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MaRussiya [10]
3 years ago
9

Which of the following diagrams involves a virtual image ?

Physics
1 answer:
sergiy2304 [10]3 years ago
7 0

Answer:

The third diagram

Explanation:

  • <u>A virtual image</u> is an image that can not be formed on a screen.
  • <u>A convex lens</u> can form both virtual and real image depending on the position of the object from the lens.
  • A virtual image in convex lens is formed when the object is placed between the focus and the optical center of the lens.
  • In the third diagram, a virtual image is formed because the position of the object is between the focus and the optical center of the convex lens.
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A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Tems11 [23]

Answer:

a) V_{wf} = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

m_{b}V{b_{i} }  + V_{wi}  = m_{w} V_{wf} + m_{b}V_{bf}

where m_{b},V{b_{i} are the mass and the initial velocity of the bullet, m_{w} and V_{wi} are the mass and the initial velocity of the wooden block, and V_{wf} and V_{bf} are the final velocities of the wooden block and the bullet

The wooden block is initial at rest (V_{wi} = 0) this yields

m_{b}V{b_{i} }  = m_{w} V_{wf} + m_{b}V_{bf}

By solving for V_{wf} adn substitute the givens

V_{wf} = \frac{m_{b}V_{bi} - m_{b} V_{bf}    }{m_{W} }

= \frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}

V_{wf} = 4.67m/s

b) The center of mass speed is defined as

V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}

substituting:

V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)

V = 8.29 m/s

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Explanation:

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