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ICE Princess25 [194]
2 years ago
15

The only planet in the solar system with the temperature that can support life is

Physics
2 answers:
EleoNora [17]2 years ago
7 0
The planet that you live in
evablogger [386]2 years ago
6 0

Answer:

earth

Explanation:

its the only planet.

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A single point charge q is located at the center of both an imaginary cube and an imaginary sphere. How does the electric flux t
saveliy_v [14]

Answer:

Explanation:

Electric flux is defined as the flow of electric field intensity through a given surface.

Mathematically:

\phi=E.A.cos\ \theta

where:

E = electric field

A = area

θ = angle between the area vector and the electric field

Electric flux through the surface of a sphere will be uniform throughout the surface area due to a charge at the center of the sphere. The distance of the surface from the center is always at a constant distance of radius of the sphere.

Electric flux through the surface of a cube will be varying as the surface area is at a varying distance from the center of the cube. The distance of the surface from the center is not at a uniform distance from the center of the cube and so the projection of solid angle changes.

7 0
3 years ago
When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
Describe how engineers designed a parachute to create the forces needed to slow down the falling person.
shepuryov [24]

Answer:

Due to the resistance of air, a drag force acts on a falling body (parachute) to slow down its motion. Without air resistance, or drag, objects would continue to increase speed until they hit the ground. The larger the object, the greater its air resistance. Parachutes use a large canopy to increase air resistance. Also, Once the parachute is opened, the air resistance overwhelms the downward force of gravity. The net force and the acceleration on the falling skydiver is upward. An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down. Sorry if not helpful.

8 0
3 years ago
Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 31.5°c.
Artemon [7]
The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant =  8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec
8 0
3 years ago
Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 13.9 rad. Du
alisha [4.7K]

Answer:

(a) Angular acceleration is 1.112 rad/s².

(b) Average angular velocity is 2.78 rad/s .

Explanation:

The equation of motion in Rotational kinematics is:

θ = θ₀ + 0.5αt²

Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.

(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:

13.9 = 0 + 0.5α(5)²

α = 1.112 rad/s²

(b) The equation of average angular velocity is:

ω = Δθ/Δt

ω = \frac{13.9}{5}

ω = 2.78 rad/s

3 0
3 years ago
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