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2 years ago

What properties of sound determine the volume of sound? Is this affected by the motion of the sound source?

1 answer:

8 0

The property of sound that determine its volume is its AMPLITUDE. The amplitude and the volume of the sound are directly varied. Which means that the larger the amplitude, the larger is the volume and the softer it is if the amplitude is smaller. It is not affected by the motion of the sound source so long as the barriers are not breached.

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An electric winch is used to raise a 40-kg package and a 10-kg package vertically up the side of a building as pictured in the d

just olya [345]

Answer:

Explanation:

40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.

3 0

2 years ago

a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio

Katena32 [7]

Answer:

a) factor $b=\sqrt{2}$

b) factor $b=\frac{1}{2}$

c) factor $b=1$

d) factor $b=1$

Explanation:

Time period of oscillating spring-mass system is given as:

$T=\frac{1}{f}$

$T={2\pi} \sqrt{\frac{m}{k} }$

where:

$f=$ frequency of oscillation

$m=$ mass of the object attached to the spring

$k=$ stiffness constant of the spring

a) On doubling the mass:

New mass, $m'=2m$

Then the new time period:

$T'=2\pi\sqrt{\frac{m'}{k} }$

$T'=2\pi\sqrt{\frac{2m}{k} }$

$T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }$

$T'=\sqrt{2} \times T$

where the factor $b=\sqrt{2}$ as asked in the question.

b) On quadrupling the stiffness constant while other factors are constant:

New stiffness constant, $k'=4k$

Then the new time period:

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

where the factor $b=\frac{1}{2}$ as asked in the question.

c) On quadrupling the stiffness constant as well as mass:

New stiffness constant, $k'=4k$

New mas, $m'=4m$

Then the new time period:

$T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T$

where factor $b=1$ as asked in the question.

d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.

so, factor $b=1$

7 0

1 year ago

A spaceship of frontal area 10 m2 moves through a large dust cloud with a speed of 1 x 106 m/s. The mass density of the dust is

Step2247 [10]

Answer:

The decelerating force is $3\times 10^{- 11}\ N$

Solution:

As per the question:

Frontal Area, A = $10\ m^{2}$

Speed of the spaceship, v = $1\times 10^{6}\ m/s$

Mass density of dust, $\rho_{d} = 3\times 10^{- 18}\ kg/m^{3}$

Now, to calculate the average decelerating force exerted by the particle:

$Mass,\ m = \rho_{d}V$ (1)

Volume, $V = A\times v\times t$

Thus substituting the value of volume, V in eqn (1):

$m = \rho_{d}(Avt)$

where

A = Area

v = velocity

t = time

$m = \rho_{d}(A\times v\times t)$ (2)

$Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t$

From Newton's second law of motion:

$F = \frac{dp}{dt}$

Thus differentiating w.r.t time 't':

$F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}$

where

$F_{avg}$ = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

$F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N$

4 0

1 year ago

Two horses, Thunder and Misty, are accelerating a wagon 1.3 m/s2. The force of friction is 75 N. Thunder is pulling with a force

sasho [114]

Assumption both thunder and misty are pulling in same direction,
Net force= 1000N+800N-75N=1725N
Mass of wagon = 1725N/1.3ms^-2 = 1327kg

7 0

1 year ago

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The best way to cool soft and thick foods (such as beans, sauce or chili) when using the refrigerator is?

AnnZ [28]

The best way to cool soft and thick foods when using the refrigerator is by having them to be placed and poured on a pan or another way is by having them to be placed in one container in which they are in a water bath, to be heated of.

3 0

1 year ago

Read 2 more answers