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2 years ago

If a 50 N force was applied to a really massive object as well as a tiny object, what can you say

Physics

1 answer:

frozen [14]2 years ago

6 0

Answer:

The tiny object's acceleration will be much greater.

Explanation:

Because of Newton's second law of motion, (which is F=ma or Force= Mass*acceleration), then if the force maintains the same (which in this case it does, because it says 50 N to both obejcts) and one mass was much greater than the other, then the ould be less for the more massive object and much greater for the lighter object.

For example: If 50 N were applied to a 500 kg object and a 50 kg object, then theformulas for each (respectively) are:

50 = 500*acceleration

and

50 = 50*acceleration

(Because of Newtons Second Law of Motion)

Then, solving for the equations, we get for equation 1:

Acceleration = .1 m/s^2

And for equation 2:

Accleration = 1 m/s^2

Thus, you can see that more massive objects (when applied he same amount of force as the smaller object) clearly have less accleration than the smaller objects.

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krek1111 [17]

Answer:

Explanation:

The heart cells must contract simultaneously to move blood.

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Therefore, the connections among heart cells are characterized by :

having many branches
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The correct option should be <u>A</u>

4 0

2 years ago

Read 2 more answers

A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.