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4 years ago

A girl hits a baseball with a bat. When the bat strikes the ball, what is the

Physics

2 answers:

torisob [31]4 years ago

8 0

Answer: Guys, the right answer is D.

Explanation: Got it right on Apex :)

Illusion [34]4 years ago

5 0

Answer:

Its simple! B The ball accelerates upward!♡♡

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Question 1 Gary is on the space shuttle. It takes off and lifts him to a height of 300 km above Earth's surface. a. How has Gary

balandron [24]

A) The mass is an intrinsic property of an object: it means it depends only on the properties of the object, so it does not depend on the location of the object. Therefore, Gary's mass at 300 km above Earth's surface is equal to his mass at the Earth's surface.

b) The weight of an object is given by
$W=mg$
where
m is the mass
$g= \frac{GM}{r^2}$ is the gravitational acceleration at the location of the object, with G being the gravitational constant, M the mass of the planet and r the distance of the object from the center of the planet.

At the Earth's surface, $g=9.81 m/s^2$ , so Gary's weight is
$W=mg=9.81 m$ (1)
where m is Gary's mass.

Then, we must calculate the value of g at 300 km above Earth's surface. the Earth's radius is
$R=6370 km$
So the distance of Gary from the Earth's center is
$r=R+h=6370 km+300 km =6670 km = 6.67 \cdot 10^6 m$

The Earth's mass is $M=5.97 \cdot 10^{24} kg$ , so the gravitational acceleration is
$g'=G \frac{M}{r^2}= (6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )\frac{5.97 \cdot 10^{24} kg}{(6.67 \cdot 10^6 m)^2}=8.95 m/s^2$

Therefore, Gary's weight at 300 km above Earth's surface is
$W' = mg' = 8.95 m$ (2)

If we compare (1) and (2), we find that Gary's weight has changed by
$\frac{W'}{W}= \frac{8.95 m}{9.81 m}=0.91$
So, Gary's weight at 300 km above Earth's surface is 91% of his weight at the surface.

6 0

3 years ago

Does anyone know what the answers are?

baherus [9]

Answer:

Option (C) is the answer

Explanation:

may be it is possible if that we stand so far

4 0

3 years ago

There are four springs stretched by the same mass.

kirill115 [55]

Answer:

That would be Spring C

Explanation:

i took the test :P

6 0

4 years ago

During the time a compact disc (CD) accelerates from rest to a constant rotational speed of 477 rev/min, it rotates through an a

atroni [7]

Answer:

126.01 rad/s^2

Explanation:

since it starts from rest, initial angular speed ω' = 0 rad/s

angular speed N = 477 rev/min

angular speed in rad/s ω = $\frac{2\pi N}{60}$ = $\frac{2*3.142* 477}{60}$ = 49.95 rad/s

angular displacement ∅ = 1.5758 rev

angular displacement in rad/s = $2\pi N$ = 2 x 3.142 x 1.5758 = 9.9 rad

angular acceleration $\alpha$ = ?

using the equation of angular motion

ω^2 = ω'^2 + 2 $\alpha$ ∅

imputing values, we have

$49.95^{2} = 0^{2} + (2 *\alpha*9.9 )$

2495 = 19.8 $\alpha$

$\alpha$ = 2495/19.8 = 126.01 rad/s^2

8 0

3 years ago

the power of a physicians eyes is 57.1 D while examinging a patient. how far from her eyes (in m) is the feature being examined.

yanalaym [24]

Answer:

14 cm

Explanation:

Power of eye = 57.1 D

The relation between the focal length and the power is

f = 1 / P = 1 / 57.1 = 0.0175 m = 1.75 cm

The distance between the image and the lens is, v = 2 cm

Let the distance between the object and the eye is u

Use the lens equation

1/ f = 1 / v - 1 / u

1 / 1.75 = 1 / 2 - 1 / u

1 / u = 1 / 2 - 1 / 1.75

1 / u = (1.75 - 2) / 3.5

u = - 14 cm

Thus, the distance between the feature and eye is 14 cm .

8 0

3 years ago