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3 years ago

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How much time would it take for an airplane to reach its destination if it traveled at an average velocity of 820 km/hr NE for a

displacement of 5,200 km NE?

1 answer:

Vladimir79 [104]3 years ago

5 0

Answer:

Time, t = 6.34 hours.

Explanation:

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

$Velocity = \frac{displacement}{time}$

Therefore, making time the subject of formula;

$Time = \frac{displacement}{velocity}$

Given the following data;

Displacement = 5200km

Average velocity = 820km/hr

Substituting into the equation, we have;

$Time = \frac{5200}{820}$

Time = 6.34 hours.

<em>Hence, it would take 6.34 hours for the airplane to reach its destination. </em>

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Answer:

The vector equation of the line is

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Parametric equations for given line are

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3 years ago

What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed

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Answer:

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Explanation:

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3 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

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Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

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divide the above equation by eq. (1)

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now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

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$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

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