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Anastaziya [24]
3 years ago
10

A ball is dropped from an aircraft flying at an altitude of 8,848 meters assuming gravity is 9.8m/s what is the total amount of

time it takes to return to earth
Physics
1 answer:
Dafna11 [192]3 years ago
7 0
In this question, you're determining the time (t) taken for an object to fall from a distance (d).

The equation to represent this is:
Time equals the square root of 2 times the distance divided by the gravitational force of earth.
In equation from it looks like this (there isn't an icon to represent square root so just pretend like there's a square root there):
t = 2d/g (square-rooted)

d = 8,848m and g = 9.8m/s

Now plug in the information we have:
t = 2 x 8,848m/9.8m/s (square-rooted)

The first step is to multiply 2 times 8,848m:
t = 17,696m/9.8m/s (square-rooted)

Now divide 9.8m/s by 17,696m (note that the two m's (meters) cancels out leaving you with only s (seconds):
t = 1805.72s (square-rooted)

Now for the last step, find the square root of the remaining number:
t = 42.5s

So the time it takes the ball to drop from the height (distance) of 8,848 meters, and falling with the gravitational pull of 9.8 meters per second is 42.5 seconds.

I hope this helps :)

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A 11 g plastic ball is moving to the left at 29 m/s. How much work must be done on the ball to cause it to move to the right at
Rasek [7]

Answer:

4.25 J

Explanation:

Given that

mass of plastic ball = 11 g

Mass of plastic ball = 0.011 kg

velocity of ball = 29 m/s

We know that from work power energy theorem

W_{all}=Change\ in\ kinetic\ energy\ of\ system

We know that kinetic energy of moving mass given as

KE=\dfrac{1}{2}mv^2

Now by pitting the values

KE=\dfrac{1}{2}mv^2

KE=\dfrac{1}{2}\times 0.011\times 29^2

KE= 4.25 J

So the work done on the ball is 4.25 J

8 0
3 years ago
A) Determine the x and y-components of the ball's velocity at t = 0.0s, 2.0, 3.0 secs.
malfutka [58]

The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7º

given parameters

  • the initial velocity of the body vₓ = 1.8m / s and v_y = 1.2 m / s
  • the movement times t = 1.0s, 2.0s and 3.0 s

to find

    a) position

    b) acceleration

    c) launch angle

Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration

b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.

         v_y = v_{oy} - g t

where v and v({oy}  are the velocities of the body, g the acceleration of the planet's gravity and t the time

          0 = v_{oy} - gt

           g = v_{oy} / t

from the graph we observe that the highest point occurs for t = 2.0 s

           g = 1.2 / 2.0

           g = 0.6 m / s²

 

a) The position is requested for several times

X axis

in this axis there is no acceleration so we can use the uniform motion relationships

          vₓ = x / t

          x = vₓ t

where x is the position, vx is the velocity and t is the time

we calculate for the time

t = 0.0 s

          x₀ = 0

           

t = 2.0 s

          x₂ = 1.8 2

          x₂ = 3.6 m

t = 3.0 s

          x₃ = 1.8 3

          x₃ = 5.4 m

Y axis

In this axis there is the acceleration of the planet, let us use for the position the relation

          y = v_{oy} t - ½ g t²

t = 0.0 s

          y₀ = 0

          y₀ = 0 m

t = 2.0 s

         y₂ = 1.2 2 - ½ 0.6 2²

         y₂ = 1.2 m

t = 3.0 s

        y₃ = 1.2  3 - ½  0.6  3²

        y₃ = 0.9 m

c) the launch angle use the trigonometry relation

        tan θ = \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{1.2}{1.8}

        θ = 33.7º

measured counterclockwise from the positive side of the x-axis

With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7ºto)

learn more about projectile launch here:

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4 0
2 years ago
Tech A says that as engines gain miles, the spark plug gap increases, which raises the ignition system’s available voltage. Tech
Ugo [173]

Answer: Tech A is correct

Explanation:

Every vehicle has ignition system and without this system,it will not work. The battery of everything vehicle contain energy that start the vehicle and ignore it to start working. Electrical current move from the vehicle's battery and get to the induction coil, the induction coil increases the voltage in it so that the plug will be ignited. The spark plugs produce fire. The spark plug is connected to the ignition system. Once voltage is produced from the induction coil, electrical impulses move from induction coil to insulated plug wires. The spark plug need a very high voltage from the small voltage battery. Once the high voltage exceed the dielectric strength of the gases, spark jump the gap between the plug's fire end.

8 0
3 years ago
The velocity time graph of an object is shown below. How far does the object travel in the time interval t =4 s to t = 6 s?
Trava [24]

The distance covered by the object between t =4 s and t = 6 s is 4 m

Explanation:

In a velocity-time graph, the distance covered by the object represented can be found by calculating the area under the curve.

Therefore, the distance covered by the object between t = 4 s and t = 6 s is the area under the curve between 4 s and 6 s.

We see that we have to calculate the area of a triangle, with:

Base:

b=6-4 = 2

And height:

h=4-0 = 4

Therefore, the area is

A=\frac{1}{2}bh=\frac{1}{2}(2)(4)=4

So, the distance covered by the object is 4 m.

Learn more about distance:

brainly.com/question/3969582

#LearnwithBrainly

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3 years ago
You are currently getting 26 sales opportunities per day and closing 64% of them. How many sale opportunists per day?
Vikki [24]

Answer: 16

Explanation:

8 0
3 years ago
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