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oee [108]
3 years ago
10

Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is

the new level of sound β2?
Physics
1 answer:
topjm [15]3 years ago
4 0

Answer:

β2= β1+10*f

Explanation:

comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

therefore if the β1 is increases by an intensity of factor f

the new intensity would be β1+ 10*f

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denis-greek [22]

Answer: B,A,C,C

Explanation:

3 0
3 years ago
Read 2 more answers
Please help me with these answers
koban [17]

Answer:

question 1 is centripetal acceleration question 2 is 4

Explanation:

centripetal acceleration basically is acceleration that only changes velocity but not speed, while for question 2, you divide 10 by 2.5

6 0
3 years ago
Chegg ) Alice owns 20 grams of a radioactive isotope that has a half-life of ln(4) years. (a) Find an equation for the mass m(t)
stepladder [879]

Answer:

m(t)=20e^{-0.5t}

Explanation:

Given:

Initial mass of isotope (m₀) = 20 g

Half life of the isotope (t_{1/2}) = (ln 4) years

The general form for the radioactive decay of a radioactive isotope is given as:

m(t)=m_0e^{-kt}

Where,

m(t)\to mass\ after\ 't'\ years\\t\to years\ passed\\k\to rate\ of\ decay\ per\ year

So, the equation is: m(t)=20e^{-kt}

At half-life, the mass is reduced to half of the initial value.

So, at t=t_{1/2},m(t)=\frac{m_0}{2}. Plug in these values and solve for 'k'. This gives,

\frac{m_0}{2}=m_0e^{-k\times\ln 4}\\\\0.5=e^{-k\times\ln 4}\\\\Taking\ natural\ log\ on\ both\ sides,we\ get:\\\\\ln(0.5)=-k\times \ln 4\\\\k=\frac{\ln 0.5}{-\ln 4}=0.5

Hence, the equation for the mass remaining is given as:

m(t)=20e^{-0.5t}

8 0
4 years ago
Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

4 0
3 years ago
Two rollerbladers face each other and stand at rest on a flat parking lot. tracey has a mass of 32 kg, and jonas has a mass of 4
galina1969 [7]
Momentum is a product of mass and the velocity of a body. The initial momentum is always equal to the final momentum during collisions between two bodies.
Therefore; M1U1 +M2U2 = M1V1+ M2V2, where m1 is the mass of tracey and m2 is the mass of jonas, while u is the initial velocity and v is the final velocity.
 (32 ×0)+ (45×0) = (32 × v1) + (45 × 0.50)
                       0 = 32V1 + 22.5
                     32 V1 = -22.5
                          V1 = - 0.703 ( negative indicates difference in direction)
There, Tracey's speed will be 0.703 m/s
8 0
3 years ago
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