Can nuclear be reused? Why or why not? Use in your own words.

Hoping for funny and stupid answers

harina [27]

Answer:

ive once eaten octopus.. i dont recommend.. ANYWAY

i would rather shoot spaghetti out of my fingers ofc, that like free food and sneezing meatballs just sounds painful

Explanation:

6 0

2 years ago

Read 2 more answers

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

2 years ago

Explain why atoms only emit certain wavelengths of light when they are excited. Check all that apply. Check all that apply. Elec

JulsSmile [24]

Answer:

Explanation:

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. FALSE. The specific lines are obseved because of the energy level transition of an electron in an specific level to another level of energy.

The energies of atoms are not quantized. FALSE. The energies of the atoms are in specific levels.

When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. FALSE. During absorption, a specific wavelength of light is absorbed, not emmited.

Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. TRUE. Again, you can observe just the transition due the change of energy of an electron in the quantized energy level

When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted. TRUE. The electron decreases its energy releasing a specific wavelength of light.

The energies of atoms are quantized. TRUE. In fact, the energy of all subatomic, atomic, and molecular particles is quantized.

7 0

2 years ago

Salmon often jump waterfalls to reach their

Y_Kistochka [10]

Answer:

5.38 m/s

Explanation:

Given (in the x direction):

Δx = 2.45 m

v₀ = v cos 42.5°

a = 0 m/s²

Δx = v₀ t + ½ at²

(2.45 m) = (v cos 42.5°) t + ½ (0 m/s²) t²

2.45 = (v cos 42.5°) t

t = 3.32 / v

Given (in the y direction):

Δy = 0.373 m

v₀ = v sin 42.5°

a = -9.8 m/s²

Δx = v₀ t + ½ at²

(0.373 m) = (v sin 42.5°) t + ½ (-9.81 m/s²) t²

0.373 = (v sin 42.5°) t − 4.905 t²

0.373 = (v sin 42.5°) (3.32 / v) − 4.905 (3.32 / v)²

0.373 = 2.25 − 54.2 / v²

v = 5.38

Graph:

desmos.com/calculator/5n30oxqmuu

4 0

2 years ago

A balloon is released from a tall building. The total mass of the balloon including the enclosed gas is 2.0 kg. Its volume is 5.

AleksandrR [38]

(B) The balloon will rise because the upward buoyant force is greater than its weight.

Explanation:

In order to evaluate what happens to the balloon, we need to compare the magnitude of the two forces acting on the balloon:

- Its weight, W, acting downward

- The buoyant force, B, acting upward

The weight of the balloon is given by:

$W=mg$

where

m = 2.0 kg is the mass of the balloon

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(2.0)(9.8)=19.6 N$

The buoyant force on the balloon is given by:

$B=\rho V g$

where

$\rho = 1.3 kg/m^3$ is the air density

$V=5.0 m^3$ is the balloon's volume

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$B=(1.3)(5.0)(9.8)=63.7 N$

We observe that the buoyant force B is larger than the weight, so the balloon will accelerate upward, and the correct answer is

(B) The balloon will rise because the upward buoyant force is greater than its weight.

#LearnwithBrainly

3 0

2 years ago