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Genrish500 [490]
3 years ago
13

baseball player hits a line drive estimated to have traveled 145 meters the ball leaves the bat with a horizontal velocity of 40

.0 m/s . how much time was the ball in the air
Physics
1 answer:
Taya2010 [7]3 years ago
8 0

Answer:

3.63 s

Explanation:

Since distance d = vt where v = velocity and t = time taken to cover the distance.

Now, the baseball traveled a distance of d = 145 m with a horizontal velocity of 40.0 m/s.

So, making t subject of the formula,

t = d/v

Substituting the values of d and v we have

t = 145 m/40.0 m/s

= 3.625 s

≅ 3.63 s

So the time the ball was in the air is 3.63 s

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A 7.5 cm object is 14 cm from a concave lens, which has a focal length of –7 cm. Its image is 5 cm in front of the lens. What is
Rama09 [41]
Given conditions:
height of object = 7.5cmdistance of object from mirror  = 14 cmfocus length = -7 cmimage distance =  ?
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Also, magnification of image is:
image height /height of object = - image distance /object distance
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3 years ago
Read 2 more answers
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Answer:

23.3 g/cm3

Explanation:

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3 0
3 years ago
Not only did Skid get shot out of a cannon when he was a clown in the circus, but he was also launched into the air by a vertica
tiny-mole [99]

Answer:

v = 16.11 m / s

Explanation:

For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation

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        Em_f = K = ½ m v²

energy is conserved

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let's calculate

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8 0
3 years ago
A 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.9-T magnetic field. The loop is rotated so that its p
Alchen [17]

Answer:

0.3405V

Explanation:

#Given a magnetic field of 1.9T, diameter= 18.5cm(r=9.25cm or 0.0925m), we find the magnetic flux of the loop as:

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\frac{\phi}{\bigtriangleup t}=\frac{5.107\times 10^-^2}{0.15}\\=3.405\times 10^-^1V

Hence, the induced emf of the loop is 0.3405V

8 0
3 years ago
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