Read the following scenario and decide if it is a hypothesis, theory or law.

An engine extracts 441.3kJ of heat from the burning of fuel each cycle, but rejects 259.8 kJ of heat (exhaust, friction,etc) dur

Lilit [14]

Answer:

The thermal efficiency is 0.4113.

Explanation:

We know that the thermal efficiency is the ratio of work done by the engine over the heat taken

$\eta = \frac{W_{made}}{Q_{in}}$

Now, how much work the engine do in a cycle?

We know that the work done in a cycle must be equal to the heat taken minus the heat rejected

$W{made} = Q_{in} - Q_{rejected}$

So, the thermal efficiency will be:

$\eta = \frac{Q_{in} - Q_{rejected}}{Q_{in}}$

$\eta = \frac{Q_{in}}{Q_{in}} - \frac{Q_{rejected}}{Q_{in}}$

$\eta = 1 - \frac{Q_{rejected}}{Q_{in}}$

Putting the values of the problem

$\eta = 1 - \frac{259.8 kJ }{441.3kJ}$

$\eta = 0.4113$

7 0

4 years ago

Find the electric flux passing through the 45° surface of an upright circular cone with a base diameter of 4 m if the electric f

alexgriva [62]

Answer:

electric flux = 177.625 N-m²/C

Explanation:

given data

angle = 45°

diameter = 4 m

electric field E = 20 N/C

to find out

electric flux

solution

we know electric flux formula that is

electric flux = E × A × cosθ ..............1

here A is area that is

Area = πd²/4 = π(4)²/4 = 12.56 m²

now put all value in equation 1

electric flux = 20 × 12.56 × cos45

electric flux = 177.625 N-m²/C

8 0

3 years ago

A soldier throws a grenade horizontally from the top of a cliff. Which of the following curves best describes the path taken by

Natasha_Volkova [10]

Answer:

4. Parabola

Explanation:

The motion of the flight of the grenade is a projectile motion, it's shape is best illustrated using a parabolic diagram.

It's not a circle as the path is not totally round.

It's not an ellipse as the path of motion is not a a completely bounded shape.

Its not a hyperbola as a hyperbola is an open curve with two branches.

Its a parabolic shape because a parabola has just one branch

7 0

4 years ago

John and mary are skating at an ice rink. john skates at a constant speed of 6.7 m/s, with respect to the ice surface, directly

iragen [17]

60.3Â° from due south and 5.89 m/s For this problem, first calculate a translation that will put John's destination directly on the origin and apply that translation to Mary's destination. Then the vector from the origin to Mary's new destination will be the relative vector of Mary as compared to John. So John is traveling due south at 6.7 m/s. After 1 second, he will be at coordinates (0,-6.7). The translation will be (0,6.7) Mary is traveling 28Â° West of due south. So her location after 1 second will be (-sin(28)*10.9, -cos(28)*10.9) = (-5.117240034, -9.624128762) After translating that coordinate up by 6.7, you get (-5.117240034, -2.924128762) The tangent of the angle will be 2.924128762/5.117240034 = 0.57142693 The arc tangent is atan(0.57142693) = 29.74481039Â° Subtract that value from 90 since you want the complement of the angle which is now 60.25518961Â° So Mary is traveling 60.3Â° relative to due south as seen from John's point of view. The magnitude of her relative speed is sqrt(-5.117240034^2 + -2.924128762^2) = 5.893783 m/s Rounding the results to 3 significant digits results in 60.3Â° and 5.89 m/s

8 0

3 years ago

These has two questions, any help would be much appreciated. Ignore the current answers.

Sati [7]

Bro sorry I don’t know these answer but keeep trying

4 0

3 years ago