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Deffense [45]
2 years ago
12

1. A brick is dropped from a high wall. a. What is its velocity after 5.0 s? b. How far does the brick fall during this time?​

Physics
1 answer:
RSB [31]2 years ago
5 0

Answer:

(I) 49m/s

(ii) 122.5m

Explanation:

Data obtained from the question include:

Time (t) = 5 s

Acceleration due to gravity (g) = 9.8 m/s²

<h3>(A) Determination of the brick's velocity.</h3>

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

<h3>Velocity (v) =?</h3>

v = gt

v = 5× 9.8

v = 49 m/s

Thus, the brick's velocity after 4 s is 49m/s

B. Determination of how far the brick fall in 5 s.

Time (t) = 5 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

h = ½gt²

h = ½ × 9.8 × 5²

h = 5× 16

h = 122.5 m

Thus, the brick fall 122.5m during the time.

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An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1
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Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

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