<span>6.67 metros por segundo
~ Haga 800/120 que equivale a 6.67 porque hay 60 segundos en un minuto y hay dos minutos, entonces 60 veces 2 es igual a 180, luego configure su problema
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Espero que esto te ayude:)
Answer:
It will take 15.55s for the police car to pass the SUV
Explanation:
We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:
1. 
2. 
Since both cars will travel the same distance x, we can equal both formulas and solve for t:

We simplify the fraction present and rearrange for our formula so that it equals 0:

In the very last step we factored a common factor t. There is two possible solutions to the equation at
and:

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at
(when the SUV passed the police car) and
(when the police car catches up to the SUV)
Answer:
As much I know the gravity on moon is 1.62m/s२.
Answer:
Height reached will be 28.35 m
Explanation:
Here we can use the work energy theorem to find the maximum height
As we know by work energy theorem
Work done by gravity + work done by friction = change in kinetic energy

now we will have



so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction
<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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