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3 years ago

Plz help i really need someone help

Physics

1 answer:

lesantik [10]3 years ago

6 0

Answer:

i can't see the picture, it is blocked off, can you write down your question?

Explanation:

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2. nakabuti ba sa mga bansa sa kanlurang asya ang sistemang mandato na ipinatupad ng mga bansang kanluranin

Vilka [71]

yes o=====/::::::::::::::::>

8 0

3 years ago

A 1.75 kg stone falls from the top of a cliff and strikes the ground at a speed of 58.8 m/s. What is the height of the cliff??

Reil [10]

Answer:

172.9m

Explanation:

h = 1/2 gt^

First calculat for t using

v = gt

t = v/g = 58.8/10

= 5.88secs

now h = 1/2 x 10 x 5.88^2

h =1/2 x 10 x 34.57

= 345.74/2

= 172.9m

6 0

3 years ago

The earth's gravity is pulling on you. Are you pulling on the earth? Explain why or why not.

mel-nik [20]

Answer:

"we both attract each other with the same force but we know that attraction between two bodies depends upon their mass, greater the mass of two bodies is the force of attraction between them"(got this off the internet).

7 0

2 years ago

Suppose that the sound level of a conversation is initially at an angry 71 dB and then drops to a soothing 54 dB. Assuming that

Goryan [66]

Answer:

Explanation:

In the decibel scale , intensity of sound changes logarithmically as follows

$10log\frac{I}{I_0} =$ Value in decibel scale , the value of I₀ = 10⁻¹² W /m².

Putting the values

$10log\frac{I}{10^{-12}} = 71$

$log\frac{I}{10^{-12}} = 7.1$

$\frac{I}{10^{-12}} = 10^{7.1}$

$I= 10^{-4.9}$ W/m²

Similarly for 54 dB sound intensity can be given as follows

I = 10⁻¹² x $10^{5.4}$

$I= 10^{-6.6 }$ W / m²

For intensity of sound the relation is as follows

I = 2π²υ²A²ρc where υ is frequency , A is amplitude , ρ is density of air and c is velocity of sound .

Putting the given values for 71 dB

$I= 10^{-4.9}$ = 2π² x 504²xA²x 1.21 x 346

A² = 60.03 x 10⁻¹⁶

A = 7.74 x 10⁻⁸ m

For 54 dB sound

$10^{-6.6}$ = 2π² x 504²xA²x 1.21 x 346

A² = 1.1978 x 10⁻¹⁶

A = 1.1 x 10⁻⁸ m

6 0

3 years ago

An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t

scoundrel [369]

Answer:

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

$C = \frac{Q}{V} = e_0\frac{A}{d}$

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

$V = \frac{Q}{e_0} *\frac{d}{A}$

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

$V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V$

So, if we double the distance between the plates, the potential difference will also be doubled.

6 0

3 years ago