What does the superscript in a chemical formula tell you?

WHICH ONE!!! ASAP FOR A RETAKE FOR SCIENCE PLS

aleksandr82 [10.1K]

I’m 95% sure it’s covalent bonds.

8 0

2 years ago

What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×

xxMikexx [17]

Answer:

9.82 × $10^{-35}$ Hz

Explanation:

De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:

λ = $\frac{h}{mv}$

where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.

Given that: h = 6.63 × $10^{-34}$ Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;

λ = $\frac{h}{mv}$

= $\frac{6.63*10^{-34} }{2.5*2.7}$

= $\frac{6.63 * 10^{-34} }{6.75}$

= 9.8222 × $10^{-35}$

The wavelength of the object is 9.82 × $10^{-35}$ Hz.

4 0

2 years ago

When energy is changed from one form to another, some of the energy is lost to the surroundings as waste thermal energy. The mor

Taya2010 [7]

The answer is B Solar cell just took the test

7 0

3 years ago

Read 2 more answers

Một xe xuất phát lúc 9 giờ 30 phút , chuyển dộng thẳng đều từ A đến B với vận tốc 40 km/h

OLEGan [10]

HUHHHHHHHHHHHHHHHHHHHHH????????

3 0

2 years ago

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 50.9 J and a maximum d

Llana [10]

(a) 2446 N/m

When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:

$E=U=\frac{1}{2}kA^2$

where

U is the elastic potential energy

k is the spring constant

A is the maximum displacement (the amplitude)

Here we have

U = E = 50.9 J

A = 0.204 m

Substituting and solving the formula for k,

$k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m$

(b) 50.9 J

The total mechanical energy of the system at any time during the motion is given by:

E = K + U

where

K is the kinetic energy

U is the elastic potential energy

We know that the total mechanical energy is constant: E = 50.9 J

We also know that at the equilibrium point, the elastic potential energy is zero:

$U=\frac{1}{2}kx^2=0$ because x (the displacement) is zero

Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:

$K=E=50.9 J$

(c) 8.55 kg

The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when

K = 50.9 J (at the equilibrium position)

Kinetic energy can be written as

$K=\frac{1}{2}mv^2$

where m is the mass

Solving the equation for m, we find the mass:

$m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg$

(d) 2.14 m/s

When the displacement is

x = 0.160 m

The elastic potential energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

So the kinetic energy is

$K=E-U=50.9 J-31.3 J=19.6 J$

And so we can find the speed through the formula of the kinetic energy:

$K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s$

(e) 19.6 J

The elastic potential energy when the displacement is x = 0.160 m is given by

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

And since the total mechanical energy E is constant:

E = 50.9 J

the kinetic energy of the block at this point is

$K=E-U=50.9 J-31.3 J=19.6 J$

(f) 31.3 J

The elastic potential energy stored in the spring at any time is

$U=\frac{1}{2}kx^2$

where

k = 2446 N/m is the spring constant

x is the displacement

Substituting

x = 0.160 m

we find the elastic potential energy:

$U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J$

(g) x = 0

The postion at that instant is x = 0, since it is given that at that instant the system passes the equilibrium position, which is zero.

4 0

2 years ago