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larisa86 [58]
3 years ago
13

How far would a jet going 155 m/s travel in 9 s?

Physics
1 answer:
docker41 [41]3 years ago
5 0

Answer:

1,395 m

Explanation:

155×9

multiply m/s by 9s

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If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th
mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

x=ut

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

y=\frac{1}{2} gt^2

Substitute 9.8 m/s² for g and 0.461 s for t.

y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by,(2.0 m)-(1.04 m)=0.96 m

The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>

8 0
3 years ago
What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a
Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

M=-\frac{f_o}{f_e}

Here, M is the magnification of an astronomical telescope, f_e is the focal length of the eyepiece lens and f_o is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put f_o=74 cm and f_e=2.4 cm in the above expression.

M=-\frac{74}{2.4}

M=-30.83

Therefore, the magnification of an astronomical telescope is -30.83.

5 0
3 years ago
URGENT!!!! WILL GIVE BRAINLIEST!!!!!! 15 POINTS
Artyom0805 [142]
It's the angle made by the incident ray when it's perpendicular to the surface. (Perpendicular lines are the lines that form a graph or like a 90-degree angle)
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3 years ago
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Which memory disorder is a common illness on television and films
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amnesia is the most common illness used in tv an films
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3 years ago
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4. A metal fork has a density of 7.8 g/cm ³. Is this fork made up of aluminum?
Morgarella [4.7K]
No, aluminum has a density near 2.7 g/cm^3 
<span>7.8 g/cm^3 is near the density of iron (or in the case of a fork, steel).
this is it

</span>
5 0
3 years ago
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