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MakcuM [25]
3 years ago
15

What is the value of acceleration that the car experiences

Physics
1 answer:
Anni [7]3 years ago
7 0
Can u show the problem ?
Acceleration= the rate and change of an object also cars go fast so I think it would be positive acceleration
Hope I helped
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A particle is confined to a one-dimensional box that is 50 pm long. What is the smallest possible uncertainty in momentum for th
nevsk [136]

Answer:

The smallest possibility is 0.01E-22kgm/s

Explanation:

Using

Momentum= h/4πx

= 6.6x 10^-34Js/ 4(3.142* 50*10-12m)

= 0.01*10^-22kgm/s

6 0
3 years ago
Marys airplane trip took 5.8 hours for one-half of that time, the airplane flew at a constant speed of 640 miles per hour and fo
Olenka [21]
Distance is speed x time.  Half of the trip is 5.8/2 = 2.9hrs.
640 x 2.9 = 1856mi
580 x 2.9 = 1682mi
1856mi+1682mi=3538mi.

You could also calculate her average speed.  This is easy since it was divided in two equal time slices.  Average Speed = (640+580)/2 = 610mi/hr
Now 610mi/hr x 5.8hrs = 3538mi
7 0
3 years ago
How did you manage to overcome the difficulties that arose?
Flauer [41]

Answer:

I just toughed it out and talked with friends

Explanation:

4 0
2 years ago
Two hockey pucks with mass 0.1 kg slide across the ice and collide. Before
RideAnS [48]

The velocity of pluck 1 is 12 m/s west.

<h3>What is the conservation of momentum?</h3>

The principle of the conservation of the linear momentum states that momentum before collision is equal to momentum after collision.

Now given that;

m1u1 + m2u2 = m1v1 + m2v2

(0.1 * 15) - (0.1 * 12) = 0.1* v + (0.1 * 15)

1.5 - 1.2 = 0.1v + 1.5

0.3 - 1.5 = 0.1v

v = -1.2/0.1

v = - 12 m/s

Hence, the velocity of pluck 1 is 12 m/s west.

Learn  more about linear momentum:brainly.com/question/27988315

#SPJ1

7 0
1 year ago
A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853
Norma-Jean [14]

Recall that

{v_f}^2-{v_i}^2=2a\Delta x

where v_i and v_f are the initial and final velocities, respecitvely; a is the acceleration; and \Delta x is the change in position.

So we have

\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)

\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

7 0
2 years ago
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