What is her velocity

A student uses a microwave oven to heat a meal. The wavelength of the radiation is 8.97 cm. What is the energy of one photon of

krek1111 [17]

Answer:

The energy of one photon is 2.21x10⁻²⁴ J. Multiplied by 10²⁵ is 22.10 J.

Explanation:

The energy (E) of a photon is:

$E = h\frac{c}{\lambda}$

Where:

h: is the Planck's constant = 6.62x10⁻³⁴ J.s

λ: is the wavelength of the radiation = 8.97 cm

c: is the speed of light = 3.00x10⁸ m/s

$E = h\frac{c}{\lambda} = 6.62 \cdot 10^{-34} J.s\frac{3.00\cdot 10^{8} m/s}{8.97 \cdot 10^{-2} m} = 2.21 \cdot 10^{-24} J$

Hence, the energy of one photon is 2.21x10⁻²⁴ J.

Now, if we multiply the answer by 10²⁵ we have:

$E = 2.21 \cdot 10^{-24} J \cdot 10^{25} = 22.10 J$

I hope it helps you!

8 0

3 years ago

Astronauts aboard the ISS move at about 8000 m/s, relative to us when we look upward.How long does an astronaut need to stay abo

Luba_88 [7]

Answer:

#_time = 7.5 10⁴ s

Explanation:

In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.

t = $\frac{t_p}{ \sqrt{1- (v/c)^2} }$

where t_p is the person's own time in an immobile reference frame,

$t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }$

let's calculate

we assume that the speed of the space station is constant

$t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }$

$t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}$

t_ = 0.99998666657 s

therefore the time change is

Δt = t - t_p

Δt = 1 - 0.9998666657

Δt = 1.3333 10⁻⁵ s

this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s

#_time = 1 / Δt

#_time = $\frac{1}{1.3333 \ 10^{-5}}$

#_time = 7.5 10⁴ s

5 0

3 years ago

a car tire is filled with air at pressure 325000 pa at 283 k. if the tire warms up to 302 k, what is the new pressure of the tir

WARRIOR [948]

Answer:

346819 Pa or ,347000 Pa in 3 significant figures

Explanation:

P1= 325000Pa , T1= 283K ,

P2=? T1= 302 K

as here volume and mass both are constant so using ratio method for pressure temperature law we have P1/T1 = P2/T2

THIS WE CAN ALSO OBTAIN BY RATIO METHOD FOR GENERAL GAS LAW AS

P1V1/(m1T1 ) = P2 V2/ (m2 T2)

IF V1 = V2 =V AND m1=m2=m then expression reduces to

P1/T1 = P2/T2

or P2 = (P1/T1)×T2

P2 = (325000/283) × 302

P2 = 1148.41×302

P2=346819

P2 = 347000 Pa in 3 significant figures

8 0

3 years ago

Many chemical reactions release energy. Suppose that at the beginning of a reaction, an electron and proton are separated by 0.1

ludmilkaskok [199]

Answer:

$1.189eV$

Explanation:

The electric potential energy is the potential energy that results from the Coulomb force and is associated with the configuration of two or more charges. For an electron in the presence of an electric field produced by a proton, the electric potential energy is defined as:

$U=\frac{kq_{e}q_{p}}{r}$

where $q_{e}$ is the electron charge, $q_{p}$ is the proton charge, r is the separation distance between the charges and k is the coulomb constant.

Knowing this, we can calculate how much electric potential energy was lost:

$\Delta U=U_{f}-U{i}\\\Delta U=\frac{kq_{e}q_{p}}{r_{f}}-\frac{kq_{e}q_{p}}{r_{i}}\\\Delta U=kq_{e}q_{p}(\frac{1}{r_{f}}-\frac{1}{r_{i}})\\\Delta U=(8.99*10^9\frac{Nm^2}{C^2})(-1.60*10^{-19}C)(1.60*10^{-19}C)(\frac{1}{0.105*10^{-9}m}-\frac{1}{0.115*10^{-9}m})\\\Delta U=1.90*10^{-19}J*\frac{6.2415*10^{18}eV}{1J}=1.189eV$

5 0

4 years ago

A potential difference of 300 volts is applied to a 2.0-µf capacitor and an 8.0-µf capacitor connected in series. (a) What are t

zubka84 [21]

Answer:

a) Q1=Q2=480μC V1=240V V2=60V

b) Q1=96μC Q2=384μC V1=V2=48V

c) Q1=Q2=0C V1=V2=0V

Explanation:

Let C1 = 2μC and C2=8μC

For part (a) of this problem, we know that charge in a series circuit, is the same in C1 and C2. Having this in mind, we can calculate equivalent capacitance first:

$C=\frac{1}{\frac{1}{C1} +\frac{1}{C2} } = 1.6\mu C$

$Q_{T} = V_{T}*C_{T} = 480\mu C$

$V1 = \frac{Q1}{C1}=240V$

$V2 = \frac{Q2}{C2}=60V$

For part (b), the capacitors are in parallel now. In this condition, the voltage is the same for both capacitors:

$V1=V2$ So, $\frac{Q1}{C1} = \frac{Q2}{C2}$

Total charge is the same calculated for part (a), so:

$\frac{Qt - Q2}{C1} = \frac{Q2}{C2}$ Solving for Q2:

Q2 = 384μC Q1 = 96μC.

Therefore:

V1=V2=48V

For part (c), both capacitors would discharge, since their total voltage of 300V would by applied to a wire (R=0Ω). There would flow a huge amount of current for a short period of time, and capacitors would be completely discharged. Q1=Q2=0C V1=V2=0V

3 0

3 years ago