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3 years ago

5

On a sunny day does a bird or an aeroplane flying high in the sky casts its shadow on the ground

1 answer:

Alex73 [517]3 years ago

3 0

Explanation:

It depends because the higher the object is, the bigger the shadow, however, the object should be close to the line directly from the sun, for you to see the shadow visible.

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Explain why ventilation is very important if there is risk of exposure to random gas in your home school

bonufazy [111]

Ventilation is very important because it helps remove the gas form people’s homes and schools and it redirects the random gas outside so it is less likely to hurt people

3 0

3 years ago

An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte

telo118 [61]

Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

$I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2$

a. The angular acceleration is Torque divided by moments of inertia:

$\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2$

b. 5 revolution would be equals to $10\pi$ rad, or 31.4 rad. Since the engine just got started

$\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5$

$\omega = \sqrt{1893.5} = 43.5 rad/s$

c. Work done during the first 5 revolution would be torque times angular displacement:

$W = T*\theta = 1930 * 31.4 = 60633 J$

d. The time it takes to spin the first 5 revolutions is

$t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s$

The average power output is work per unit time

$P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W$ or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

$P_i = T*\omega = 1930*43.5=83983 W$ or 84 kW

7 0

3 years ago

A 50.0-turn circular coil of radius 5.00 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

Blababa [14]

Answer:

The maximum torque in the coil is $4.9\times 10^{-3}\ N-m$ .

Explanation:

Given that,

Number of turns in the circular coil, N = 50

Radius of coil, r = 5 cm

Magnetic field, B = 0.5 T

Current in coil, I = 25 mA

We need to find the magnitude of the maximum possible torque exerted on the coil. The magnetic torque is given by :

$\tau=NIAB\ \sin\theta$

For maximum torque, $\theta=90^{\circ}$

$\tau=NIAB\\\\\tau=50\times 25\times 10^{-3}\times \pi (0.05)^2\times 0.5\\\\\tau=4.9\times 10^{-3}\ N-m$

So, the maximum torque in the coil is $4.9\times 10^{-3}\ N-m$ .

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3 years ago

HELP ASAP!!!! SCIENCE LOVERS !!! What do the lines making a circle around the LOW-PRESSURE area indicate?

Degger [83]

1 is D.
2 is A.

hope i helped

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3 years ago

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What are 3D glasses made of?

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<span>two filters with vertical polarization</span>

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3 years ago