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iren2701 [21]
3 years ago
14

Que capacidade física consiste em deslocar o corpo no espaço o mais rápido possível, mudando o centro de gravidade de posição, s

em perder o equilíbrio e a coordenação dos movimentos?
Physics
1 answer:
grin007 [14]3 years ago
6 0

Answer:ur mom

Explanation:because

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A shell is fired with 500km/h on a target half kilometer away.In what time in which shell will hit the target.
Drupady [299]

It depends on the direction in which the shell is launched.  The time can be anything from 3.6 seconds to never.

6 0
3 years ago
IS
jarptica [38.1K]

Answer:

5.03

Explanation:

trust me

7 0
3 years ago
how much water is needed to produce 1kwh of electricity at a power plant that is 30% efficient if the temperature increase 10 C
Dimas [21]

The amount of water needed is 287 kg

Explanation:

The amount of energy that we need to produce with the power plant is

E=1 kWh = (1000W)(1h)=(1000W)(3600s)=3.6\cdot 10^6 J

We also know that the power plant is only 30% efficient, so the energy produced in input must be:

E_{in}=\frac{E}{0.30}=\frac{3.6\cdot 10^6}{0.3}=1.2\cdot 10^7 J

The amount of water that is needed to produce this energy can be found using the equation

E_{in}=mC\Delta T

where:

m is the amount of water

C=4186 J/kg^{\circ}C is the specific heat capacity of water

\Delta T=10^{\circ}C is the increase in temperature

And solving for m, we find:

m=\frac{E_{in}}{C\Delta T}=\frac{1.2\cdot 10^7}{(4186)(10)}=287 kg

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 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
3 years ago
A 10.0-kg box starts at rest and slides 3.5 m down a ramp inclined at an angle of 10° with the horizontal. If there is no fricti
Alenkinab [10]

Answer:

v_f = 3.45 m/s

Explanation:

As we know that box was initially at rest

so here work done by all forces on the box = change in its kinetic energy

so we will have

mg sin\theta L = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

now we have

m = 10 kg

\theta = 10^0

L = 3.5 m

so we will have

10(9.81)sin10 \times 3.5 = \frac{1}{2}(10)(v_f^2 - 0)

2(9.81) sin10 \times 3.5 = v_f^2

so final speed is given as

v_f = 3.45 m/s

6 0
3 years ago
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