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ch4aika [34]
3 years ago
11

When a plane flies with the​ wind, it can travel 1575 miles in 3.5 hours. when the plane flies in the opposite​ direction, again

st the​ wind, it takes 4.5 hours to fly the same distance. find the average velocity of the plane in still air and the average velocity of the wind?
Physics
1 answer:
Zepler [3.9K]3 years ago
3 0

When plane is flying along the wind then we can say

V_{plane} + v_{wind} = \frac{distance}{time}

V_{plane} + v_{wind} = \frac{1575}{3.5}

V_{plane} + v_{wind} = 450 mph

Now when its going against the wind the speed is given by

V_{plane} - v_{wind} = \frac{distance}{time}

V_{plane} - v_{wind} = \frac{1575}{4.5}

V_{plane} - v_{wind} = 350 mph

Now by the above two equations we will have

V_{plane} = 400 mph

v_{wind} = 50 mph

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You push on the top edge of a 1.8 m tall solid circular cylinder of iron which is 4.00 cm in diameter. You push with a horizonta
Citrus2011 [14]

Answer:

\triangle x=3.2*10^-^5 m

Explanation:

From the question we are told that

Height of circular cylinder is h= 1.8m

Diameter of cylinderD=4cm=>0.04m

Horizontal Force  F=900N

Y = 10.0 *10^1^0 N/m^2 \\B = 9.0 * 10^1^0 N/m^2\\S = 4.0 * 10^1^0 N/m^2\\

Generally the formula for shear modulus is mathematically represented by

G=\frac{\tau}{\gamma}

Where

G=Shear modulus\\\tau= shear\ stress\\\gamma=shear strain

G=\frac{F/A}{\triangle x/L}

G=\frac{900/\pi r^2}{\triangle x/1.8}

4.0*10^1^0=\frac{900/\pi r^2}{\triangle x/1.8}

4.0*10^1^0=\frac{900}{\pi r^2}  *\frac{1.8}{\triangle x}

{\triangle x} =\frac{1620}{\pi r^2* 4.0*10^1^0}

\triangle x=3.2*10^-^5 m

5 0
3 years ago
758 j of heat are added to 0.750 kg of copper. how much does its temperature change?(unit=degrees c) PLEASE HELPPPPPPP MEEEEEE
DiKsa [7]

Answer:

2.6^{\circ}C

Explanation:

When a substance is supplied with a certain amount of heat energy, the temperature of the substance increases according to the equation

Q=mC\Delta T

where

m is the mass of the substance

Q is the amount of energy supplied

C is the specific heat of the substance

\Delta T is the temperature change

In this problem:

Q = 758 J is the energy supplied

m = 0.750 kg is the mass of the sample

C=385 J/kg^{\circ}C is the specific heat of copper

Re-arranging the equation, we can find the increase in temperature:

\Delta T=\frac{Q}{mC}=\frac{758}{(0.750)(385)}=2.6^{\circ}C

5 0
3 years ago
What is the new pressure of 150 ml of a gas that is compressed to 50 ml when the original pressure was 3.0 atm and the temperatu
Firlakuza [10]

Answer: 9.0 atm

Explanation:

To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=3.0atm\\V_1=150mL\\P_2=?mmHg\\V_2=50mL

Putting values in above equation, we get:

3.0\times 150mL=P_2\times 50mL\\\\P_2=9.0atm

Thus new pressure of 150 ml of a gas that is compressed to 50 ml is 9.0 atm

5 0
3 years ago
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