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3 years ago

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If the activation energy required for a chemical reaction were reduced, what would happen to the rate of the reaction?The rate w

ould increase.The rate would decrease.The rate would remain the same.The rate would go up and down.

1 answer:

VikaD [51]3 years ago

8 0

Answer:the rate of the reaction will decrease

Explanation:the lower the activation energy the lower the rate at which products would be forked because there won't be effective collision to give product

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A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its

gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

$d = 2\pi r$ ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

$T = \frac{2\pi r}{v}$

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
$v = \sqrt{\frac{Gm}{r}}$

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
$v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s$

Now, we can solve for the period:

$T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}$

7 0

2 years ago

At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?

sergey [27]

1) Initial upward acceleration: $6.0 m/s^2$

2) Mass of burned fuel: $0.10\cdot 10^4 kg$

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude $F_g = mg$ , in the downward direction, where

$m=1.9\cdot 10^4 kg$ is the rocket's mass

$g=9.8 m/s^2$ is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

$T=3.0\cdot 10^5 N$

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

$T-mg=ma$ (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

$a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2$

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

$a'=6.9 m/s^2$

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

$T-m'g=m'a'$

where

$m'$ is the new mass of the rocket

Re-arranging the equation and solving for m', we find

$m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg$

And since the initial mass of the rocket was

$m=1.9 \cdot 10^4 kg$

This means that the mass of fuel burned is

$\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg$

3 0

3 years ago

How does what happens to the particles in a substance during melting differ from what happens during freezing?

sasho [114]

When in the melting process particles start to move more freely when in the freezing process particles tend to slow and vibrate in place

8 0

3 years ago

what happens to the volume when temperature is held constant and the pressure is increased to 125 kPa

Semenov [28]

The volume decreases, by a factor of

(the original pressure/(125 kPa).

5 0

3 years ago

When 1,250^3/4 is written in simplest radical form, which value remains under the radical?

GaryK [48]

Answer:

$125\sqrt[4]{8}$

Explanation:

A number of the form

$a^{\frac{m}{n}}$

can be re-written in the radical form as follows:

$\sqrt[n]{a^m}$

In this problem, we have:

a = 1,250

m = 3

n = 4

So, if we apply the formula, we get

$1,250^{\frac{3}{4}}=\sqrt[4]{(1,250)^3}$

Then, we can rewrite 1250 as

$1250 = 2\cdot 5^4$

So we can rewrite the expression as

$=\sqrt[4]{(2\cdot 5^4)^3}=5^3 \sqrt[4]{2^3}=125\sqrt[4]{8}$

7 0

3 years ago

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