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Ad libitum [116K]

3 years ago

7

Is there is only one tool used to heat equipment in the laboratory.

2 answers:

yaroslaw [1]3 years ago

7 0

This is false there are many tools used to heat equipment in the laboratory such as a Fuel burner.

Dafna1 [17]3 years ago

6 0

No there is not only one tool used. They are many. Examples are:

Buchner funnel
Bunsen burner
Buret.

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If F(theta)=tan theta=3, find F(theta+pi)

Alex_Xolod [135]

The period of the tan function is π so (∅ + π) would yield the same value as ∅
F(∅ + π) = 3

4 0

3 years ago

I need to show my work as well but on the computer so, please show work for how you got the answers. Thank you!

balandron [24]

Answer:

1) $1H_2 + 1Cl_2$ => 2HCl

2) $2Al + 6HCl$ => $2AlCl_3 + 3H_2$

3) $1Ca_2Br_2 + 2NaCO_3$ => $2CaCO_3 + 2NaBr$

4) $3NaOH + 1FeCl_3$ => $3NaCl + 1Fe(OH)_3$

Explanation:

4 0

3 years ago

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

A)

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is: $w_{c} =m_{c} g$ and maximum tension is $T=2.50 m_{c} g=1.41*10^4N$

total mass and weight is :

$M =m_{c}+ m_{b} =740kg+550kg=1290 kg$

$w_{M} =1.2650*10^4N$

∑ $F_{y} =ma_{y}$

$T-M_{g} =Ma_{y}$

$a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg$

$=0.645m/s^2$

B)

maximum acceleration

$a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0$

using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

to solve for t

$t=\sqrt{2(y-y_{0} )/a_{y} }$

$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

6 0

3 years ago

A baseball is thrown with an initial velocity of 45.4 m/s at an angle of 31.2 ∘ .

sveticcg [70]

Answer:

V (initial vertical velocity) = 45.4 sin 31.2 = 23.52 m/s

1/2 m V^2 = m g h conservation of energy

h = V^2 / (2 g) = 23.52^2 / 19.6 = 28.2 m max height

Check:

t = 28.2 / 9.8 = 2.88 sec time to reach max height

h = 23.52 * 2.88 - 1/2 g 2.88^2 = 27.1 m

8 0

2 years ago

How much time will it take to do 33j of work with 11 w of power

Yuri [45]

Time required : 3 s

<h3>Further explanation </h3>

Power is the work done/second.

$\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s$

To do 33 J of work with 11 W of power

P = 11 W

W = 33 J

$\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}$

8 0

3 years ago