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spin [16.1K]
2 years ago
15

A novice pilot sets a plane’s controls, thinking the plane will fly at 2.50 × 102 km/h to the north. if the wind blows at 75 km/

h toward the southeast, what is the plane’s resultant velocity? use graphical techniques.
Physics
1 answer:
serg [7]2 years ago
3 0

Velocity of plane is given as

v_1 = 2.50 * 10^2 km/h towards north

v_2 = 75 km/h towards south east

we can write the two velocity in vector form

v_1 = 2.50 * 10^2 \hat j

v_2 = 75 cos45 \hat i - 75 sin45 \hat j

now the net velocity will be given as

v_{net} = v_1 + v_2

v_{net} = 250\hat j + 53.03 \hat i - 53.03 \hat j

v_{net} = 53.03\hat i + 196.97 \hat j

so the resultant velocity will be given as

v_{net} = \sqrt{53.03^2 + 196.97^2}

v_{net} = 203.98 m/s

and the angle of velocity will be

\theta = tan^{-1}\frac{v_y}{v_x}

\theta = tan^{-1}\frac{196.97}{53.03} = 74.9 degree

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Answer:

When a an object is been rotated its resistance capacity to that rotational force is know as rotational inertia  and this mathematically given as

          I = mr^2

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For the spinning of the lamp as a baton to work the location of the center of mass of the floor lamp needs to be located

This is more likely to be located closer to base of the lamp as compared to the top, so success of spinning a floor lamp like a baton is highly likely if the lamp is grabbed closer to the base because that is where the position of its center of mass is likely to be.

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A na+ ion moves from inside a cell, where the electric potential is -72 mv, to outside the cell, where the potential is 0 v. wha
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The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:
\Delta U = q \Delta V = q (V_f - V_i)
where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge: q=1.6 \cdot 10^{-19} C, and Vf and Vi are the final and initial voltages.

Substituting the numbers, we find:
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This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating
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Answer:

rod end A is strongly attracted towards the balls

rod end B is weakly repelled by the ball as it is at a greater distance

Explanation:

When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.

Therefore rod end A is strongly attracted towards the balls and

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2 years ago
Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
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Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

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The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

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