A stone is thrown straight up. when it reaches it’s highest point, ____

An aircraft engine takes in 9000 j of heat and discards 6400 j each cycle. (a) what is the mechanical work output of the engine

charle [14.2K]

Work done = heat in - heat out
= 9000 -6400 = 3600 J per cycle. 

And the efficiency = work out / heat in 
= 3600/9000 
= 0.4 or 40%

Sorry if this does not help.

4 0

3 years ago

The moon sun is directly overhead at the equator during ( a solstice or an equinox)?.

mina [271]

I think it equinox but I’m not 100% sure

4 0

2 years ago

A flowerpot falls from a window sill 36.5 m

hram777 [196]

When a flowerpot falls from a window sill 36.5 m

above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.

From Newton's third equation of motion,

v^2 = u^2 + 2gh

where,

h is the height of the object or body from ground

u is the initial velocity of the body or object

v is the final velocity of the body or object

g is the acceleration due to gravity

Now, as we know that

Flowerpot is at rest. So, u = 0

g = 9.81m/s^2

h = 36.5m

By substituting all the values, we get

v^2 = 2 × 9.81 × 36.5

= 716.13

v = 26.7m/s

Thus, we concluded that when a flowerpot falls from a window sill 36.5 m

above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.

learn more about Newton's equation of law of motion:

brainly.com/question/8898885

#SPJ9

7 0

1 year ago

The flow of air from an ocean or lake to the land is called a ___________________________.

slava [35]

Maybe sea or onshore breeze

It develops due to the difference of air pressure greater by the different temperature of the water and land

8 0

3 years ago

What distance will a vehicle travel before coming to a complete stop from a speed of 70 mph, (a) When the vehicle is traveling o

OLga [1]

Answer:

(a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be $723.91>SSD>620.2$

(c), The SSD will be 910.5 ft.

Explanation:

Given that,

Speed = 70 mph

Suppose, a perception reaction time of 2.5 sec and the coefficient of friction is 0.35

We need to calculate the stopping sight distance

Using formula of SSD

$SSD=1.47\times v\times t+\dfrac{v^2}{30\times(f\pm g)}$

Where, v = speed of vehicle

t = perception reaction time

f = coefficient of friction

g = gradient of road

(a). If the gradient of road is zero.

Then, the stopping sight distance will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35)}$

$SSD=723.9\ ft$

(b-1). If the gradient of road is 0.1

Then, the stopping sight distance will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35+0.1)}$

$SSD=620.2\ ft$

(b-2). If the grade continuously decrease then the SSD will be increase.

But if the grade is increase then the SSD will be decrease and for flat grade the SSD will be more.

So, The SSD will be $723.91>SSD>620.2$

(c). When the vehicle is traveling downhill on a roadway of constant grade then the vehicle take will be more SSD

So, The SSD will be

$SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35-0.1)}$

$SSD=910.5\ ft$

Hence, (a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be $723.91>SSD>620.2$

(c), The SSD will be 910.5 ft.

7 0

3 years ago

A stone is thrown straight up. when it reaches it’s highest point, _____