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3 years ago

Sound travels at about 360 meters/sec. What is the wavelength of sound with a

Physics

1 answer:

ollegr [7]3 years ago

6 0

<u>Answer:</u>

0.24 m

<u>Explanation:</u>

Given:

Wave velocity ( v ) = 360 m / sec

Frequency ( f ) = 1500 Hz

We have to calculate wavelength ( λ ):

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > λ = v / f

Putting values here we get:

= > λ = 360 / 1500 m

= > λ = 36 / 150 m

= > λ = 0.24 m

Hence, wavelength of sound is 0.24 m.

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A cube of wood having an edge dimension of 20.0cm and a density of 650 kg /m³ floats on water. (b) What mass of lead should be p

schepotkina [342]

The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.

<h3>What mass of lead should be placed on the cube?</h3>

Given: Side of the cube (a) = 20cm

The density of the cube (ρc) = $$$650 kg/m^3$

a) Applying the force balance, the buoyant force must be equal to the weight of the cube

ρcgV = ρg × (Ax)

Substituting the values in the above equation, we get

$(650*(0.2 m)^3)=1000*(0.2 m)^2*x$

x = 0.13

where x is the height of the cube in the water

$$$A = a^2$ is the area of the cross-section

ρ is the density of the water

V is the volume of the cube

Now, the height above the surface of the water would be

h = a − x

Substituting the values, then we get

h = 0.2 − 0.13

h = 0.07 m

b) The mass added is "m" so the complete cube is submerged in the water, therefore

ρcgV + mg = ρg × (V)

$$(650*(0.2 m)^3)+m=1000*(0.2 m)^3$

m = 2.8 kg

The mass added is "m" so the complete cube is submerged in the water is 2.8 kg.

To learn more about buoyant force refer to:

brainly.com/question/11884584

#SPJ4

7 0

1 year ago

A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves o

aniked [119]

Correct question is;

A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers a damping force numerically equal to √2 times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 7 ft/s. (Use g = 32 ft/s²)

Answer:

x(t) = 7te^(-2t√2)

Explanation:

We are given;

Weight; W = 8 lbs

mass; m = W/g

g = 32 ft/s²

Thus;

m = 8/32

m = ¼ slugs

From Newton's second law we can write the equation as;

m(d²x/dt²) = -kx - β(dx/dt)

Rearranging this, we have;

(d²x/dt²) + (β/m)(dx/dt) + (k/m)x = 0

Where;

β is damping constant = √2

k is spring constant = W/s

Where s = 8ft - 4ft = 4ft

k = 8/4

k = 2

Thus,we now have;

(d²x/dt²) + (√2/(¼))(dx/dt) + (2/(¼))x = 0

>> (d²x/dt²) + (4√2)dx/dt + 8x = 0

The auxiliary equation of this is;

m² + (4√2)m + 8 = 0

Using quadratic formula, we have;

m1 = m2 = -2√2

The general solution will be gotten from;

x_t = c1•e^(mt) + c2•t•e^(mt)

Plugging in the relevant values gives;

x_t = c1•e^(mt) + c2•t•e^(mt)

At initial condition of t = 0, x_t = 0 and thus; c1 = 0

Also at initial condition of t = 0, x'(0) = 7 and thus;

Since c1 = 0, then c2 = 7

Thus,equation of motion is;

x(t) = 7te^(-2t√2)

8 0

3 years ago

I need 4 answers for Earth's layers and 4 answers for the Descriptions for Earth's layers graphic organizer.

Arada [10]

Ok well have a nice day

3 0

3 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

3 years ago

Kristen is running a half marathon. By her tenth minute of running, her breathing rate has increased from her resting rate.

pashok25 [27]

They need more oxygen to function, so gas exchange needs to increase

7 0

3 years ago

Read 2 more answers