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3 years ago

Which battery will last the longest Superlast or energy star? IV: DV: CV:

Physics

2 answers:

Bas_tet [7]3 years ago

7 0

Ig will because it’s the best battery

docker41 [41]3 years ago

4 0

Answer:

If I

Explanation:

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If it is known that a non-zero net force is acting on an object, then which of the

Anna [14]

Answer:

I think it is other B,C I think not fur sure though

7 0

3 years ago

At a distance of 8 m, the sound intensity of one speaker is 66 dB. If we were to place 3 speakers in a circle of radius 8 m, wha

ludmilkaskok [199]

Answer:

dβ = 70. 77 dβ

Explanation:

The intensity of sound in decibels is

dβ = 10 log I/I₀

let's look for the intensity of this signal

I / I₀ = 10 dβ/10

I / I₀ = 3.981 10⁶

the threshold intensity of sound for humans is I₀ = 1 10⁻¹² W / m²

I = 3.981 10 ⁶ 1 10⁻¹²

I = 3,981 10⁻⁶ W / m²

It is indicated that 3 cornets are placed in the circle, for which total intensity is

I_total - 3 I

I_total = 3 3,981 10⁻⁶

I_total = 11,943 10⁻⁶ W / m²

let's reduce to decibels

dβ = 10 log (11,943 10⁻⁶/1 10⁻¹²)

dβ = 10 7.077

dβ = 70. 77 dβ

3 0

3 years ago

How far will a free falling object fall in 8.7 secs if it started from rest? Remember acceleration is negative for free fall. Do

sleet_krkn [62]

Answer:

h~=371.26m

Explanation:

when an object falls we use the equations of accelerated motion. There is only one that gives distance.

$x = ut + \frac{1}{2} a {t}^{2}$

Since we have no initial velocity (started from rest) we can get rid of the (ut) term

where a we substitute g (gravitational acceleration, constant for given heights and almost 9.81m/s^2).

$h = \frac{1}{2} g {t}^{2} = \frac{1}{2} \times 9.81 \times {8.7}^{2} = 371.26m$

4 0

3 years ago

HOW FAR IS THE MOON?<br><br><br> [COVER UP QUESTION BORED}

Bad White [126]

238,900 miles exactly

4 0

3 years ago

Read 2 more answers

We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes

Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

$1.56421\times 10^{-6}\ T$

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = $\frac{d}{2}=\frac{7}{2}=3.5\ cm$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

The intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2$

5% of this energy goes to the visible light so the intensity is

$I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2$

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

$u=\frac{1}{2}\epsilon_0E^2$

Energy density is also given by

$\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m$

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

$B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T$

The amplitude of the magnetic field at this surface is $1.56421\times 10^{-6}\ T$

7 0

3 years ago