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Oksanka [162]
3 years ago
9

À A car moves with an initial velocity of 18 m/s due north. Find the velocity of the car after 7 Os if

Physics
1 answer:
Nonamiya [84]3 years ago
3 0

Answer:

(a) v_f=28.5m/s

(b) v_f=7.5m/s

Explanation:

Hello.

(a) In this case since the car is moving at an initial velocity of 18 m/s due north, the final velocity is computed considering the acceleration as positive since it is due north as well:

v_f=v_0+at=18m/s+1.5m/s^2*7s\\\\v_f=28.5m/s

(b) In this case, since the car is moving due north by the acceleration is due south it is undergoing a slowing down process, thereby the acceleration is negative therefore the final velocity turns out:

v_f=v_0+at=18m/s-1.5m/s^2*7s\\\\v_f=7.5m/s

Best regards.

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Hello!

This is a matter of superposition.
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3 years ago
D section a only still need help on this
SIZIF [17.4K]

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Scientists are studying a moving glacier. To monitor the flow of the glacier, they place a series of five markers, A, B, C, D, a
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4 0
3 years ago
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

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3 years ago
For an extended object, the weight force can be considered to act at which point?
JulsSmile [24]

Answer:

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At the center of gravity of the object

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Explanation:

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