An object travels 20 m in 4 s heading south. What is it’s velocity?

A Carnot engine is operated between two heat reservoirs at temperatures of 550 K and 300 K . 1.) If the engine receives 6.70 kJ

Goshia [24]

Answer:

a) 3654.5 joules per cycle is rejected to the reservoir at 300 K by the Carnot Engine.

b) Mechanical work performed by the engine during each cycle = 3045.5 J = 3.046 KJ

Explanation:

For a Carmot engine,

(Q꜀/T꜀) + (Qₕ/Tₕ) = 0

(Q꜀/T꜀) = - (Qₕ/Tₕ)

Q꜀ = - T꜀ × (Qₕ/Tₕ)

Qₕ = 6.70 KJ

Tₕ = 550 K

Q꜀ = ?

T꜀ = 300 K

Q꜀ = - 300 (6700/550) = - 3654.5 J = - 3.65 KJ

b) Mechanical work done by the Carnot engine is given by

W = |Qₕ| - |Q꜀| = 6700 - 3654.5 = 3045.5 J

6 0

3 years ago

A sled plus passenger with total mass m = 53.1 kg is pulled a distance d = 25.3 m across a horizontal, snow-packed surface for w

Alex Ar [27]

Answer:

Explanation:

Force of friction

F = μ mg

μ is coefficient of friction , m is mass and g is acceleration due to gravity .

If f be the force applied to pull the sled , the horizontal component of force should be equal to frictional force

The vertical component of applied force will reduce the normal force or reaction force from the ground

Reaction force R = mg - f sin28.3

frictional force = μ R where μ is coefficient of friction

frictional force = μ x (mg - f sin28.3 )

This force should be equal to horizontal component of f

μ x (mg - f sin28.3 ) = f cos 28.3

μ x mg = f μsin28.3 + f cos 28.3

f = μ x mg / (μsin28.3 + cos 28.3 )

a )

work done by pulling force = force x displacement

f cos28.3 x d

μ x mg d cos28.3 / (μsin28.3 + cos 28.3 )

b ) Putting the given values

= .155 x 53.1 x 9.8 x 25.3 cos28.3 / ( .155 x sin28.3 + cos 28.3 )

= 1796.76 / (.073 + .88 )

= 1885.37 J

c )

Work done by frictional force

= frictional force x displacement

= - μ x (mg - f sin28.3 ) x d

= - μ x mgd + f μsin28.3 x d

= - μ x mgd + μsin28.3 x d x μ x mg / (μsin28.3 + cos 28.3 )

d )

Putting the values in the equation above

- .155 x 53.1 x 9.8 x 25.3 +

.155 x .474 x 25.3 x .155 x 53.1 x 9.8 /( .155 x .474 + .88)

= -2040.67 + 149.92 / .95347

= -2040.67 + 157.23

= -1883.44 J .

6 0

3 years ago

The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and

Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

$\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m$

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

$d=\sqrt{(x-x_o)^2+(y-y_o)^2}$ (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

$d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m$

hence, the displacement of the airplane is 3.45*10^4 m

6 0

3 years ago

A straight wire carries a 12-A current eastward and a second straight wire carries a 14-A current westward. The wires are separa

USPshnik [31]

Answer:

$5.12\cdot 10^{-4} N$

Explanation:

The force exerted between two current-carrying wires is given by

$F=\frac{\mu_0 I_1 I_2 L}{2 \pi r}$

where

$\mu_0$ is the vacuum permeability

I1 and I2 are the two currents

L is the length of the segment of wire on which we want to calculate the force

r is the distance between the wires

In this problem we have:

$I_1 = 12 A\\I_2 = 14 A\\r = 42 cm = 0.42 m\\L = 6.4 m$

Substituting into the formula, we find:

$F=\frac{(4\pi \cdot 10^{-7})(12)(14)(6.4)}{2 \pi (0.42)}=5.12\cdot 10^{-4} N$

And since the direction of the two currents is opposite, the force between the wires is repulsive.

7 0

4 years ago

What is the pressure inside such a balloon (in atm) if it starts out at sea level with a temperature of 12.8°C and rises to an a

tatiyna

Answer:

The final pressure inside the balloon will be $P_2=0.0359\ atm$ .

Explanation:

Given initial temperature of the balloon is $T_1=12.8\°C$

And the final temperature of the balloon is $T_2=-47.7\°C$

Initially, the balloon is at atmospheric pressure $P_1=1\ atm$ and volume be $V_1$

And the final pressure in the balloon is $P_2$ and the volume be $V_2$

Also, it is given in the question that the final volume became twenty-two times the original volume.

We can write $V_2=22V_1$

Now, using ideal gas equation.

$\frac{P_2V_2}{P_1V_1}=\frac{nRT_2}{nRT_2}$

Where, $R$ is the gas constant. And $n$ is moles of substance inside the balloon.

In our problem the value of $n$ will be the same for both cases. Also, $R$ is the gas constant.

$\frac{P_2V_2}{P_1V_1}=\frac{T_2}{T_1}$

Also, we have $V_2=22V_1$ from the question.

$\frac{P_2\times 22V_1}{P_1V_1}=\frac{T_2}{T_1}\\\\\frac{P_2\times 22}{P_1}=\frac{T_2}{T_1}$

We need to convert the temperature from $\°C$ to Kelvin.

$T_1=12.8\°C\\T_1=12.8+273.15=285.95\\T_2=-47.7\°C\\T_2=-47.7+273.15=225.45$

Plug these values we get,

$\frac{P_2\times 22}{1}=\frac{225.45}{285.95}\\\\22\times P_2=0.789\\P_2=\frac{0.789}{22}\\P_2=0.0359\ atm.$

So, the final pressure inside the balloon will be $P_2=0.0359\ atm$ .

6 0

4 years ago