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KiRa [710]
3 years ago
12

The MSDS for glacial acetic acid says that it is a flammable liquid that can severely burn any human tissue it comes in contact

with. It reacts with bases, various metals, and strong oxidizing agents. Its vapors can form explosive mixtures with air. Based on this information, which statement is applicable to glacial acetic acid?
Physics
1 answer:
Alex777 [14]3 years ago
3 0

Answer: D) It should be handled in a fume hood, away from open flames.

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A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
disa [49]
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, \mu m g. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
ma=-\mu m g
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
2aS=v_f^2-v_i^2
where v_f=0 is the final speed of the skier and v_i=4.75 m/s is the initial speed. Substituting numbers, we find:
S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m
5 0
3 years ago
A block of aluminum occupies a volume of 15.0 ml and weighs 40.5 what is its density
AVprozaik [17]
D = 40.5 g / 15.0 mL<span>d = 2.70 g/mL</span>
5 0
3 years ago
Read 2 more answers
__________ is the most common type of stretching.
Verizon [17]

Answer: static stretching

Explanation:

e.g rubberband

6 0
3 years ago
Read 2 more answers
How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer:
zimovet [89]

According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:

W=\Delta K

Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:

K=\frac{1}{2}mv^2

Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:

\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\  \\ \therefore K\approx532,346J \end{gathered}

Therefore, the answer is: 532,346 J.

5 0
1 year ago
A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding t
Tom [10]

Answer: Yes, he is exceeding the speed limit

Explanation:

Hi!

This is problem about unit conversion

1 mile = 1,609.344 m

Then the speed limit v is:

v = 75 mi/h = 120,700.8 m/h

1 hour = 60 min = 60*60 s = 3,600 s

v = (120,700.8/3,600) m/s = 33.52 m/s

38 m/s is higher than the speed limit v.

5 0
3 years ago
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