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Pani-rosa [81]
3 years ago
14

You hang a heavy ball with a mass of 30 kg from a tungsten rod 2.8 m long by 1.5 mm by 2.6 mm. You measure the stretch of the ro

d, and find that the rod stretched 0.000594 m. Using these experimental data, what value of Young's modulus do you get? Y = N/m2 The atomic mass of tungsten is 184 g/mole, and the density of tungsten is 18.7 g/cm3. Using this information along with the measured value of Young's modulus, calculate the speed of sound in tungsten.

Physics
2 answers:
guajiro [1.7K]3 years ago
7 0

Answer:

Young's modulus (Y) = 3.56×10^11 N/m^2

The speed of sound in tungsten = 6166.4 m/s

Explanation:

Young's modulus (Y) = stress/strain

Stress = force/area

Force = mg = 30×9.8 = 294 N

Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2

Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2

Strain = extension/length

Extension = 0.000594 m

Length = 2.8 m

Strain = 0.000594/2.8 = 2.12×10^-4

Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2

Y = h × rho × g

rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3

h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m

From the equations of motion

v^2 = u^2 + 2gh =

Initial speed (u) = 0 m/s

v = sqrt (2×9.8×1.94×10^6)

v = 6166.4 m/s

egoroff_w [7]3 years ago
6 0

Explanation:

Below is an attachment containing the solution.

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D. Advances in technology were needed to gather more evidence.

Explanation:

New technologies had to be developed to ascertain Wegener's claims.

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His theory needed more scientific backing also.

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6 0
2 years ago
A4 kg bowling ball begins rolling down a at bowling alloy at 6 m/s . When it strikes the pins, it is estimated to be moving at 5
Paul [167]

Answer:

Energy lost due to friction is 22 J      

Explanation:

Mass of the ball m = 4 kg

Initially velocity of ball v = 6 m/sec

So kinetic energy of the ball KE=\frac{1}{2}mv^2

KE=\frac{1}{2}\times 4\times 6^2=72J

Now due to friction velocity decreases to 5 m/sec

Kinetic energy become

KE=\frac{1}{2}\times 4\times 5^2=50J

Therefore energy lost due to friction = 72 -50 = 22 J

8 0
3 years ago
g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn
krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

            T² = (\frac{4\pi }{G M_s} ) r³             (1)

           

in this case the period of the season is

            T₁ = 93 min (60 s / 1 min) = 5580 s

            r₁ = 410 + 6370 = 6780 km

            r₁ = 6.780 10⁶ m

for the satellite

           T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

            T² = K r³

            K = T₁²/r₁³

            K = \frac{ 5580^2}{ (6.780 10^6)^2}

            K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

            r³ = T² / K

            r³ = 86400² / 9.99 10⁻¹⁴

            r = ∛(7.4724 10²²)

            r = 4.21 10⁷ m

this distance is from the center of the earth

7 0
2 years ago
Thermal escape of an atmosphere is most pronounced on worlds where the gravity is low and the temperature is high.
Alona [7]
The answer is false
7 0
3 years ago
Not sure if it went through last time. Please help asap!
Olin [163]
The equation for force is F=ma. Because we have the value of mass (0.42 kg) and the acceleration (14.8 m/s^2), simply plug them into the equation for force to get
0.42 \times 14.8 = 6.22
The answer is 6.22 N because newtons are the unit used to measure force.
8 0
3 years ago
Read 2 more answers
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