Which of these have a potential energy?

PLEASE HELP <3

aliya0001 [1]

1) The landing spot of the projectile is given by $d=v_x \sqrt{\frac{2h}{g}}$

2) The common variable is the time

Explanation:

1)

The motion of a projectile consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

The landing spot can be determined in the following way:

- First of all, we analyze the vertical motion to find the time of flight of the projectile. This can be done by using the suvat equation

$h=ut+\frac{1}{2}at^2$

where

h is the vertical displacement of the projectile, which corresponds to the height from which the projectile has been fired, above the ground

u = 0 is the initial vertical velocity

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

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2h}{g}}$

- After finding the time of flight, we analyze the horizontal motion, which is a uniform motion with constant horizontal velocity $v_x$ . Therefore, the horizontal distance covered is given by

$d=v_x t$

And substituting the time of flight,

$d=v_x \sqrt{\frac{2h}{g}}$

2)

Since the horizontal motion is uniform, the horizontal component of the displacement of the projectile is given by

$x=v_x t$

where $v_x$ is the horizontal velocity and t is the time.

The vertical motion is accelerated, so the vertical component of the displacement is given by

$y=\frac{1}{2}gt^2$

where g is the acceleration of gravity and t is the time.

Therefore, from the two equations we see that the common variable is t, the time.

Learn more about projectile:

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6 0

3 years ago

Which characteristics do temperate marine and temperate continental climates have in common? Check all that apply.

katrin2010 [14]

Answer:

moderate amount of precipitation

two seasons with temperature differences

Explanation:

Temperate marine climates is also known as oceanic climate and areas who experience this climate are closer to oceans. There is relative precipitation all round the year due to the close proximity to the water body and have the hot and cold season.

The temperate continental climate also has two season and relative rainfall but found in the areas along the Northern hemispheres.

Both climates have moderate amount of precipitation and two seasons with temperature differences.

6 0

3 years ago

Ayden and Steven are playing catch with a football. Ayden throws the football at a velocity of 15 m/s with an angle of 60° above

Pachacha [2.7K]

1) 1.33 s

2) 8.6 m

3) 19.9 m

Explanation:

1)

The motion of the football is a projectile motion, which consists of two independent motions:

- A uniform motion (=constant velocity) along the horizontal direction

- A uniformly accelerated motion (=constant acceleration) along the vertical direction

The hang time of the football is the time it takes for the football to reach its maximum height. It is given by:

$t=\frac{u_y}{g}$

where

$u_y = u sin \theta$ is the initial vertical velocity of the ball, where

u = 15 m/s is the initial velocity

$\theta=60^{\circ}$ is the angle of projection of the ball

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

Substituting, we find the hang time:

$t=\frac{u sin \theta}{g}=\frac{(15)(sin 60^{\circ})}{9.8}=1.33 s$

2)

The motion of the ball along the vertical direction is a uniformly accelerated motion, so we can use the suvat equation:

$s=u_y t - \frac{1}{2}gt^2$

where:

s is the vertical displacement

$u_y = u sin \theta$ is the initial vertical velocity

t is the time

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

In this part, we want to find the maximum height, so the height reached by the ball when the time is

t = 1.33 s

Therefore, by substituting the values in the equation, we can find the maximum height:

$s=usin \theta t-\frac{1}{2}gt^2=(15)(sin 60^{\circ})(1.33)-\frac{1}{2}(9.8)(1.33)^2=8.6 m$

3)

Here we want to find the horizontal range covered by the ball during its flight.

The horizontal range for a projectile is given by the equation

$d=\frac{u^2 sin(2\theta)}{g}$

where

u is the initial velocity

$\theta$ is the angle of projection

g is the acceleration due to gravity

For the ball in this problem we have:

u = 15 m/s

$\theta=60^{\circ}$

$g=9.8 m/s^2$

Substituting into the equation, we find the horizontal distance covered by the ball:

$d=\frac{(15)^2sin(2\cdot 60^{\circ})}{9.8}=19.9 m$

6 0

4 years ago

The weight of the space shuttle is about 4.5 million pounds. How many people would it take to move it? Answer in units of people

larisa [96]

Assuming that they are pulling the space shuttle straight up, and that this is done on the surface of the earth, where gravity is equal to 9.8 m/s^2. So taking the average pull force of a person to be 110 kg, then: 

110 kg/person * 9.8 m/s^2 = 1078 kg*m/s^2, or 1078 N/person (Newtons).

1 Newton is about 0.225 lb. So, 1078 N/person * 0.225 lb/N = 242.34 lb/person. 

So if the space shuttle weighs 4,500,000 lb, then it would take 4,500,000 lb / (242.34 lb/person) = 18,569 people.

Answer:

18,569 people

7 0

4 years ago

84 ounces to milliliters

Arte-miy333 [17]

Answer:

2484.18

Explanation:

Hope this helps :)

6 0

4 years ago

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