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2 years ago

Calculate the average velocity in m/y of a tectonic plate that has travelled 9000 km to the south in 60 million years

1 answer:

8 0

The distance covered by the tectonic plate, in meters, is
 $d=9000km=9\cdot 10^6 m$ 
The time taken for the tectonic plate to cover this distance is equal to
 $t=60 mil.y=60 \cdot 10^6 y$ 
Therefore, the average velocity of the tectonic plate is the ratio between the distance covered and the time taken:
 $v= \frac{S}{t} = \frac{9\cdot 10^6m}{60 \cdot 10^6y} =0.15 m/y.$

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a 0.0215m diameter coin rolls up a 20 degree inclined plane. the coin starts with an initial angular speed of 55.2rad/s and roll

marissa [1.9K]

Answer:

$h = 0.0362\,m$

Explanation:

Given the absence of non-conservative force, the motion of the coin is modelled after the Principle of Energy Conservation solely.

$U_{g,A} + K_{A} = U_{g,B} + K_{B}$

$U_{g,B} - U_{g,A} = K_{A} - K_{B}$

$m\cdot g \cdot h = \frac{1}{2}\cdot I \cdot \omega_{o}^{2}$

The moment of inertia of the coin is:

$I = \frac{1}{2}\cdot m \cdot r^{2}$

After some algebraic handling, an expression for the maximum vertical height is derived:

$m\cdot g \cdot h = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega_{o}^{2}$

$h = \frac{r^{2}\cdot \omega_{o}^{2}}{g}$

$h = \frac{(0.0108\,m)^{2}\cdot (55.2\,\frac{rad}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }$

$h = 0.0362\,m$

3 0

2 years ago

The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a

navik [9.2K]

Answer:

230

Explanation:

$\omega$ = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

$K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J$

Energy used in one acceleration

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J$

Number of accelerations would be given by

$n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4$

So the number of complete accelerations is 230

8 0

2 years ago

Tyler stands at rest on a skateboard. He has a mass of 120 kg. His friend (m = 60 kg) jumps into his arms at a speed of 2 m/s. I

Andrews [41]

Momentum question. This is an inelastic collision, so

m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
Vf=120kg m/s / 180kg
Vf=0.67m/s

0.67m/s

5 0

2 years ago

A diver in the pike position (legs straight, hands on ankles) usually makes only one or one-and-a-half rotations. To make two or

Korolek [52]

Answer:

from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

Explanation:

This can be explained on the basis of conservation of angular momentum.

This means the initial and the final angular velocity is conserved. Consider initial position (1)in the pike and final position in the be truck position. So there inertia's will also be different.

⇒ $I_1\omega_1 = I_2\omega_2$

$\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}$

also,

$I_1= mr_1^2$

$I_2= mr_2^2$

since, $r_2^2$

$I_2^2$

therefore,

$\omega_1^2$

So, from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

6 0

2 years ago

Calculate the longest wavelength visible to the human eye. express the wavelength in nanometers to three significant figures.

slega [8]

NOTE: The given question is incomplete.

The complete question is given below.

The human eye contains a molecule called 11-cis-retinal that changes conformation when struck with light of sufficient energy. The change in conformation triggers a series of events that results in an electrical signal being sent to the brain. The minimum energy required to change the conformation of 11-cis-retinal within the eye is about 164 kJ/mole. Calculate the longest wavelength visible to the human eye.

Solution:

Energy (E) = 164 kJ/mole

E = 164 kJ/mole = 164 kJ /6.023 x 10²³

= 2.72 x 10⁻²² kJ = 2.72 x 10⁻¹⁹J

Planck's constant = 6.6 x 10⁻³⁴ J s,

Speed of light = 3.00 x 10⁸ m/s

Let the required wavelength be λ.

Formula Used: E = hc / λ

or, λ = hc / E

or, λ = (6.6 x 10⁻³⁴ J s)× (3.00 x 10⁸ m/s) / (2.72 x 10⁻¹⁹J)

or, λ = 7.28 x 10⁻⁷ m

or, λ = (7.28 x 10⁻⁷ m) ×( 1.0 x 10⁹ nm / 1.0 m)

or, λ = (7.28 x 10² nm)

or, λ = 728 nm

Hence, the required wavelength will be 728 nm.

6 0

2 years ago