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Crank
3 years ago
11

Find the velocity of a baseball thrown 78 m from third base to first base in 30 sec​

Physics
1 answer:
ankoles [38]3 years ago
3 0

Answer:

V=2.6m/s

Explanation:

velocity=Distance/time

V=78/30=2.6m/s

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How are newtons third law and his law of universal gravitation are connected
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Explanation:

Gravitational force keeps it attracted to the Earth, and centripetal force keeps it moving in a circle. ... Explain how Newton's third law and his law of universal gravitation are connected. His third law states that every force has an equal opposite force attracted to it, and that force is caused by gravity.

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25.16 what is the reaction product of acetic acid and ethylamine at room temperature?
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The ammonium salt of acetic acid is the reaction product of acetic acid and ethylamine at room temperature

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Acetic acid is a monofunctional carboxylic acid containing two carbon atoms. It acts as a protein solvent,  food acidity regulator, antibacterial food preservative. It is a conjugate acid of an acetate.

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learn more about acetic acid, visit;

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7 0
1 year ago
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis
Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

Y= VoT+0.5gt^{2}

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s

c)

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t    

Vf=0+(9.81 )(3.5)=34.335m/S          

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S

7 0
3 years ago
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