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Bài tập ôn tập - Vật lý đại cương | Trường Đại Học Duy Tân
Bài tập ôn tập - Vật lý đại cương | Trường Đại Học Duy Tân được sưu tầm và soạn thảo dưới dạng file PDF để gửi tới các bạn sinh viên cùng tham khảo, ôn tập đầy đủ kiến thức, chuẩn bị cho các buổi học thật tốt. Mời bạn đọc đón xem!
Vật lý đại cương (Phy 101) 68 tài liệu
Đại học Duy Tân 1.8 K tài liệu
Bài tập ôn tập - Vật lý đại cương | Trường Đại Học Duy Tân
Bài tập ôn tập - Vật lý đại cương | Trường Đại Học Duy Tân được sưu tầm và soạn thảo dưới dạng file PDF để gửi tới các bạn sinh viên cùng tham khảo, ôn tập đầy đủ kiến thức, chuẩn bị cho các buổi học thật tốt. Mời bạn đọc đón xem!
Môn: Vật lý đại cương (Phy 101) 68 tài liệu
Trường: Đại học Duy Tân 1.8 K tài liệu
Thông tin:
Tác giả:
Preview text:
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s7),!+τ -N12 1 2 1 ϕ = β τ . = 1 . 2 , 2 6 . 02 = 700π (rad) 2 2
EAB0!3Ad), -N12 ϕ 70 π 0 = = = ]R Z /f 8 *π π 2
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Ad_-N3 g^Ad_-N8F36cA2!3Ad26cC)
,-N8
Fc'xA3771P372 ω − ω 180 2
. π / 60 − 300 2 . π / 60 0 β = = = 0 − 2 , 1 ( 2 rad / s )8 τ 60
s7),(MA23)MAL)A..!
J6<tN 2 2 ω − ω . π / − . π / 0 ( 2 2 180 2 60 ) (300 2 60) ϕ = = = 24 0 1/ f# 8 2β − 2.0 2 , 1
fa(MA2P>AB376 ω + ω 180 + 300 0 ϕ = .τ = .1 = 240 1/ f# 2 2
R6c76/Ow ^1NI>40!7))
7AL376;S0 T(_!%8fQ0!V>
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$s7@32-IA26/O6c">ALq4 A26`
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s3<-<7'/PtA23--< α 1N2
f
a = β. R = 3,14. 01 , = 0,314 ( 2 m/ s t ) 2 2 a = ω R . = 3 1 , 4 .01 , = 0 9 , 86 m / s n ( 2 )
od32-I6; 2 2 a = a + a = 1 0
, 3 m / s 8 t n ( 2 )
$s7@32-IA26/O12 α=C a , t 0 314 sin α = = α [^TYj8 a 1 0 , 3
o/)6c6/OX^12^0 V8F
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6$s3--<@6/O8
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6$s3--<{3LV@6/O 2 2 2 a = ω r = ω R . / 2 = 62 8 , 0 . 5
, / 2 = 986 m / s 8 n ( 2 )
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HB3I3+Ad 1
v = v + a t = 15+ .30 = 25 (m / s) = 90 (km / h )8 0 t 3
s3--<{3LVI 2 2 2 v 25 a = ω R = = = 0 6 , 25 m / s n ( 2 ) R 1000
od32-I12 2 2 2 2 1 5 a = a + a = + = 0 7 , 08 m / s t n ( 2 ) 3 8
s372I
a 1/ 3 t −4 β = = ≈ 3,3. 10 ( 2 rad/ s ) R 1000
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6$HB3A3AL(dL5 Z o
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FG]<o<+ ^-N0]<Z<+ ]
%0X-N0Zo7(2!Zo %^8 f .T %
V 12 62 t ,- AB 38 F W Z o →
+ q AL (d L AL AB 3 u A2 1 →
!AL(dL"( +•$ALAB 3 v 8 H A γ
ot,-O123AL → → →
6+!PALAB3 ] V = v+ u 8
F+,->62>AL .T0+ f .T6
,->>AL .T68
FcAb07-!
!8 51A8 51"A8! 5A8! γ$8 % γ5 s 120 u = = = 0 2 , (m / s)8 t 600 1 t 10 4
l = v.t = v.cosγ .t → cos 1 γ = = = → γ = 360 5 ' 3 1 2 t2 12 ,5 5 3 u u 0 2 , 1 sinγ = = → v = = = = 0 3 , 3 (m / s)8 5 v sinγ 3 / 5 3
o(d!P l= v.t = 0 3
, 3.( 10. 60) = 200 m8 1
U+•)!PcLAP7AL6+!PAL
AB3[0%/_8UL=CA-OP(d/= X^8
F$HB3(dL3AL6+!P5
6$ F+ I ) , !P8 o 6< !P 6; ^0X/8
Z(d!P1^0X/X^^8! X^0H[0%/_%_!8
FGAb Z o u s s 150 ! = → u = v . = 2 . = 0 6 , 0 (m / s)8 v l l 500 A 1
F+<!!P H AC AB l 500 t = = = = = 250 (s )8 ] V v v 2
-!3$^0Y^_!56$%X^!8
f62 .%[
R66GA'O]LA'OZ8]Z;cLFVPA2
/=S^^/8&'+6<
$nP775
6$o77tcLUZe5
$o77tcLFVP8
o6<AB376;A %^_!0AB363AL/P/OA % Y^^/_8
]ZS^^/07A %^_![%/_0A Y^^/_8 %
$F+66M<-G]<Z l 300 t = = = 0 5 , (h )= 3 0 1 B# 8 v 600 2
6$FM62 .%Y063LA'OZ07-=6<A-O
7α!AL-]Z8F7
V = v 2 − v 2 = 6002 − 722 = 596 / 8 2 1 (km h) ] Z
F+66G]<Z12 α A s 300 t = = = 0 5
, 03 (h) = 30, 2 B 8 A% V 596
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R201M!(14c
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R2„ /8U4 !
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FP!3CαTX^0A%_!0!SY0T^0SYT,/≈^0%8
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%.g8 $RPP/31,4+6;0!3!
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Ft,-1M(14PPW1M/r F PP0m1M P 0-=
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f62%.g → → → → →
'1BiiUcƒPP12 F + P + N + f = m a ms
om(12c8o<-214-
,
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F!3 ^^^/5/^0 5u0g_!%5A2
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%$ nc4+(3
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F = ma + f + Psin α = ma + kmgcosα +mgsinα ms ms
F70!α^0^T12(3(3!α 1 − 0 042 , ≈ 1 0 , F = 1000× 0 + 0 1 , .1000 9 . 8 , .1+ 1000.9 8 , .0 04 , = ( 1372 N )
6$ˆP3(3„"/!α.! α$8 F− f + P si α n = ma
F = ma + f − P sin α = ma + kmgcosα − mg sinα ms ms F!3 F = 1000 ×0 + 0,1.1000. , 9 8.1 −1000.9,8. , 0 04 = 58 ( 8 N)