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chubhunter [2.5K]
3 years ago
6

8. (7.MS-PS3-5.) When energy changes form, energy is

Physics
1 answer:
Furkat [3]3 years ago
5 0

Answer:

Not lost or created

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Arrange the phases of the Moon in order of increasing rising time, from the phase with the earliest rising time at 12:00 a.m. to
ZanzabumX [31]

Answer:

Lets assume the Sun rise time to be 06:00 AM. The rise time of different phases of Moon will be as follows:

12:00 AM : Waning Half

01:00 AM - 05:00 AM : Waning Crescent

06:00 AM : New Moon

07:00 AM - 11:00 AM : Waxing Crescent

12:00 PM :  Waxing Half

01:00 PM - 06:00 PM : Waxing Gibbous

06:00 PM : Full Moon

07:00 PM - 09:00 PM : Waning Gibbous

Explanation:

The Moon is the only celestial object which shows visible changes in its shape and rise and set time over a very short period of time i.e. just one day. One can observe it by observing the Moon daily. One will notice the change easily. This happens because of the geometry of the Sun, Earth and Moon. The Moon doesn't have its own light and shines because of the light of Sun.

At any given time half of the Moon would be illuminated by the Sun but how much of this illuminated portion is facing the Earth decides the phase of the Moon visible from the Earth. Due to this the Moon shows us various phases namely: New, Waxing Crescent, Waxing Half, Waxing Gibbous, Full, Waning Gibbous, Waning Half, Waning Crescent.

Also, the Moon revolves around the Earth completing the orbit in 29.5 Days. Everyday the Moon will change its position in the orbit. Due to this the rising time of Moon shifts by approximately 52 minutes daily. So, the New Moon rises with the Sun and Full Moon rises just after the sunset.

Lets assume the Sun rise time to be 06:00 AM. The rise time of different phases of Moon will be as follows:

12:00 AM : Waning Half

01:00 AM - 05:00 AM : Waning Crescent

06:00 AM : New Moon

07:00 AM - 11:00 AM : Waxing Crescent

12:00 PM :  Waxing Half

01:00 PM - 06:00 PM : Waxing Gibbous

06:00 PM : Full Moon

07:00 PM - 09:00 PM : Waning Gibbous

5 0
3 years ago
Which of the following is not a part of dalton s atomic theory?
Alex777 [14]
<span>c. atoms are always in motion..............</span>
6 0
3 years ago
Read 2 more answers
An object is thrown upwards with the speed of 40.0 km/hr in 5 second, calculate the distance in millimeters (mm).
sertanlavr [38]
V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s

v=d/t
d=vt
d=(11111.1mm/s)(5s)
d=55555mm
d=5.56x10^4mm
4 0
3 years ago
Which one of the following temperatures is equal to 5°C?
natali 33 [55]

Answer : The correct option is, (D) 278 K

Explanation :

We are given temperature 5^oC.

Now the conversion factor used for the temperature is,

K=^oC+273

where, K is kelvin and ^oC is Celsius.

Now put the value of temperature, we get

K=5^oC+273=278K

Therefore, the temperature 278 K is equal to the 5^oC


7 0
3 years ago
A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio
Fed [463]

Answer:

Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g

Explanation:

The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

N = mg cos θ

m = 100 g = 0.10 kg

g = acceleration due to gravity = 9.8 m/s²

θ = 15°

N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

Frictional force = Tension + F

Frictional force = μN

where μ = coefficient of kinetic friction = 0.60

N = normal reaction = 0.9466 N

Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

F = (0.10 × 9.8 × sin 15°) = 0.253624 N

Tension in the rope = T = ?

Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

8 0
4 years ago
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