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3 years ago

5. A sled with no initial velocity accelerates at a rate of 5.0 m/s2 down a hill. How

Physics

1 answer:

ad-work [718]3 years ago

6 0

Answer:

60 m/s

Explanation:

$a = \frac{vf - vi}{t}$

$5.0 = \frac{vf - 0}{12}$

$5.0 \times 12 = vf$

$60 = vf$

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What factors affect the gravitational pull of two massive objects?

Mrrafil [7]

Answer:Mass and distance

Explanation:The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.

Hope this helps!!

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3 years ago

Resonance occurs when an object vibrating at or near the resonant frequency of a second object to vibrate. What form of waves ar

Ivanshal [37]

Answer:

Resonance depends on objects, this may happen for example when you play guitar in a given room, you may find that for some notes the walls or some object vibrate more than for others. This is because those notes are near the frequency of resonance of the walls.

So waves involved are waves that can move or affect objects (in this case the pressure waves of the sound, and the waves that are moving the wall).

this means that the waves are mechanic waves.

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3 years ago

What causes an impact crater to form?

Scrat [10]

Answer:

it is the impact of a meteorite, volcanic activity, or an explosion.

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3 years ago

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

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3 years ago

Imagine two billiard balls on a pool table. Ball A has a mass of 2 kilograms and ball B has a mass of 3 kilograms. The initial v

ratelena [41]

1. The balls move to the opposite direction but the same speed. This represents Newton's third law of motion.
2. The total momentum before and after the collision stays constant or is conserved.
3. If the masses were the same, the velocities of both balls after the collision would exchange.
4 and 5. Use momentum balance to solve for the final velocities.

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3 years ago

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