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harkovskaia [24]

2 years ago

15

A car initially miving at 0.5m/s along a track.the car come to rest after travelling 1m.the car is repeated on the same of track

,but car initial moving at 1m/s.How far does the car travel before come to rest?

1 answer:

cluponka [151]2 years ago

6 0

Answer:

it will travel 69 m long consuming. it is totally wrong

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The sediment deposited by glaciers is called Glacial deposition.

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Explanation:

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3 years ago

) a wrench 0.6 meters long lies along the positive y-axis, and grips a bolt at the origin. a force is applied in the direction o

Lera25 [3.4K]

<span>Applied Force Direction vector = [0,1,3] force F = j + 4k torque is t = 100 Newton-meters = r x F Wrench is 0.5 meters long on positive side of the y-axis, r = 0.6 = [0,0.6,0] We know torque |t| = |r x F| = |r| x |F| sin theta r x F = |r| x |F| cos theta r x (j + 3k) = |r| x |j + 3k| cos theta => [0,0.6,0] [0,1,3] = 0.6 x squareroot of ((0)^2 + (1)^2 + (3)^2) cos theta => 0.6 = 0.6 x squareroot of (1 + 9) cos theta => cos theta = 1 / squareroot of (10) Calculationg the sin theta, sin theta = squareroot of (1 - (1 / squareroot of (10))^2) = squareroot of (9/10) sin theta = 3 / squareroot of (10) Substituting the values, |T| = |r| x |F| sin theta => 100 = |0.6| x |F| x 3 / squareroot of (10) |F| = (100 x squareroot of (10)) / 1.8 |F| = (1000 / 18) x squareroot of (10) Magnitude of force |F| = 55.55 x squareroot of (10)</span>

6 0

3 years ago

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

shutvik [7]

Answer:

Answer is explained in the explanation section.

Explanation:

A)

Solution:

For this to find, we need to calculate the centripetal acceleration on the equator.

The centripetal acceleration of the equator:

a = 4 $\pi ^{2}$ RcosФ/ $T^{2}$ ,

where,

R is the radius of the earth

R = 6378 KM = 6.3 x $10^{6}$ and

T is the time period

T = 24 h = 86164.1 s

At Equator, Ф = 0°

So, CosФ = 1

Hence,

a = 4 $\pi ^{2}$ R/ $T^{2}$

By plugging in the values, we get:

a = 4 x ( $3.14^{2}$ ) x (6.3 x $10^{6}$ ) / $86164.1^{2}$

a = 0.03 m/ $s^{2}$

Hence, this is the centripetal acceleration on the equator. And we also know that, acceleration due to gravity is 9.8 m/ $s^{2}$ which is very higher than the centripetal acceleration on the equator.

B) Normal force exerted by chair will always be equal and opposite to the mass times gravitational acceleration (F = mg). Otherwise, I would be thrown away from chair in case the normal force is not equal and opposite or I would be drag down to the earth due to greater mass times gravitational acceleration. Hence, both are equal and opposite.

C) Of course, this is not a lie, it is true because the acceleration due to gravity is 9.8 m/ $s^{2}$ and as we calculated the acceleration on the equator is 0.03 m/ $s^{2}$ which way too low to experience.

For percentage difference,

9.8 - 0.03 = 9.77

So, % diff = (9.8 - 9.77)/9.8 x 100

% diff = 0.00306 x 100

% diff = 0.306%

Obviously, this is way too low to experience.

D) With the help of same formula as discussed above, we have:

a = 4 $\pi ^{2}$ RcosФ/ $T^{2}$ ,

Here, Ф = 44.4°

Just putting the values. we get

a = 0.0241 m/ $s^{2}$

Acceleration while sitting in my chair.

6 0

2 years ago