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Basile [38]
4 years ago
11

What effect does air resistance have on the projectile

Physics
2 answers:
Vesnalui [34]4 years ago
7 0

more deceleration.

in vertical motion downwards => terminal velocity ... raindrops etc

tino4ka555 [31]4 years ago
7 0

Answer:

Decreases its maximum acceleration

Explanation:

Air resistance which is also known as drag force occurs with an object in air while in motion, it encounters air resistance which reduces its speed and reduces its maximum height. Example of bodies in which air resistance affects are rockets, bullets, a ball in air etc

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One Newton is expressed in ?
ra1l [238]
The newton is the SI unit for force; it is equal to the amount of net force required to accelerate a mass of one kilogram at a rate of one meter per second squared.
3 0
3 years ago
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Un resorte se alarga 5 cm bajo la acción de una fuerza de 39,2 N. ¿Cuál es la constante del resorte? Si ahora la fuerza es 68,6
Lorico [155]

Answer:

k=784 N/m

\Delta x=8,8 cm

Explanation:

Usando la ley de Hook tenemos:

F=k\Delta x

Solving it for k we have:

k=\frac{F}{\Delta x}

k=\frac{39,2}{0,05}

k=784 N/m

Usando la misma ecuación y sabiendo k tenemos:

\Delta x=\frac{F}{k}

\Delta x=\frac{68,6}{784}

\Delta x=8,8 cm

Espero esto te ayude!

6 0
3 years ago
According to einstein's theory of simple relativity (_E + mc(2)_). BLANK is converted into BLANK.
True [87]

Answer:

energy is converted into mass

Explanation:

8 0
3 years ago
A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.
DanielleElmas [232]

Answer:

39.7 m

Explanation:

First, we conside only the last second of fall of the body. We can apply the following suvat equation:

s=ut+\frac{1}{2}at^2

where, taking downward as positive direction:

s = 23 m is the displacement of the body

t = 1 s is the time interval considered

a=g=9.8 m/s^2 is the acceleration

u is the velocity of the body at the beginning of that second

Solving for u, we find:

ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s

Now we can call this velocity that we found v,

v = 18 m/s

And we can now consider the first part of the fall, where we can apply the following suvat equation:

v^2-u^2 = 2as'

where

v = 18 m/s

u = 0 (the body falls from rest)

s' is the displacement of the body before the last second

Solving for s',

s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m

Therefore, the total heigth of the building is the sum of s and s':

h = s + s' = 23 m + 16.7 m = 39.7 m

7 0
3 years ago
A 18.0-kg rock is sliding on a rough, horizontal surface at 7.10 m/s and eventually stops due to friction. the coefficient of ki
Bond [772]
A = .3*g = 2.94 m/s² 

<span>t = v/a = 9/2.94 = 3.061 sec </span>

<span>W = E/t = ½mv²/t = ½*40*9²/3.061 = 529.2 watts</span>
4 0
3 years ago
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