what is the largest speed this projectile can have without causing the cable to break
PHY131�������� ������������� Ch half dozen � 9 Test����������� Name
| A force F = (2x i - 5y j) | The x component is perpendicular; must omit (4/4) ; no work in the vertical direction | Work = ∫F∙dr � (6/6) Work = -5∫ydy� Work = -5 �y2 � (14/xiv) ���� (0 to 4) Piece of work = -40 Joules |
| A10.0 kg wood ball hangs from a 2.00 m-long wire. The maximum tension the wire tin can withstand without breaking is 500 N.�� A 1.00 kg projectile traveling horizontally hits and embeds itself in the wood brawl.� What is the largest speed this projectile can have without causing the cable to break? | ����������� (2/two) ���� (x/x) ��� FT ��� ��� mg�� =�� m� vf two /r 500 �11(x) = (10+1)fivef 2/2 780/11 = 5f 2 | pinit ��� =��� pfinal � ��� �(ten/ten) ���� mp vp � = (mp+mb )5f (1)5p �� = (i + 10)vf 5f ������� =� vp / xi | |
| (3/3) ���� 780/11 = fivep two / 112 5p = 92.six m/s | |||
| 3 blocks are connected by a massless string that passes over frictionless pulleys as shown. The 60� incline is frictionless but the table has a frictional coef of 0.2. Find the tension at signal �A� in the string. ����������� (5/5) ��� (5/5) � ������ (five/5) ���� (4/4) ���� ��������� FNet ������������ ������ Ff ��� =�� mT ���� a ��� | | |
| [iii(k)sin60� - ane(g)] � [�(2)g] = (3+1+two) a a = ii k/due south2 | FT = grand (gsinθ � a) �� �� (6/6) ���� FT = 20 North | |
| You identify your water bottle on acme of your apartment dashboard. The coefficient of static friction between your water bottle and the panel is μs = 0.300. How fast can yous drive on a horizontal roadway around a gradual correct turn of radius 40.0 grand earlier the bottle slides? | Ff = μFN Fc = mvii/r Logic: The frictional force provides the centripetal forcefulness towards the middle | Ff = Fc μFNorth = mvii/r (14/xiv) ��� (10/10) ���� μmg = mv2/r | v2 = μ g r v2 = 0.5(x)(lxxx) v = xx m/s (1/one) ���� |
PHY131�������� ������������� Ch 6 � 9 Exam����������� Proper name
| An odd jump with a forcefulness of F = -grand x2 is compressed from three cm to i cm.� What was the free energy required to compress the spring if k = 100 N/cm2? | Work =�� F�� dx ������� (6/6) ���� Work = -k ∫ten2 dx ��� (14/14) ���� Work = k/iii (x0 3 -� xf 3)���� (from 3 to 1) Piece of work = 800/3 N●cm �� (5/five) ���� |
| A brawl is launched upwardly a semicircular chute in such a way that at the summit of the chute, only before information technology goes into free fall, the brawl has a centripetal acceleration of magnitude 3g.� How far from the bottom of the chute does the ball land? | | ||
| (8/8) ���� �m3g = mv2/R vx 2 = (3Rg)� | (8/8) ���� y�� = � g t2 2R = � g t2 t2 = (4R/g)� | � fivex =��� Δx �� /� Δt ���� (8/eight) ���� Δx =����� v10 �������� t Δx = (3Rg) � � (4R/g) � Δx = R(12)� ��� � (i/i) ���� | |
| An elastic cord has a force of F = 8t + 4 | ���� F��� dt ��� �����= k dv �������������� (6/6) ���� 8∫tdt + 4∫dt� = m��� Δv (14/14) ���� iv t2 + 4t���� �����= m��� Δv i N + 2 N�� ������= 1 kg Δv ���������� (4/4) ���� Δv = iii one thousand/s������������������������������� (ane/ane) ���� |
| Two blocks both of mass 10 kg are connected by a massless string that passes over a frictionless pulley. The 60� and 30� inclines both have a frictional coef of 0.1. Find the tension in the string. | | ||
| Fin dir = sin lx� mg Fin dir = 0.866(100N) Fin dir-left = 86.6 N FT-lx� =� yard�� a����� - Ff FT-60� = ten(sin60�grand-a) - 5 FT-sixty� = 70.one N | FNet � -���������������������� Ff����������� = chiliadfull a (86.6 � 50) - (.1*cos60mg +.one*sin60mg) = (ten+10) a (4/4) �� (4/4) � (vii/7) ���� (ii/two) ���� ��������� (2/two) ���� ������ (4/4) 36.6���������� � ( v��������������� +�� 8.66�� )���� = 20 a a = ane.15 1000/s2 (1/1) ���� | Fin dir = sin 30� mg Fin dir = � (100 North) Fin dir-right = 50 N FT-thirty� =� m��� a������� -�� Ff FT-30� = 10 (sin30�g+a) + 8.66 FT-30� = 70.1 Northward | |
PHY131�������� ������������� Ch 6 � nine Exam����������� Name
| A 4 kg object is moving in a aeroplane with its y-coordinate given past y = (2tthree + 8t +ane) meters. Notice the magnitude of the velocity and the net force acting on this object at t = iii sec. | y = 2t3 + 8t + 1 vy = 6t2 + 8 v = dx / dt ay = 12t����� a = dv / dt | ||
| vy = 6t2 + 8 5y = vi*iii2 + 8 vy = 62 m/s | Fy = m ay Fy = 4 (12t) Fy = 144 North | ||
| Ii blocks both of mass 10 kg are connected by a massless string that passes over a frictionless pulley. The 60� and xxx� inclines are frictionless. Detect the tension in the cord. | | | |
| sin threescore � mg Fin dir = 0.866(100N) Fin dir-left = 86.half dozen Northward FT-sixty� ������� = m�� a FT-60� ������� = 10 (sin 60 � g - a) ������� FT-60� ������� = x (seven.82) ������ FT-lx� ������� = 68.3 N | Fin dir = sin 30 � mg Fin dir = � (100 Northward) Fin dir-correct = l North FT-30� ������� = m����� a FT-30� ������� = x (sin thirty � g + a) FT-thirty� ������� = 10 (half-dozen.83) ������ FT-30� ������� = 68.iii N | ||
| FInternet ����������������� =�� thoutotal � a (86.6 � 50) ����� = (ten+10) a a = 1.83 m/s2 �� |
| Two basketballs are on a collision course. 1 is big sized (150 g) and the other is a junior ball of 100 grams. The large brawl is traveling 2 m/s; the pocket-size ball is traveling at -four k/s. What are their final velocities? Initial:� v1f - 52f = half-dozen m/south������������ Final:� v2f - v1f = six one thousand/south;���� v2f = 6 m/southward + 51f | m151 ����� � + 1000iiv2 ����������� = k1f 51f ����������� + ��������� thousand2f five2f 0.15(2) + 0.one(-4) ���������� = 0.15 v1f ���������� + ��������� 0.3 v2f 0.3 ����� - 0.4 ��������������� = 0.15v1f ���������� + 0.3(vi + v1f ) -0.1������������������������������ = 0.15v1f ���������� + i.8 + 0.3v1f v1f ������� = -four.22 m/s������������������ ����������� v2f ������ = 2.44 m/due south |
| Mass m1 on the frictionless table and is connected by a string through a hole in the tabular array to a hanging mass yardii.� With what speed must mane rotate in a circle of radius r if m2 is to remain hanging at rest? m2 g = m1 v2 / r v = (m2 g r / 1000one)� | |
Source: https://www.cpp.edu/~skboddeker/131/quiz/09-4fall-e3.htm
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