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Lady bird [3.3K]
3 years ago
15

Can u help me on the first one: a ship going out of sea goes out of sight

Physics
2 answers:
k0ka [10]3 years ago
7 0

A ship sailing away from you gradually goes out of sight,
starting from the bottom and moving up the sides of the ship. 
The Earth is a sphere.  As the ship gets farther away from
you, it's working its way around the curve of the sphere. 
Since you can't see around the curve of the Earth, the
lowest point on the ship that you can see moves slowly
up the ship.
 
kow [346]3 years ago
7 0
The earth is a sphere
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What is the reaction force if a girl pulls on a cow?
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Answer : B) The cow pulls back on the girl.

From newton’s third law we know that every action has a reaction force pushing back. So when the girl pulls on a cow, the cow is pulling back on her.
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A ______ is a closed loop containing a source of electrical energy and a load.
gavmur [86]

Answer:

d. circuit

Explanation:

the parts of the circuit consists of a load or resistance

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3 years ago
Which is an example of someone responding from a personal perspective?
Anika [276]

Answer:

C

Explanation:

If Ami is saying she likes it then it it personal. If you are speaking from statistics and studies it is impersonal and technically not from there perspective. All of these do this except C.

5 0
3 years ago
Read 2 more answers
You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh
Alexandra [31]

Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m  , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

\beta =10\ log\dfrac{I}{10^{-12}}

\beta _1=10\ log\dfrac{I_1}{10^{-12}}

20=10\ log\dfrac{I_1}{10^{-12}}

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

\beta _2=10\ log\dfrac{I_2}{10^{-12}}

60=10\ log\dfrac{I_1}{10^{-12}}

10⁶ x  10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity  ,A=Area

\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}

\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}

r₂=0.1 m

4 0
3 years ago
A kayakeris paddling 2.50 m/s at an angle of 45° (northeast) and the current is moving 1.25 m/s at an angle of 315° (southeast).
PIT_PIT [208]

The kayaker has velocity vector

<em>v</em> = (2.50 m/s) (cos(45º) <em>i</em> + sin(45º) <em>j</em> )

<em>v</em> ≈ (1.77 m/s) (<em>i</em> + <em>j</em> )

and the current has velocity vector

<em>w</em> = (1.25 m/s) (cos(315º) <em>i</em> + sin(315º) <em>j</em> )

<em>w</em> ≈ (0.884 m/s) (<em>i</em> - <em>j</em> )

The kayaker's total velocity is the sum of these:

<em>v</em> + <em>w</em> ≈ (2.65 m/s) <em>i</em> + (0.884 m/s) <em>j</em>

That is, the kayaker has a velocity of about ||<em>v</em> + <em>w</em>|| ≈ 2.80 m/s in a direction <em>θ</em> such that

tan(<em>θ</em>) = (0.884 m/s) / (2.65 m/s)   →   <em>θ</em> ≈ 18.4º

or about 18.4º north of east.

5 0
3 years ago
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