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3 years ago

Which phrase describes how Jupiter and Saturn are similar?

Physics

2 answers:

Readme [11.4K]3 years ago

5 0

Answer:

Many Moons

Explanation:

DedPeter [7]3 years ago

4 0

The phase many moons .

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A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

I hope it helps you!

3 0

3 years ago

Light propagating in the glass n1= 1.65 wall of an aquarium tank strikes the interior edge of the wall with incidence angle 19.0

jonny [76]

We can solve this using Snell's Law which is represented by the equation:
sin θ₁ / sin θ₂ = n₂ / n₁

From the problem, we can substitute values and solve for the angle of refraction.
sin 19 / sin θ₂ = 1.65 / 1
θ₂ = 11.38°

The angle of refraction would be 11.38°.

3 0

2 years ago

A weightlifter lifts a set of weights a vertical distance of 2.00m.If a constant net force of 350 N is exerted on the weights,wh

MrMuchimi

Answer:

W=700 Joule

Explanation:

Physics Work

Is the dot product of the force vector by the displacement vector

$W=\vec F \cdot \vec r$

When both the force and the displacements are pointed in the same direction, the formula reduces to its scalar version

$W=F.d$

The weightlifter is applying a net force of 350 N to lift the weights a distance of 2 m, thus the net work done is

$W=350\ N\ .\ 2\ m=700\ Joule$

4 0

3 years ago

Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3

cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

$J = - D \frac{\Delta c}{\Delta x}$

Where $\frac{\Delta c}{\Delta x}$ is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

$D= D_0 e^{\frac{E_d}{RT}}$

Where $D_0$ is the proportionality constant,

R is the gas constant, T the temperature and $E_d$ is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

$J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}$

Rearrange the equation to get the value of temperature,

$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

We have all the values in our equation.

$\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}$

$\Delta x = 9.7*10^{-3}m$

$E_d = 82000J$

$D_0 = 6.5*10^{-7}m^2/s$

$J = 3.2*10^{-9}m^2/s$

$R= 8.31Jmol^{-1}K$

Substituting,

$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

$T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}$

$T=1118.07K=844\°C$

4 0

3 years ago

Suppose that you are standing on a train accelerating at 0.20g. What minimum coefficient of static friction must exist between y

Ilia_Sergeevich [38]

Acceleration = (0.2 x g) = 1.96m/sec^2.
Accelerating force on 1kg. = (ma) = 1.96N. 
1kg. has a weight (normal force) of 9.8N. 
Coefficient µ = 1.96/9.8 = 0.2 minimum. 

Coefficient is a ratio, so holds true for any value of mass to find accelerating force acting. 
e.g. 75kg = (75 x g) = 735N. 
Accelerating force = (735 x 0.2) = 147N

5 0

3 years ago

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