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3 years ago

8

A net force of 1.0N acts on a 4.0 kg object, initially at rest, for 4.0 seconds. What is the distance the object moves during th

e same time?

1 answer:

Eva8 [605]3 years ago

8 0

Use F=ma to solve for a=0.25m/s^2. Now to get distance you need to integrate this twice to get
x=(1/2)at^2. Plug in a and t to get
x=2m

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Explanation:

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1 year ago

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3 years ago

Read 2 more answers

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a

ValentinkaMS [17]

Answer:

a. 11.5kv/m

b.102nC/m^2

c.3.363pF

d. 77.3pC

Explanation:

Data given

$area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v$

to calculate the electric field, we use the equation below

V=Ed

where v=voltage, d= distance and E=electric field.

Hence we have

$E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m$

b.the expression for the charge density is expressed as

σ=ξE

where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2

If we insert the values we have

$8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2} \\102nC/m^{2}$

c.

from the expression for the capacitance

$C=eA/d$

if we substitute values we arrive at

$C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF$

d. To calculate the charge on each plate, we use the formula below

$Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC$

8 0

3 years ago