Ask question

Ask question

All categories

3 years ago

5

Why won’t anyone help me please anybody help me I really need help .

1 answer:

irga5000 [103]3 years ago

8 0

Answer:

1➡️ this is the method of decomposition

2➡️ H2 and O2

3➡️ b

sorry if I am wrong

You might be interested in

A current of 3.60A flows for 15.3s through a conductor. Calculate the number of electrons that pass through a point in the condu

Artyom0805 [142]

3.60 A = 3.60 coulombs of charge per second

(3.60 Coul/sec) x (15.3 sec) = 55.08 coulombs of charge

1 coulomb of charge is carried by 6.25 x 10^18 electrons

Number of electrons =

(55.08 Coul) x (6.25 x 10^18 e/coul) = <em>3.4425 x 10^20 electrons</em>

7 0

4 years ago

The gravitational constant G was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance

belka [17]

The gravitational force between the spheres is

$F_{\rm g}=\dfrac{G(188\,\mathrm{kg})(0.93\,\mathrm{kg})}{(0.27\,\mathrm m)^2}\approx1.6\times10^{-7}\,\mathrm N$

where <em>G</em> = 6.674 x 10⁻¹¹ N m²/kg².

The weight of the lighter sphere is

$F_{\rm w}=(0.93\,\mathrm{kg})g\approx9.1\,\mathrm N$

where <em>g</em> = 9.80 m/s².

The ratio between the two forces is then

$\dfrac{F_{\rm g}}{F_{\rm w}}\approx1.8\times10^{-8}$

7 0

3 years ago

How do l calculate e<br> nergy

aivan3 [116]

The formula for energy of motion is KE = .5 x m x v^2

Ke= Kinetic Energy in Joules

m = Mass in Kilograms

v = Velocity in Meters per Second

4 0

3 years ago

Often called velocity this is the velocity of an object at a particular moment in time

Yuliya22 [10]

Instantaneous speed?

8 0

3 years ago

Hanging from a horizontal beam are nine simple pendulums of the following lengths:

LenaWriter [7]

Answer:

Options d and e

Explanation:

The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.

We can get the length of the pendulums likely to oscillate with the formula;

$L =\frac{g}{w^{2} }$

where g=9.8m/s

ω= 2rad/s to 4rad/sec

when ω= 2rad/sec

$L= \frac{9.8}{2^{2} }$

L = 2.45m

when ω= 4rad/sec

$L=\frac{9.8}{4^{2} }$

L = 9.8/16

L=0.6125m

L is between 0.6125m and 2.45m.

This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.

Have a great day ahead

8 0

4 years ago