As in problem 80, an 76-kg man plans to tow a 128000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of 6.7° above the horizontal. The coefficient of static friction between his shoes and the runway is 0.87. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.

Answers

Answer 1

Answer:

In order to solve this problem it is necessary to apply the concepts related to Newton's second law and the respective representation of the Forces in their vector components.

The horizontal component of this force is given as

F_x = Fcos(6.7)

While the vertical component of this force would be

F_y = Fsin(6.7)

In the vertical component, the sum of Force indicates that:

$\sum F_y= 0$

The Normal Force would therefore be equivalent to the weight and vertical component of the applied force, therefore:

$N = mg+Fsin(6.7)$

In the horizontal component we have that the Force of tension in its horizontal component is equivalent to the Force of friction:

$\sum F_x = 0$

$F_x = F_(friction)$

$Fcos (6.7) = N\mu$

Using the previously found expression of the Normal Force and replacing it we have to,

$Fcos(6.7)= \mu (mg+Fsin(6.7))$

Replacing,

$Fcos(6.7)= (0.87) (mg+Fsin(6.7))$

$Fcos(6.7) = (0.87)(mg) + (0.87)(Fsin(6.7))$

$Fcos(6.7) -(0.87)(Fsin(6.7)) = 0.87 (mg)$

$F(cos(6.7)-0.87sin(6.7)) = 0.87 (mg)$

$F = (0.87 (mg))/((cos(6.7)-0.87sin(6.7)))$

$F = (0.87(128000*9.8))/((cos(6.7)-0.87sin(6.7)))$

$F = 1.95*10^6N$

Finally the acceleration would be by Newton's second law:

$F = ma$

$a = (F)/(m)$

$a = ( 1.95*10^6)/(128000)$

$a = 15.234m/s^2$

Therefore the greatest acceleration the man can give the airplane is $15.234m/s^2$

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The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact with the ground. Determine the weight of the automobile.

Answers

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, $P=2.2* 10^5\ Pa$

Area of each tire, $A=0.023\ m^2$

Area of 4 tires, $A=0.092\ m^2$

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

$P=(F)/(A)$

$F=P* A$

$F=2.2* 10^5* 0.092$

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

Final answer:

The vehicle's weight can be calculated by rearranging the definition of pressure (Pressure = Force/Area) to solve for Force (Force = Pressure * Area), then multiplying by four to account for all four tires. Remember that the result will be in newtons, so to convert it to kilograms, divide by gravitational acceleration.

Explanation:

Your question revolves around the concept of pressure. Tire pressure is a type of air pressure which is a part of physics. To determine the weight of the automobile in which each of the four tires has an area of 0.023 m2 and is inflated to a gauge pressure of 2.2 x 105 Pa, we need to utilize the fundamental equation of pressure:

P = F/A

Where P is the pressure, F is the force (which in this case will be the weight of the car), and A is the area of each of the tires where they are in contact with the ground. Solving for the weight (F) results in:

F = P * A

In your case, because there are four tires we multiply the result by four, therefore:

F = 4 * (2.2 x 105 Pa) * (0.023 m2)

We have to multiply this by 4 to account for all four tires. Finally, your weight will be in newtons, to convert it to kg you will divide by gravitational acceleration (approx 9.8 m/s2).

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How many times will the temperature of oxygen with a mass of 1 kg increase if its volume is increased by 4 times, and the pressure is decreased by 2 times?Round off the answer to the nearest whole number.

Answers

Answer:

9.2 Relating Pressure, Volume,

Figure 1. In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.

Explanation:

hope its help :)

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Diethyl ether has a H°vap of 29.1 kJ/mol and a vapor pressure of 0.703 atm at 25.0°C. What is its vapor pressure at 60.0°C?

Answers

Answer:

The vapor pressure at 60.0°C is 2.416 atm

Explanation:

To solve this problem, we use Clausius-Clapeyron equation

$ln(P_2)/(P_1) = (-\delta H)/(R)[(1)/(T_2)-(1)/(T_1)]= (\delta H)/(R)[(1)/(T_1)-(1)/(T_2)]$

where;

Initial pressure P₁ = 0.703 atm

Initial Temperature T₁ = 25+273 = 298K

Final temperature T₂ = 60+273 = 333K

Change in enthalpy of vaporization ΔH = 29.1 KJ/mol = 29100J/mol

R is Boltzman constant = 8.314 J/K.mol

$ln(P_2)/(P_1) = (29100)/(8.314)[(1)/(298)-(1)/(333)] =1.23449$

$(P_2)/(P_1) = e^(1.23449)$ ⇒ $(P_2)/(P_1) = 3.43663$

P₂ = P₁ (3.43663) = (0.703 atm)(3.43663) = 2.416 atm

P₂ = 2.416 atm

Therefore, the vapor pressure at 60.0°C is 2.416 atm.

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 V. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answers

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

The additional resistance is $R_z = 4.4 \Omega$

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = (E)/(R_T)$

Substituting values

$4 = (28)/(R_z + 2.6)$

Making $R_z$ the subject of the formula

So $R_z = (28 - 10.4)/(4)$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

When you blow some air above a paper strip, the paper, rises. This is because 1.) the air above the paper moves faster and the pressure is lower 2.) the air above moves faster and the pressure is higher 3.) the air above the paper moves faster and the pressure remains constant 4.) the air above the paper moves slower and the pressure is higher 5.) the air above the paper moves slower and the pressure is lower

Answers

The correct option is option (1)

The faster movement of air on the upper surface of the paper creates lower pressure above the paper.

Pressure difference:

The movement of air is always from a region of higher pressure to a region of lower pressure.

As we blow air above the paper strip a low pressure is created above the strip due to the fast movement or high speed of the air. And the pressure below the strip is higher in comparison to the pressure above since the air below is not moving.

So, due to the pressure difference, a force is generated on the paper strip by the air from the lower surface to the upper surface.

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This is happened because "the air" above "moves faster" and "the pressure" is "lower".

Option: 1

Explanation:

Air movement take place from the region where air pressure is more than the region where the pressure is low. When we "blow" air above the "paper strip" paper rises because "low pressure" is created above the strip as high speed winds always travel with reduced air pressure. Hence due to higher air pressure below the strip, it is pushed upwards. This difference in pressure results into fast air moves. This happen because "speed" of the wind depends on "the difference between the pressures" of the air in the two regions.

Find the force on a 5 pС charge in a place where the electric field is 400 N/C.

Answers

Answer:

Electric force, $F=2* 10^(-9)\ N$

Explanation:

It is given that,

Charge on the particle, $q=5\ pC=5* 10^(-12)\ C$

Electric field, $E=400\ N/C$

Let F is the electric force acting on the charged particle. The electric force per unit electric charge is called electric field. Mathematically, it is given by :

$F=qE$

$F=5* 10^(-12)\ C* 400\ N/C$

$F=2* 10^(-9)\ N$

So, the force acting on the charged particle is $2* 10^(-9)\ N$ . Hence, this is the required solution.

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