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11 hours ago

A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

r atmosphere is ozone, O3O3. In particular, ozone absorbs radiation with frequencies around 9.38×1014 HzHz . What is the wavelength λλlambda of the radiation absorbed by ozone? Express your answer in nanometers.
λ = ___.

Physics

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A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn

Ostrovityanka [3204]

The alteration in potential energy is $\Delta PE = - 3.8*10^{-16} \ J$

In the query, it is stated that

The intensity of the uniform electric field equals $E = 950 \ N/C$

The distance the electron covers is $x = 2.50 \ m$

Typically, the force exerted on this electron is expressed mathematically as

$F = qE$

Where F signifies the force and q represents the charge of the electron, which is a fixed value of $q = 1.60*10^{-19} \ C$

Thus

$F = 950 * 1.60 **10^{-19}$

$F = 1.52 *10^{-16} \ N$

Generally, the work-energy theorem is mathematically framed as

$W = \Delta KE$

Where W denotes the work done on the electron by the electric field and $\Delta KE$ is the change in kinetic energy

Additionally, work done on the electron can also be described as

$W = F* x *cos( \theta )$

Where $\theta = 0 ^o$ assuming that the electron's movement aligns with the x-axis

$\Delta KE = F * x cos (0)$

Inserting values

$\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)$

$\Delta KE = 3.8*10^{-16} J$

According to the conservation of energy

$\Delta PE = - \Delta KE$

Where $\Delta PE$ signifies the change in potential energy

Thus

$\Delta PE = - 3.8*10^{-16} \ J$

7 0

2 months ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

ValentinkaMS [3465]

No one is going to handle that for a mere 5 points lol.

8 0

3 months ago

A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential o

Maru [3345]

Answer:

The excess charge is $Q = 3.5 *10^{-7} \ C$

Explanation:

According to the question, we are informed that

The diameter is $d = 45 \ cm = 0.45 \ m$

The potential of the surface is $V = 14 \ kV = 14 *10^{3} \ V$

The radius of the sphere is

$r = \frac{d}{2}$

by plugging in given values

$r = \frac{0.45}{2}$

$r = 0.225 \ m$

The potential at the surface is mathematically expressed as

$V = \frac{k * Q }{r }$

Where k is Coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

Based on the question stating there are no other charges, Q represents the excess charge

Therefore

$Q = \frac{V* r}{ k}$

inserting the numerical values

$Q = \frac{14 *10^{3} 0.225}{ 9*10^9}$

$Q = 3.5 *10^{-7} \ C$

7 0

1 month ago

Select the areas that would receive snowfall because of the lake effect.

Keith_Richards [3271]

Result:

- Grand Marais

- Two Harbors

- Duluth

Explanation:

The locations likely to receive snowfall due to lake effect include Grand Marais, Two Harbors, and Duluth. These areas are situated directly along the shores of Lake Superior, one of the world’s largest lakes. Its substantial water volume significantly influences the local climate, generating a lot of humidity in the air and considerable evaporation, both of which lead to cloud formation and, when temperatures drop adequately, result in significant snowfall rather than rain.

8 0

2 months ago

A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of

Sav [3153]

Answer:

A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.

Explanation:

τchild=τrock

We will utilize the formula for torque:

(F)child(d)child)=(F)rock(d)rock)

The gravitational force acts equally on both objects.

(m)childg(d)child)=(m)rockg(d)rock)

We can eliminate gravity from both sides of the equation for simplification.

(m)child(d)child)=(m)rock(d)rock)

Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.

(25kg)(1m)=(50kg)drock

Solve for the distance where the rock should be positioned in relation to the seesaw's center.

drock=25kg⋅m50kg

drock=0.5m

6 0

3 months ago