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3 years ago

Differences Between light year and astronomical unit in two points .

Physics

2 answers:

Marina CMI [18]3 years ago

7 0

Answer:

A light year is the distance light travels in a year. ... And an astronomical unit is the average distance between the earth and the sun. So the distance to the sun is by definition one AU. A parsec is the distance at which one astronomical unit subtends an angle of one second of arc.

Maslowich3 years ago

5 0

Answer: A light year is the distance light travels in a year. ... And an astronomical unit is the average distance between the earth and the sun. So the distance to the sun is by definition one AU. A parsec is the distance at which one astronomical unit subtends an angle of one second of arc

Explanation:

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A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after bein

mars1129 [50]

<span>(M G H)=(0.5 x 9.8 x 10) = 49 joules.</span>

4 0

3 years ago

Which sentence from the article supports the idea that the visible spectrum is more limited in marine animals than humans? A Mos

lidiya [134]

Most marine bioluminescence is blue-green, which is easier to see in the deep ocean

Explanation:

As per science, Emission and production of light by a living organism is defined as Bioluminescence. Bioluminescence occurs widely in marine animals whereas it is triggered by a physical disturbance is seen by humans, such as a moving boat hull or waves.

Throughout the water column bioluminescent organisms live and bioluminescence is extremely common in deep sea which shows that visible spectrum is more limited to marine animals than humans.

8 0

3 years ago

A pan hangs from a 50 cm spring. When a 10 kg mass is placed in the pan, it stretches the spring 6 cm. What is a function rule l

MrRissso [65]

Answer:

Explanation:

A Spring stretches / compresses when force is applied on them and they are governed by the Hookes Law which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched.

$F = -kx$

F is the force applied and x is the elongation of the spring

k is the spring constant.

negative sign indicates the change in direction from equilibrium position.

In the given question, we dont have force but we know that the pan is hanging. We also know from the Newton's second law of motion that

$F=mg$

Inserting this into Hooke's Law

$mg=-kx$

computing it for x,

$-x=mg/k$

This is the model which will tell the length of the spring against change in the mass located in the pan.

3 0

2 years ago

a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame

natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

$\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}$

where;

L = length of the bed

$\mu$ = viscosity

U = superficial velocity

$\epsilon$ = void fraction

dp = equivalent spherical diameter of bed material (m)

$\rho$ = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6 -------------- equation (1)

24A + 576B = 24.1 --------------- equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

B = 0.017014

From equation (1)

12A + 144B = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

$A = \frac{9.6 -144(0.017014}{12}$

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

3 0

2 years ago

Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary

kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

$I=mr^2$

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

$I_A=1 \times 0.5^2\ kg.m^2$

$I_A=0.25\ kg.m^2$

T= I α

0.5= 0.25 α

$\alpha = 2\ rad/s^2$

For Wheel B

m= 1 kg

d= 2 m,r=1 m

$I_B=1 \times 1^2\ kg.m^2$

$I_B=1 \ kg.m^2$

Given that angular acceleration is same for both the wheel

$\alpha = 2\ rad/s^2$

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0

3 years ago