Which type of cloud would you expect to be involved in some form of precipitation?

A space shuttle with a mass of approximately 7.08 E5 kg is sitting on the launch pad. What would be the weight of the space shut

Afina-wow [57]

The weight of the shuttle is $6.94\cdot 10^6 N$

Explanation:

The weight of an object on Earth is the gravitational force exerted by the Earth on the object.

The magnitude of the weight of an object is given by:

$W=mg$

where

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity on Earth's surface

And the direction is downward (towards the Earth's centre).

For the shuttle in this problem, its mass is

$m=7.08\cdot 10^5 kg$

So, its weight is

$W=(7.08\cdot 10^5)(9.8)=6.94\cdot 10^6 N$

Note that while the mass of an object (m) does not change, its weight (W) changes according to the location, since the value of g can be different at different location (for example, on the Moon, the value of g is about 1/6 the value of g on the Earth).

Learn more about weight and forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

6 0

2 years ago

A car of mass m push = 1200 kg is capable of a maximum acceleration of 6.00 m / s 2 . If this car is required to push a stalled

motikmotik

1)

first you find the maxium force that the car can produce.

f=ma

Fmax=(1100kg)(6m/s^2)

then use f = ma again to find the accel with the passengers

Fmax=(1100kg +1650kg)(a)

=> a = (1100kg)(6m/s^2)/( 1100kg +1650kg)

= 2.4 m/s^2

4 0

2 years ago

What is the momentum of an 18-kg object moving at 0.1 m/s ?

Anit [1.1K]

Answer:

1.8 m/s

Explanation:

here's the solution : -

momentum = mass × velocity

=》18 × 0.1

=》1.8 m/s

6 0

3 years ago

A high voltage transmission line with a resistance of 0.51 Ω/km carries a current of 1099 A. The line is at a potential of 1300

Illusion [34]

<h2>Answer:</h2>

$\boxed{P=96.09MW}$

<h2>Explanation:</h2>

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

$R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega$

So we can calculate the power loss as:

$P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}$

Finally, the power loss due to resistance in the line is 96.09MW

5 0

3 years ago

Please help me no links or I will report I really need the answer and I’ll give brainly if it’s right!

Annette [7]

Just put a full picture clearly we can’t really see

3 0

3 years ago