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den301095 [7]
2 years ago
14

Liquids, when heated,

Physics
1 answer:
iren2701 [21]2 years ago
6 0

Answer:

d

Explanation:

According to me answer is d but gas expand more than others

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What’s the answer anyone??
saul85 [17]

Answer:

5.2L == 5200 mL

3.75 kg == 3750 g

500 mm == 0.5 m

Explanation:

For liters to milliliters, simply multiply by 1000.

For kilograms to grams, simply multiply by 1000

For millimeters to meters, simply divide by 1000.

Cheers.

5 0
3 years ago
Read 2 more answers
If the temperature of the developing solution is slightly above normal, radiographic images of the required density may be produ
OLEGan [10]

Answer: shorter/longer developing time.

Explanation:

Ideally, the time to process a radiographic image in the developer is 680 F for 5 minutes. But films may be removed from the fixing solution after 5 minutes (and fixed back) for viewing, only at emergencies like if the temperature of the developing solution is slightly above normal.

If the temperature of the developing solution is slightly above normal, the radiographic images of the required density may be produced by permitting a shorter/longer developing time.

8 0
3 years ago
A group of scientists decide to repeat the muon decay experiment (TR Section 2.7) at the Mauna Kea telescope site in Hawaii, whi
m_a_m_a [10]

Answer:They decided to count 10000 muons because of classical predictions. When you count 1000 initial muons.they would expect only 1.5 muons

Explanation:

t= 4205/0.98 = 1.43×10^-5sec

Classically,number of surviving muons= No exp(-0.693t/t)

N=10^4exp[(-0.694×(1.43×10^-5)/1.52×10-6)]

N= 14.7=15

Relativistic time t'=t/y

t'=(1.43×10^-5)×sqrt(1-(0.98c)^2/c^2

t'=2.84×10^-6secs

N= 10^4exp[(-0.693×(3.84×10^-6)/1.52×10^-6]

N relativistic=2739

4 0
3 years ago
A load of bricks is lifted to the second floor of a building. How do work and power relate to this job?
Nady [450]
The faster the job is done, the greater the power
5 0
2 years ago
Read 2 more answers
A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along
storchak [24]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear density of charge as:

\lambda = \frac{Q}{L}

     Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

dv = \frac{kdq}{r}

          where k is the coulomb's constant

                     r is the distance from dq to center of the circle

Thus.

v = \int_{}^{}\frac{kdq}{a}  

v = \frac{k}{a}\int_{}^{}dq

v = \frac{kQ}{a}     Potential at the center of the semicircle

4 0
2 years ago
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