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Blizzard [7]
3 years ago
13

If an air traffic controller is describing the speed of an airplane as well as the direction in which it should begin its decent

, then the air traffic controller is describing the airplane's
a. magnitude.
b. velocity.
c. speed.
d. acceleration.
Physics
2 answers:
Sever21 [200]3 years ago
8 0
I think the correct answer from the choices listed above is option B. The air traffic controller is describing the airplane's velocity. Velocity<span> is a physical vector quantity; both magnitude and direction are needed to define it. Hope this answers the question. Have a nice day.</span>
liberstina [14]3 years ago
5 0

The correct answer for this is B. Velocity. To find velocity you need the speed and direction, what we have. Sorry for the wait, Good luck :)

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While standing atop a building 49.6 m tall, you spot a friend standing on a street corner. Using a protractor and a dangling plu
olga nikolaevna [1]

Answer:

75degree don't forget wind and gravity force pulling down

6 0
2 years ago
A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it
Degger [83]

Answer:

a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

d)\alpha_2=4.1454\,rad.s^{-2}

e)t_2=1.061\,s

Explanation:

Given that:

  • initial speed of turntable, N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

4 0
3 years ago
A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in
Kruka [31]

Answer:

1020g

Explanation:

Volume of can=1100cm^3=1100\times 10^{-6}m^3

1cm^3=10^{-6}m^3

Mass of can=80g=\frac{80}{1000}=0.08kg

1Kg=1000g

Density of lead=11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3

By using 1g/cm^3=10^3kg/m^3

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can  and lead inside it they are floating in the water.

Buoyancy force =F_b=Weight of can+weight of lead

\rho_wV_cg=m_cg+m_lg

Where \rho_w=10^3kg/m^3=Density of water

m_c=Mass of can

m_l=Mass of lead

V_c=Volume of can

Substitute the values then we get

1000\times 1100\times 10^{-6}=0.08+m_l

1.1-0.08=m_l

m_l=1.02 kg=1.02\times 1000=1020g

1 kg=1000g

Hence, 1020 grams of lead shot can it carry without sinking water.

4 0
3 years ago
Inertia is ______ to mass.
UkoKoshka [18]

Answer:

Inertia is directly proportional to mass of an object. Therefore, when the force of inertia increases the mass also increases, and when it decreases the mass also decreases.

Explanation:

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3 years ago
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A swimmer is swimming to the left with a speed of 1.0\,\dfrac{\text m}{\text s}1.0 s m ​ 1, point, 0, start fraction, start text
bulgar [2K]

Answer:

2.9s

Explanation:

3 0
3 years ago
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