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Natali5045456 [20]
3 years ago
15

A 1490-kg car is moving due east with an initial speed of 29.5 m/s. After 9.14 s the car has slowed down to 18.8 m/s. Find the m

agnitude of the net force that produces the deceleration.
Physics
1 answer:
zalisa [80]3 years ago
8 0

Answer:

The magnitude of the net force that produces the deceleration = 1744 N

Explanation:

Using the equations of motion, we can find the deceleration of the car and thereby find the magnitude of the met force that produces such deceleration

v = u + at

where v = final velocity = 18.8 m/s

u = initial velocity = 29.5 m/s

t = 9.14 s

a = ?

v = u + at

18.8 = 29.5 + 9.14a

9.14a = 18.8 - 29.5 = -10.7

a = - 1.171 m/s²

Magnitude of Force causing deceleration = ma = 1490 × 1.171 = 1744 N

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Answer:

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Explanation:

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Distance between the plates, d = 6 cm = 0.06 m

The electric field between the plates is given by

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Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio
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Answer:

The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

Explanation:

Given that,

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We need to calculate the electric field

For first charge,

Using formula of electric field

E_{1}= \dfrac{kq_{1}}{r^2}

Put the valueinto the formula

E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}

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Direction of electric field along AB

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For second charge,

Using formula of electric field

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Put the valueinto the formula

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E_{net}=E_{1}-E_{2}

E_{net}=(18-7.2)\times10^{5}\ N/C

E_{net}=10.8\times10^{5}\ N/C

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Hence, The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

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Consider the forces acting on left cube, from the force diagram, we have

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