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Alexxandr [17]
3 years ago
7

Was the object accelerating during the time interval labeled “B”? A) Yes B) No

Physics
2 answers:
Oxana [17]3 years ago
7 0
Yes it was accelerating
Alla [95]3 years ago
5 0

There's not enough information in the graph to determine whether it was or it wasn't.  It could have been, but we can't tell.

Before I explain this answer, it's VERY important to know exactly what "accelerating" means:

-- "Accelerating" does NOT always mean "speeding up".

-- "Accelerating" means ANY change in velocity.

-- That can be ==> <u>speeding up</u>, or ==> <u>slowing down</u>, or ==> <u>turning</u>.

This graph is labeled "Velocity vs Time", but it isn't really. "Velocity" means <u>speed</u> AND <u>direction</u>. The graph only shows us the object's speed, but nothing about its direction.

During time-interval-B, the speed isn't changing, but the object could still be changing <u>direction</u>, like turning a corner, or on a curved path. We don't know. The graph doesn't show us that information.

The object could be accelerating during time-interval-B. It's true that the speed wasn't changing. But maybe the direction was. If the direction was changing, then we have to say the object was 'accelerating' during the interval labeled B .  

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An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag
Afina-wow [57]

Answer:

The induced emf in the loop is 7.35\times 10^{-4}\ V

Explanation:

Given that,

Length of the wire, L = 1.22 m

It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L

4a = 1.22

a = 0.305 m

Area of square, A=a^2=(0.305)^2=0.0930\ m^2

Circumference of the loop,

C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m

Area of circle,

A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2

The induced emf is given by :

\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V

So, the induced emf in the loop is 7.35\times 10^{-4}\ V

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