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3 years ago

A plane has a take off speed of 1 point

Physics

1 answer:

oksian1 [2.3K]3 years ago

4 0

Answer:

Acceleration = 1.85m/s²

Explanation:

Given the following data;

Final velocity = 300km/h to m/s = 300*1000/3600 = 83.33m/s

Time = 45 seconds

Since the plane started from rest, initial velocity is equal to 0m/s.

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Where,

a is acceleration measured in $ms^{-2}$

v and u is final and initial velocity respectively, measured in $ms^{-1}$

t is time measured in seconds.

Substituting into the equation, we have;

Acceleration = (83.33 - 0)/45

Acceleration = 83.33/45

Acceleration = 1.85m/s²

Therefore, the acceleration of the plane is 1.85m/s².

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Answer:

1400 N

Explanation:

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Equating this to impulse formula then

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3 years ago

When the saw slices wood, the wood exerts a 104-N force on the blade, 0.128 m from the blade’s axis of rotation. If that force i

Soloha48 [4]

Answer:

Explanation:

When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.

So torque exerted by wood on the blade

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2 years ago

1. Determine the magnitude of two equal but opposite charges if they attract one another with a force of 0.7N when at distance o

adell [148]

Answer:

q = 2.65 10⁻⁶ C

Explanation:

For this exercise we use Coulomb's law

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In this case they indicate that the load is of equal magnitude

q₁ = q₂ = q

the force is attractive because the signs of the charges are opposite

F = $k \ \frac{q^2}{r^2}$

q = $\sqrt{\frac{F \ r^2}{k} }$

we calculate

q = $\sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9} }$

q = $\sqrt{7 \ 10^{-12} }$ Ra 7 10-12

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3 years ago

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3 years ago

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padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

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Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

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3 years ago