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2 years ago

What is the acceleration?

Physics

1 answer:

Andrews [41]2 years ago

4 0

0.2m/s^2 is you answer because acceleration is (Vf-Vi) divided by (Tf-Ti) so it's (22-2)/(0-100) which is 0.2m/s^2

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One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau

suter [353]

Answer:

actual efficiency is 47.78 %

Carnot efficiency is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy = 7.0 × 10^9 cal (4.184 J/1 cal) = 29.29 × 10^9 J

actual efficiency = 1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency = 0.4778

actual efficiency is 47.78 %

and

Carnot efficiency is = 1 - ( 703 / 2173 )

so Carnot efficiency is = 0.67648

Carnot efficiency is 67.65 %

and

power output = work / time

power output = 1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0

2 years ago

Two objects gravitationally attract with a force of 100 N. If the mass of one object is doubled and the mass of the other object

barxatty [35]

Hello!

Recall the equation for gravitational force:
$F_g = \frac{Gm_1m_2}{r^2}$

Fg = Force of gravity (N)
G = Gravitational constant

m1, m2 = masses of objects (kg)
r = distance between the objects' center of masses (m)

There is a DIRECT relationship between mass and gravitational force.

We are given:
$F_g = 100N$

If we were to double one mass and triple another, according to the equation:
$F'_g = \frac{G(2m_1)(3m_2)}{r^2} = 6(\frac{G(m_1)(m_2)}{r^2}) = 6F_g$

Thus:
$6 * F_g = 6 * 100 = \boxed{600N}$

5 0

2 years ago

Help!!! If anyone could do one of these, i'm confused on how I should write the equation down.

balu736 [363]

F=ma=m(change in velocity/change in time)

Number 1

F=ma

F=55kg(1.1ms^-1/1.6s)=37.8N

Number 2

F=ma

F=0.440kg(10ms^-1/0.02s)=220N

Number 3

F=ma

F=1400kg(15ms^-1/0.73s)=2.88*10^3N or 28,767N

Any questions please feel free to ask.

4 0

3 years ago

3. Maverick and Goose are flying a training mission in their F-14. They are

Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of $y_0=1500m$ , and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

$y=y_0 - \frac{gt^2}{2}$ [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

$x = v.t$ [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

$t=\sqrt{\frac{2y_0}{g}}$

Since we have $y_0=1500m$

$t=\sqrt{\frac{2(1500)}{9.8}}$

$t=17.5 sec$

Replacing in [2]

$x = 688\ m/sec \ (17.5sec)$

$x = 12040\ m$

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0

2 years ago

On hot sunny days, tire should not be over-inflated,why?

Alenkinab [10]

The adds pressure to the tire which would cause it to burst

4 0

2 years ago