Which will have the lowest density A) Ice B) Steam C) water

The critical angle for refraction from glass to air is 30° and that from water to air is 37°.Find the critical angle for refract

erik [133]

Answer:

Å

1,55

5 400 Å

n

=

193548,4 km/s

1,55

3 10 m/s

n

c

Vidrio v: =

3 724,14 Å

1,45

5 400 Å

n

=

206 896,5 km/s

1,45

3 10 m/s

n

c

Aceite = v:

07,1

45,1

55,1

n

I

8 0

2 years ago

Does gender affect your aerobic capacity?

alekssr [168]

To a degree yes, the mass and volume of a male is far greater than that of the opposite gender. As well as the type of aerobics , but with practice and daily aerobics activities one can become very flexible

8 0

2 years ago

A 2.0 g particle moving at 5.2 m/s makes a perfectly elastic head-on collision with a resting 1.0 g object.

sesenic [268]

Answer:

(a) The speed of the first particle is 1.75 m/s. The speed of the second particle is 6.9 m/s after the collision.

(b) The speed of the first particle is 3.45 m/s in the negative direction. The speed of the second particle is 1.73 m/s.

(c) The final kinetic energy of the incident particle in part (a) and part(b) is 0.0031 J and 0.011 J, respectively.

Explanation:

(a)

In an elastic collision, both momentum and energy is conserved.

$\vec{P}_{initial} = \vec{P}_{final}\\m_1v_1 = m_1v_1' + m_2v_2'\\K_{initial} = K_{final}\\\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2$

Combining these equations will give the speed of the second particle.

$v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2*2}{2+1}(5.2) = 6.9~m/s$

We can use this to find the speed of the first particle.

$m_1v_1 = m_1v_1' + m_2v_2'\\2(5.2) = 2v_1' + (1)(6.9)\\v_1' = 1.75~m/s$

(b)

If m_2 = 10g.

$v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2*2}{2+10}(5.2) = 1.73~m/s$

$m_1v_1 = m_1v_1' + m_2v_2'\\2(5.2) = 2v_1' + (10)(1.73)\\v_1' = -3.45~m/s$

The minus sign indicates that the first particle turns back after the collision.

(c)

The final kinetic energy of the particle in part (a) and part (b) is

$K_a = \frac{1}{2}m_1v_1'^2 = \frac{1}{2}(2\times10^{-3})(1.75)^2 = 0.0031 ~J\\K_b = \frac{1}{2}m_2v_1'^2 = \frac{1}{2}(2\times10^{-3})(3.45)^2 = 0.011~J$

8 0

2 years ago

How many electrons point in 2 second in a wire carrying a current of 20mA ?

Afina-wow [57]

Answer and working shown on photo, final answer for number electrons=2.5*10^17 electrons

6 0

2 years ago

Coulomb is a very large unit for practical use. Justify your answer if 10^10 electrons are transferred from a body/second

zlopas [31]

Given:

10^10 electrons per second

To justify that coulomb is a very large unit for practical use, we need to convert the quantity of electron given to Coulombs:

From literature,

1 Coulomb is equivalent to 6.242×10^18 electrons.

So,

= 10^10 electrons * (1 coulomb/6.242×10^18 electrons) / second
= 1.602 x 10^-9 coulumbs

This value is too small to be used in an actual setting.

3 0

2 years ago

Which will have the lowest density A) IceB) SteamC) water​

Which will have the lowest density A) IceB) SteamC) water