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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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This is because it steps up or steps down electrical voltage. It multiplies either voltage (if it is a voltage transformer )or current (if it is a current transformer), but it does not multiply electrical power.

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A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A pie
liubo4ka [24]

Answer:

Explanation:

Capacitance of the capacitor = 13.5μF

Voltage across plate is 24V

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a. Energy in capacitor is given by

E=1/2CV^2

We want to calculate energy without the dielectric substance

Given that C=13.5 μF and V=24V

The capacitance give is with dielectric so we need to remove it

C=kCo

Co=C/k

Then the Co=13.5μF/3.55

Co=3.803μF

Then

E=(1/2)×3.803×10^-6×24^2

E=1.1×10^-3J

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b. Energy in capacitor is given by

E=1/2CV^2

The capacitance given is with a dielectric, so we are going to apply it direct.

Given that C=13.5 μF and V=24V

Then

E=(1/2)×13.5×10^-6×24^2

E=3.89×10^-3J

E=3.9mJ

c. The energy without dielectric is 1.1mJ and the energy with dielectric is 3.9mJ

The energy increase when the dielectric material is added

d. Dielectrics in capacitors serve three purposes: to keep the conducting plates from coming in contact, allowing for smaller plate separations and therefore higher capacitances;

Therefore, Since dielectric allow higher capacitance, and energy of a capacitor is directly proportional to the capacitance, then the higher the capacitance the higher the energy.

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Answer:

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