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4 years ago

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A rocket launching is an example of newtons _____ law

1 answer:

Karolina [17]4 years ago

8 0

Its an example of newton's 2nd law

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A negative slope on the velocity vs. time graph indicates that the object is not accelerating

liubo4ka [24]

AnswerNo

However, the acceleration is the opposite of the movement:

Explanation:

8 0

3 years ago

A ball rolls up a ramp with a velocity of 8.0 m/s. How high up the ramp does it travel?

Flura [38]

The greatest height the ball will attain is 3.27 m

<h3>Data obtained from the question</h3>

Initial velocity (u) = 8 m/s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =?

The maximum height to which the ball can attain can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = 8² – (2 × 9.8 × h)

0 = 64 – 19.6h

Collect like terms

0 – 64 = –19.6h

–64 = –19.6h

Divide both side by –19.6

h = –64 / –19.6h

h = 3.27 m

Thus, the greatest height the ball can attain is 3.27 m

Learn more about motion under gravity:

brainly.com/question/13914606

5 0

3 years ago

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it

KatRina [158]

Answer:

$t = 1.82$

Explanation:

Given

$u = 7.70m/s$ -- initial velocity

$s = 30.2m$ --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

$s = ut + \frac{1}{2}gt^2$

Where

$g = 9,8m/s^2$

So, we have:

$30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2$

$30.2 = 7.70t + 4.9 * t^2$

Subtract 30.2 from both sides

$30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9t^2 - 30.2$

$7.70t + 4.9t^2 - 30.2 = 0$

$4.9t^2 + 7.70t - 30.2 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 4.9;\ b = 7.70;\ c = -30.2$

$t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}$

$t = \frac{-7.70\±\sqrt{651.21}}{9.8}$

$t = \frac{-7.70\±25.52}{9.8}$

Split the expression

$t = \frac{-7.70+25.52}{9.8}$ or $t = \frac{-7.70-25.52}{9.8}$

$t = \frac{17.82}{9.8}$ or $t = -\frac{33.22}{9.8}$

Time can't be negative; So, we have:

$t = \frac{17.82}{9.8}$

$t = 1.82$

Hence, the time to hit the ground is 1.82 seconds

7 0

3 years ago

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit

Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

$\Delta KE = \Delta PE$

$\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1$

Where,

m = mass

$v_{f,i}$ = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

$\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2$

$\frac{1}{2} (v_f^2-v_0^2) = gh_2$

$(v_f^2-v_0^2) = 2gh_2$

$v_f = \sqrt{2gh_2+v_0^2}$

$v_f = \sqrt{2(9.8)(23.5)+13.6^2}$

$v_f = 25.4m/s$

7 0

3 years ago

PLEASE HELP ME I FONT HAVE THAT MUCH TIME LEFT!!! THE ALL CAPS ARE TO CATCH YOUR ATTENTION SO NOW THAT I’VE GOT IT, PLEASE HELP

podryga [215]

Answer:

hmm if it were up to me i would say gravity potential energy and sorry I don't really have a third one hope this helps though.

8 0

3 years ago

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