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Illusion [34]
2 years ago
7

Mark and Balthazar are preparing to conduct neutralization reactions in which they add a base to two different solutions, citric

acid and sodium carbonate. These solutions may irritate the skin or eyes. What safety equipment should Mark and Balthazar be familiar with before they begin?
Physics
2 answers:
jarptica [38.1K]2 years ago
6 0

Mark and Balthazar should be wearing lab coats, gloves, and protective eyewear. If chemicals get into the eyes, an eye wash station can be used to flush the eyes with water. Also, a safety shower can be used if chemicals get on the skin or clothing.

zhuklara [117]2 years ago
5 0
Lab safety equipment prevents damage from accidents and helps keep the people working in the lab safe. The equipment goes hand in hand with the clothing of the person. The first step would be to wear closed shoes and a lab coat.
The equipment that must be worn are goggles to protect the eyes from irritants and latex gloves to protect the skin on the hands.
You might be interested in
Two students are playing paddle ball with a 5 kg spongy ball. If the ball is thrown at the batter with a speed of 5 m/s and boun
fenix001 [56]

Answer:

75 kgm/s

Explanation:

Impulse: This can be defined as the product of mass and change in velocity. The S.I unit is kgm/s.

From the question,

I = m(v-u)................... Equation 1

Where I = impulse, m = mass, v = final velocity, u = initial velocity.

Let the direction of the initial velocity be the positive direction.

Given: m = 5 kg, v = -10 m/s (bounce off), u = 5 m/s.

Substitute into equation 1

I = 5(-10-5)

I = 5(-15)

I = -75 kgm/s.

The negative sign tells that the impulse act in the same direction as the final velocity of the ball

Hence,

I = 75 kgm/s

3 0
2 years ago
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

6 0
2 years ago
This is really urgent
hodyreva [135]

20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is n=\frac{c}{v}

26) Critical angle: [tex]sin \theta_c = \frac{n_2}{n_1}[/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

with

n_1, n_2 are the refractive index of 1st and 2nd medium

\theta_1, \theta_2 are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

  • A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
  • A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

  • A material with refractive index of 1.1
  • A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1 (1)

For light moving from air to water:

n_1 \sim 1.00 is the index of refraction of air

n_2 = 1.33 is the index of refraction ofwater

In this case, \frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75

For light moving from air to glass,

n_2 = 1.51 is the index of refraction of glass

And so

\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66

From eq.(1), we see that the angle of refraction \theta_2 is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

n=\frac{c}{v}

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

  • The higher the index of refraction, the slower the light
  • The lower the index of refraction, the faster the light

26)

From Snell's law,

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, n_1 > n_2, so \frac{n_1}{n_2}>1, and since sin \theta_2 cannot be larger than 1, there exists a maximum value of the angle of incidence \theta_c (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

sin \theta_c = \frac{n_2}{n_1}

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

sin \theta_c = \frac{n_2}{n_1}

For the interface glass-air,

n_1 \sim 1.51\\n_2 = 1.00

The critical angle is

\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}

For the interface glass-water,

n_1 \sim 1.51\\n_2 = 1.33

The critical angle is

\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}

So, the critical angle is larger for the glass-water interface.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

7 0
2 years ago
If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 12m/s, how far from the base of the dorm should
Hitman42 [59]

Answer:

3.6 m

Explanation:

let x = horizontal distance between emily and allison should be for allison to catch the ball

Find horizontal speed of the ball

vx = 12 sin 30 = 12 x 0.5 = 6 m/s

To find time taken, we will use vertical values of the ball motion

Initial velocity in vertical direction

u = 12 cos 30 = 10.392 m/s

let a = g = 9.8m/s2

Use equation of motion

s = ut +1/2at^2

s = vertical distance = 8

8 = (10.392)t + (1/2)(9.8)t^2

8 = (10.392)t + (4.9)t^2

4.9t^2 + 10.392t - 8 = 0

Using formula of quadratic or calculator, we'll find

t = 0.6 and t = -2.72

We pick t=0.6s since it's not logical time in negative

Assuming no air resistance or external forces, the ball will move 6m/s horizontally. Hence using the formula of speed

speed vx = distance x / time

x = (vx)(t)

  = 6 x 0.6

  = 3.6 m

6 0
2 years ago
A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by
Delicious77 [7]

Answer:

W = -510.98J

Explanation:

Force = 43N, 61° SW

Displacement = 12m, 22° NE

Work done is given as:

W = F*d*cosA

where A = angle between force and displacement.

Angle between force and displacement, A = 61 + 90 + 22 = 172°

W = 43 * 12 * cos172

W = -510.98J

The negative sign shows that the work done is in the opposite direction of the force applied to it.

6 0
2 years ago
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