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icang [17]
3 years ago
7

A particle is uncharged and is thrown vertically upward from ground level with a speed of 23.0 m/s. As a result, it attains a ma

ximum height h. The particle is then given a positive charge q and reaches the same maximum height h when thrown vertically upward with a speed of 29.0 m/s. The electric potential at the height h exceeds the electric potential at ground level. Finally, the particle is given a negative charge -q. Ignoring air resistance, determine the speed with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the maximum height h. In all three situations, be sure to include the effect of gravity.1_________m/s
Physics
1 answer:
Andreyy893 years ago
5 0

Answer:

J Two objects carry initial charges that are qi and q2, respectively, where |q2| > |qi|. ... [45 ptsj A particle is uncharged and is thrown vertically upward from ground level with a ... The particle is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of 40.0 m/s.

Explanation:

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4 0
3 years ago
An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo
REY [17]

Answer:

a) The tangential component of acceleration at the edge of the motor at  t = 1.5\,s is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor (\alpha), measured in radians per square second, as a function of time (t), measured in seconds, is determined by the following formula:

\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right] (1)

The function for the angular velocity of the electric motor (\omega), measured in radians per second, is found by integration:

\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right] (2)

Where \omega_{o} is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration (a_{t}), measured in meters per square second, is defined by the following formula:

a_{t} = R\cdot \alpha (3)

Where R is the radius of the electric motor, measured in meters.

If we know that R = 7.165\times 10^{-2}\,m, \alpha = 10\cdot t and t = 1.5\,s, then the tangential component of the acceleration at the edge of the motor is:

a_{t} = (7.165\times 10^{-2}\,m)\cdot (-10)\cdot (1.5\,s)

a_{t} = -1.075\, \frac{m}{s^{2}}

The tangential component of acceleration at the edge of the motor at  t = 1.5\,s is -1.075 meters per square second.

b) If we know that \omega_{o} = 104.720\,\frac{rad}{s} and \omega = 26.180\,\frac{rad}{s}, then the time needed is:

26.180\,\frac{rad}{s} = 104.720\,\frac{rad}{s}-5\cdot t^{2}

5\cdot t^{2} = 104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}

t^{2} = \frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}  }{5}

t = \sqrt{\frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}  }{5} }

t \approx 3.963\,s

The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

8 0
3 years ago
Please help me on due today
Bad White [126]

Subject-Matter Achievement Goal

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4 0
3 years ago
changes in the number of turns of the coil affects the deflection in the galvanometer. True or false?
Vadim26 [7]

Answer:True

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6 0
3 years ago
To test the performance of its tires, a car
yarga [219]

Answer:

0.43

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∑F = ma

N − mg = 0

N = mg

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ac = v² / r

ac = (35 m/s)² / 564 m

ac = 2.17 m/s²

The total acceleration is:

a = √(ac² + at²)

a = √((2.17 m/s²)² + (3.62 m/s²)²)

a = 4.22 m/s²

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μ = a / g

μ = 4.22 m/s² / 9.8 m/s²

μ = 0.43

5 0
4 years ago
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