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3 years ago

All objects in the universe radiate some form of energy. a. True b. False

2 answers:

5 0

True . but also depends on which kind of object because if you have metal it will not really form energy in my opinion it will just attract another object.

hope i helped

Harrizon [31]3 years ago

3 0

That's true. The only way to stop an object from radiating energy
is to cool it to absolute zero. Since the temperature of space is
roughly 3 degrees above absolute zero, the atoms or molecules
of every object have some kinetic energy, and the object radiates
some heat.

Of course it also absorbs heat at the same time, mostly from the
huge number of stars shining on it.

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An electron is trapped in an infinite square-well potential of width 0.6 nm. If the electron is initially in the n = 4 state, wh

grandymaker [24]

Answer:

E₁ = 1.042 eV

E₄₋₃= 7.29 eV

E₄₋₂= 12.50 eV

E₄₋₁= 15.63 eV

E₃₋₂= 5.21eV

E₃₋₁= 8.34eV

E₂₋₁= 3.13eV

Explanation:

The energy in an infinite square-well potential is giving by:

$E = \frac {h^{2} n^{2}}{8mL^{2}}$

where, h: Planck constant = 6.62x10⁻³⁴J.s, n: is the energy state, m: mass of the electron and L: widht of the square-well potential

The energy of the electron in the ground state, n = 1, is:

$E_{1} = \frac {(6.62 \cdot 10^{-34})^{2} (1)^{2}}{(8) (9.11 \cdot 10^{-31}) (0.6 \cdot 10^{-9} m)^{2}}$

$E_{1} = 1.67 \cdot 10^{-19} J = 1.042 eV$

The photon energies that are emitted as the electron jumps to the ground state is the difference between the states:

$E_{\Delta n} = \Delta n^{2} E_{1}$

$E_{(4 - 3)} = (4^{2} - 3^{2}) 1.042 eV = 7.29eV$

$E_{(4 - 2)} = (4^{2} - 2^{2}) 1.042 eV = 12.50eV$

$E_{(4 - 1)} = (4^{2} - 1^{2}) 1.042 eV = 15.63eV$

$E_{(3 - 2)} = (3^{2} - 2^{2}) 1.042 eV = 5.21eV$

$E_{(3 - 1)} = (3^{2} - 1^{2}) 1.042 eV = 8.34eV$

$E_{(2 - 1)} = (2^{2} - 1^{2}) 1.042 eV = 3.13eV$

Have a nice day!

7 0

2 years ago

What kind of motion is the motion of the “flying bully”? How much and what is

lesantik [10]

The flying bully is a move used in the Superhero Movie "Hancock", it is not a real motion in our universe. However, the direction would be towards the target object and the acceleration would be maximal.

7 0

3 years ago

An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acce

ankoles [38]

Answer:

d=510.2m

t=10.2s

Explanation:

The formulas for accelerated motion are:

$v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}$

From them we can get $v^2=v_0^2+2ad$ .

We have:

$v-v_0=at\\t=\frac{v-v_0}{a}$

And substitute:

$x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}$

We multiply both sides by 2a, and continue:

$2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2$

Being d the displacement $x-x_0$ , we have $v^2=v_0^2+2ad$

For our exercise, we will write this as:

$d=\frac{v^2-v_0^2}{2a}$

And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:

$d=\frac{-v_0^2}{2a}=\frac{-(100m/s)^2}{2(-9,8m/s^2)}=510.2m$

For the time we use:

$t=\frac{v-v_0}{a}=\frac{-v_0}{a}=\frac{-(100m/s)}{(-9.8m/s^2)}=10.2s$

6 0

3 years ago

A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse

Elanso [62]

Answer:

$v_{f} = 0.51 \frac{m}{s}$

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg : crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)

μk: coefficient of kinetic friction

N : Normal force (N)

Problem development

We apply the formula (1)

∑Fy = m*ay , ay=0

N-W = 0

N = W

N = 882 N

We replace the data in the formula (2)

Ff= μk*N = 0.351* 882 N

Ff= 309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax , ax=0

282 N - 309.58 N = 90*a

a= (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

$v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}$

$v_{f} = 0.51 \frac{m}{s}$

8 0

3 years ago

PLEASE HELP!!!!

VashaNatasha [74]

In my opinion I would say all of the above; because people's job depends on where they live or how their environment is. The way they travel is also affected because we can't travel by car through water and can't use a boat in land. Free time as well because people like to travel and or just stay at home depending on the weather which is part of earths features.

7 0

3 years ago