabin is descending down with constant speed

so in that case the force balance is given as

$F + (150 * 9.8) = (400 * 9.8)$

$F = 2450 N$

now power required is

$P = F.v$

$P = (2450)(1.2)$

$P = 2.94 * 10^3 W$

Part c)

If no counter weight is used here then for part a)

$F = 7848 N$

now power required is

$P = F.v$

$P = 7848 (1.2)$

$P = 9.4 * 10^3 W$

Part d)

Now in part b) if friction force of 800 N act in opposite direction

then we have

$F + (150 * 9.8) = 800 +(400 * 9.8)$

$F = 3250 N$

now power is

$P = (3250)(1.2)$

$P = 3.9 * 10^3 W$

If the balloon takes 0.19 s to cross the 1.6-m-high window, from what height above the top of the window was it dropped?

Answers

Answer:

$heigth=2.86m$

Explanation:

Given data

time=0.19 s

distance=1.6 m

To find

height

Solution

First we need to find average velocity

$V_(avg)=(distance)/(time)\nV_(avg)=(1.6m)/(0.19s)\nV_(avg)=8.42m/s$

Also we know that average velocity

$V_(avg)=(V_(i)+V_(f))/2\n$

Where

Vi is top of window speed

Vf is bottom of window speed

Also we now that

$V_(f)=V_(i)+gt\nV_(f)=V_(i)+(9.8)(0.19)\nV_(f)=V_(i)+1.862$

Substitute value of Vf in average velocity

$V_(avg)=(V_(i)+V_(f))/2\nwhere\nV_(f)=V_(i)+1.862\nand\nV_(avg)=8.42m/s\nSo\n8.42m/s=(V_(i)+V_(i)+1.862)/2\n2V_(i)+1.862=16.84\nV_(i)=(16.84-1.862)/2\nV_(i)=7.489m/s\n$

Vi is speed of balloon at top of the window

Now we need to find time

$V_(i)=gt\nt=V_(i)/g\nt=7.489/9.8\nt=0.764s$

So the distance can be found as

$distance=(1/2)gt^(2)\n distance=(1/2)(9.8)(0.764)^(2)\n distance=2.86m$

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