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Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d at time t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet each other ?
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Answers
Given :-
Three particles A, B and C are situated at the vertices of an equilateral triangle ABC.The side of the triangle is d .A always has its velocity along AB, B along BC and C along CA.To Find :-
Time at which the particles will meet .Solution :-
See attachment .
Firstly break the velocity of A , into it's vertical components . That will be ,
[tex] v_A = v_A cos60^o + v_A sin60^o [/tex]
Now the component , [tex]v_A cos60^o [/tex] coincides on side AC and is directed from A to C . Along CA, the velocity of particle C is directed . So both are in opposite directions.
So , the relative velocity of C with respect to A will be ,
[tex] v_{CA}= v_C - v_A [/tex]
Substitute ,
[tex] v_{CA}= v - ( - v cos 60^o) [/tex]
Value of cos 60° is 1/2 ,
[tex] v_{CA}= v + v\bigg(\dfrac{1}{2}\bigg)=v +\dfrac{v}{2}[/tex]
Add the terms,
[tex] v_{CA}= \dfrac{2v + v}{2}=\dfrac{3v}{2}[/tex]
Hence , now we can find the time taken by A and C ,to meet each other will be ,
Since the speed is constant , we can use distance = speed*time , as ;[tex]s = v t [/tex]
Substitute ,
[tex]d = \dfrac{3v}{2} t [/tex]
Multiplying both sides by 2 ,
[tex] 2d = 3vt [/tex]
Dividing both sides by 3v ,
[tex] t = \dfrac{2d}{3v} [/tex]
Similarly , taking any two particles you will get the same time .
Hence the particles meet after 2d/3v time .
I hope this helps .
![[tex]\bold{\huge{\underline{ Hola \: Brainlians!!! }}}[/tex]Three Particles A, B And C Are Situated At](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/Z5g8PLR1Ud9OrJslRW96o3k0rFIbOsT0.jpeg)
Answer:
never
Explanation:
Particles moving on paths that don't meet at a point will be particles that never meet. Even if the paths meet at a point, which these don't, the particles may not meet if they arrive at different times.
__
Consider the diagrams in the attachments. In order, they show the positions of the particles shortly after takeoff, midway along the line, and after they have gone farther than the length of the line. The point of closest approach is shown in the second attachment, when the particles are halfway along the line.
Consider particles A and C. Their north/south direction is 'north' for each of them. They both have the same velocity in the 'north' direction, but have started at different points, so can never meet. Those two particles have opposite velocities in the east/west direction, so can pass each other (which they do in the second attachment), but cannot meet.
![[tex]\bold{\huge{\underline{ Hola \: Brainlians!!! }}}[/tex]Three Particles A, B And C Are Situated At](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/zTC2NCzcFLmVstKZDxpzyeqOt2E09c5i.png)
![[tex]\bold{\huge{\underline{ Hola \: Brainlians!!! }}}[/tex]Three Particles A, B And C Are Situated At](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/7ZocH6SVq9vsgplLVSFQAO6Uv1BN5mH7.png)
![[tex]\bold{\huge{\underline{ Hola \: Brainlians!!! }}}[/tex]Three Particles A, B And C Are Situated At](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/Jb0lRj55AhWznP42LcU76e4pvAFx8niG.png)
Related Questions
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Answers
Answer:
Kangaroo should be the answer.
Answer:
Kangaroo So there's your answer
Explanation:
Their numbers became so large that they destroyed crops and land, leading to soil erosion. They also negatively affected agriculture and plants by overgrazing. Not only did the rabbits wreak havoc on Australian croplands, but they also contributed to the decline of native plant and animal species.
Cart A has mass M and is released from rest at a height 2H on a ramp making an angle 2 with the horizontal, as shown above. Cart B has mass 2M and is released from rest at a height H on a ramp making an angle with the horizontal. The carts roll toward each other, have a head- on collision on the horizontal portion of the ramp, and stick together. The masses of the carts’ wheels are negligible, as are any frictional or drag forces.
a.) Derive an expression to determine the velocity of Cart A in terms of the variables given in the prompt.
b.) Derive an expression to determine the velocity of Cart B in terms of the variables given in the prompt.
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d.) Is the collision elastic or inelastic? Justify your answer.
*Answers should not include numbers, only variables.
Answers
a) The expression to determine the velocity of cart A is [tex]v_{A} = 2\cdot \sqrt{g\cdot H}[/tex].
b) The expression to determine the velocity of cart B is
[tex]v_{B} =\sqrt{2\cdot g\cdot H }[/tex].
