2. A person drives 120 km [N 32° W] in a car to a friend's place to visit and then drives 150 km [W 24° NJ to visit family. Determine the total displacement of the trip. A

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Chapter14: Nuclear Physics Applications
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Please answer question and just send me the paper solutions asap dont type the answer question 2 And show the picture and all the work and use this formula and 9.81 if needed not 9.8 please faster asap
Equations and Constants
Fr = µFn
sinA
sinB
sinc
Fnet
= ma
a? = b2 + c2 – 2bccosA
a
b.
C
Ad
Vav =
Ad = v;At + %aA? 4d = v,At - %adť
V2 = V1 + gAt
Ad = %(v, + v2)At
(v2) = (v:) + 2aAd
v2
a. =
4n2r
4n rf2
g = 9.81
T2
W = FAd
W = AE
1
PE = mgAy
%3D
KE =
mv2
2.
F = kx
1
Elastic Energy
kx2
2.
FAt = Ap
p= mv
GMm
mearth = 5.972 x 1024kg
mmoon =
7.348 x 1022kg
%3D
r²
Nm2
Te = 6378 km
Tтоon - 1737 m
G = 6.67x10-11
kg?
kq192
mproton
1.67 x 10-27kg
melectron =
9.11 x 10-31kg
Fe =
r2
Nm2
k = 9.00x10°
C2
lelectron/proton
= +1.60x10-19C
%3D
Good to go
Transcribed Image Text:Equations and Constants Fr = µFn sinA sinB sinc Fnet = ma a? = b2 + c2 – 2bccosA a b. C Ad Vav = Ad = v;At + %aA? 4d = v,At - %adť V2 = V1 + gAt Ad = %(v, + v2)At (v2) = (v:) + 2aAd v2 a. = 4n2r 4n rf2 g = 9.81 T2 W = FAd W = AE 1 PE = mgAy %3D KE = mv2 2. F = kx 1 Elastic Energy kx2 2. FAt = Ap p= mv GMm mearth = 5.972 x 1024kg mmoon = 7.348 x 1022kg %3D r² Nm2 Te = 6378 km Tтоon - 1737 m G = 6.67x10-11 kg? kq192 mproton 1.67 x 10-27kg melectron = 9.11 x 10-31kg Fe = r2 Nm2 k = 9.00x10° C2 lelectron/proton = +1.60x10-19C %3D Good to go
Total dispiacement ans: 830.0 km [W 29.09 S
2. A person drives 120 km [N 32° W] in a car to a friend's place to visit and then drives 150 km
[W 24° NJ to visit family. Determine the total displacement of the trip.
Transcribed Image Text:Total dispiacement ans: 830.0 km [W 29.09 S 2. A person drives 120 km [N 32° W] in a car to a friend's place to visit and then drives 150 km [W 24° NJ to visit family. Determine the total displacement of the trip.
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