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Zanzabum
3 years ago
6

Say you dropped a cannonball from the 17–meter mast of a ship sailing at 2 meters/second. How far from the base of the mast will

the ball land?
Physics
1 answer:
vichka [17]3 years ago
3 0
It should land at the base of the mast. the cannonball, whilst in your hand, is travelling at the same rate as the boat in the horizontal plane ie at 2m/s. as it leaves you hand it should retain that velocity and therefore land right underneath where it was dropped. this assumes all other things are equal, eg there is no air resistance etc that might cause the ball to accelerate or decelerate.
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A resistor with resistance R is connected to a battery that has emf 13.0 V and internal resistance r = 0.350 Ω . For what two va
makvit [3.9K]

Answer:

R = 1,746 Ω

Explanation:

The power dissipated in the circuit is

   P = V I = V² / R

Let's find the current

   R = V² / P

Let's calculate

  R = 13²/81

   R = 2,096 Ω

This is total resistance

  R_total = R + r

   R = R_total - r

   R = 2,096 -0,350

   R = 1,746 Ω

8 0
3 years ago
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wlad13 [49]
Technological advances clearly.

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3 years ago
Calculate how much work you need to move the 130N trunk to a ledge 2 n above
arsen [322]
260 joules is hopefully right.
3 0
3 years ago
Read 2 more answers
Which explains why more energy is released in nuclear reactions then in chemical reactions
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A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x ax
alexdok [17]

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

so

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p = - .444 i -  .748 j  

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

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if v be its velocity

.024 v = .87

v = 36.25 m / s .

6 0
3 years ago
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