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Veseljchak [2.6K]
2 years ago
6

Jessica does not like how her BestFriend has been treating her around others. When confronted, the friend says Jessica is being

emotional. Which beat describes the relationship between Jessica’s emotional and mental health
Physics
1 answer:
Eva8 [605]2 years ago
7 0
She is sad and feels left out because they are treating her badly. Hope this helps :)
You might be interested in
A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel
trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

1\ mile=1609.34\ m

1\ h=3600\ seconds

18.03\times \frac{3600}{1609.34}=40.33\ mph

Speed of the boat by when it will hit the dock is 40.33 mph

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m

The distance at which the boat will have to start decelerating is 312.5 m

5 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
Which of the following statements is true?A. Radiant energy is the same as sound waves. B. Radiant energy is a form of potential
svetoff [14.1K]
D.Radiant energy does not require a medium through which to travel.
5 0
3 years ago
Read 2 more answers
About where is our solar system located within the milky way galaxy?
sashaice [31]
Solar system is nested nearly 2/3 of the way from the center of the galaxy to the outskirt of the galactic disc.
6 0
3 years ago
A solid weighs 200N in air, 150N in water and 170N in a liquid. Find relative density of solid, relative density of liquid and d
fomenos

Answer:

\rho_{s} = 4

\rho_{l} = 0.6

\rho{liq} = 600 kg/m^{3}

Given:

Weight of solid in air, w_{sa} = 200 N

Weight of solid in water, w_{sw} = 150 N

Weight of solid in liquid, w_{sl} = 170 N

Solution:

Calculation of:

1. Relative density of solid, \rho_{s}

\rho_{s} = \frac{w_{sa}}{w_{sa} - w_{sw}}

\rho_{s} = \frac{200}{200 - 150} = 4

2. Relative density of liquid, \rho_{l}

\rho_{l} = \frac{w_{sa} - w_{sl}}{w_{sa} - w_{sw}}

\rho_{l} = \frac{200 - 170}{200 - 150} = 0.6

3. Density of liquid in S.I units:

Also, we know:

\rho{l} = \frac{\rho_{liq}}{\rho_{w}}

where

= {\rho_{liq}} = density of liquid

= {\rho_{w}} = 1000 kg/m^{3} = density of water

Now, from the above formula:

0.6 = \frac{\rho_{liq}}{1000}

\rho{liq} = 600 kg/m^{3}

3 0
2 years ago
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