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3 years ago

What is the total energy of a 850 kg sea plane flying at 38 m/s and 20 m above the ocean?

Physics

1 answer:

Nikolay [14]3 years ago

3 0

Answer: 7.8×10^5 J

Explanation:

Total energy = kinetic energy of plane + gravitational potential energy of plane.

Total energy = ((1/2)×m×v²) + (m×g×h)

Here, m is mass of plane, v is velocity of plane, g is acceleration due to gravity and h is height above the ocean.

Now,

Total energy = ((1/2)×m×v²) + (m×g×h)

Total energy = m((v²/2) + (g×h))

Total energy = 850((38²/2) + (9.81×20))

Total energy = 850(722 + 196.2)

Total energy = 7.8×10^5 J

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3 years ago

10. A kicked soccer ball has an initial velocity of 30 m/s at an angle of 42° above the horizontal, level

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2 years ago

1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa

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Answer:

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$given \to \\ f_{r} = 350 \: \\ v_{r} = {3 \times 10}^{8} \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{r} = \frac{v_{r}}{f_{r}} = \frac{3 \times 10^{8} }{350} = \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\ f_{w} = water \: frequency = \: \boxed{ ?}\: \\ v_{w} = 14 50 \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{w} = \frac{v_{w}}{f_{w}} = \frac{1}{100} \times \gamma _{r} = \frac{1}{100} \times 857,142.85714 \\\gamma _{w} = \boxed{8,571.4285714 \: m} : hence \to \: \\ f_{w} = \frac{v_{w}}{ \gamma _{w}} = \frac{1450}{8,571.4285714} = \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) = \frac{f_{w}}{f_{r}} = \frac{0.1691666667}{350} = 0.0004833333 \\ hence \to \\ the \: frequency \: of \: the \: radio \: wave \: is \to \: \boxed{{4.8 \times 10}^{ - 4} }\: \\ that \: of \: the \: wave \: created \: in \: the \: water.$

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3 years ago

Which event happens in a solar cell when the photoelectric effect occurs?

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Answer:

current forms when electromagnetic waves strike a semiconductor, removing some of its electrons.

Explanation:

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2 years ago

A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside

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Answer:

Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

Explanation:

Given;

number of turns of solenoid, N = 269 turn

length of the solenoid, L = 102 cm = 1.02 m

radius of the solenoid, r = 2.3 cm = 0.023 m

current in the solenoid, I = 3.9 A

Magnitude of the magnetic field inside the solenoid near its centre is calculated as;

$B = \frac{\mu_o NI}{l} \\\\$

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

$B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T$

Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

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3 years ago