All categories

3 years ago

In order to safely conduct any experiment in the laboratory, it is crucial that you

1 answer:

3 0

Lab Safety Rules:

Report all accidents, injuries, and breakage of glass or equipment to instructor immediately. Keep pathways clear by placing extra items (books, bags, etc.) on the shelves or under the work tables. If under the tables, make sure that these items can not be stepped on. Long hair (chin-length or longer) must be tied back to avoid catching fire. Wear sensible clothing including footwear. Loose clothing should be secured so they do not get caught in a flame or chemicals.Work quietly — know what you are doing by reading the assigned experiment before you start to work. Pay close attention to any cautions described in the laboratory exercises Do not taste or smell chemicals. Wear safety goggles to protect your eyes when heating substances, dissecting, etc. Do not attempt to change the position of glass tubing in a stopper. Never point a test tube being heated at another student or yourself. Never look into a test tube while you are heating it.Unauthorized experiments or procedures must not be attempted.Keep solids out of the sink. Leave your work station clean and in good order before leaving the laboratory. Do not lean, hang over or sit on the laboratory tables. Do not leave your assigned laboratory station without permission of the teacher. Learn the location of the fire extinguisher, eye wash station, first aid kit and safety shower. Fooling around or "horse play" in the laboratory is absolutely forbidden. Students found in violation of this safety rule will be barred from participating in future labs and could result in suspension. Anyone wearing acrylic nails will not be allowed to work with matches, lighted splints, Bunsen burners, etc. Do not lift any solutions, glassware or other types of apparatus above eye level. Follow all instructions given by your teacher.Learn how to transport all materials and equipment safely. No eating or drinking in the lab at any time!

You might be interested in

A ball of 0.1kg is thrown straight up into the air with

Contact [7]

a) The momentum of the ball at the maximum height is zero

b) The momentum of the ball at halfway is 1.06 kg m/s

Explanation:

There is only one force acting on the ball during its motion: the force of gravity, acting downward. As a result, the ball hasa constant acceleration downward ( $g=9.8 m/s^2$ , acceleration of gravity), and therefore it is in free fall motion.

This means that the velocity of the ball is given by the equation:

$v=u-gt$

where

v is the velocity at time t

u = 15 m/s is the initial velocity

$g=9.8 m/s^2$

As we see from the equation, the term $-gt$ is negative: this means that as the ball goes higher and higher, its velocity decreases. The point of maximum height is reached when the velocity has become zero (after that point, the velocity changes direction and points downward), so when

$v=0$

The momentum of the ball at any point is given by

$p=mv$

where m = 0.1 kg is its mass. Therefore, at the maximum height, since $v=0$ , the momentum is also zero:

$p=0$

We can use the following suvat equation to find the maximum height reached by the ball:

$v^2-u^2=2as$

where

v = 0 is the velocity at the point of maximum height

u = 15 m/s is the initial velocity

$a=-g=-9.8 m/s^2$ is the acceleration (downward)

s is the vertical displacement (the maximum height)

Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{0-(15)^2}{2(-9.8)}=11.48 m$

This means that the ball is halfway when its height is

$h'=\frac{s}{2}=\frac{11.48}{2}=5.74 m$

We can find the velocity of the ball v' when it is at h' by using again the same equation:

$v'^2-u^2=2ah'\\v=\sqrt{u^2+2ah'}=\sqrt{(15)^2+2(-9.8)(5.74)}=10.6 m/s$

And therefore, the momentum of the ball here is

$p'=mv'=(0.1)(10.6)=1.06 kg m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

7 0

3 years ago

Si un cuerpo se encuentra en reposo tiene energía cinética?¿por qué?

posledela

Answer:

This is because, when an object which has possessed kinetic energy under its being in motion hits another object, which is stationary the object at rest receives part of the energy inherent in the body which has kinetic energy and expresses the received energy in the form of motion or some forms different from the initial stationary position.

Explanation:

7 0

4 years ago

3. A block falls from a table 0.6 m high. It lands on a vertical spring which has a spring constant of k= 2400 N/m.

goldfiish [28.3K]

<h2>SOLUTION :-</h2>

PLEASE CHECK THE ATTACHED FILE

8 0

2 years ago

What is the acceleration of a ball traveling horizontally with an initial velocity os 20 meters/second and ,2.0 seconds later, a

ch4aika [34]

Acceleration is the rate of change of velocity, so can be calculated by:
a = v2 - v1 / t where v2 and v1 are the final and initial velocities of the object respectively and t is the time taken. Therefore the acceleration is:
a = 30 m/s - 20 m/s / 2.0 s = 10 m/s / 2.0 s = 5 m/(s^2)
Hope this helps!

5 0

4 years ago

) A steel guitar string with a diameter of 1.00 mm is stretched between supports 80.0 cm apart. The temperature is 0.0°C. (a) Fi

ladessa [460]

Answer: a. Mass per unit length =0.0245kg/m

b. Tension =2.45x10^-8N

C. Tension = 2.45 x10^-8N

Fundamental frequency =200Hz

Explanation:

7 0

4 years ago