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3 years ago

Two common terms for a decrease in velocity are

Physics

2 answers:

Colt1911 [192]3 years ago

6 0

deceleration or rėtardation i’m pretty sure (it won’t let me say the second word but it’s correct)

polet [3.4K]3 years ago

5 0

-- negative acceleration

-- deceleration

You might be interested in

In your own words, describe why melting ice with salt freezes cream. Compare your descriptions to your classmates and describe h

gizmo_the_mogwai [7]

Answer:

Water normally freezes at 0°C (32°F). Salt lowers the freezing temperature. (That is, it can remain a liquid at much lower temperatures.)

When sprinkled on ice, the salt lowers the freezing temperature of the water which effectively melts the ice when the salt dissolves into it. There is a limit to how low it can reduce the temperature, though. If the temperature drops below -9°C (15°F), it's too cold for the salt to dissolve into the ice.

When making ice cream, the salt lowers the temperature of the ice and water sufficiently enough to freeze the cream.

3 0

2 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c

e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = $e^{kt}$

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = $e^{-0.01t}$

V = V₀ $e^{-0.01t}$

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀ $e^{-0.01t}$

0.1V₀ = V₀ $e^{-0.01t}$

0.1V₀/V₀ = $e^{-0.01t}$

0.1 = $e^{-0.01t}$

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0

2 years ago

4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p

Komok [63]

Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A , n 1 = 4 .2 mol

Number of moles mono atomic gas B , n 2 = 3.2mol

Initial energy of gas A , K A = 9500 J

Thermal energy given by gas A to gas B , Δ K = 600 J

Gas constant R = 8.314 J / molK

Let K B be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

6 0

3 years ago

Rank these significant figures numbers from the least to the most

Mumz [18]

Answer:

0.006<357<700.003<6010<9256.0<9520.00

8 0

2 years ago