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svetoff [14.1K]
3 years ago
11

Mr. Fuller assigned his science class a lab comparing the masses of objects to the nearest gram using a balance. The mass of a p

aper clip is about 1 gram. How many paper clips will it take to balance the scale, if it is determined that a 1 centimeter cube of aluminum has the mass of 2.7 grams?
Physics
2 answers:
galben [10]3 years ago
6 0

Answer:

We know that the mass of one papper clip is 1 gram.

now, on the other side of the balance there is a cube centimeter of aluminium, that has a mass of 2.7 grams.

Then, if we want that the mass is equal in both sides, we must have:

X*1g = 2.7g

X=2.7g/1g = 2.7

So we need 2.7 paper clips, rounded to the next whole number, 3 papper clips.

telo118 [61]3 years ago
5 0
2.7 Paper clips, or if rounded, 3 whole paper clips.
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Answer:

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3 years ago
The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic
Drupady [299]

a) Density at 100 degrees: 1.34\cdot 10^4 kg/m^3

Explanation:

The density of mercury at 0 degrees is d=1.36\cdot 10^4 kg/m^3

Let's take 1 kg of mercury. Its volume at 0 degrees is

V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3

The formula to calculate the volumetric expansion of the mercury is:

\Delta V= \alpha V \Delta T

where

\alpha=180\cdot 10^{-6} K^{-1} is the cubic expansivity of mercury

V is the initial volume

\Delta T is the increase in temperature

In this part of the problem, \Delta T=100 C-0 C=100 C=100 K

So, the expansion is

\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3

So, the new density is

d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3


b) Density at 22 degrees: 1.355\cdot 10^4 kg/m^3

We can apply the same formula we used before, the only difference here is that the increase in temperature is

\Delta T=22 C-0 C=22 C=22 K

And the volumetric expansion is

\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3

So, the new density is

d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3


8 0
3 years ago
A 2.0 kg stone is tied to a 0.30 m string and swung around a circle at a constant angular velocity of 12.0 rad/s. The net torque
saul85 [17]

Answer: τ = 0

Explanation:

At constant angular velocity there is no angular acceleration therefore no torque.

τ = Iα

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3 years ago
Under federal and state law, boats must carry certain equipment. to determine the types of equipment your boat must carry, you n
shepuryov [24]
So we want to know what measurement do we need to make in order to determine what equipment can the boat carry. The boat overall length dictates what equipment should a boat have to comply with federal and state law. So the correct answer is the LENGTH measurement. 
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3 years ago
Read 2 more answers
The work function for cesium is 1.96 eV. (a) Find the cutoff wavelength for the metal, (b) what is the maximum kinetic energy fo
romanna [79]

Answer:

Explanation:

λ = hc/¢

Where

h = the Plank constant 6.63 x 10-34 Is

C = 3.0×10^8

¢= 1.96eV

= (6.63×10^-34Js)×(3×10^8)÷( 1.96eV) × 1eV/1.6×10^-19J

= (1.989×10^-25)÷( 1.96eV)×1eV/1.6×10^-19J

= 6.342×10^-7m

B) maximum kinetic energy

= K=hf−ϕ ........1

ϕ = hc

Where

​ h = constant 6.63 x 10^-34Js

ϕ= 1.96eV

Recall

λ =425×10^-9m

f = frequency in Hz

f = c / λ

C = 3.0×10^8

f = 3.0×10^8 / 425×10^-9m

f = 0.000705Hz

From equation 1

K = (6.63 x 10^-34Js×0.000705Hz )- 6.63 x 10^-34Js×3.0×10^8

= 4.68×10^-37 - 1.989×10^-25

= - 1.98×10^-25J

7 0
3 years ago
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