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vfiekz [6]
3 years ago
5

The decibel level of an orchestra is 90 db, and the violin section achieves a level of 80 dB. How does the sound intensity from

the full orchestra compare to that from the violin section alone?
Physics
1 answer:
skad [1K]3 years ago
4 0

Answer:

The difference in the decibel corresponses to a constant difference in the loudness perceived.

The refore the sound intensity from the orchestra is like 100 times that of the violin.

Explanation:

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HELP ASAP TIMED TEST
balu736 [363]

Answer:

<em>Correct choice: b 4H</em>

Explanation:

<u>Conservation of the mechanical energy</u>

The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):

E = U + K

The GPE is calculated as:

U = mgh

And the kinetic energy is:

\displaystyle K=\frac{1}{2}mv^2

Where:

m = mass of the object

g = gravitational acceleration

h = height of the object

v = speed at which the object moves

When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

U_1 = mgH

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

\displaystyle U_2=\frac{1}{2}mv^2

Since the energy is conserved, U1=U2

\displaystyle mgH=\frac{1}{2}mv^2    \qquad\qquad [1]

For the speed to be double, we need to drop the snowball from a height H', and:

\displaystyle mgH'=\frac{1}{2}m(2v)^2

Operating:

\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]

Dividing [2] by [1]

\displaystyle \frac{mgH'}{mgH}=\frac{4\frac{1}{2}m(v)^2}{\frac{1}{2}m(v)^2}

Simplifying:

\displaystyle \frac{H'}{H}=4

Thus:

H' = 4H

Correct choice: b 4H

4 0
2 years ago
Science help please!
OLga [1]

Answer:

104 N

Explanation:

m = 1300 kg

a = 0.08m/s^2

F = 1300*0.08

F = 104 N

Newtons is the unit of force.

3 0
2 years ago
e force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If the
anastassius [24]
The force acting between the particles is

F=k \frac{Q_{1}Q_{2}}{r^2}
Then
Q_{2}= \frac{5.2 \times 10^-^5 \times 0.024^2}{ 9.0 \times 10^9 7.2 \times 10^-^8} =4.622 \times 10^-^1^1C




7 0
2 years ago
Read 2 more answers
A capacitor of capacitance C has charge Q and stored energy is U if the charge is increased to 2Q then energy will be A)4U B)2U
Lubov Fominskaja [6]

Answer:

2u

Explanation:

2u

W = Vq

q = CV

W = V.CV

W = CV²

W/C = V²

V = √(W/C)

√(W1/C1) = √(W2/C2)

√(u/c) = √(x/2c)

x = 2u

8 0
3 years ago
Calculate the angular velocity of the neptune (orbital period of 165 earth period) in its orbit around the sun.
ludmilkaskok [199]
The correct answer is: Angular velocity =  1.208 * 10^{-9} rad/s

Explanation:
The angular velocity is given as:
ω = \frac{2\pi}{T} --- (1)

Where T = 165 * (365 days) * (24 hours/day) * (60 minutes/hour) * (60 seconds/minute) = 5203440000 s

Plug in the value in (1):
ω = \frac{2\pi}{5203440000} = 1.208 * 10^{-9} rad/s
7 0
2 years ago
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