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3 years ago

An object is dropped from 42m tall building. How long does It take to reach the ground?

Physics

1 answer:

yKpoI14uk [10]3 years ago

4 0

Answer:

It takes 2.93 s for the object to reach ground from the height of 42 m.

Explanation:

Given:

Displacement of the object is, $d=-42\ m$

As it is dropped, initial velocity of the object is, $u=0$

Acceleration of the object is equal to acceleration due to gravity and is equal to, $a=g=-9.8\ m/s^2$

Now, using Newton's equation of motion and plugging in the values, we get:

$d=ut+\frac{1}{2}at^2\\-42=0+\frac{1}{2}(-9.8)t^2\\-42=-4.9t^2\\t^2=\frac{-42}{-4.9}\\t^2=8.571\\t=\sqrt{8.571}=2.93\ s$

So, it takes 2.93 s for the object to reach ground from the height of 42 m.

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What happens when a star exhausts its core hydrogen supply?

Allushta [10]

Answer:

It becomes a giant or supergiant.

Explanation:

Once all the hydrogen supply is gone, fusion of hydrogen into helium stops. The core starts to contract and liberates energy, which heats the superior layer until it becomes hot enough to start the fusion of hydrogen into helium.

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2 years ago

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True [87]

Answer:

The last one " They are equal and act in opposite directions"

Explanation:

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2 years ago

Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statemen

muminat

Answer:

Option A

Explanation:

From the question we are told that:

Mass $m=0.20kg$

Velocity $v=4m/s$

Generally the equation for momentum for Ball A is mathematically given by

Initial Momentum

$M_{a1}=mV$

$M_{a1}=0.2*4$

$M_{a1}=0.8$

Final Momentum

$M_{a2}=-0.8kgm/s$

Therefore

$\triangle M_a=-1.6kgm/s$

Generally the equation for momentum for Ball B is mathematically given by

Initial Momentum

$M_{b1}=mV$

$M_{b1}=0.2*4$

$M_{b1}=0.8$

Final Momentum

$M_{b2}=-0 kgm/s$

Therefore

$|\triangle M_a|>|\triangle Mb|$

Option A

4 0

2 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the x-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its y-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the x- and y-components of the copter's position vector after t = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

2 years ago

Which is one way scientists indicate how precise and accurate there experimental measurements are

kherson [118]

They do the method 3 times to be sure. Because if you do it once, that could mean anything. If you do it twice, it may or may not have the same result. If you do it 3 times and it matches one of the previous answers, then it's likely that it's correct.

8 0

2 years ago