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3 years ago

Which formula can be used to calculate the horizontal displacement not of a horizontally launched projectile

1 answer:

8 0

<span>If you are looking to get an object up the highest, shoot it straight up. If you want to go for a specific horizontal displacement, use the range equation. R = v2sin(twice the launch angle)/ g. g is the gravitaional constant, 9.8 meters per second. Use degrees for the angle. v is the launch velocity. R is the horizontal displacement. This formula only works if your start altitude and end altitude are the same, i.e. you must shoot over a level field. it depends on the gravitational force of attraction of earth and air resistance. if we are neglecting air resistance, the max.horizontal distance is according to this formulae, V0/2 * sin (2theta) where V0 is the initial velocity theta is the angle with x axis and the projection. There are a number of ways that you could find a horizontally displaced object. You could for example just look.</span>

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1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as

telo118 [61]

Explanation:

The tangential speed of Andrea is given by :

$v=r\omega$

Where

r is radius of the circular path

ω is angular speed

The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'

r' = 2r

New angular speed,

$v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v$

New angular speed is twice that of the Chuck's speed.

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3 years ago

Whats A Controlled Experiment? Whats A Controlled Experiment? Like what the definition is please!

frozen [14]

A controlled experiment is when scientests go throw thests to make an experiment

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3 years ago

A pendulum makes 60 vibrations in 15 secs what’s its frequency

o-na [289]

Answer:

4 hertz

Explanation:

The defination of freqyency = the total no of cycle made by a wave in one second .

so,

cycle or vibrations=60

tame taken = 15

now,

frequency = no. of cycle / time taken

= 60/15

=4 hertz

hence, the its frequency = 4hertz

8 0

3 years ago

If a football player collides with a goal post, what forces are at work?

AlekseyPX

When a footballer collides with the goal post, the forces at work are the action and reaction forces. The player will exert an action force on the goal post, and then a reaction force from the goal post will stop the player. The reaction force call will cause pain and even injury to the player.

7 0

3 years ago

Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m

mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0

3 years ago