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Leni [432]
3 years ago
14

If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the initial displacement is 2.25 m, and the

change in momentum is 0.58 kg*m/s, how far will it coast back up the ramp before changing directions? Please express your answers with 4 decimal place. _________________________ m
Physics
1 answer:
gtnhenbr [62]3 years ago
5 0

Answer:

L_f = 0.0196

Explanation:

given,

mass of the car,m = 0.2 Kg

height to width of ramp, y/x = 12/75

initial displacement, L_i = 2.25 m

change in momentum, Δp = 0.58 kg.m/s

distance of change in direction of the ramp, L_f = ?

Using equation

\Delta P = m \sqrt{\dfrac{2gy}{x}} (\sqrt{L_i}+\sqrt{L_f})

inserting all the values

0.58 = 0.2\times \sqrt{\dfrac{2\times 9.8\times 12}{75}} (\sqrt{2.25}+\sqrt{L_f})

1.5 + \sqrt{L_f} = 1.64

\sqrt{L_f}=0.14

L_f = 0.0196

The car will cost back up to the distance of 0.0196 m

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Answer:

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3 years ago
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A stone takes 5.4 seconds to fall from the top of a cliff. The cliff is
Allushta [10]

Answer:

143

Explanation:

Using one of the 3 fundamental equations in physics, y=vo*t+1/2gt^2, we can use this equation to find the total distance that was traveled.

Acceleration due to gravity is always 9.8m/s^2 and time is 5.4s, we also have no initial velocity.

Given this, we can plug in the known variables.

y=0t+1/2*9.8*5^2

simplify,

y=4.9*5.4^2

y=4.9*29.16

y=142.884m which we can round up to 143 meters

Final Answer: 143 meters

4 0
3 years ago
the moon revolves around the earth in a nearly circular orbit kept by gravitational force exerted by the earth work done will be
rodikova [14]

Answer:

Zero because the applied force is perpendicular to the motion of the object.

No work is done on an object moving is a circular path about a central attractive force.

Any work done in such a case would result in a change in the orbit.

3 0
2 years ago
At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She
Citrus2011 [14]

Answer:

a

The height is   H  = 6.74 \  m

b

The horizontal distance is  D  = 23.74 \  m

Explanation:

From the question we are told that

  The speed is  v  =  15 \  m/s

  The angle is  \theta  =  40^o

   The height of the cannon from the ground is  h  =  2 m

  The distance of the net from the ground is k  =  1 m

 

Generally the maximum height she reaches is mathematically represented as  

     H  =  \frac{v^2 sin^2 \theta }{2 *  g }  +  h

=>    H  =  \frac{(15)^2 [sin (40)]^2 }{2 * 9.8}  +  2

=>    H  = 6.74 \  m

Generally from kinematic equation  

    s = ut + \frac{1}{2} at^2

Here s is the displacement which is mathematically represented as

         s  =  [-(h-k)]  

    =>  s =  -(2-1)

    =>  s  = -1 m

There reason why s =  -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

     a =  -g = -9.8

    u  =  v sin (\theta)

So

    -1 = (vsin 40 )t + \frac{1}{2} * (-9.8) t^2

=>  -4.9t^2 + 9.6418t + 1 = 0

using  quadratic formula to solve the equation we have

    t  =  2.07 \  s

Generally distance covered along the horizontal is  

   D  =  v cos (40) *  2.07

=>   D  =  15 cos (40) *  2.07

=>   D  = 23.74 \  m

7 0
3 years ago
A ball is kicked from level ground at an angle 60° with initial velocity 10 m/s. The distance the ball travels, in meters, is:
RUDIKE [14]

Answer:

<em>the ball travels a distance of 8.84 m</em>

Explanation:

Range: Range is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again.

R = (U²sin2∅)/g.............................. Equation 1

Where R = range, U = initial velocity, ∅ = angle of projection, g = acceleration due to gravity.

<em>Given: U = 10 m/s, ∅ = 60°</em>

<em>Constant: g = 9.8 m/s²</em>

Substituting these values into equation 1

R = [10²×sin(2×60)]/9.8

R = (100sin120)/9.8

R = 100×0.8660/9.8

R = 86.60/9.8

R = 8.84 m

<em>Therefore the ball travels a distance of 8.84 m</em>

4 0
3 years ago
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