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3 years ago

5

according to the law of conservation of energy the total amount of energy in the universe does not change but remains constant t

rue or false ?

1 answer:

allsm [11]3 years ago

7 0

True hope this helps

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What are the signs of a chemical change

Rom4ik [11]

Gas or bubbles are formed
Heat, light, or sound is given off
Change in color
Different taste or smell (never taste or smell anything in a chemical lab, cause if u do bad reactions may occur to your body, unless directed to do so.)
New substance is formed (a precipitate)
<span>Temperature (exothermic or endothermic reactions) ( exothermic RELEASES heat energy) (endothermic ABSORBS energy)</span>

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3 years ago

If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how high above the earth's surface

Alecsey [184]

Answer: h = 3R

Explanation:

Using the law of conservation of energy,

Total energy at the beginning of the launch would be equal to total energy at any point.

kinetic energy + gravitational potential energy = constant

Initial energy of the projectile = $\frac{1}{2}mv_e^2-\frac{GMm}{R}$ ... (1)

where R is the radius of the Earth, M is the mass ofthe Earth, m is the mass of the projectile.

escape velocity, $v_e=\sqrt{\frac{2GM}{R}}$

Total energy at height h above the Earth where speed of the projectile is half the escape velocity:

$\frac{1}{2}m(v_e/2)^2-G\frac{Mm}{R+h}$ ...(2)

(1)=(2)

⇒ $\frac{1}{2}m(v_e/2)^2-G\frac{Mm}{R+h} = \frac{1}{2}mv_e^2-\frac{GMm}{R}$

⇒ $G\frac{Mm}{R}-G\frac{Mm}{R+h}= \frac{1}{2}m \frac{3}{4}v_e^2 =\frac{1}{2}m \frac{3}{4} (\sqrt{\frac{2GM}{R}})^2$

⇒ $\frac{GM(R+h-R)}{R(R+h)} = \frac{3}{4}(\frac{GM}{R})$

⇒ $\frac{h}{R+h} = \frac{3}{4}$

⇒h = 3R

Thus, at height equal to thrice radius of Earth, the speed of the projectile would reduce to half of escape velocity.

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3 years ago

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Convection currents move tectonic plates in two locations as described below:

Verizon [17]

The answer to this is C.

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4 years ago

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A trumpeter plays at a sound level of 75dB. three equally loud trumpet players join in. what is the new sound level?

Serjik [45]

The answer is approx 87 decibels.

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3 years ago

Water traveling along a straight portion of a river normally flows fastest in the middle, and the speed slows to almost zero at

miv72 [106K]

Answer:

The angle at which the boat must head is $- 22.47^{\circ}$

Solution:

As per the solution:

Distance between the parallel banks, d = 40 m

The maximum speed of water, v' = 3 m/s

constant speed, u' = 5 m/s

Also,

The speed of water of the river at a distance of 'x' units from the west bank is given as a sine function:

$f(x) = 3sin(\frac{\pi x}{40})$ (2)

Now, to determine the angel at which the boat must head:

The velocity of the engine of the boat:

v = $u'cos\theta\hat{i} + u'sin\theta\hat{j}$

v = $5cos\theta \hat{i} + 5sin\theta\hat{j}$

The abscissa of the boat at time t:

v = $5cos\theta t\hat{i}$

Now, from above and eqn(1) , we can write:

$f(5cos\theta t) = 3sin(\frac{\pi \times 5cos\theta t}{40})$

Now, boat's velocity at time t:

v = $5cos\theta \hat{i} + (5sin\theta + 3sin(\frac{\pi \times 5cos\theta t}{40})\hat{j}$

In order to obtain the position of the boat, we integrate both the sides, we get:

r = $5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40})\hat{j}$ + C (3)

Now, at r = 0:

0 = $5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40})\hat{j}$ + C

C = $\frac{24}{\pi cos\theta}\hat{j}$

Now, from eqn (3)

r = $5sin\theta t\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta t}{40} + \frac{24}{\pi cos\theta})\hat{j}$ (4)

the baot will reach the point at y = 0 and x = 40

Now,

40 = $5cos\theta t$

$t = \frac{8}{cos\theta}$

Substituting the above value of 't' in eqn (4):

r = $5sin\theta \frac{8}{cos\theta}\hat{i} + (5sin\theta - \frac{3\times 40}{5\pi cos\theta}cos(\frac{\pi \times 5cos\theta \frac{8}{cos\theta}}{40} + \frac{24}{\pi cos\theta})\hat{j}$

We get:

$48 + 40\pi sin\theta = 0$

$\theta = sin^{- 1}(\frac{- 48}{40\pi}) = - 22.47^{\circ}$

8 0

3 years ago