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Deffense [45]
4 years ago
6

A 0.060 kg ball hits the ground with a speed of -32m/s. The ball is in contact with the ground for 45 milliseconds and the groun

d exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?
A. 9.3m/s
B.12m/s
C.41m/s
D.73m/s
Physics
2 answers:
Dominik [7]4 years ago
9 0
Using the kinematic equation V=Vo + at, the initial velocity (Vo) is -32, the acceleration (a) is found by dividing force by mass 55N/.06kg=916.66 m/s/s, and the time is 0.045 sec.

V=Vo+at
V=-32+916.66*.045
V=9.25

The answer is A.
aleksklad [387]4 years ago
7 0

it's A. i took the test right now

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A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia
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Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

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8 0
3 years ago
A 0.55-μF capacitor is connected to a 3.5-V battery. How much charge is on each plate of the capacitor?
yan [13]

Answer:

1.925 μC

Explanation:

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The formula for the charge stored in a capacitor is given as,

Q = CV ................... Equation 1

Where Q = charge, C = Capacitor, V = Voltage.

Note: 1 μF  = 10⁻⁶  F

Given: C = 0.55 μF = 0.55×10⁻⁶ F, V = 3.5 V.

Substitute into equation 1

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Q = 1.925 μC

Hence the charge on the plate = 1.925 μC

6 0
4 years ago
Read 2 more answers
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