25 Points PLEASE I need Help

A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ

Svetradugi [14.3K]

Answer:

d = 4180.3m

wavelengt of sound is 0.251m

Explanation:

Given that

frequency of the sound is 5920 Hz

v=1485m/s

t=5.63s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

$velocity=\frac{distance}{time}\\\\ v=\frac{2d}{t} \\\\vt=2d\\\\d=\frac{vt}{2}$

$d=\frac{1485*5.63}{2}\\d= 4180.3m$

wavelengt of sound is $\lambda$ = v/f

= (1485)/(5920)

= 0.251 m

7 0

2 years ago

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

= $4.59\times 10^{-19} \ J$

or,

= $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

= $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒ $\Phi_0=2.3\times 10^{-19} \ J$

or,

⇒ $=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}$

⇒ $=1.4375 \ eV$

5 0

2 years ago

A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a

klemol [59]

Answer:

15.4 kg.

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m').................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.

Given: m = 7.7 kg, u' = 0 m/s (at rest)

Let: u = x m/s, and V = 1/3x m/s

Substitute into equation 1

7.7(x)+m'(0) = 1/3x(7.7+m')

7.7x = 1/3x(7.7+m')

7.7 = 1/3(7.7+m')

23.1 = 7.7+m'

m' = 23.1-7.7

m' = 15.4 kg.

Hence the mass of the second sphere = 15.4 kg

7 0

2 years ago

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Please help me! Some people have proposed a new way to build houses in areas that are likely to experience tsunamis. In this des

Alla [95]

Answer:

house wouldn't have solid walls on all four sides. Instead, some of the wall areas would be replaced by substances that

water can travel through quickly, as shown in the diagram. How would this design help a house survive a tsunami? What

drawbacks might there be to this design?

Explanation:

3 0

2 years ago

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The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.10. The mass of the object i

denis-greek [22]

Answer:

Explanation:

1) Force Friction = Normal Force * Coefficient of Friction

Force Friction = Mass * Gravity * Coefficient of Friction

2) F = ma

Force = mass * acceleration

Force Friction (from #1) = mass * acceleration

acceleration = Force Friction / Mass

6 0

2 years ago