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3 years ago

Please help me i neeed help with all threeas soon as possible thank you

Physics

1 answer:

liberstina [14]3 years ago

7 0

Answer:

4b: comets

5a: supercluster

5b: they just changed 4b's solar system for milky way. I think it is still comets. if not, then just say black holes.

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What is the potential energy of two charges of +4.6 μC and +1.0 μC that are separated by a distance of 10.0 cm?

Artist 52 [7]

Answer:

U = 0.413 J

Explanation:

the potential energy between two charges q1 and q2 is given by the following formula:

$U=k\frac{q_1q_2}{r}$ (1)

k: Coulomb's constant = 8.98*10^9 NM^2/C^2

q1: first charge = 4.6 μC = 4.6*10^-6 C

q2: second charge = 1.0 μC*10^-6 C

r: distance between charges = 10.0 cm = 0.10 m

You replace the values of all variables in the equation (1):

$U=(8.98*10^9Nm^2/C^2)\frac{(4.6*10^{-6}C)(1.0*10^{-6}C)}{0.10m}=0.413\ J$

Hence, the energy between charges is 0.413 J

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2 years ago

What is the frequency if a wave that pases a given pount 22 times in 2 seconds

34kurt

The answer is B. 11 Hz

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3 years ago

James took two pea plants, placing one in a dark closet and the other on a sunny window sill. Both are located in air-conditione

Fudgin [204]

The constant is the temperature of the air that the plants get.

The independent variable is the thing that YOU control. That's the amount of sunlight each plant gets.

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5 0

3 years ago

Read 2 more answers

A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows: