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3 years ago

7

What can you observe to know that when clear methane gas is burned it is a chemical change?

1 answer:

aalyn [17]3 years ago

6 0

<span>4) Formation of a gas

When methane burns, it produces the gases water vapor and carbon dioxide.
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I need help with question 6

Irina-Kira [14]

Answer:

She will have to ride her scooter approximately 10 minutes.

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3 years ago

Read 2 more answers

X-rays with an energy of 86 keV scatter off of stationary electrons. How much energy do the photons scattering at 60° have? A. 7

AveGali [126]

Answer:

A. 79.3 keV

Explanation:

Because the procedure involves many steps for its resolution and it works faster on paper and pencil, the detailed solution of this exercise is attached as a scanned image of the procedure for review.

In the procedure, the initial values of the problem and the replacement of these values with the correct formulas for this process are taken into account.

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3 years ago

Can someone help me its A,B,C,D

Tatiana [17]

I think C i’m so sorry if i’m wrong tho.

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3 years ago

The mass of Tesla Model S is 2000 kg and its engine produces

yulyashka [42]

Answer:

The ratio of the power generated by the engines are $\dfrac{2}{3}$

(b) is correct option.

Explanation:

Given that,

Mass of Tesla = 2000 Kg

Torque = 400 Nm

Mass of Mercedes = 1.5 M Tesla

The Benz engine produces 50% more torque than the Tesla.

We need to calculate the power generated by the engines 6 seconds after a start from rest

Using formula of power

$P=F\cdot v$

We know that,

$v=at$

$v=\dfrac{F}{m}t$

$v=\dfrac{\tau}{mr}t$

Put the value of F and v in equation (I)

$P=\dfrac{\tau}{r}\times\dfrac{\tau}{mr}t$

$P=\dfrac{\tau^2t}{mr^2}$

Here, r and t are same for both bodies.

$P\propto\dfrac{\tau^2}{m}$

We need to calculate the ratio of the power generated by the engines

Using formula of power

$\dfrac{P_{T}}{P_{B}}=\dfrac{\tau_{T}^2}{m_{T}}\times\dfrac{m_{B}}{\tau_{B}}$

Put the value into the formula

$\dfrac{P_{T}}{P_{B}}=\dfrac{400^2}{2000}\times\dfrac{3000}{600^2}$

$\dfrac{P_{T}}{P_{B}}=\dfrac{2}{3}$

Hence, The ratio of the power generated by the engines are $\dfrac{2}{3}$

3 0

3 years ago

A force of 50n is applied to an object at an angle of 50 degrees measured relative to the horizontal. Determine the horizontal a

kobusy [5.1K]

Answer:

Fx = 32.14 [N]

Fy = 38.3 [N]

Explanation:

To solve this problem we must decompose the force vector, for this we will use the angle of 50 degrees measured from the horizontal component.

F = 50 [N]

Fx = 50*cos(50) = 32.14 [N]

Fy = 50*sin(50) = 38.3 [N]

We can verify this result using the Pythagorean theorem.

$F = \sqrt{(32.14)^{2}+ (38.3)^{2}} \\F = 50 [N]$

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3 years ago