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12345 [234]
3 years ago
15

How far does a plane fly in 15 s while its velocity is changing from 145 m/s to 75 m/s at a uniform rate of acceleration?

Physics
2 answers:
kirza4 [7]3 years ago
6 0

Answer:

The plane travels 1650m.

Explanation:

Information Given:

  • The initial velocity of the plane v = 75 m/s
  • Duration of change in velocity t = 15s
  • The final velocity of the plane after 15 s, vf = 75 m/s

v_{f} =v_{0} +a*t\\75m/s=145m/s+a*15s\\a=(75m/s-145m/s)/15s\\a=-4.66m/s^{2}

Note that acceleration is negative, because the plane is in constant desacceleration.

To know how far does the plane fly, we use following formula:

s=v_{0} *t+1/2*a*t^{2} \\s=145m/s*15s+1/2*-4.66m/s^{2} *225s^{2} \\s=1650 m

MAVERICK [17]3 years ago
4 0

The solution would be like this for this specific problem:

Given:<span>
- The initial velocity of the plane </span><span>u = 75<span><span> m</span>/s</span><span>
</span></span>- Duration of change in velocity <span><span><span>t = 15</span>s</span><span>
</span></span>- The final velocity of the plane after 15 s, <span>v = 145<span><span> m</span>/s</span><span>
</span></span>- Let its uniform acceleration be <span>am<span>s<span>^−2</span></span></span>

<span>
Now we know that </span><span>v = u + at ⇒ .145 = 75 + a × 15</span>

<span>
</span><span>⇒ a =<span> (145) / 15</span> =<span> 14/3</span>m<span>s<span>^−2</span></span></span>

<span>
Now distance traversed during 15s is:</span>

<span>
</span><span>s = u × t +<span> 1/2</span> × a ×<span><span> t</span><span>^2</span></span> = 75 × 15 +<span><span> ½ ×<span> 14/3</span> ×<span> 15<span>^2</span></span> = 1650</span><span>m</span></span></span>

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A building made with a steel structure is 565 m high on a winter day when the temperature is 0◦F. How much taller is the buildin
anyanavicka [17]

To solve this problem we apply the thermodynamic equations of linear expansion in bodies.

Mathematically the change in the length of a body is subject to the mathematical expression

\Delta L = L_0 \alpha \Delta T

Where,

L_0 = Initial Length

\alpha = Thermal expansion coefficient

\Delta T = Change in temperature

Since we have values in different units we proceed to transform the temperature to degrees Celsius so

0\°F \Rightarrow (0-32)*\frac{5}{9} = -17.77\°C

103\°F \Rightarrow (103-32)*(\frac{5}{9})= 39.44\°C

The coefficient of thermal expansion given is

\alpha = 1.1*10^{-5}/\°C

The initial length would be,

L_0 = 565m

Replacing we have to,

\Delta L = L_0 \alpha \Delta T

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = 0.355m

This means that the building will be 35.5cm taller

3 0
2 years ago
You have a string with a mass of 0.0133 kg. You stretch the string with a force of 8.89 N, giving it a length of 1.97 m. Then, y
Kazeer [188]

Answer:

(i) The wavelength is 0.985 m

(ii) The frequency of the wave is 36.84 Hz

Explanation:

Given;

mass of the string, m = 0.0133 kg

tensional force on the string, T = 8.89 N

length of the string, L = 1.97 m

Velocity of the wave is:

V = \sqrt{\frac{F_T}{M/L} } \\\\V = \sqrt{\frac{8.89}{0.0133/1.97} } \ = 36.29 \ m/s

(i) The wavelength:

Fourth harmonic of a string with two nodes, the wavelength is given as,

L = 2λ

λ = L/2

λ = 1.97 / 2

λ = 0.985 m

(ii) Frequency of the wave is:

v = fλ

f = v / λ

f = 36.29 / 0.985

f = 36.84 Hz

3 0
3 years ago
an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?
muminat

Answer:

4°C

Explanation:

Water is densest at 4°C.  Since dense water sinks, the bottom of the lake will be 4°C.

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Hope this helps

5 0
2 years ago
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