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2 years ago

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What is the electric field 4.0 m from the center of the terminal of a Van de Graaff with a 7.10 mC charge, noting that the field

is equivalent to that of a point charge at the center of the terminal?

1 answer:

Jobisdone [24]2 years ago

5 0

Answer:

3984875 N/C

Explanation:

Applying,

E = kq/r².................. Equation 1

Where E = Electric field, q = charge, r = distance, k = coulomb's constant.

From the question,

Given: q = 7.10 mC = 0.0071 C, r = 4.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation 1

E = 0.0071(8.98×10⁹)/4²

E = 3984875 N/C

Hence the electric field is 3984875 N/C

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022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati

svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

$v_f = 24.5 m/s$

Explanation:

Part 1)

initial speed of the ball upwards

$v_i = 24.5 m/s$

so maximum height of the ball is given by

$H = \frac{v_i^2}{2g}$

$H = \frac{24.5^2}{2(9.80)}$

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Part 2)

As we know that final speed will be zero at maximum height

so we will have

$v_f - v_i = at$

$0 - 24.5 = (-9.8)t$

$t = 2.5 s$

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

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$v_f = 24.5 m/s$

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

$y = \frac{1}{2}at^2$

$4.88 = \frac{1}{2}a(1^2)$

$a = 9.76 m/s^2$

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

$a = 90 m/s^2$

time taken by the rocket

t = 0.33 min

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4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

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Part 1)

As it took t = 2.3 s to hit the water surface

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$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.81)(2.3^2)$

$y = 25.95 m$

Part 2)

Distance traveled by it in horizontal direction is given as

$d = v_x t$

$d = 2.92 \times 2.3$

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