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Sliva [168]
3 years ago
8

Which words describe the composition of the inner planets ? Select three options,

Physics
2 answers:
IgorLugansk [536]3 years ago
7 0

Answer:a b c

Explanation: I’m not sure tho

Molodets [167]3 years ago
6 0

Answer:

b,a

Explanation:

bc u just past the lesson

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URGENT. Please help.
exis [7]

1). c ... 2). d ... 3). a ... 4). d ... 5). c ... 6). a

7). b-mass ... c-m/s ... d-Newton's 1st ... e-Newton's 2nd

6 0
3 years ago
How are light waves used to bring far away objects into view and how does the eye translate them?
adelina 88 [10]

Answer:

When focused light is projected onto the retina, it stimulates the rods and cones. The retina then sends nerve signals are sent through the back of the eye to the optic nerve. The optic nerve carries these signals to the brain, which interprets them as visual images.

Explanation:

Hope it will help u

5 0
2 years ago
Why does friction cause a spinning stop to eventually stop?
Genrish500 [490]
Mechanical Energy transforms into Thermal due to the moving parts rubbing on eachother creating heat and friction.
7 0
3 years ago
Show all your work clearly on the scratch paper) Two in-phase loudspeakers that emit sound with the same frequency are placed al
Darya [45]

Answer:

Explanation:

The speed of sound in air to be 343 m/s.

Given:

distance 'd' = 5 m

L = 12 m

It can be concluded that path difference must be equal to half of the wavelength when person is observing destructive interference'y' at 1 m distance from the equidistant position

Since

λ/2 = yd/L

λ/2 = (1 x 5)/12

λ = 0.833m

Frequency of the sound is given by,

f = v / λ => 343 / 0.833

f=411.6 Hz

8 0
3 years ago
Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl
tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s

For Train 2 Velocity of the Source,

v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s

Frequency of Source,

f_{s}=500\ Hz

To Find:

Frequency of Observer,

f_{o}=?  (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source v_{s} and observer v_{o}, the original frequency of the sound waves f_{s} and the observed frequency of the sound waves f_{o},

The Equation is

f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

7 0
3 years ago
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