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Gre4nikov [31]
3 years ago
5

1. A record with a radius of 0.3m spins in a clockwise circle with a centripetal

Physics
1 answer:
Hitman42 [59]3 years ago
6 0

Solve for the linear/tangential speed:

<em>a</em> = <em>v</em>²/<em>r</em>

where <em>a</em> = centripetal acceleration, <em>v</em> = speed, and <em>r</em> = radius.

4.7 m/s² = <em>v</em>²/(0.3 m)

<em>v</em>² = (0.3 m) (4.7 m/s²)

<em>v</em> ≈ 3.96 m/s

For every time the record completes one revolution, a fixed point on the edge of the record travels a distance equal to its circumference, which is 2<em>π</em> (0.3 m) ≈ 1.88 m. So if 1 rev ≈ 1.88 m, then the angular speed of the record is

(3.96 m/s) (1/1.88 rev/m) ≈ 7.46 rev/s

Take the reciprocal of this to get the period:

1 / (7.46 rev/s) ≈ 0.134 s/rev

So it takes the record about 0.134 seconds to complete one revolution.

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2 years ago
How can you use distance and displacement to describe an object's motion​
andrew11 [14]

Answer:

distance can describe the total distance moved and displacement shows how far something has moved from its starting position (in a straight line from point a to point b) the object doesn't have to move in a straight line, but that is how displacement is measured

8 0
3 years ago
If we connect a third bulb in our series circuit, say one with 4
Alexus [3.1K]

Answer:

The current will decrease.

Explanation:

When another bulb is added, the resistance is going to increase. Keep in mind that the current is inversely proportional to the resistance (<em>Ohm's law: R= </em><em>V</em><em>/</em><em>I</em><em> </em><em>).</em> Therefore when the resistance increase, the current running in the circuit will decrease.

6 0
2 years ago
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

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3 years ago
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Vesnalui [34]

Answer:

<em>The comoving distance and the proper distance scale</em>

<em></em>

Explanation:

The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster.  Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.

4 0
3 years ago
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