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11Alexandr11 [23.1K]
3 years ago
14

The car in the figure travels at a constant speed along the road shown. Draw vector showing its acceleration at the point C if t

he acceleration at point A is drawn. The lengths of your vectors should correspond to the magnitudes of the accelerations.
Draw the vector Ac with its tail at point C. The orientation and exact length of your vector will be graded.

Physics
1 answer:
yuradex [85]3 years ago
6 0

Answer:

The vector form is as shown in the attachment

Explanation:

The figure as shown in the diagram, indicates that the car is moving along the road at a constant speed. Centripetal acceleration comes into play for an object moving in a circular motion at uniform speed. The centripetal acceleration is the acceleration experienced by an object while in uniform circular motion.

Mathematically from centripetal acceleration; a = v2/r

The equation shows that there is an inverse relationship between the acceleration and the radius of curvature as such the radius of curvature at the point A will be more than the radius of curvature at the point C, this shows that the centripetal acceleration at point C will be more than the centripetal acceleration at point A.

The attachment shows the figure and the representation in vectorial form.

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If Vector A is (3, 0) and Vector B is (-3, 3), what is the resultant?
Archy [21]

Answer:resultant vector R = (0, 3)

Explanation: vector A = (3, 0)

vector B =(-3, 3)

Vectors are added such that those in same directions are added together. The resultant vector R is the given by R = (3-3, 0+3)

= (0, 3)

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When light reflects from a surface, there is a change in its
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The answer is: none of the above.

Explanation:

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3 years ago
A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/
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First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

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An electron and a 0.033 0-kg bullet each have a velocity of magnitude 495 m/s, accurate to within 0.010 0%. Within what lower li
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Answer:

1.170*10^-3 m

3.23*10^-32 m

Explanation:

To solve this, we apply Heisenberg's uncertainty principle.

the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"

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If we make Δx the subject of formula, by rearranging, we have

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on substituting the values, we have

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Δx = 6.63*10^-34 / 5.67*10^-31

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for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2

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Change in position of object = Displacment
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