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3 years ago

15

The Car That Ran On Chocolate Achieve 3000 - Go On Achieve go to physical science scroll down untill you see The Car That Runs O

n Chocolate. give me all the 8 answers and questions.

1 answer:

worty [1.4K]3 years ago

5 0

Explanation:

Mobil Yang Berlari Di Atas Coklat Mencapai 3000 - Terus Mencapai pergi ke ilmu fisika gulir ke bawah sampai Anda melihat Mobil Yang Berjalan Di Atas Coklat. beri saya semua 8 jawaban dan pertanyaan.

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Natural causes are the reason for the recent warming trend of the average global temperature.

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The reasoning for this is false

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Hold a pencil in front of your eye at a position where its blunt end just blocks out the Moon.

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Answer:

$D = 3.55 \times 10^6 m$

Explanation:

Light rays coming from moon is blocked by the pencil

so as per figure we know that angle subtended by pencil and angle subtended by moon must be same

so we have

$Angle = \frac{Arc}{Radius}$

so we have

$\frac{D}{3.8 \times 10^8 m} = \frac{0.7 cm}{75 cm}$

so we have

$D = 3.55 \times 10^6 m$

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The _______ is the time required for one complete wave oscillation to occur.

aleksandrvk [35]

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3 years ago

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A train locomotive is pulling two cars of the same mass behind it. Determine the ratio of the tension in the coupling (think of

Anna007 [38]

Answer:

The ratio is $\frac{F_{T1}}{F_{T2}} = 2$

Explanation:

The diagram for this question is shown on the first uploaded image

Here we are assume the acceleration of the train is a

which makes the acceleration of each car a

From the question we are told that

Considering the second car

The force causing it s movement is mathematically represented as

$F_{T2} = ma$

Considering the first car

The force causing it s movement is mathematically represented as

$F = F_{T1} -F_{T2} = ma$

=> $F_{T1} -ma = ma$

=> $F_{T1} = 2 ma$

=> $\frac{F_{T1}}{ma} = 2$

=> $\frac{F_{T1}}{F_{T2}} = 2$

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3 years ago

You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an el

Lapatulllka [165]

Answer:

$a=0.5418\ m.s^{-2}$ upwards

$a=1.283\ m.s^{-2}$ downwards

Explanation:

Given:

weight of the person, $w=688\ N$

So, the mass of the person:

$m=\frac{w}{g}$

$m=\frac{688}{9.81}$

$m=70.132\ kg$

Now if the apparent weight in the elevator, $w_a= 726\ N$

<u>Then the difference between the two weights is :</u>

$\Delta w=w_a-w$

$\Delta w=726-688$

$\Delta w=38\ N$ is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

$a=\frac{\Delta w}{m}$

$a=\frac{38}{70.132}$

$a=0.5418\ m.s^{-2}$ upwards, because the normal reaction that due to the weight of the body is increased here.

Now if the apparent weight in the elevator, $w_a= 598\ N$

<u>Then the difference between the two weights is :</u>

$\Delta w=w-w_a$

$\Delta w=688-598$

$\Delta w=90\ N$ is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

$a=\frac{\Delta w}{m}$

$a=\frac{90}{70.132}$

$a=1.283\ m.s^{-2}$ downwards, because the normal reaction that due to the weight of the body is decreased here.

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3 years ago

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