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3 years ago

How do catalysts speed up reactions please help quickly

2 answers:

8 0

How do catalysts speed up reactions ?

by reducing the amount of energy required for the reaction to occur. other factors such as <span>catalysts' ability to speed a process are temperature, concentration </span>

nikitadnepr [17]3 years ago

7 0

i think its because They lower the energy barrier of a reaction.

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An circuit has a section AB as shown in the fig. with E= 10V. (sf C_1) = 1.0 F, (sf C_2) = 2.0 F and the potential difference (s

bagirrra123 [75]

Without the fig, we know nothing about AB, C_1, C_2, V_a, or V_b.
The probability of getting any kind of answer from anybody who hasn't
seen or tasted the fig is vanishingly small.

4 0

4 years ago

A source emits monochromatic light of wavelength 495 nm in air. when the light passes through a liquid, its wavelength reduces t

Katyanochek1 [597]

By definition, the refractive index is
n = c/v
where c = 3 x 10⁸ m/s, the speed of light in vacuum
v = the speed of light in the medium (the liquid).

The frequency of the light source is
f = (3 x 10⁸ m/s)/(495 x 10⁻⁹ m) = 6.0606 x 10¹⁴ Hz

Because the wavelength in the liquid is 434 nm = 434 x 10⁻⁹ m,
v = (6.0606 x 10¹⁴ 1/s)*(434 x 10⁻⁹ m) = 2.6303 x 10⁸ m/s

The refractive index is (3 x 10⁸)/(2.6303 x 10⁸) = 1.1406

Answer: a. 1.14

6 0

3 years ago

Dave throws a 10 kg bowling ball straight up in the air. At the very tippy-top of its path, what is it's momentum?

svetoff [14.1K]

Answer:Zero

Explanation:

Given

mass of ball $m=10\ kg$

If the ball is thrown upward then at maximum point velocity of ball is zero because ball is no longer able to move upward

Momentum(P) of a particle is given by

$P=mass\times velocity$

$P=10\times 0$

$P=0$

Therefore at the highest point momentum is zero .

8 0

3 years ago

What is hooke's law? Also what does it say about how elastic materials respond to force? Explain.

gogolik [260]

It <span>states that the force F needed to extend or compress a spring by some distance X is proportional to that distance.

For elastic materials, they extend more in same amount of force, (as they are directly proportional), due to it's elastic nature (presence of large deforming force)

Hope this helps!</span>

3 0

4 years ago

baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the

Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0

3 years ago