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3 years ago

Explain the qualifications that must be met for any vehicle to go in space.

Physics

2 answers:

Lubov Fominskaja [6]3 years ago

8 0

Answer: It must be able to handle a lot of pressure and heat

Explanation:

Shkiper50 [21]3 years ago

5 0

Ais 1+1is2 so you put 9x9 in 2+2 what the answer is your answer hope it help

You might be interested in

What is solid matter and can we covert matter into something else

lbvjy [14]

Answer: Solid matter is one of the three main states of matter, along with liquid and gas. Matter is the "stuff" of the universe, the atoms, molecules and ions that make up all physical substances. In a solid, these particles are packed closely together and are not free to move about within the substance.

All matter exists as solids, liquids, or gases. These are called the states of matter. Matter can change from one state to another if heated or cooled. If ice (a solid) is heated it changes to water (a liquid).

Explanation:

6 0

3 years ago

Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1

Nonamiya [84]

Answer:

$a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

Magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$

Explanation:

$t_1=2\ \text{s}$

$v_x=1\ \text{m/s}$

$v_y=3\ \text{m/s}$

$t_2=2.5\ \text{s}$

$v_x=4\ \text{m/s}$

$v_y=3\ \text{m/s}$

Average acceleration in the different axes

$a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2$

$a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2$

The components of the acceleration is $a_x=6\ \text{m/s}^2$ and $a_y=0\ \text{m/s}^2$

The magnitude of acceleration

$a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2$

Direction

$\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}$

The magnitude of accleration is $6\ \text{m/s}^2$ and the direction is $0^{\circ}$ .

7 0

2 years ago

During a hard sneeze your eyes might shut for 0.5sec .if you are driving a car at 90kmhr during such a sneeze, how far does the

Lana71 [14]

Eyes shut = 0.5 sec

Speed = 90 km/hr

In 1 hour, it will travel at 90 km.

Let us divide 90 to 60 since 1 hour has 60 minutes.

The distance would be 1.5km/min.

Let us again divide it by 60 because 1 minute is equals to 60 seconds, so it will be 0.025km/second.

0.025 divided by 2 equals 0.0125km.

So the car’s distance is 0.0125km or 12.5 meters per 0.5 second.

6 0

2 years ago

One side of a triangle is increasing at a rate of 5 cm/sec and a second side is increasing at a rate of 7 cm/sec. If the area of

Vanyuwa [196]

Answer:

The rate of angle is 26.25 rad/sec.

Explanation:

Given that,

First side of triangle a= 20 cm

Second side of triangle b= 50 cm

One side of a triangle is increasing at a rate = 5 cm/sec

Second side is increasing at a rate = 7 cm/s

Angle $\theta=\dfrac{\pi}{3}=60^{\circ}$

If the area of the triangle remains constant,

We need to calculate rate of angle

Using formula of area of triangle

$A=\dfrac{1}{2}ab\sin\theta$

On differentiating

$\dfrac{dA}{dt}=\dfrac{1}{2}ab\cos\theta\dfrac{d\theta}{dt}+\dfrac{1}{2}a\sin\theta\dfrac{db}{dt}+\dfrac{1}{2}b\sin\theta\dfrac{da}{dt}$

Put the value into the formula

$0=\dfrac{1}{2}\times20\times50\cos60\dfrac{d\theta}{dt}+\dfrac{1}{2}\times20\sin60\times7+\dfrac{1}{2}\times50\sin60\times5$

$\dfrac{d\theta}{dt}=\dfrac{35\sqrt{3}\times\dfrac{25}{2}\sqrt{3}\times5}{250}$

$\dfrac{d\theta}{dt}=26.25\ rad/sec$

Hence, The rate of angle is 26.25 rad/sec.

8 0

3 years ago

A 50.0 mg sample of an unknown radioactive substance was placed in storage and its mass measured periodically. After 19.7 days,

PilotLPTM [1.2K]

<h2>Answer: 4.928 days </h2><h2 />

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula: </u>

<u></u>

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=3.13mg$ is the final amount of the material

$A_{o}=50mg$ is the initial amount of the material

$t=19.7days$ is the time elapsed

$h$ is the half life of the material (the quantity we are asked to find)

Knowing this, let's substitute the values and find $h$ from (1):

$3.13mg=(50mg)2^{\frac{-19.7days}{h}}$ (2)

$\frac{3.13mg}{50mg}=2^{\frac{-19.7days}{h}}$ (3)

Applying natural logarithm in both sides:

$ln(\frac{3.13mg}{50mg})=ln(2^{\frac{-19.7days}{h}})$ (4)

$-2.77=-\frac{19.7days}{h}ln(2)$ (5)

Clearing $h$ :

$h=\frac{-19.7days}{-2.77}(0.693)$ (6)

Finally:

$h=4.928days$

6 0

3 years ago