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Zarrin [17]
3 years ago
12

This is the question with the options

Physics
1 answer:
melamori03 [73]3 years ago
4 0

Answer:

(c) \frac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}

Explanation:

Average speed is calculated as (total distance)/(total time). We have three segments in the journey, indexed by 0, 1, and 2:

v = \frac{s}{t}=\frac{v_0t_0+v_1t_1+v_2t_2}{t_0+t_1+t_2}\\t_1=t_2\implies\\v=\frac{v_0t_0+v_1t_1+v_2t_1}{t_0+t_1+t_1}=\frac{v_0t_0+v_1t_1+v_2t_1}{\frac{v_0t_0}{v_0}+\frac{v_1t_1}{v_1}+\frac{v_2t_1}{v_2}}

We also know that the distance of the first segment is the same as one of segment 2 and 3 together:

v_0t_0=v_1t_1+v_2t_1\\v_0\frac{t_0}{t_1}=v_1+v_2

Going back to the average speed expression, divide by t_1:

\frac{v_0t_0+v_1t_1+v_2t_1}{\frac{v_0t_0}{v_0}+\frac{v_1t_1}{v_1}+\frac{v_2t_1}{v_2}}=\frac{\frac{v_0t_0}{t_1}+v_1+v_2}{\frac{v_0t_0}{v_0t_1}+2}

and combine the two equations:

\frac{\frac{v_0t_0}{t_1}+v_1+v_2}{\frac{v_0t_0}{v_0t_1}+2}=\frac{2(v_1+v_2)}{\frac{v_1+v_2}{v_0}+2} = \frac{2v_0(v_1+v_2)}{v_1+v_2+2v_0}

The last form matches your choice (c).

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Suppose we hang a heavy ball with a mass 13 kg (so the weight is ) from a steel wire 3.9 m long that is 3.1 mm in diameter (radi
Katena32 [7]

Answer:

1.635×10^-3m

Explanation:

Young modulus is the ratio of the tensile stress of a material to its tensile strain.

Young modulus = Tensile stress/tensile strain

Tensile stress = Force/Area

Given force = 130N

Area = Πr² = Π×(1.55×10^-3)²

Area = 4.87×10^-6m²

Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²

Tensile strain = extension/original length

Tensile strain = e/3.9

Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²

2×10¹¹N/m² = 8.39×10^7/{e/3.9)}

2×10¹¹ = (8.39×10^7×3.9)/e

2×10¹¹e = 3.27×10^8

e = 3.27×10^8/2×10¹¹

e = 1.635×10^-3m

The stretch of the steel wire will be

1.635×10^-3m

7 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
3 years ago
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100 seconds is the right thing
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You are 12 miles north of your base camp when you begin walking north at a speed of 2 miles per hour. What is your location, rel
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If you walk at a pace of 2 miles per hour for 5 hours, you should have walked 10 miles. You would be 2 miles away from your base camp. 
8 0
4 years ago
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