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DaniilM [7]
2 years ago
6

Bodies A and B have equal mass. Body B is initially at rest. Body A collides with body B in a one-dimensional elastic collision.

Which statement is true?
A. BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A

B. THE MOMENTUM OF THE SYSTEM IS NOT CONSERVED

C.THE KINETIC ENERGY OF THE SYSTEM IS NOT CONSERVED

D.THE FINAL VELOCITY
Physics
2 answers:
inn [45]2 years ago
6 0
After looking on the scaleattached with the question I recognized the right answer in the first option represented above. A. BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A - is definitely the only correct answer which perfectly concides with the given task.Hope you still need the answer it because this one I gave you is really useful.
jek_recluse [69]2 years ago
4 0
According to the statement " Collision <span>between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision."
The best answer is :
Option A " </span><span>BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A "</span>
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Two long, straight wires are separated by a distance of 32.2 cm. One wire carries a current of 2.75 A, the other carries a curre
Igoryamba

Answer:

a)\frac{F_1}{L}=1.95*10^-^5N

b)\frac{F_2}{L}=1.95*10^-^5N

Explanation:

From the question we are told that:

Distance between wires d=32.2

Wire 1 current I_1=2.75

Wire 2 current I_2=4.33

a)

Generally the equation for Force on l_1 due to I_2 is mathematically given by

F_1=I_1B_2L

Where

B_2=Magnetic field current by I_2

B_2=\frac{\mu *i_2}{2\pi d}

Therefore

F_1=I_1B_2L

F_1=I_1(\frac{\mu *i_2*l_1}{2\pi d})L

\frac{F_1}{L} =\frac{4*\pi*10^{-7}*2.75*4.33*100 }{2*\pi*12.2 }

\frac{F_1}{L}=1.95*10^-^5N

b)

Generally the equation for Force on I_2 due to I_1 is mathematically given by

F_2=I_2B_1L

Where

B_1=Magnetic field current by I_2

B_1=\frac{\mu *I_1}{2\pi d}

Therefore

\frac{F_2}{L} =I_2(\frac{\mu *I_1*I_2}{2\pi d})

\frac{F_2}{L}=1.95*10^-^5N

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to remove a tight-fitting jar, megan runs the lid under hot water. What happends to the jar lid when its temperature increases?
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66. Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown st
kozerog [31]

Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:  

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t  Equation 2

y: The vertical distance the ball moves at time t  

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t  

Known information

We know the following data:

Vi=15 m / s

g =9.8 \frac{m}{s^{2} }

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)  

a) t=1s

y = 15*1 - ½ 9.8*1^{2}=15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

y = 15*1.5 - ½ 9.8*1.5^{2}=22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

y = 15*2 - ½ 9.8*2^{2}= 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

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Calculate the density of a solid cube that
tia_tia [17]
Ok so if each side is 4.53 cm, we can multiply 4.53 x 4.53 x 4.53 to get the volume (since v= l x w x h). Density equals mass/volume, so

519 g/4.53 cm 
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4 0
3 years ago
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