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3 years ago

9. A plane starts at rest & accelerates along the ground before takeoff. It

Physics

1 answer:

Phoenix [80]3 years ago

6 0

Answer:

9.877 m/s^2

Explanation:

The acceleration can be computed from ...

d = (1/2)at^2

(1600 m) = (1/2)a(18 s)^2

a = (1600/162) m/s^2 ≈ 9.877 m/s^2

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Predict whether an element with a valence of 8 is mostly likely form ionic bonds ,covalent bonds or no chemical bonds with other

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An element with a valence of 8 are most likely noble gases (He, Ne, Ar, etc) which means they're already stable so they dont need to undergo chemical bonds with another atom because they already have a full valence shell.

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3 years ago

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2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de

kow [346]

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

u = 0

Therefore, the equation becomes

S = 1/2 gt²

Substituting the given values in the above equation

S = 0.5 x 9.8 x 1.5²

= 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

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3 years ago

What part of the atom takes up most of the atoms space?

Ahat [919]

Answer:

The electrons

Explanation:

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2 years ago

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Irina18 [472]

Answer:

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2 years ago

The 60.0 kg skier shown below is skiing down a 35.0 degree incline where the magnitude of the friction force is 38.5N

Sunny_sXe [5.5K]

Answer:

a) 4.98m/s²

b) 481.66N

Explanation:

a) Using the Newtons second law of motion

$\sum F_x = ma_x\\F_m - F_f = ma_x\\Wsin \theta - F_f = ma_x\\mgsin \theta - F_f = ma_x\\$

m is the mass of the object

g is the acceleration due to gravity

Fm is the moving force acting along the plane

Ff is the frictional force opposing the moving froce

a is the acceleration of the skier

Given

m = 60kg

g = 9.8m/s²

$\theta$ = 35°

Ff = 38.5N

Required

acceleration of the skier a

Substituting into the formula;

$60(9.8)sin 35^0 - 38.5 = 60a\\588sin35^0 - 38.5 = 60a\\337.26 - 38.5 = 60a\\298.76 = 60a\\a = 298.76/60\\a = 4.98m/s^2\\$

Hence the acceleration of the skier is 4.98m/s²

b) The normal force on the skier is expressed as;

N = Wcosθ

N = mgcosθ

N = 60(9.8)cos 35°

N = 588cos 35°

N = 481.66N

Hence the normal force on the skier is 481.66N

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2 years ago