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Bas_tet [7]
4 years ago
8

The nose of an ultralight plane is pointed south, and its airspeed indicator shows 28 m/s . the plane is in a 18 m/s wind blowin

g toward the southwest relative to the earth.
Physics
1 answer:
leva [86]4 years ago
6 0
<span>Here I think you have to find the velocity in x and y components where x is east and y is north
 So as air speed indicator shows the negative speed in y component and adding it in
  air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
 v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
 solving
  v= -12.7 m/s x-40.7 m/s y
 if magnitude of velocity or speed is required then
  speed= sqrt(12.7^2 + 40.7^2)
 speed= 42.63 m/s
 if angle is asked
  angle = arctan (40.7/12.7)
 angle = 72.67 degrees south of west</span>
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Winnie the Pooh pushes a jug of with a force of 60 Newtons 3.5 meters. How much work did he do?
KonstantinChe [14]

Answer:

For him being such a fat bear, it would have taken alot of work

Explanation:

7 0
3 years ago
1. A 1.32 x 104 meter steel railroad track with a coefficient of linear expansion of 12 x 10-6 per degree Celsius changes temper
GarryVolchara [31]

Answer:

1) 0.1584 m

2) To allow for expansion without derailment

3) 0.101376 m

4) 213.675 °C

5) 266.67 m

6) 8.33 × 10⁻⁶ /°C

7) The alloy meets the requirement

8) 1.95 × 10⁻³ /°C

9) 32.095 m

10) -12157.72°C

Explanation:

1) Equation for the coefficient of linear expansion = \frac{\Delta L}{L} = \alpha _L \Delta T

Where:

ΔL = Change in length = Required

L = Initial length = 1.32 × 10⁴ m

\alpha _L = Coefficient of linear expansion of steel = 12 × 10⁻⁶ /°C

ΔT = Change in temperature = 37°C - 27°C = 10°C

Plugging the values in the equation for the temperature expansion of steel, we have m;

ΔL = L × \alpha _L ×ΔT = 1.32 × 10⁴ × 12 × 10⁻⁶ × 10  = 0.1584 m

2. Here we have that by segmenting railroad tracks into short pieces, the expansion of the metal tracks with temperature can be absorbed by the gaps between the segment without distorting the shape and direction (pattern) of the tracks

3. Here we have;

\alpha _L = Coefficient of linear expansion of iron = 12 × 10⁻⁶ /°C

ΔT = Temperature change = 27°C - 3°C = 24°C

L = Height of the Eiffel Tower = 352 meters

∴ ΔL = L × \alpha _L ×ΔT = 352 × 12 × 10⁻⁶ × 24 = 0.101376 m

Therefore, the height of the Eiffel Tower changes from 352 m to about 352.101376 m each year, with an average change in height experienced each year = 0.101376 m

4. Here, we have

L = 13.0 ft

ΔL = 1 in.

\alpha _L = 30 × 10⁻⁶ /°C

ΔT = Required temperature change

From  \frac{\Delta L}{L} = \alpha _L \Delta T

\Delta T =\frac{\Delta L}{L \times  \alpha _L} = \frac{1}{156 \times 30 \times 10^{-6}} = 213.675^{\circ}C

5. Here, we have;

L = \frac{\Delta L}{\alpha _L \Delta T}

∴ L = 1/(150×25 × 10⁻⁶) = 266.67 m

The bars original length = 266.67 m

6. Here we have;

\alpha _L = \frac{\Delta L}{L \times \Delta T}

Where:

ΔL = 3.00 - 3.002 = 0.002 m

L = 3.00 m

ΔT = 110°C - 30°C = 80°C

∴ \alpha _L = 0.002/(3.00 × 80) = 8.33 × 10⁻⁶ /°C

7. Here we have;

ΔL = L × \alpha _L ×ΔT = 3 × 8.33 × 10⁻⁶ × 210 = 0.00525 m

Therefore, final length = 3.00 m + 0.00525 m = 3.00525 m

Since 3.00525 m < 3.017 m hence the alloy meets the requirement.

8. Here, we have

L = 3.2 m

ΔL = 0.5 m

ΔT = 84°C - 24°C = 60°C

∴ \alpha _L = 0.5/(3.2 × 60) = 1.95 × 10⁻³ /°C

The coefficient of linear expansion of the material from which the rod is made = 1.95 × 10⁻³ /°C

9. Here, we have

Length of steel girder, L = 32.10 m

ΔT = 8°C - 22°C = -14°C

\alpha _L = 12 × 10⁻⁶ /°C

ΔL = L × \alpha _L ×ΔT

Hence ΔL = 32.1 × 12 × 10⁻⁶× -14 = -0.0054 m

New length = 32.1 - 0.0054 = 32.095 m

10. Here we have;

ΔL = 92.6 cm - 123 cm  = -30.4 cm

\alpha _L = 2.0 × 10⁻⁵ /°C

L = 123 cm

∴ \Delta T =\frac{\Delta L}{L \times  \alpha _L} = \frac{-30.4}{123 \times 2.0 \times 10^{-5}} = -12357.724^{\circ}C

Therefore, the temperature will be 200 - 12357.724 = -12157.72°C.

3 0
3 years ago
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3 years ago
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You've been called in to investigate a car accident by the Bluffs. A car went over a 7.93 meter high cliff
valina [46]

Explanation:

Use the height of the cliff to determine how long it took the car to land.

Take down to be positive.  Given:

Δy = 7.93 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

7.93 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 1.27 s

Use the time to calculate the horizontal velocity.

Given:

Δx = 26.7 m

a = 0 m/s²

Find: v₀

Δx = v₀ t + ½ at²

26.7 m = v₀ (1.27 s) + ½ (0 m/s²) (1.27 s)²

v₀ = 21.0 m/s

The driver was going 21.0 m/s, faster than the speed limit of 9.72 m/s.

4 0
4 years ago
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How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?
podryga [215]

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J

Final velocity = 40 m/s

\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

4 0
3 years ago
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