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kirza4 [7]
3 years ago
15

Se lanza una pelota de hule elástica de masa mA con una velocidad VAi contra la puerta trasera de un camión muy pesado de masa m

B que se encuentra en reposo ¿Cuáles son las velocidades de ambos objetos después de la colisión? Suponga un choque elástico.
Physics
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

v_{A}=v_{Ai}

Explanation:

The total momentum of the system must conserve by the law of momentum conservation.

p_b=p_a\\\\m_Av_{Ai}+m_Bv_{Bi}=m_Av_{A}+m_Bv_{B}    ( 1 )

If you consider the shock between the ball and the truck as totally elastic, and also if the truck is much more massive than the ball, you can assume that the truck does not move when the ball hits it.

Then the truck is always at rest

Hence, you have in the expression ( 1 ):

m_Av_{Ai}+0=m_Av_{A}+0\\\\v_{Ai}=v_{A}

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2. In these diagrams, the sunlight is coming from the left, as shown by the arrows. Which
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Answer:

Explanation:

a) A is accurate because the half of the Moon that is facing the sun is it by the sun, and the other half is dark.

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3 years ago
A 15-µF capacitor and a 25-µF capacitor are connected in parallel, and charged to a potential difference of 60 V. How much energ
Alborosie

Answer:

Energy stored, E = 0.072 J

Explanation:

Given that,

Capacitance, C_1=15\ \mu F

Capacitance, C_2=25\ \mu F

These two capacitor are connected in parallel, and charged to a potential difference of, V = 60 volts

We know that in parallel combination of capacitor, the equivalent capacitance is given by :

C=C_1+C_2\\\\C=(15+25)\ \mu F\\\\C=40\times 10^{-6}\ F

The energy stored in the capacitor is given by :

E=\dfrac{1}{2}CV^2\\\\E=\dfrac{1}{2}\times 40\times 10^{-6}\times (60)^2\\\\E=0.072\ J

So, the energy stored in the capacitor in this capacitor combination is 0.072 J.

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3 years ago
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Read 2 more answers
A long wire carrying a 5.0 A current perpendicular to the (xy)-plane intersects the x-axis at x = -2.00 cm. A second parallel wi
zimovet [89]

Answer:

a. 05cm from x axis

b. 8cm from x axis

Explanation:

If the net magnetic field is zero and the currents are in the same direction then the thanks point is between the currents i1 and i2 as show in the attachment below

a. Given that i1= 5A and i2=3A

Let assume the null point is xcm from current i1, then the null point will be (4-x)cm from current i2 since the total length is 4cm.

Now the magnetic field of the current i1 from the null point= to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(4-x)

i1/x=i2/(4-x)

5/x=3/(4-x)

20-5x=3x

8x=20

8x=2.5cm

since from the left of x axis is 2cm, then the null point is 2.5-2 which 0.5cm from the origin x axis.

The null point is 0.5cm from the origin x axis

b. If both current are flowing in opposite direction, the null point lies outside of the current.

Then with same analysis let assume the first current i1 is xcm from the null point and since the total length is 4cm the second current i2 will be (x-4)cm from the null point.

Also the magnetic field of the current i1 from the null point = to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(x-4)

i1/x=i2/(x-4)

5/x=3/(x-4)

5x-20=3x

2x=20

x=10cm.

This shows that the distance of the null point from current i1 is 10cm and the current i1 is 2cm from the x axis, then the null point is 10-2=8cm from the origin x axis.

The null point is 8cm from the x axis.

Check the attachment to see the diagram of the current and the null points

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