Which energy transformation converts energy from the sun’s core to light energy needed by plants?

The second-order decomposition of hi has a rate constant of 1.80 x 10−3 m−1 s−1. How much hi remains after 45.6 s if the initial

zlopas [31]

Answer:

2.9 M

Explanation:

The concentration-time equation for a second order reaction is:

1/[A] = kt + 1/[A°]

Where,

A = concentration remaining at time, t

A° = initial concentration

k = rate constant

1/[A] = (1.80 x 10^-3) * (45.6) + 1/3.81

1/[A] = 0.345

= 1/0.345

= 2.9 M.

6 0

3 years ago

Complete the equation???

soldi70 [24.7K]

Here in nuclear reaction we can say that sum of neutrons and protons in reactant side and product side will be same always

Here mass number on the product side is given to us

so sum of mass number is given as

$A_1 + A_2 = 265 + 1 = 266$

now on the reactant side also the number must be same

$A_1' + A_2' = 58 + x$

now we will have

$58 + x = 266$

$x = 208$

Now number of protons on product side is given as

$P_1 + P_2 = 108 + 0$

now we also know that atomic number of Fe is 26

so now we will have

$P_1' + P_2' = 108$

$26 + P_2' = 108$

$P_2' = 82$

now the equation is given as

$_{26}^{58}Fe + _{82}^{208}Pb = _{108}^{266}Hs + _0^1X$

8 0

2 years ago

A 1,200 kg car travels at 20 m/s. what is it’s momentum ?

WINSTONCH [101]

Answer:c vx v xv xv x

Explanation:

mlmcmcm

6 0

3 years ago

Why am alive??????????????

Alex777 [14]

Answer:

because i love you <3

Explanation:

7 0

2 years ago

Read 2 more answers

The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the

tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.

A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.

B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

6 0

3 years ago