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OleMash [197]
3 years ago
5

If a substance has a density of 2.7g/cm3 and a mass of 86.4g, what is its volume?

Physics
1 answer:
Gennadij [26K]3 years ago
7 0

Answer:

32cm³

Explanation:

Given parameters:

  Density of substance  = 2.7g/cm³

   Mass of substance  = 86.4g

Unknown:

Volume of substance  = ?

Solution:

Density is the mass per unit volume of a substance.

    Density  = \frac{mass}{volume}

Since the unknown is volume we solve for it;

   mass  = density x volume

   86.4 = 2.7 x volume

    volume  = \frac{86.4}{2.7}   = 32cm³

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Which measuring tool should be used to measure the diameter of a soda can?
polet [3.4K]

Answer:

A ruler in centimeters and/or inches.

Explanation:

This is the most convenient option to measure a soda can.

4 0
3 years ago
Older railroad tracks in the U.S. are made of 12 m-long pieces of steel. When the tracks are laid, gaps are left between the sec
Shkiper50 [21]

Answer:

0.005 m

Explanation:

length of steel (L°) = 12 m

initial temperature (T) = 16 degrees

expected temperature (T') = 50 degrees

We can find how large the gaps should be if the track is not to buckle when the temperature is as high as 50 degrees from the formula below

ΔL = ∝L°ΔT where

  • ΔL = expansion / gap
  • ∝ = linear expansion coefficient of steel = 12x10^{-6} C^{-1}
  • L° = initial length
  • ΔT = change in temperature

ΔL = 12x10^{-6} x 12 x (50-16) = 0.005 m

3 0
4 years ago
A stereo speaker produces a pure \"c\" tone, with a frequency of 262 hz. what is the period of the sound wave produced by the sp
jasenka [17]

Answer:

3.82 ms

Explanation:

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

where f is the frequency.

In this problem, f = 262 Hz, so the period if this sound wave is

T=\frac{1}{262 Hz}=3.82\cdot 10^{-3} s=3.82ms

4 0
3 years ago
A person pushes a 25kg box across a horizontal frictionless surface with a force of 200 Newtons.
inna [77]
Second law of physics. F = MA
200 = 25A
I believe you are capable of doing the rest
6 0
3 years ago
A car leaves an intersection traveling west. Its position 4 sec later is 21 ft from the intersection. At the same time, another
Alex_Xolod [135]

Answer:

15.8 ft/s

Explanation:

\frac{da}{dt} = Velocity of car A = 9 ft/s

a = Distance car A travels = 21 ft

\frac{db}{dt} = Velocity of car B = 13 ft/s

b = Distance car B travels = ft

c = Distance between A and B after 4 seconds = √(a²+b²) = √(21²+28²) = √1225 ft

From Pythagoras theorem

a²+b² = c²

Now, differentiating with respect to time

2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\\\Rightarrow \frac{dc}{dt}=\frac{21\times 9+28\times 13}{\sqrt{1225}}\\\Rightarrow \frac{dc}{dt}=15.8\ ft/s

∴ Rate at which distance between the cars is increasing three hours later is 15.8 ft/s

6 0
4 years ago
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