College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Please answer question and just send me the paper solutions asap dont type the answer question 3abcd only And use the formula that is given dont use any other formula and use 9.81 for acceleration not 9.8 if required please solve it faster asap and dont give wrong answer
Equations and Constants
F = µPm
sinA
sinB
sinc
Fnet = ma
a? = b2 + c? -
2bccosA
a
b.
C
V2 = V1 + aAt
Ad
Ad = v;At + ½aAť
Ad = v2At - ½aAť
Vav
Ad = %(v1 + v2)At
(vz) = (v1} + 2aAd
v2
a. =
g = 9.81
4n rf2
T2
W FAd
W= ΔΕ
PE = mgAy
1
KE =
mv2
2.
F = kx
Elastic Energy
kx2
P= mv
FAt = Ap
GMm
mearth = 5.972 x 1024kg
mmoon = 7.348 x 1022kg
Fg
Nm2
G = 6.67x10-11
kg2
re = 6378 km
Tmoon = 1737 m
kq,92
F. =
mproton = 1.67 x 10-27kg
melectron = 9.11 x 10-31kg
Nm2
k = 9.00x109
C2
lelectron/proton
+1.60x10-19C
Fmagnetic = qvBsin®
expand button
Transcribed Image Text:Equations and Constants F = µPm sinA sinB sinc Fnet = ma a? = b2 + c? - 2bccosA a b. C V2 = V1 + aAt Ad Ad = v;At + ½aAť Ad = v2At - ½aAť Vav Ad = %(v1 + v2)At (vz) = (v1} + 2aAd v2 a. = g = 9.81 4n rf2 T2 W FAd W= ΔΕ PE = mgAy 1 KE = mv2 2. F = kx Elastic Energy kx2 P= mv FAt = Ap GMm mearth = 5.972 x 1024kg mmoon = 7.348 x 1022kg Fg Nm2 G = 6.67x10-11 kg2 re = 6378 km Tmoon = 1737 m kq,92 F. = mproton = 1.67 x 10-27kg melectron = 9.11 x 10-31kg Nm2 k = 9.00x109 C2 lelectron/proton +1.60x10-19C Fmagnetic = qvBsin®
gun, Caiculatc no W Tai uiT II IS away HO uc guuT (0.04A1TOʻ III)
3. A ball is fired from a cannon inclined at 36.9° above the horizontal. The speed with which it
leaves the barrel is 200 m/s and the mouth of the cannon is 2.00 m above ground level.
Assume level ground.
What are the horizontal and vertical components of the initial velocity of the
cannon ball?
Determine the velocity of the cannon ball at t = 6.0 s.
What is the ball's velocity when it reaches its maximum height?
How far down range (from the cannon mouth) does the ball land?
(b)
(d)
expand button
Transcribed Image Text:gun, Caiculatc no W Tai uiT II IS away HO uc guuT (0.04A1TOʻ III) 3. A ball is fired from a cannon inclined at 36.9° above the horizontal. The speed with which it leaves the barrel is 200 m/s and the mouth of the cannon is 2.00 m above ground level. Assume level ground. What are the horizontal and vertical components of the initial velocity of the cannon ball? Determine the velocity of the cannon ball at t = 6.0 s. What is the ball's velocity when it reaches its maximum height? How far down range (from the cannon mouth) does the ball land? (b) (d)
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