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Veseljchak [2.6K]

3 years ago

12

(ANSWER NOW PLS) A diver with a mass of 90kg is at a height of 10m, and he has not jumped off of the board yet (v=0m/s) what's h

is kinetic energy?

1 answer:

Korolek [52]3 years ago

3 0

Answer: there is zero kinetic energy but there is Gravitational Potential Energy (GPE) and GPE = 8826.3 J

Explanation:

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In the writing of ionic chemical formulas, what factor is "crossed over" in the crossover rule?

Aleks04 [339]

In the writing of ionic chemical formulas the value of each ion's charge is crossed over in the crossover rule.

Rules for naming Ionic compounds

Frist Rule
The cation (element with a negative charge) is written first in the name then the anion(element with a positive charge) is written second in the name.
Second rule
When the formula unit contains two or more of the same polyatomic ion, that ion is written in parentheses with the subscript written outside the parentheses.
Example: Sodium carbonate is written as Na₂CO₃ not Na₂(CO)₃
Third rule
If the cation is a metal ion with a fixed charge then the name of the cation will remain the same as the (neutral) element from which it is derived (Example: Na+ will be sodium).
If the cation is a metal ion with a variable charge, the charge on the cation is indicated using a Roman numeral, in parentheses, immediately following the name of the cation (example: Fe³⁺ = iron(III)).

Fourth rule
If the anion is a monatomic ion, the anion is named by adding the suffix <em>-ide</em> to the root of the element name (example: F = Fluoride).

The oxidation state of each ion is also important, thus in the crossover rule, the value of each ion's charge is crossed over.

Learn more about chemical formulas here:

<u>brainly.com/question/11995171</u>

#SPJ4

3 0

2 years ago

If wheel turning at a constant rate completes 100 revolutions in 10 s its angular speed is:

GenaCL600 [577]

The frequency of the wheel is given by:
$f= \frac{N}{t}$
where N is the number of revolutions and t is the time taken. By using N=100 and t=10 s, we find the frequency of the wheel:
$f= \frac{100}{10 s}=10 s^{-1}$

And now we can find the angular speed of the wheel, which is related to the frequency by:
$\omega=2 \pi f=2 \pi (10 s^{-1})=62.8 s^{-1}$

6 0

4 years ago

AYUDAAA! <br> Hola, alguien podría ayudarme en esto la C), me ha complicado demasiado!!

Softa [21]

Answer:

I dont know spanish... and i live in Cali... wow....

Explanation:

6 0

3 years ago

Two groups of students were tested to compare their speed working math problems. Each group was given the same problems. One gro

oee [108]

Answer:

The people with caculators will probably answer faster due to thier ablitiy to use a device of technology

3 0

4 years ago

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

0

0

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

3 years ago