If for every action force an equal and opposite reaction force exists, how can anything ever be accelerated?

1 answer:

8 0

If there are two equal and opposite forces on the SAME THING, then the thing doesn't accelerate. You're right about that. But the action and reaction forces act on two different things. The bullet and the rifle. The ball and the bat. The airplane and the air. etc. So BOTH can accelerate.

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A rollerblader is blading along the sidewalk. Which forms of measurement would be the best to use to determine the rollerblader'

Greeley [361]

You would use distance an time formula to mathmaticly solve

3 0

2 years ago

A baseball is thrown horizontally at a rate of 40m/s toward a home plate 18.4 m away. How far below the launch height is the bal

sammy [17]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g where

R is the distance moves in horizontal direction = 18.4m

H is the height

U is the velocity of the baseball = 40m/s

g is the acceleration due to gravity = 9.8m/s²

Substitute the given parameters into the formula and calculate H as shown;

18.4 = 40√2H/9.8

18.4/40 = √2H/9.8

0.46 = √2H/9.8

square both sides;

(0.46)² = (√2H/9.8)²

0.2116 = 2H/9.8

2H = 9.8*0.2116

2H = 2.07368

H = 2.07368/2

H = 1.03684m

Hence the ball is 1.03684m below the launch height when it reached home plate.

8 0

2 years ago

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C = 50 - x cm

We know that ,electric field is given as

$E=K\dfrac{q}{r^2}$

$K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\$