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3 years ago

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You are trying to decide between two new stereo amplifiers. One is rated at 130 W per channel and the other is rated at 200 W pe

r channel. Part A In terms of dB, how much louder will the more powerful amplifier be when both are producing sound at their maximum levels?

1 answer:

NARA [144]3 years ago

7 0

Answer:

The sound level will be 1.870 dB louder.

Explanation:

Given that,

Power = 130 W

Power = 200 W

We need to calculate the sound level

Using formula of sound level

$I_{dB}=10\log(\dfrac{I}{I_{0}})$

For one amplifier,

$I_{1}=10\log(\dfrac{130}{I_{0}})$ ...(I)

For other amplifier,

$I_{2}=10\log(\dfrac{200}{I_{0}})$ ...(II)

For difference in dB levels

$I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}})-10\log(\dfrac{130}{I_{0}})$

$I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}}\times\dfrac{I_{0}}{130})$

$I_{2}-I_{1}=10\log(\dfrac{200}{130})$

$I_{2}-I_{1}=1.870\ dB$

Hence, The sound level will be 1.870 dB louder.

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10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr

12345 [234]

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

$F_{g0}=G \frac{mM_{E0}}{r^{2}}$

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=2M_{E0}$

so:

$F_{gf}=G \frac{2mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=2$

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=3M_{E0}$

so:

$F_{gf}=G \frac{3mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=3$

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

$M_{Ef}=\frac{M_{E0}}{4}$

so:

$F_{gf}=G \frac{mM_{E0}}{4r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=\frac{1}{4}$

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

4 0

3 years ago

*8–52. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the to

denis23 [38]

Answer:

μ₁ = 0.1048

μ₂ = 0.1375

Explanation:

Using static equation can find in both point the moment and the forces so:

∑ M = F *d , ∑ F = 0

∑ M A = 0

N₁ * 3 - 200 * 9.81 * 1.5 = 0

N₁ = 981

∑ M y = 0

N₂ + 300 * ³/₅ - 981 - 20 * 9.81 = 0

N₂ = 997.2 N

∑ M C = 0

F₁ * 1.75 - 300 * ⁴/₅ * 0.75 = 0

F₁ = 102.86

∑ M B = 0

300 * ⁴/₅ * 1 - F₂ * 1.75 = 0

F₂ = 137.14 N

The Force F1 and F2 related the coefficients of static friction

F₁ = μ₁ * N₁ ⇒ 102.86 N = μ₁ * 981 ⇒ μ₁ = 0.1048

F₂= μ₂ * N₂ ⇒ 137.14 N = μ₂ * 997 ⇒ μ₂ = 0.1375

8 0

4 years ago

A shot-putter puts a shot (weight = 71.5 N) that leaves his hand at a distance of 1.84 m above the ground. (a) Find the work don

givi [52]

Answer:

a)W= - 47.19 J

b)47.19 J

Explanation:

Given that

Weight ,mg= 71.5 N

y= 1.84 m

H= 2.5

a)

The distance above the hand ,h= 2.5 - 1.84 m

h= 0.66 m

We know that gravitaional force act in the downward direction but the displacement is in upward direction that is why work done will be negative.

W= - m g h

W= -71.5 x 0.66 J

W= - 47.19 J

b)

The potential energy at initial position = m g y

The potential energy at final position = m g H

So change in the potential energy = m g H - m g y

= mg ( H- y)

=71.5 ( 2.5 - 1.84 ) = 47.19 J

Therefore change in the potential energy = 47.19 J

5 0

4 years ago

What’s the total system momentum before collision?

Soloha48 [4]

The momentum would be 9 kg

3 0

3 years ago

Read 2 more answers

URGENT HELP !! The coefficients of static and kinetic frictions for plastic on wood are 0.53 and 0.40, respectively. How much ho

IRISSAK [1]

Answer:

43.83 N

Explanation:

Given that,

The mass of an object, m = 34.4 kg

The coefficients of static and kinetic frictions for plastic on wood are 0.53 and 0.40, respectively.

The force of static friction,

$F_s=\mu_smg\\\\F_s=0.53\times 34.4\times 9.8\\\\F_s=178.67\ N$

The force of kinetic friction,

$F_k=\mu_kmg\\\\F_k=0.40\times 34.4\times 9.8\\\\F_k=134.84\ N$

Net force acting on the object is :

F = 178.67-134.84

= 43.83 N

Hence, this is the required solution.

7 0

3 years ago