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marshall27 [118]
3 years ago
5

Which metal is liquid at room temperature?

Physics
2 answers:
White raven [17]3 years ago
7 0
D. mercury.............
Paul [167]3 years ago
5 0
Mercury will be our answer.

As we know, our lovely thermometers contain mercury (mostly old fashioned kinds now), and the only way that we could see the temperature through a thermometer was by seeing the liquid inside of the thermometer rise and fall. If mercury were to be a solid, then it wouldn't be able to rise or fall.
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A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force
Oliga [24]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

6 0
4 years ago
Read 2 more answers
If you begin with 40 grams of a radioactive isotope and end with 10 grams, how many half-lifes of the radioactive isotope have p
Rom4ik [11]
30 grams of radioactive isotope have passed.
8 0
3 years ago
If you walk 6mph how far would you get in 10 seconds?
abruzzese [7]

Answer:

0.017

Explanation:

6÷3600×10=

6÷360=0.00166666666

=0.017(rounded)

3 0
3 years ago
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Calculate the monentun pf 75 kg bicycle and boy who has ghe velocitg of 3m/s
Grace [21]

Explanation:

n=75 kg

v=3m/s

m=n×v

m=75x3

m= 225n

6 0
3 years ago
How far apart are two conducting plates that have an electric field strength of 4.5 × 103V/m between them, if their potential
nalin [4]

Answer:

Explanation:

Given

Electric Field Strength E=4.5\times 10^{3}\ V/m

Potential Difference between Plates is given by V=12.5\ kV

In conducting plates a Potential difference exist between two plate which accelerate the charge when put between the conducting plates

The potential difference is given by

\Delta V=Ed

where E=Electric Field strength

d=distance between Plates

d=\frac{\Delta V}{E}

d=\frac{12.5\times 10^3}{4.5\times 10^{3}}

d=2.78\ m

     

8 0
3 years ago
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