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3 months ago

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2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind

1 answer:

Yuliya22 [3.3K]3 months ago

5 0

The speed resulting from the plane is (3) 226 m/s

Reasoning:

We can determine the plane's resultant speed using the Pythagorean Theorem because the two speeds form a right angle (creating a right triangle).

Thus, the computation is as follows:

$ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}$

$ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}$

Consequently, the plane's resultant speed is (3) 226 m/s

Have a wonderful day!

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Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change $(\Delta h)=-50 m$

initial speed $(v_1)=15 m/s$

Final Speed $(v_2)=30 m/s$

Specific internal Energy $(\Delta U)=-5 kJ/kg$

according to the first Law of thermodynamics

$Q-W=\Delta H$

$\Delta H=\Delta KE+ \Delta PE +\Delta U$

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

$\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U$

$Q=W+\Delta KE+ \Delta PE +\Delta U$

$Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10$

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

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4 months ago

A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or

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Answer:

Baseball mass: $m_b=3.992*10^{14}kg$

Explanation:

Circumference of a baseball is calculated using 2πr = 23 cm

Thus, the radius comes out to be 3.66 cm, which equals $3.66*10^{-2}$ m

The mass density of the baseball matches that of a neutron or proton.

Proton mass = $10^{-27}$ kg

Proton diameter = $10^{-15}$ m

Proton radius = $5*10^{-16}$ m

Volume of the baseball is $\frac{4}{3} \pi r^3$

Now by substituting all values into the mass per unit volume equation for the baseball, we get:

$\frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}$

$\frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}$

$m_b=3.992*10^{14}kg$

Therefore, the baseball mass amounts to $m_b=3.992*10^{14}kg$

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The force that the car applies on the truck will be 6000 N in the reverse direction.

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