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3 years ago

An object is dropped from a height of 25 meters. At what velocity will it hit the ground? A. 7.0 meters/second B. 11 meters/seco

nd C. 22 meters/second D. 49 meters/second E. 70 meters/second

Physics

1 answer:

Basile [38]3 years ago

4 0

$Final^{2}=Initial^{2}+2ad \\ x^{2}=0^{2}+2(9.8)(25) \\ x^{2}=490 \\ x=22.13 \\ C$

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Otrada [13]

It would be 1. B 2. A 3. A

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3 years ago

An object is initially moving at an unknown velocity. It accelerates at a rate of 1.5m/s2 to a new velocity of 60 m/s in 25 s. W

Alexandra [31]

<h3>Answer: 22.5 m/s</h3>

=====================================================

Work Shown:

acceleration = ( finalVelocity - initialVelocity )/(change in time)

1.5 = (60 - x)/(25)

1.5*25 = 60-x

37.5 = 60-x

x = 60-37.5

x = 22.5

The initial velocity is 22.5 m/s

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3 years ago

Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st

scoray [572]

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

$x(t) = v_0 cos \theta t$

where

$v_0$ is the initial speed

t is the time

$\theta=32.0^{\circ}$ is the angle below the horizontal

We can rewrite this equation as

$t=\frac{x(t)}{v_0 cos \theta}$ (1)

The vertical displacement instead is given by

$y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2$ (2)

where

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting (1) into (2),

$y(t) = -x(t) tan \theta - \frac{1}{2}gt^2$

We know that for t = time of flight, the horizontal displacement is

$x(t) =50.8 m$

We also know that the vertical displacement is

$y(t) = -45 m$

Substituting everything into the equation, we can find the time of flight:

$\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s$

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

$x(t) = v_0 cos \theta t$

where we have

x = 50.8 m

$\theta=32.0^{\circ}$

Substituting the time of flight,

t = 1.64 s

We find:

$v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s$

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

$v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s$

The initial vertical velocity is instead

$u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s$

And it changes according to the equation

$v_y = u_y -gt$

So at t = 1.64 s (when the ball hits the ground),

$v_y = -19.3 - (9.8)(1.64)=-35.4 m/s$

So the impact speed is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s$

While the direction is:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}$

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3 years ago

Sea Food is a rich source of ______. *

k0ka [10]

Answer:

Sea Food is a rich source of ______. *

Explanation:

Seafood is a rich source of minerals, such as iron, zinc, iodine, magnesium, and potassium.

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2 years ago

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Electrically charged sunspot gases which escape the sun's chromosphere and enter the earth's atmosphere near the magnetic north

Kisachek [45]

Answer:

Northern Lights ( Aurora Borealis)

Explanation:

When the electricaly charged sunspot gases (they are named a solar wind) escape the sun's chromosphere and penetrates from the earth magnetic sheild which is called earth's magnetosphere then upon there interaction with atoms and molecules of our atmosphere there are little bursts of photons in the form of light which made up these northern lights.

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