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Anna [14]
3 years ago
6

The trajectory of a projectile always ________________. The trajectory of a projectile always ________________. is a straight li

ne, in the same direction as the initial velocity vector follows a circular path curves downward, below the initial velocity vector at first goes up above the initial velocity vector and then curves down
Physics
2 answers:
tiny-mole [99]3 years ago
8 0

Answer:

<h2>Curves downward. </h2><h2>At first goes up above the initial velocity and then curves down. </h2>

Explanation:

The trajectoy of a projectile is always a parabola concave down, beacuse it's movement is modeled by

y=v_{0} t-\frac{1}{2}gt^{2} (vertical motiion)

So, we know that quadratic equations are represented by parabolas.

So, the first answer is curves downward.

A simpler reason why it curves down is gravity, this acceleration makes the object to change is direction and velocity.

An important fact is that the inital and final point are at the same level, which means the initial velocity vector and the final velocity vector are equal.

However, notice that a projectile trajectory first moves up the initial vector and then moves down.

So, the second answer is at first goes up above the initial velocity and then curves down. However, it's important to notice that we are referring to the trajectory here, because higher positions of the projectile means a least velocity vector, because gravity is reducing the velocity.

jenyasd209 [6]3 years ago
6 0

Answer:

curves downward, below the initial velocity vector.

Explanation:

Projectile launches are generally divided into two types: the oblique throw and the free fall. The free fall of bodies consists of throwing or abandoning projectiles from a hill or any unevenness that has a height in relation to a frame (usually the ground), while the oblique launch consists of launching a projectile at an angle. any relative to a frame (usually the ground).

Regardless of the type, when reading the paragraph above, we can say that the trajectory of a projectile will always be curved down and below the initial velocity vector.

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A 60.0-kg person is in an elevator with a mass of 2000 kg. The elevator moves vertically up through a distance of 24.0 m with a
kati45 [8]

The mass of the (elevator + person) is (2,000 kg + 60 kg) = 2,060 kg .

The weight is (mass x gravity) = (2,060 x 10) = 20,600 newtons

Work = (force x distance) = (20,600 newtons x 24m) = <em>494,400 joules</em>

The speed, acceleration, and how much time it takes don't make any
difference, unless you want the average power during the lift.

         Power = (work)/(time) =
           494,000 J / (24/4 m) =
           494,000 J / 6 sec = <em><u>82,400 watts</u></em>      wow !
6 0
3 years ago
a merry go round has a radius of 8 meters makes 2 revolutions every 2.5 minutes. A. express the angular speed of the merry go ar
pantera1 [17]

Answer:

a).v=83.77x10^{-3} rad/s

b).v=0.8rpm

c).v=0.5865 m/sec.

Explanation:

Given:

r=8m

v=\frac{2rev}{2.5minutes}

a).

2 rev*\frac{2\pirad}{1rev}=4\pi  rad

t=2.5minutes*\frac{60s}{1minute}=150s

The angular speed in radians per seconds is

v=\frac{4\pi}{150s}=83.77x10^{-3} rad/s

b).

v=\frac{2rev}{2.5minute}rpm

v=0.8rpm

c)

Child's distance per revolution

(pi*2r) = 43.988 metres.  

v=(43.988 x 0.0133333) = 0.5865 m/sec.

4 0
3 years ago
2.486 L is equal to:
swat32
Is equal to 2486 milliliters
8 0
2 years ago
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What is the sound intensity level in decibels? Use the usual reference level of I0 = 1.0×10−12 W/m2.
jek_recluse [69]

Answer:

L = 130 decibels

Explanation:

The computation of the sound intensity level in decibels is shown below:

According to the question, data provided is as follows

I = sound intensity = 10 W/m^2

I0 = reference level = 1 \times 10-12 W/m^2

Now

Intensity level ( or Loudness)is

L = log10 \frac{I}{10}

L = log10 \frac{10}{1\times 10^{-12}}

L = log10 \times 1013

= 13 \times 1 ( log10(10) = 1)

Therefore  

L = 13 bel

And as we know that

1 bel = 10 decibels

So,

The  Sound intensity level is

L = 130 decibels

5 0
3 years ago
A 3.9 kg ball traveling towards a soccer player at a velocity of -3.5 m/s rebounds off the soccer player's foot at a velocity of
Tresset [83]

Answer: 2.92 s

Explanation:

Given

Mass of ball is m=3.9\ kg

The initial velocity of the ball is u=-3.5\ m/s

Velocity after the rebound is v=15.9\ m/s

Force during the contact is F=25.9\ N

We know, change in momentum is Impulse

\Rightarrow F\cdot \Delta t=m(\Delta v)

\Rightarrow 25.9\cdot \Delta t=3.9(15.9-(-3.5))\\\\\Rightarrow \Delta t=\dfrac{3.9\times 19.4}{25.9}=2.92\ s

Thus, the force is applied for 2.92 s

4 0
3 years ago
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