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3 years ago

A 2.5-kg rock is dropped

Physics

2 answers:

Butoxors [25]3 years ago

7 0

Answer:

Okay it dropped now what..........................................

Explanation:

ipn [44]3 years ago

3 0

Is the rock okay I’m concerned

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The sun is approximately 27,000 light years away from the center of our galaxy.

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3 years ago

The chart shows data for a moving object. A 2-column table with 3 rows. Column 1 is labeled time in seconds with entries 2, 4, 6

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The object is not accelerating. I think I got this question right

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2 years ago

If the car speeds up at a steady 1.6 m/s2 , how long after starting is the magnitude of its centripetal acceleration equal to th

Rufina [12.5K]

Based on internet sources, the basic formulas are: v^2/r = (at)^2/r = a ==> at^2 = r ==> t = sqrt(r/a).

Assuming the missing units are mutually compatible, as in the following example, they don't need to be known. 
Acceleration = 1.6 cramwells/s^2 
Radius = 150 cramwells 
t = sqrt(150/1.6) = 9.68 s.

I hope this helps.

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3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

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3 years ago