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oksano4ka [1.4K]
2 years ago
12

A 900 kg car is traveling at 20 m/s along the road. What force must be applied to the car to stop it in a distance of 30 m2 Assu

me a
constant change in velocity.
A. 6000 N
B. 12,000 N
C. 7000 N
D. 18,000 N
h​
Physics
1 answer:
serg [7]2 years ago
5 0

Answer:

A. 6000 N

Explanation:

v²=u²+2as

0²=20²+2x30xa

-400=60a

a=-400/60

a =-6.667m/s²

f =ma

f = 900 x 6.667 = 6003N

F = 6000N

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Ayuda por favor (archivo adjunto) con un ejercicio de expresión sobre periodo de oscilación de esta figura:
vodka [1.7K]

Answer:

nolo se

Explanation:

no lo se

8 0
2 years ago
System A has masses m and m separated by a distance r; system B has masses m and 2m separated by a distance 2r; system C has mas
Anna [14]

Answer:

System D --> System C --> System A --> System B

Explanation:

The gravitational force between two masses m1, m2 separated by a distance r is given by:

F=G \frac{m_1 m_2}{r^2}

where G is the gravitational constant. Let's apply this formula to each case now to calculate the relative force for each system:

System A has masses m and m separated by a distance r:

F=G\frac{m \cdot m}{r^2}=G \frac{m^2}{r^2}

system B has masses m and 2m separated by a distance 2r:

F=G\frac{m \cdot 2m}{(2r)^2}=G \frac{2m^2}{4r^2}=\frac{1}{2} G \frac{m^2}{r^2}

system C has masses 2m and 3m separated by a distance 2r:

F=G\frac{2m \cdot 3m}{(2r)^2}=G \frac{6m^2}{4r^2}=\frac{3}{2} G \frac{m^2}{r^2}

system D has masses 4m and 5m separated by a distance 3r:

F=G\frac{4m \cdot 5m}{(3r)^2}=G \frac{20m^2}{9r^2}=\frac{20}{9} G \frac{m^2}{r^2}

Now, by looking at the 4 different forces, we can rank them from the greatest to the smallest force, and we find:

System D --> System C --> System A --> System B

5 0
3 years ago
if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?
madreJ [45]

Answer:

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

a_{c} =\frac{(velocity)^{2} }{radius}

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

a_{c} = ?

Solution:

We have,

a_{c} =\frac{(velocity)^{2} }{radius}

Substituting  given value in it we get

a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

7 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
Define kinetic energy and derive its relation.
IRINA_888 [86]

Answer:

The kinetic energy of a body is the energy that it possessed due to its motion. Kinetic energy can be defined as the work needed to accelerate an object of a given mass from rest to its stated velocity. Kinetic energy depends upon the velocity and the mass of the body.

3 0
3 years ago
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