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3 years ago

In projectile motion, why don't we add acceleration due to gravity on the formula of horizontal motion

Physics

2 answers:

Basile [38]3 years ago

8 0

Answer:

Horizontal motion is free from gravity which is zero

Vika [28.1K]3 years ago

7 0

Answer:

• For horizontal motion, the projectile is not under a free fall. So, it is not experienced by gravitational force hence acceleration due to gravity is negligible.

${ \tt{s _{x} = u _{x} t}} \\$

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an athlete sprints from 150 m south of the finish line to 65 m south of the finish line in 5.0s what is his average velocity

Ilya [14]

Total distance=150m-65m=85m
Time=5s

$\\ \rm\longrightarrow Avg\:Velocity=\dfrac{Total\:Distance}{Total\:Time}$

$\\ \rm\longrightarrow Avg\:Velocity=\dfrac{85}{5}$

$\\ \rm\longrightarrow Avg\:Velocity= 17m/s$

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3 years ago

Read 2 more answers

There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

$F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})$

$F_{net}=0.208 N$

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3 years ago

A box rests on the floor. If you pull sideways on the box with a 300 N force, but don't move it, what is the magnitude of the fr

Dahasolnce [82]

The magnitude of the frictional force is B) 300 N

Explanation:

For this problem, we just have to consider the forces acting on the box in the horizontal direction.

There are only two forces in this direction:

- The pull applied by the man, of magnitude F=300 N, forward

- The frictional force, $F_f$ , acting backward

So the equation of motion for the box is

$F-F_f = ma$

where

m is the mass of the box

a is its acceleration

However, the box is at rest, so its acceleration is zero:

a = 0

This means therefore that

$F-F_f=0$

And so the magnitude of the frictional force is equal to the pulling force:

$F_f = F = 300 N$

Learn more about friction:

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3 years ago

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Darya [45]

for example

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2.picking up tomatoes for salads which contains oxalic acid

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2 years ago

Describe in your own words, What is insulation?

serg [7]

Insulators are the materials that are bad conductors of heat/electricity.insulation is tge pricess of covering any thing with such materials in order to prevent heat or electricity.

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3 years ago

In projectile motion, why don't we add acceleration due to gravity on the formula of horizontal motion​

In projectile motion, why don't we add acceleration due to gravity on the formula of horizontal motion