Most stars spend the majority of their lifetimes as which type of star?

Which one of the following statements about sulfuric acid is correct?

kherson [118]

The answer is C. Sulfuric Acid is dangerous to living organism

Direct contact with your skin and it will cause Chemical Burns , Direct contact with your eyes and it could cause a permanent Blindness.

This chemical also suck the water out of everything, so you can guess what will happen if it get into your body.

6 0

2 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

3 years ago

Not only did Skid get shot out of a cannon when he was a clown in the circus, but he was also launched into the air by a vertica

tiny-mole [99]

Answer:

v = 16.11 m / s

Explanation:

For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation

starting point. When the spring is compressed

Em₀ = K_e + U = ½ k x² + m g x ’

final point. The point where it leaves the platform

Em_f = K = ½ m v²

energy is conserved

Em₀ = Em_f

½ k x² + m g x ’= ½ m v²

v² = $\frac{k}{m}$ x² + g x

let's calculate

v² = $\frac{8700}{55}$ 1.25² + 9.8 1.25

v² = 247.159 + 12.25 = 259.409

v = 16.11 m / s

8 0

3 years ago

Which of the following is not a state of matter?

zzz [600]

Water doesn't belong on that list.

If it's removed, then there's a space to add GAS, which does belong.

8 0

3 years ago

Two narrow slits are illuminated by a laser with a wavelength of 593 nm. The interference pattern on a screen located x = 4.80 m

masya89 [10]

Answer:

The distance is $d = 1.39 *10^{-4} \ m$

Explanation:

From the question we are told that

The wavelength is $\lambda = 593 \ nm = 593 *10^{-9} \ m$

The distance of the screen is x = 4.80 m

The location of the fourth order bright fringe is y = 8.20 cm = 0.082 m

The order of the fringe is n = 4

Generally the position of a fringe with respect to the central fringe is mathematically represented as

$y = \frac{ n * x * \lambda }{d}$

Where d is the distance between the slits, so making d the subject

$d = \frac{\lambda * x * n }{ y }$

substituting values

$d = \frac{ 593 *10^{-9} * 4.80 * 4 }{ 0.082 }$

$d = 1.39 *10^{-4} \ m$

7 0

3 years ago