All categories

3 years ago

When using a micron gauge, if the pressure gauge falls frim 1,300 to 1,000 instantly, the vehicle needs to?

1 answer:

5 0

If the micron gauge pressure falls instantly, it means that the moisture of the air-conditioning is frozen, therefore you need to move the vehicle to a warmer place in order to unfreeze the moisture.

You might be interested in

Is it possible for a nazi style government to exist in the modern world? Explain why or why not.

nevsk [136]

I think only if they were too overpowered maybe, but the modern world doesn't except this kind of dictatorship. Most armies are much more powerful than in the past.

7 0

3 years ago

Which of the following must be true about the object labeled X in the circuit below?

Alex787 [66]

the answer is c and if I help you thank me

4 0

2 years ago

The space shuttle travels at a speed of about 7.6times10^3 m/s. The blink of an astronaut's eye lasts about 110 ms. How many foo

sveta [45]

Answer:

It covers distance of 9.15 football fields in the said time.

Explanation:

We know that

$Distance=Speed\times Time$

Thus distance covered in blinking of eye =

$Distance=7.6\times 10^{3}m/s\times 110\times 10^{-3}s\\\\Distance=836 meters$

Thus no of football fields= $\frac{936}{91.4}=9.15Fields$

7 0

2 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$