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7 days ago

What is the least possible initial kinetic energy in the oxygen atom could have and still excite the cesium atom?

Physics

1 answer:

inna [2.2K]7 days ago

8 0

K=E[(m+M)/M] Kmin=4.4

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An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte

Ostrovityanka [2204]

Answer:

35.79 meters

Explanation:

We have an archer, and there is a target. Denote the distance between them as d.

The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

$v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d$

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time $t_{sound}$ . Thus, we can derive:

$v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d$ .

Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:

$t_{arrow} + t _{sound} = 1 s$ .

Using this relationship in the distance formula for sound allows us to write:

$340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d$ .

Substituting the value of d from the first equation yields:

$40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow})$ .

Now, after some calculations, we can proceed further:

$40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s$ .

Finally, the value is inserted into the initial equation:

$40 \frac{m}{s} * t_{arrow} = d$

$40 \frac{m}{s} * 340/380 s = 35,79 s = d$

6 0

1 day ago

Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g

Maru [2355]

Answer:

a) $X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

b) D does not influence the long-term results.

Explanation:

Given that

$\dfrac{dx}{dt}=-k(x-A)$

A = A0 cos(ωt)

$\dfrac{dx}{dt}=-k(x-A_o cos(\omega t))$

$\dfrac{dx}{dt}+kx=kA_o cos(\omega t)$ This is a linear equation hence the integration factor, I

$I=e^{\int kdt}$

$I=e^{kt}$ Now using the characteristics of linear equations

$e^{kt} X=\int e^{kt} kA_o cos(\omega t) dt +C$

$e^{kt} X= kA_o \dfrac{e^{kt}}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+C$

b) At t= 0

$X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

Thus, the initial condition

does not affect the long-term outcome.

$X(0)=\dfrac{k^2A_o}{\omega^2+k^2}+C$

5 0

8 days ago

A cave diver enters a long underwater tunnel, when her displacement with respect to the entry point is 20m,she accidentally drop

Maru [2355]

Answer:

3x864/y488bjehdksuwiieirjr

4 0

27 days ago

Why must the height of the meniscus in the graduated cylinder match the height of the water in the tub when measuring volume?

serg [2593]

Answer:

Refer to the explanation.

Explanation:

To grasp the concept, start with the fundamental idea of meniscus in a graduated cylinder.

"The meniscus refers to the curve noticed at the surface of a liquid due to its container. Depending on liquid surface tension and its adherence to the container’s walls, this meniscus may be concave or convex".

Given this definition, when measuring the volume of water, it's crucial to read the measurement at the bottom of the meniscus curve, as water forms a concave shape.

Taking the reading at eye level ensures an accurate volume, as it aligns with this method.

Additionally, this way of reading balances total pressure with atmospheric pressure by adjusting the cylinder's height to match the water level.

8 0

17 days ago

A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft

Sav [2226]

Answer:

The water level increases more when the cube is above the raft before it sinks.

Explanation:

The principle involved is based on Archimedes' theory, which states that immersing an object in water will raise the initial water level. This means that the height of the water in the container increases. The increase in water volume corresponds to the volume of the submerged object.

We can consider two scenarios: when the steel cube rests on the raft and when it settles at the pool's bottom.

When the cube rests at the pool’s bottom, the volume will indeed rise, and we can ascertain this using the cube's volume.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

When an object floats, it's because the densities of the object and water are in equilibrium.

$Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]$

The formula for density is:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyant force can be calculated with this equation:

$F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]$

Vs > Vc indicates that the combined volume of the raft and the cube exceeds that of the cube alone resting at the bottom of the pool. Hence, when the cube is positioned above the raft, the water level rises more before it becomes submerged.

7 0

1 month ago