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gizmo_the_mogwai [7]
3 years ago
5

A train travels 65 kilometers in 5 hours, and then 61 kilometers in 4 hours. What is its average speed? _______km/hr

Physics
2 answers:
olasank [31]3 years ago
4 0
The trains average speed is 13 km/hr.
Galina-37 [17]3 years ago
4 0
Yes, Average speed equals total distance divided by total time.

Total distance= 65km + 61km= 126 km

Total time= 5h + 4h = 9 h


126km/9h= 14km/h

The trains average speed is 14 km/h
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A 700 kg car makes a turn going at 30 m/s with radius of
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3 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
2 years ago
A uniformly charged, straight filament 5.10 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder
Ratling [72]

Answer:

70509.8039216 N/C

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q = Charge = 2.00 µC

l = Length of filament = 5.1 m

r = Radius of cylinder = 10 cm

\lambda=\dfrac{q}{l}

Electric field is given by

E=\dfrac{2k\lambda}{r}\\\Rightarrow E=\dfrac{2\times 8.99\times 10^9\times \dfrac{2\times 10^{-6}}{5.1}}{10\times 10^{-2}}\\\Rightarrow E=70509.8039216\ N/C

The electric field at the surface of the cylinder is 70509.8039216 N/C

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3 years ago
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