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aleksandrvk [35]
3 years ago
13

A stone fell from the top of a cliff into the ocean. In the air, it had an average speed of 161616 \text{m/s}m/sstart text, m, s

lash, s, end text. In the water, it had an average speed of 333 \text{m/s}m/sstart text, m, slash, s, end text before hitting the seabed. The total distance from the top of the cliff to the seabed is 127127127 meters, and the stone's entire fall took 12 seconds. How long did the stone fall in the air and how long did it fall in the water?
Physics
1 answer:
AfilCa [17]3 years ago
4 0

Answer:

Explanation:

Initial velocity in air, Vo = 0 m/s

Final velocity in air, Vi = 16 m/s

Initial velocity in water, Vf = 3 m/s

Total distance, S = 127 m

Total time, T = 12 s

Using the equation of motion,

(V - U)t = s

S = s1 + s2

Let T = t1 + t2

127 = (16 × t1) + 3 × (12 - t1)

127 = 16t1 + 36 - 3t1

91 = 13t1

t1 = 91/13

= 7 seconds

Time taken in air, t1 = 7 seconds

t2 = 12 - 7

= 5 seconds.

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Beginning at the NW corner of the intersection of Pine & 675, thence north 950 feet, thence west 380 feet, thence south 950
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4 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

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The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
You have a tungsten sphere (emissivity ε = 0.35) of radius 25 cm at a temperature of 25°C. If the sphere is enclosed in a room w
egoroff_w [7]

Answer:

Explanation:

Stefan's formula for emission of radiation is

E = e σ A  ( T⁴ - T₀⁴ )

E is energy radiated , e is emissivity , σ is stefan's constant , T is temperature of object and T₀ is temperature of surrounding. A is area of surface .

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= 1.9845 x 10⁻⁸ ( 78.86 - 51.58 ) x 10⁸ x .0625

= 3.38  J /s

8 0
3 years ago
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