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Study of an inelastic collisionIn this question we shall apply principle of energy conservation and principle of linear momentum conservation to model an inelastic collision between two carts.
a) The combination of cart and ramp can be considered a conservative system as there are no non-conservative forces (i.e. friction), the final velocity of cart A ([tex]v_{A}[/tex]) is related to the change in gravitational potential energy:
[tex]\frac{1}{2}\cdot M\cdot v_{A}^{2} = M\cdot g \cdot 2\cdot H[/tex] (1)
Now we clear [tex]v_{A}[/tex] and simplify the resulting expression:
[tex]v_{A} = 2\cdot \sqrt{g\cdot H}[/tex]
The expression to determine the velocity of cart A is [tex]v_{A} = 2\cdot \sqrt{g\cdot H}[/tex]. [tex]\blacksquare[/tex]
b) We apply the same approach used in part b) to find the final velocity:
[tex]\frac{1}{2}\cdot (2\cdot M) \cdot v_{B}^{2} = (2\cdot M)\cdot g \cdot H[/tex] (2)
Now we clear [tex]v_{B}[/tex] and simplify the resulting expression:
[tex]v_{B} =\sqrt{2\cdot g\cdot H }[/tex]
The expression to determine the velocity of cart B is
[tex]v_{B} =\sqrt{2\cdot g\cdot H }[/tex]. [tex]\blacksquare[/tex]
c) The final velocity ([tex]v[/tex]system is determined by principle of linear momentum conservation:
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[tex]3\cdot v = 2\cdot (1-\sqrt{2})\cdot \sqrt{g\cdot H}[/tex]
[tex]v = \left(\frac{1-\sqrt{2}}{3} \right)\cdot \sqrt{g\cdot H}[/tex]
The final velocity of the carts after the collision is [tex]v = \left(\frac{1-\sqrt{2}}{3} \right)\cdot \sqrt{g\cdot H}[/tex]. [tex]\blacksquare[/tex]
d) The collision is inelastic since a part of the energy of the entire system is lost when they stick together. [tex]\blacksquare[/tex]
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How far does the plane need to travel during take-off (in given units)?
ft
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Answers
Answer:
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The [tex]10,\!000\; {\rm ft}[/tex] runway should be sufficient.
Explanation:
Convert unit of the the take-off velocity of this plane to [tex]\rm ft\cdot s^{-1}[/tex]:
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Take-off velocity of the plane [tex]v =264\; {\rm ft\cdot s^{-1}}[/tex].
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Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration [tex]t[/tex] times average velocity [tex](u + v) / 2[/tex].
The distance that the plane need to cover would be:
[tex]\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}[/tex].
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Answers
Answer:
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Answer:
No.
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Answers
Answer:
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÷4
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Helppp
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Answer:
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Answer:
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[tex]kai6417[/tex]
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Just got right
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Answers
Answer:
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PLEASE ANSWER ILL GIVE BRAINLIEST !!!!!
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Answers
Answer:
where is the problem u need help with
Explanation:
(: pls mark brainliest
Answer: math is an area of knowledge that includes the study of such topics. as numbers and formulas related structures, algebra and geometry also calculus and analysis
Explanation:
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Answers
Answer:
11km
Explanation:
5 + 4 + 2 = 11 (The whale traveled for 11km even if it changes direction)
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Motion of Car
Speed
Distance
0 meters
50 meters
100 meters
150 meters
meters
250 meters
Time
0.0 seconds
4.3 seconds
8.6 seconds
12.9 seconds
17.2 seconds
21.5 seconds
200
Which of the following statements best describe the motion of the car?
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Answers
Answer:
(B) electromagnetic radiation
Explanation:
Answer: B electromagnetic radiation
Explanation: hope this helps ik im a bit late though
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Answers
Answer:
True
Explanation:
Answer:
The answer is T (True)
Explanation:
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Answers
Answer:
about 13.8 billion years old
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Answers
Answer:
A
Explanation:
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Answers
Answer:
The first part of the question is asking about BUOYANT FORCE or UPTHRUST.
Upthrust =TRUE WEIGHT-APPARENT WEIGHT
TRUE WEIGHT=mg
TRUE weight=50kg×10m/s²
=500N
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please mark brainliest, hope it helped
when two forces on the same object are equal and oppostie, these forces are called
Answers
Answer:
Balanced forces are equal and opposite in direction, they act on the same body. The net force due to these forces is zero.
Explanation:
Answer:
Balanced forces are equal and opposite in direction, they act on the same body. The net force due to these forces is zero.
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Please. help me with this question.
Answers
Answer:
When you push it hard enough, then force for the Bush is created than the fictional force. So that means that is net force Equus. If Bush minus the fiction force is very Dan zero, so that means that is mass times acceleration. So that's why it starts moving in the direction off the bush Boxee.
Explanation: