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Over [174]
3 years ago
11

Explain what it means to view a moving object from a frame of reference

Physics
1 answer:
GaryK [48]3 years ago
5 0
When you look at an object in motion, the frame of reference is different depending on where you are when looking at this object.

For example, if you are in a train you can say that from your frame of reference the ground outside is moving compared to you and you are not moving or the train is not moving. From someone on the outside of the train, they can say that the train is moving compared to the ground and they are not moving.

Another example could be if you are driving in a car. You can say that the road is moving backwards compared to you and you are stationary. Someone standing on the sidewalk can say you are moving compared to the ground. You can also say that you are not moving since you are stationary in the car.
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If your weight is 120 pounds and your mass is 54 kilograms how would those values change if you were on the moon
BlackZzzverrR [31]
The gravitation acceleration on the moon is different than on Earth. It is 1.6 m/s^2. If you weigh 120 lbs, then you would multiply 120 pounds by the gravitational acceleration on the moon and then divide by the acceleration on Earth.

(120 lbs * 1.6) / 9.8 = 20 pounds.

The mass will always be the same no matter what planet you’re on, so it’s still 54 kg.
4 0
3 years ago
what are the common elements that make up series and parallel circuits , and how are circuits diagrammed?​
Ahat [919]
<h2>Answer:</h2>

A series circuit occurs when the elements are connected along a simple path so the same current flows through all the elements. On the other hand, a parallel circuit occurs when there are two or more paths for the electricity to flow. The diagram are shown in the Figure below. We have chosen a source and resistors to illustrate this problem.

3 0
4 years ago
In Einstein’s analysis, a photon striking the surface of a conductor is absorbed by an electron. Which statement describes the s
puteri [66]

Answer:

the correct answer is A

Explanation:

In an Einstein-type analysis, the photon is absorbed, it loses all its energy, therefore the electron must receive all or none of the energy of the incident photon. In a type of inelastic shock.

Let's analyze the different answers

A) true. In photon it is completely absorbed or passes without interaction

B) False. The photon must change energy, but in this case there is no absorption of the photon

C) False. In the insistent analyzes, the quantization of the electron in discrete states is not mentioned.

Therefore the correct answer is A

7 0
2 years ago
When dogs see police dogs do they think "oh no its the cops"?
ira [324]

No they say "Watch out it's the fuzz"

4 0
3 years ago
Read 2 more answers
A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
3 years ago
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