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dalvyx [7]
3 years ago
10

How many different types of atoms are present in one molecule of aluminum hydroxide, Al(OH)3?

Physics
2 answers:
tatyana61 [14]3 years ago
8 0
There are approximately 3 different types of atoms that are present in one molecule of aluminum hydroxide, AI(OH)3.
Dafna11 [192]3 years ago
6 0
3 is the answer hope this helps
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How high does a rocket have to go above the Earth’s surface so that its weight is reduced to 58.8 % of its weight at the Earth’s
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Explanation:

Below is an attachment containing the solution.

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3 years ago
Which is the correct formula for finding the frequency of an electromagnetic wave? f = c – f = f = + c f =
liubo4ka [24]
The answer for the online test is going to be the last one, D
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The newton is defined as the:
Orlov [11]

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b.

Explanation:

because of fhe speed an object when under the influence of earth gravitational field

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Salmon often jump waterfalls to reach their
natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

v_x = u cos \theta

where u is the initial speed and \theta=37.7^{\circ}. The horizontal distance travelled by the salmon is

d=v_x t = (ucos \theta)t

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

t=\frac{d}{v_x}=\frac{d}{u cos \theta} (1)

Along the vertical direction, the equation of motion is

y=h+u_y t -\frac{1}{2}gt^2

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

u_y = u sin \theta is the vertical component of the initial velocity of the salmon

g=9.81 m/s^2 is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}

and we can solve this formula for u, the initial speed of the salmon:

y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s

5 0
4 years ago
Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals,
Andre45 [30]

Answer:

9241.6 W or 12.39318 hp

Explanation:

u = Initial velocity = 0

v = Final velocity

m = Mass

t = Time taken

Energy

KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules

Power

P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W

Converting to hp

1\ W=\frac{1}{745.7}\ hp

\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp

The power developed by the cheetah is 9241.6 W or 12.39318 hp

7 0
4 years ago
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