Full moon: All of the moon or sun appears covers by the shadow.
When the moon is waxing, it means that the lit part is getting larger (going towards a full moon)
When the moon is waning , it means that the lit part is getting smaller (going towards a new moon)
The new moon represents the start of a new lunar cycle and occurs approximately every 29 days.

Hope that help:)

5 0

2 years ago

What happens as water evaporates from a lake

NARA [144]

Answer:

As that liquid water is further heated, it evaporates and becomes a gas—water vapor. So C... Need branliest pls

Explanation:

4 0

2 years ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

IRINA_888 [86]

Answer:

Time : 7.96 s

Distance Traveled : 357.8 m

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

t₂ = 7.96 s

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

S = 357.8 m

7 0

3 years ago

What is the passenger's apparent weight at t=1.0s?

Fittoniya [83]

Answer:

For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.

Explanation:

The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.

From the slope of graph it is clear that acceleration at t = 1 sec is given as:

Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2

Now, there are two cases:

1- Elevator moving up

2- Elevator moving down

For upward motion:

Apparent Weight = m(g + a)

Apparent Weight = (75 kg)(9.8 + 4)m/s^2

Apparent Weight = 1035 N

For downward motion:

Apparent Weight = m(g - a)

Apparent Weight = (75 kg)(9.8 - 4)m/s^2

Apparent Weight = 435 N

4 0

3 years ago

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