Answer:
1st – Place the film canister on the <u>scale</u>.
2nd – Slide the large <u>weight </u>to the right until the arm drops below the line and then move it back one notch.
3rd – Repeat this process with the <u>top</u> weight. When the arm moves below the line, back it up one groove.
4th – Slide the <u>small </u>weight on the front beam until the <u>lines</u> match up.
5th – Add the amounts on each beam to find the total <u>mass </u>to the nearest tenth of a gram.
Explanation:
The triple beam balance is an instrument that is used in measuring the mass of substances to a very high degree of precision. The reading error is given by ±0.05 grams. The triple beam balance as the name implies has three beams that measure substances of different mass levels.
The beams are categorized as small, medium, and large. There is a balance on which the substance to be weighed is placed directly upon. To use this measuring device, the procedures mentioned above are followed.
Acceleration = (change in speed) / (time for the change)
change in speed = (speed at the end) minus (speed at the beginning)
change in speed = (zero) minus (28 m/s) = -28 m/s
Acceleration = (-28 m/s) / (13 sec)
Acceleration = -2.15 m/s²
I think that’s the correct answer
Answer:
Explanation:
First of all we shall find the velocity at equilibrium point of mass 1.2 kg .
It will be ω A , where ω is angular frequency and A is amplitude .
ω = √ ( k / m )
= √ (170 / 1.2 )
= 11.90 rad /s
amplitude A = .045 m
velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s
= .5355 m /s
At middle point , no force acts so we can apply law of conservation of momentum
m₁ v₁ = ( m₁ + m₂ ) v
1.2 x .5355 = ( 1.2 + .48 ) x v
v = .3825 m /s
= 38.25 cm /s
Let new amplitude be A₁ .
1/2 m v² = 1/2 k A₁²
( 1.2 + .48 ) x v² = 170 x A₁²
( 1.2 + .48 ) x .3825² = 170 x A₁²
A₁ = .0379 m
New amplitude is .0379 m
