Prove the identity <br>Trigonometry grade 10

What is ham, I'm totally serious?

Sergeeva-Olga [200]

Ham is meat made from a pig =]

3 0

3 years ago

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A race car accelerates at a constant rate from 20 m/s to 60 m/s in a time of 2 seconds. What is the car’s acceleration?

Vlada [557]

$\huge\fbox{Answer ☘}$

Given ,

initial velocity , u = 20 m/s

final velocity , v = 60 m/s

time taken = 2 seconds

Now ,

$\bold{acceleration = \frac{v - u}{t} } \\ \\ = > \bold{ \frac{60 - 20}{2} } \\ \\ = > \bold{\frac{40}{2} } \\ \\ \bold{ => 20 \: ms {}^{ - 2}}$

hope helpful~

5 0

2 years ago

A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0

viktelen [127]

Hi there!

Initially, we have gravitational potential energy and kinetic energy. If we set the zero-line at H2 (12.0m), then the ball at the second building only has kinetic energy.

We also know there was work done on the ball by air resistance that decreased the ball's total energy.

Let's do a summation using the equations:
$KE = \frac{1}{2}mv^2 \\\\PE = mgh$

Our initial energy consists of both kinetic and potential energy (relative to the final height of the ball)

$E_i = \frac{1}{2}mv_i^2 + mg(H_1 - H_2)$

Our final energy, since we set the zero-line to be at H2, is just kinetic energy.

$E_f = \frac{1}{2}mv_f^2$

And:
$W_A = E_i - E_f$

The work done by air resistance is equal to the difference between the initial energy and the final energy of the soccer ball.

Therefore:
$W_A = \frac{1}{2}mv_i^2 + mg(H_1 - H_2) - \frac{1}{2}mv_f^2$

Solving for the work done by air resistance:
$W_A = \frac{1}{2}(.450)(15.1^2)+ (.450)(9.8)(30.2 - 12) - \frac{1}{2}(.450)(19.89^2)$

$W_A = \boxed{42.552 J}$

8 0

2 years ago

You and your friends are having a discussion about weight. He/she claims that he/she weighs less on the 100th floor of a buildin

Viktor [21]

Answer:

if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

Explanation:

The weight of a person in the force with which the Earth attracts the person, therefore can be calculated using the law of universal attraction

F = G m M / r²

Where m is the mass of the person, M the masses of the earth

Let's call the person's weight at ground level as Wo and suppose the distance to the center of the Earth is Re

W₀ = G m M / Re²

In the calculation of the weight of the person on the 100th floor the only thing that changes is the distance

r = Re + 100 r₀

Where r₀ is the distance between the floors, which is approximately 2.5 m, so the distance is

r = Re + 250

We substitute

W = G m M / r²

W = G m M / (Re + 250)²

The value of Re is 6.37 10⁶ m, so we can take it out as a factor and perform a serial expansion of the remaining fraction

W = G m M / Re² (1+ 250 / Re)²

(1 + 250 / Re)⁻² = 1 + (-2) 250 / Re + (-2 (-2-1)) / 2 (250 / Re)² +….

The value of the expression is

(1 + 250 / Re)⁻² = 1 -2 250 / 6.37 10⁶ -30 (250 / 6.37)² 10⁻¹² + ...

We can see that the quadratic term is very small, which is why we despise it, we substitute in the weight equation

W = G m M / Re² (1 - 78.5 10⁻⁶)

Remains

W = Wo (1 - 7.85 10⁻⁵)

We can see that if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

4 0

3 years ago

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How many items are present in the compound (NH4)2Cr2O7

SVEN [57.7K]

In 2Cr2O7 there’s 2 items; Cr and O

5 0

3 years ago

Prove the identity Trigonometry grade 10​

Prove the identity Trigonometry grade 10