Answer:
An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of the jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?
A.
We’ll see the bell move, but we won’t hear it ring.
B.
We won’t see the bell move, but we’ll hear it ring.
C.
We’ll see the bell move and hear it ring.
D.
We won’t see the bell move or hear it ring.
E.
We’ll see the sound waves exit the vacuum pump.
Explanation:
so, the answer to the question is
A.
We'll see the bell move, but we won’t hear it ring.
Answer:
453 gm
Explanation:
<u>Immersed </u>objects are buoyed up by force equal to mass of displaced liquid
400 + 53 = 453 gm in air
Answer:
what
ITS BLANK FOR ME I WISH I COULD HELP YOU
(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
<h3>
Work done in the spring</h3>
The work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
W(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
Learn more about work done here: brainly.com/question/25573309
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