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2 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

What evidence do you that suggest water waves are transverse wave

svp [43]

Answer:

If you throw a pebble into a pond, ripples

spread out from where it went in. These

ripples are waves travelling through the

water. The waves move with a transverse

motion.

Explanation:

5 0

3 years ago

If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a

PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s $w_o$ = 12 rad/s

Angular Velocity at time 0.15s $w_f$ = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( $w_f$ - $w_o$ ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

8 0

2 years ago

What is the volume of a bar of soap that is 9 cm long, 5 cm wide, and 2 cm high?

GalinKa [24]

Answer:

90cm

Explanation:

2x5=10

10x9=90

7 0

3 years ago

Read 2 more answers

Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.

andreev551 [17]

Answer:

the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = $m_A$

then, the mass of object B = $m_B = \frac{m_A}{4}$

work done on object A = -18 J

work done on object B = -18 J

let $v_i$ be the initial speed

let $v_f$ be the final speed

For object A;

$K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\$

$v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\$

Thus, the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

To obtain the change in the final speed of object B, apply the following equations.

$K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B = \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\$

$\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i$

Thus, the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

3 0

2 years ago

a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por q

Tatiana [17]

Answer:

a) the correct answer is 1 , b) x=0 F=0, a=0

x= 0.050 F= -7.5 N, a= -15 m/s²

x= 0.150 F= 22.5 N, a=- 45 m/s²

Explanation:

a) In a mass - spring system the force is given by the Hooke force,

F = - k x

Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position

the correct answer is 1

b) we assume that the given values are from the equilibrium position of the spring.

Let's calculate the force

x = 0

F = 0

x = 0.050

F = - 150 0.050

F = - 7.50 N

x = 0.150

F = - 150 0.150

F = - 22.5 N

let's use Newton's second law to find the acceleration

F = m a

a = F / m

x = 0 m

a = 0

x = 0.050 m

a = -7.50 / 0.50

a = - 15 m / s²

x = 0.150 m

a = - 22.5 / 0.50

a = - 45 m/s²

TRASLATE

a) En un sistema masa – resorte la fuerza es dada por la fuerza de Hoke,

F= - k x

analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio

la respuesta correcta es 1

b)suponemos que los valores dados son desde la posición de equilibrio del resorte.

Calculemos la fuerza

x=0

F= 0

x=0.050

F = - 150 0.050

F= - 7.50 N

x= 0.150

F= - 150 0.150

F= - 22.5 N

usemos la segunda ley de Newton para encontrar la aceleración

F = m a

a = F/m

x =0 m

a = 0

x= 0.050 m

a = -7.50/ 0.50

a =- 15 m/s²

x= 0.150 m

a= - 22.5 / 0.50

a= - 45 m/s²

7 0

3 years ago