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denpristay [2]
3 years ago
15

One of the Lady Spartans was falling to the ground after

Physics
1 answer:
dem82 [27]3 years ago
0 0

Answer:

Explanation:

Given that,

A lady falling has a final velocity of 4m/s

v = 4m/s

Mass of the lady is 60kg.

m = 60kg

Using conservation of energy, the potential energy of the body from the point where the lady is dropping is converted to the final kinetic energy of the lady.

Therefore,

P.E = K.E(final) = ½mv²

P.E = ½ × 60 × 4²

P.E = 480 J.

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The Outlaw Run roller coaster in Branson, Missouri, features a track that is inclined at 84 ∘ below the horizontal and that span
harina [27]

Answer:

Explanation:

a)

Ff = μmgcosθ

Ff = 0.28(1600)(9.8)cos(-84)

Ff = 458.9217...

Ff = 460 N

b)  ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.

W = Ffd

W = 458.9217(-49.4/sin(-84)

W = 22,795.6119...

W = 23 kJ

c) same assumptions as part b

The change in potential energy minus the work of friction will be kinetic energy.

KE = PE - W

½mv² = mgh - (μmgcosθ)d

v² = 2(gh - (μgcosθ)(h/sinθ))

v = √(2gh(1 - μcotθ))

v = √(2(9.8)(49.4)(1 - 0.28cot84))

v = 30.6552...

v = 31 m/s

5 0
2 years ago
Scientists use laser range-finding to measure the distance to the moon with great accuracy. A brief laser pulse is fired at the
Fofino [41]

Answer:

d = 2,042 10-3 m

Explanation:

The laser diffracts in the circular slit, so the process equation is

      d sin θ= m λ

The first diffraction minimum occurs for m = 1

We can use trigonometry in the mirror

        tan θ = Y / L

Where L is the distance from the Moon to Earth

Since the angle is extremely small

           tan θ = sin θ / cos θ

           Cos θ = 1

           tant θ = sin θ = y / L

We replace

           d y / L = λ

           d = λ L / y

Let's calculate

           d = 532 10⁻⁹ 3.84 10⁶/1 10³

           d = 2,042 10-3 m

5 0
3 years ago
A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3
kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, a=25.4\ m/s^2

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

v=u+at

v=25.4\times 3.39=86.10\ m/s    

Let x is the initial position of the rocket. Using third equation of kinematics as :

v^2=u^2+2ax_o

x_o=\dfrac{v^2}{2a}

x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m  

Let x_o is the position at the maximum height. Again using equation of motion as :

v^2-u^2=2a(x-x_o)

Now a=-g and v and u will interchange

u^2=2g(x-x_o)

x=x_o+\dfrac{u^2}{2g}

x=145.92+\dfrac{(86.10)^2}{2\times 9.8}

x = 524.14 meters

Hence, this is the required solution.

5 0
3 years ago
Using the right amount of significant figures, calculate the answer to the following problem, 215.5+101.02555
cestrela7 [59]

Answer:

This is how I figured it out:

  1. 215.5 rounded to one significant figure is 200
  2. 101.02555 rounded to one significant figure is 100.
  3. 200 + 100 = 300.

Hope this helps!

Explanation:

7 0
3 years ago
An object has a kinetic energy of 175 J and a momentum of magnitude 25.0 kg m/s. Find the
DedPeter [7]

Answer:14 m/s

Explanation:

Kinetic energy(ke)=175J

Momentum(M)=25kgm/s

Speed=v

Mass=m

Ke=(m x v x v)/2

175=(mv^2)/2

Cross multiply

175 x 2=mv^2

350=mv^2

Momentum=mass x velocity

25=mv

m=25/v

Substitute m=25/v in 350=mv^2

350=25/v x v^2

350=25v^2/v

v^2/v=v

350=25v

v=350/25

v=14 m/s

5 0
4 years ago
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