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pishuonlain [190]
3 years ago
6

During a particular time interval, the displacement of an object is equal to zero. Must the distance traveled by this object als

o equal to zero during this time interval? Group of answer choices
Physics
1 answer:
Gnom [1K]3 years ago
3 0

Answer: No, we can have a displacement equal to 0 while the distance traveled is different than zero.

Explanation:

Ok, let's write the definitions:

Displacement: The displacement is equal to the difference between the final position and the initial position.

Distance traveled: Total distance that you moved.

So, for example, if at t = 0s, you are in your house, then you go to the store, and then you return to your house, we have:

The displacement is equal to zero, because the initial position is your house and the final position is also your house, so the displacement is zero.

But the distance traveled is not zero, because you went from you traveled the distance from your house to the store two times.

So no, we can have a displacement equal to zero, but a distance traveled different than zero.

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Ondrea could drive a Jetson's flying car at a constant speed of 540.0 km/hr across oceans and space, approximately how long woul
Tcecarenko [31]

Answer:

The time taken in years is   x = 125 \  years

Explanation:

From the question we are told that

   The  speed is  v  = 540.00 \ km /hr  =  \frac{540 *1000}{3600} =  150\  m/s

    The distance from the sun to Pluto is  d =  5.9*10^{9} \  k m =  5.9*10^9 * 1000 =  5.9*10^{12} \  m

Generally the time  taken is mathematically represented as

     t =  \frac{d}{v}

=>   t = \frac{5.9*10^{11}}{150}

=>   t =  3.933*10^{9}

Converting to years

   1 year  \to  3.154*10^7 \  s

    x \  years  \to 3.933*10^{9}

=>  x = \frac{ 3.933*10^{9} * 1 }{ 3.154 *10^7}

=>    x = 125 \  years

7 0
3 years ago
A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from
ivann1987 [24]

Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f} (1)

Where:

I_{T} - Moment of inertia of the turntable, in kilogram-square meters.

r - Distance of the block regarding the center of the turntable, in meters.

m - Mass of the object, in kilograms.

\omega_{o} - Initial angular speed of the turntable, in radians per second.

\omega_{f} - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

I_{T} = \frac{1}{2}\cdot M\cdot r^{2} (2)

Where M is the mass of the turntable, in kilograms.

If we know that \omega_{o} \approx 7.330\,\frac{rad}{s}, M = 1.5\,kg, m = 0.54\,kg and r = 0.1\,m, then the angular speed of the turntable after the event is:

I_{T} = \frac{1}{2}\cdot M\cdot r^{2}

I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}

I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}

\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}

\omega_{T} = 3.004\,\frac{rad}{s} (28.687\,\frac{rev}{min})

The turntable's angular speed after the event is 28.687 revolutions per minute.

3 0
2 years ago
A 0.420 kg block of wood rests on a horizontal frictionless surface and is attached to a spring (also horizontal) with a 20.5 N/
Aneli [31]

Answer:

leon

Explanation:

leom ofjfjbfbfdnns

3 0
3 years ago
The heart working with the blood vessels to pump blood is which body system?
LiRa [457]
I think it’s the cardiovascular system
6 0
3 years ago
A particle is moving with velocity v(t) = t2 _ 9t + 18 with distance, s measured in meters, left or right of zero, and t measure
Alex787 [66]
V = t^2 - 9t + 18

position, s
s = t^3 /3 - 4.5t^2 +18t + C

       t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1

Average velocity: distance / time

   distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
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t = 5 s

     v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
     speed = |-2| m/s = 2 m/s
 
Moving right
     V > 0 => t^2 - 9t + 18 > 0
     (t - 6)(t - 3) > 0

     => t > 6 and t > 3 => t > 6 s => Interval (6,8)

    => t < 6 and t <3 => t <3 s => interval (0,3)

    

Going faster and slowing dowm

acceleration, a = v' = 2t - 9
     a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
     Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
     


4 0
3 years ago
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