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USPshnik [31]
3 years ago
13

What brand is this please no links

Physics
2 answers:
makvit [3.9K]3 years ago
6 0

Answer: YAHOO

Explanation: HAVE A GOOD DAY HOPE IT HELPS!

Readme [11.4K]3 years ago
3 0

Answer:

goo-gle did back then

Explanation:

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Two objects attract each other gravitationally. If the distance between their centers decreases by a factor of 2, how does the g
kramer

Answer:

The gravitational force between them increases by a factor of 4

Explanation:

Gravitational force is a force of attraction between two objects with masses M and m which are separated by a distance R. It is given mathematically as:

Fg = GMm/R²

Where G = Gravitational constant.

If the distance between their centers, R, decreases by a factor of 2, then it means the new distance between their centers is:

r = R/2

Hence,the gravitational force becomes:

Fg = GMm/r²

Fg = GMm/(R/2)²

Fg = GMm/(R²/4)

Fg = 4GMm/R²

Hence,the gravitational force increases by a factor of 4.

6 0
3 years ago
A banked curve is designed for a roadway. The curve is designed so that a car traveling at 19.2 m/s can negotiate it without rel
egoroff_w [7]
I don’t know wait for someone else
8 0
3 years ago
I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT.
Tems11 [23]
Ultraviolet light with a wavelength shorter than visible light and a higher radiant energy than visible light.

The shorter the wavelength, the higher the frequency and energy.
6 0
3 years ago
Read 2 more answers
You wish to watch TV at exactly 85 dB and no louder to avoid long term damage to your hearing. You record the sound intensity le
BigorU [14]

Answer:

1) the new power coming from the amplifier is 19.02 W

2) The distance away from the amplifier now is 5.50 m

3) u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

Explanation:

Lets say that I am at a distance "u" from the TV,

Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB

SO

S(indB) = 10log (I₁/1₀)

we substitute

125 = 10(I₁/10⁻¹²)

12.5 = log (I₁/10⁻¹²)

10^12.5 = I₁/10^-12

I₁ = 10^12.5 × 10^-12

I₁ = 10^0.5 W/m²

Now I₂ will be intensity of sound when corresponding sound level is 107 dB

107 = 10log(I₂/10⁻²)

10.7 = log(I₂/10⁻¹²)

10^10.7 = I₂ / 10^-12

I₂ = 10^10.7  ×  10^-12

I₂ = 10^-1.3 W/m²

Now since we know that

I = P/4πu² ⇒ p = 4πu²I

THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂

Therefore

P₁/P₂ = I₁/I₂

WE substitute

P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)

P₂ = 19.02 W

the new power coming from the amplifier is 19.02 W

2)

P₁ = 4πu²I₁

u =√(p₁/4πI₁)

u = √(1200/4π × 10^0.5)

u = 5.50 m

The distance away from the amplifier now is 5.50 m

3)

Let I₃ be the intensity corresponding to required sound level 85 dB

85 = 10log(I₃/10⁻¹²)

8.5 = log (I₃/10⁻¹²)

10^8.5 = I₃ / 10^-12

I₃ = 10^8.5  × 10^-12

I₃ = 10^-3.5 w/m²

Now, I ∝ 1/u²

so I₂/I₃ = u₁²/u²

u₁ = √(I₂/I₃) × u

u₁ = √(10^-1.3 / 10^-3.5) ×  5.50

u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

8 0
3 years ago
How far does light travel in the time it takes sound to travel 1 cm in air at 20°c?
kykrilka [37]
The speed of sound at 20^{\circ}C is approximately v=343 m/s. The distance covered by the sound wave is
s=1 cm=0.01 m
And the time it takes is
t= \frac{S}{v}= \frac{0.01 m}{343 m/s}=2.9 \cdot 10^{-5} s

Now we want to find how far the light travels during this time. Light travels at speed c=3 \cdot 10^8 m/s, therefore the distance it covers during this time is
S=ct = (3 \cdot 10^8 m/s)(2.9 \cdot 10^{-5} s)=8700 m
8 0
3 years ago
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