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2 years ago

I need help with this

Physics

2 answers:

Fed [463]2 years ago

3 0

HEY THERE !!

It is called SUBLIMATION.

so the correct answer is (D).

Salsk061 [2.6K]2 years ago

3 0

Your answer would be D.

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A light bulb whose resistance is 240 ohms is connected to a 120 voltage source. What is the current through the bulb?

Sedaia [141]

Answer:

0.5A

Explanation:

Using $R = \frac{V}{I}$ ,

R is the resistance (in Ohms)

V is the voltage (in V)

I is the current (in A)

$240=\frac{120}{I}$

I = 0.5A

7 0

2 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

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3 years ago

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Answer:

the answer is

Mahatma Gandhi

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2 years ago

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( 45 points)

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kogti [31]

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EXPLANATION :

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Hence, the correct answer is that atom will be positively charged.

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