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jekas [21]
3 years ago
14

A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between

the Earth and the satellite, yet no work is done on the satellite. How is this possible?
a. No work is done if there is no contact between objects.
b. No work is done because there is no gravity in space.
c. No work is done if the direction of motion is perpendicular to the force.
d. No work is done if objects move in a circle.
Physics
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

C. No work is done if the direction of motion is perpendicular to the force.

Explanation:

We know that work is the dot product of force and displacement.

Let's take the angle between the force and the displacement = θ

W= F . d cosθ

F=Force  , d=Displacement  

If θ = 0° then W= F.d

If θ = 90° then W= 0

So we can say that when force is perpendicular to the displacement then the work done by force will be zero.

Therefore the answer is C.

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photoshop1234 [79]

Answer:

It conserves both energy and momentum in the collision at the same time. By design, when the balls collide the strings that hold them up are vertical (assuming balls are only swung from one side).

Explanation:

Hope This Helps!!

7 0
3 years ago
A homeowner uses a snowblower to clear his driveway. knowing that the snow is discharged at an average angle of 40° with the hor
Nesterboy [21]

given that snow is projected at an angle of 40 degree

It range is given as a = 19 ft

a = 19 * 0.3048 = 5.8 m

now we can use the formula of horizontal range

R = \frac{v_o^2 sin2\theta}{g}

5.8 =\frac{ v_o^2 sin(2*40)}{9.8}

5.8 = \frac{v_o^2 * sin80}{9.8}

v_o^2 = 57.7

v_o = 7.6 m/s

<u>so its initial speed must be 7.6 m/s</u>

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:

brainly.com/question/10903823

4 0
2 years ago
A car accelerates from 6.0 m/s to 18 m/s at a rate of 3.0 m/s2. How far does it travel while accelerating
AlekseyPX

Answer:

48m, hope this helps :)

4 0
3 years ago
A p-wave arrives at 3:00:00 and the s-wave arrives at 3:07:20, what is the exact distance that the seismic station is away from
maxonik [38]

The exact distance from the seismic station to the epicenter is 6000 km.

<h3>Epicenter of earthquake</h3>

The earthquake's epicenter is the point above the fault location on the earth's surface.

Given that:

A p-wave arrives at 3:00:00 and the s-wave arrives at 3:07:20.

Difference in arrival time = 3:07:20 - 3:00:00 = 7 minutes 20 seconds

From the earthquake time travel graph, a time difference of 7 min 20 sec is at x = 6

The exact distance from the seismic station to the epicenter is 6000 km.

Find out more on Epicenter of earthquake at: brainly.com/question/1969968

4 0
2 years ago
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