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prisoha [69]
3 years ago
14

Ms. Sparkle bought 12 cans of diet soda. Each can

Physics
1 answer:
Anit [1.1K]3 years ago
6 0
4.2 liters..... there are 1,000 mL in a liter and there is a total of 4200 mL in this case which is divided by 1000 which gives you 4.2 liters.
You might be interested in
For your answer to this problem, just type in the numerical magnitude of the momentum - no units.
stepan [7]

Answer:

120 kg•m/s.

Explanation:

From the question given above, the following data were obtained:

Case 1

Mass of object = M

Velocity of object = V

Momentum = 15 kg•m/s

Case 2

Mass of object = 2M

Velocity of object = 4V

Momentum = ?

Momentum is defined as follow:

Momentum = mass × velocity

The momentum of object in case 2 can be obtained as follow:

From case 1

Momentum = mass × velocity

15 = M × V

15 = MV ....... (1)

From case 2:

Momentum = mass × velocity

Momentum = 2M × 4V

Momentum = 8MV ....... (2)

Finally , substitute the value of MV in equation 1 into equation 2.

Momentum = 8MV

MV = 15

Momentum = 8 × 15

Momentum = 120 kg•m/s

Therefore, an object with a mass of 2M and 4V would have a momentum of 120 kg•m/s

3 0
3 years ago
Which is the correct formula for cellular respiration?
Tanya [424]

D (Glucose +Oxygen --> Carbon Dioxide + Water + Energy)

8 0
3 years ago
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. She finds the pendulum m
forsale [732]

Answer:

Acceleration due to gravity will be g=5.718m/sec^2

Explanation:

We have given length of pendulum l = 55 cm = 0.55 m

It is given that pendulum completed 100 swings in 145 sec

So time taken by pendulum for 1 swing =\frac{145}{100}=1.45sec

We have to find the acceleration due to gravity at that point

We know that time period of pendulum;um is given by

T=2\pi \sqrt{\frac{l}{g}}

So 1.45=2\times 3.14\times \sqrt{\frac{0.55}{g}}

\sqrt{\frac{0.55}{g}}=0.230

Squaring both side

{\frac{0.3025}{g}}=0.0529

g=5.718m/sec^2

So acceleration due to gravity will be g=5.718m/sec^2

3 0
3 years ago
In order to walk barefoot on hot coals without hurting your feet
siniylev [52]

Before a person walks through burning coal, the person will make sure their feet are very wet. When they start walking on the coal, this moisture will evaporate and form a protective gas layer underneath the person's feet. You can see examples of this if you happen to drip some water on a hot stove or any very hot surface. The water will very easily glide around on top of a newly formed layer of air underneath it -- like air hockey pucks on an air hockey table. Note that when someone walks through burning coal, typically this is also done very quickly to prevent a great deal of exposure to possible harm. By walking quickly, thinking positively, and letting the water cushion you from immediate danger over a short distance, such a task is possible. You may have also heard of physics teachers demonstrating how this principle works by sticking their hand first in a bucket of water and then quickly in a bucket of boiling molten lead. In the lead, their hand is protected briefly by a layer of gas from the evaporated water (the water vapor). I'm fairly sure that there is a name for this particular layer of gas, but I'm afraid the name is beyond me at the moment. In other words, water vapor has a low heat capacity and poor thermal conduction. Very often, the coals or wood embers that are used in fire walking also have a low heat capacity. Sweat produced on the bottom of people's feet also helps form a protective water vapor. All of this together makes it possible, if moving quickly enough, to walk across hot coals without getting burned. WARNING: Do not attempt to perform any of the actions described above. You can seriously injure yourself. Answered by: Ted Pavlic, Electrical Engineering Undergrad Student, Ohio St.  (citing my source)

5 0
3 years ago
Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i
spayn [35]

Answer:

v = 15.56 m/s

v = 56 km/h

Explanation:

When coefficient of friction is approximately zero then we have

F_ncos\theta = mg

F_n sin\theta = \frac{mv^2}{R}

tan\theta = \frac{v^2}{Rg}

here we know that

v = 40 km/h = 11.11 m/s

R = 30 m

tan\theta = \frac{11.11^2}{30\times 9.81}

\theta = 22.75 degree

now when friction coefficient is 0.30 then we have

F_n cos\theta = mg + F_f sin\theta

F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}

now we have

v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}

v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}

v = 15.56 m/s

v = 56 km/h

3 0
3 years ago
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