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3 years ago

A ball is kicked from level ground at an angle 60° with initial velocity 10 m/s. The distance the ball travels, in meters, is:

Physics

1 answer:

RUDIKE [14]3 years ago

4 0

Answer:

the ball travels a distance of 8.84 m

Explanation:

Range: Range is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again.

R = (U²sin2∅)/g.............................. Equation 1

Where R = range, U = initial velocity, ∅ = angle of projection, g = acceleration due to gravity.

Given: U = 10 m/s, ∅ = 60°

Constant: g = 9.8 m/s²

Substituting these values into equation 1

R = [10²×sin(2×60)]/9.8

R = (100sin120)/9.8

R = 100×0.8660/9.8

R = 86.60/9.8

R = 8.84 m

Therefore the ball travels a distance of 8.84 m

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A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener

trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

$U=mgh$

where

m is its mass

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

$U=(m)(9.8)(15)=147 m$ (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

$U=K=\frac{1}{2}mv^2$

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

$147 m = \frac{1}{2}mv^2$

And re-arranging for v, we find the velocity:

$v=\sqrt{2\cdot 147}=17.1 m/s$

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3 years ago

An air filter can remove dust particles from air but will reach capacity (saturation) at 50.0 mg. Of air containing 225 µg dust

zubka84 [21]

Answer:

Time to Reach Saturation = 0.0146 day

Explanation:

In order to solve this problem, we first need to calculate the dust filtered by the filter per cubic meter of air:

Filtered Dust per m³ = Dust Particles Entering per m³ - Dust Particles Leaving per m³

Filtered Dust per m³ = 225 μg/m³ - 15 μg/m³

Filtered Dust per m³ = 210 μg/m³ = 210 x 10⁻³ mg/m³

Now, we find volume flow rate of air through filter:

Volume Flow Rate of Air = (400 ft³/min)(0.3048 m/1 ft)³

Volume Flow Rate of Air = 11.33 m³/min

Now, we calculate rate of dust filtered:

Rate of Dust Filtered = (Filtered Dust per m³)(Volume Flow Rate of Dust)

Rate of Dust Filtered = (210 x 10⁻³ mg/m³)(11.33 m³/min)

Rate of Dust Filtered = 2.38 mg/min

Now, for the time required to reach saturation:

Time to Reach Saturation = (Saturation Capacity)/(Rate of Dust Filtered)

Time to Reach Saturation = (50 mg)/(2.38 mg/min)

Time to Reach Saturation = (21.02 min)(1 day/24 h)(1 h/60 min)

Time to Reach Saturation = 0.0146 day

3 0

3 years ago