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aniked [119]
2 years ago
4

A girl on a swing varies in height from 60 cm above the ground to 180 cm. the girl's greatest velocity ???

Physics
1 answer:
hoa [83]2 years ago
4 0
Use mechanical energy conservation

[Ek + Ep] at lower point = [Ek + Ep] at higher point


Ek at lower point = Ek at higher point + Ep at higher point - Ep at lower point

Ek at lower point = 1/2 m v^2

Ek at higher point = 0

Ep at higher point - Ep at lower point = mg(1.80m -0.60m)

1/2 mv^2 = mg (1.20m)

v^2 = 2g(1.20m)

use g = 10m/s^2

v = √[(2)(10m/s^2)(1.20m)] = 4.90 m/s^2


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A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
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Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e
Blizzard [7]
<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity V_{esc} is given by the following equation:

V_{esc}=\sqrt{\frac{2GM}{R}}   (1)

Where:

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M  is the mass of the asteroid

R  is the radius of the asteroid

On the other hand, we know the density of the asteroid is \rho=3.84(10)^{8}g/m^{3} and its volume is V=2.17(10)^{12}m^{3}.

The density of a body is given by:

\rho=\frac{M}{V}  (2)

Finding M:

M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})  (3)

M=8.33(10)^{20}g=8.33(10)^{17}kg  (4)  This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

V=\frac{4}{3}\piR^{3}   (5)

Finding R:

R=\sqrt[3]{\frac{3V}{4\pi}}   (6)

R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}   (7)

R=8031.38m   (8)  This is the radius of the asteroid

Now we have all the necessary elements to calculate the escape velocity from (1):

V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}   (9)

Finally:

V_{esc}=117.626m/s This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

6 0
2 years ago
_____ discovered that light also showed properties commonly found in waves.
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