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3 years ago

Not a question but need a lot of tips

Physics

1 answer:

skad [1K]3 years ago

8 0

DibujaZos da buenos tips e incluso el pinta con colores faber castel son económicos y el dibujo queda epico

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What gases can CFC and HCFC refrigerants decompose into at high temperatures

nekit [7.7K]

Answer:

Hydrochloric and Hydrofluoric Acids.

3 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.

BARSIC [14]

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

$y_{i}=0m$

$v_i=80\frac{m}{s}$

$a=4\frac{m}{s^2}$

$y_{e}=1000m$

$g=9.8\frac{m}{s^2}$

From Kinematics, the position $y$ as a function of time when the engine still works will be:

$y(t)=v_it+\frac{1}{2}at^2$

At what time the altitud will be $y_{e}=1000m$ ?

$v_it+\frac{1}{2}at^2=y_{e}$ ⇒ $\frac{1}{2}at^2+v_it-y_{e}=0$

Using the quadratic formula: $t_1=10s$ .

How much time does it take for the rocket to touch the ground? No the function of position is:

$y(t)=y_{e}+v_et-\frac{1}{2}gt^2$

Where our new initial position is $y_{max}$ , the velocity when the engine breaks is $v_e=v_i+at=120\frac{m}{s}$ and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

$y_{e}+v_et-\frac{1}{2}gt^2=0$

Again, using the quadratic ecuation:

$t_2=24.49s$

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

$t_T=t_1+t_2=34.49s$ .

6 0

3 years ago

What happens to the Force due to gravity between 2 masses if one

creativ13 [48]

There is half the force that there was before it was split in half

7 0

3 years ago

An object is initially moving at an unknown velocity. It accelerates at a rate of 1.5m/s2 to a new velocity of 60 m/s in 25 s. W

Alexandra [31]

<h3>Answer: 22.5 m/s</h3>

=====================================================

Work Shown:

acceleration = ( finalVelocity - initialVelocity )/(change in time)

1.5 = (60 - x)/(25)

1.5*25 = 60-x

37.5 = 60-x

x = 60-37.5

x = 22.5

The initial velocity is 22.5 m/s

6 0

3 years ago

You do 30 J of work to load a toy dart gun. However, the dart is 10 cm long and feels a frictional force of 10 N while going thr

shepuryov [24]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

$K.E_{1}$ = 30 J, l = 10 cm = 0.1 m (as 1 m = 100 cm)

F = 10 N

The formula to calculate the work done is as follows.

W = $F \times l$

Hence, putting the given values into the above formula as follows.

W = $F \times l$

= $10 N \times 0.1 m$

= 1 J

As the dart loses 1 joule of energy when it goes through the barrel. Therefore, the kinetic energy it has when it comes out is calculated as follows.

K.E = 30 - 1

= 29 J

Thus, we can conclude that the kinetic energy of the fired dart is 29 J.

6 0

3 years ago