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uysha [10]
3 years ago
5

A 0.50-kg bomb is sliding along an icy pond (frictionless surface) with a velocity of 2.0 m/s to the west. The bomb explodes int

o two pieces. After the explosion, a 0.20-kg piece moves south at 4.0 m/s. What are the components of the velocity of the 0.30-kg piece?
Physics
2 answers:
nevsk [136]3 years ago
5 0

Answer:

2.667m/s to the north and 3.333 m/s to the west

Explanation:

According to law of momentum conservation, the total momentum should be conserved before and after the explosion.

Before the explosion, the momentum was

0.5*2 = 1 kg m/s to the west

Therefore the total momentum after the explosion should be the same horizontally and vertically.

Vertically speaking, it was 0 before the explosion. After the explosion:

0.2*4 + 0.3v = 0

0.3v = -0.8

v = -0.8/0.3 = -2.667 m/s

So the vertical component of the 0.3kg piece is 2.667m/s to the north

Horizontally speaking, since the 0.2kg-piece doesn't move west or east post-explosion:

0.2*0 + 0.3V = 1

0.3V = 1

V = 1/0.3 = 3.333 m/s

So the horizontal component of the 0.3kg piece is 3.333 m/s to the west

marishachu [46]3 years ago
3 0

Answer:

x = - 3.33 i

y = 2.667 j

Explanation:

Given that;

mass of bomb m1 = 0.50 kg

After explosion , 0.20 kg piece moves south m2 = 0.20 kg

At v2 = 4.0 m/s

By using conservation of momentum

m1 v1 + m2 v2 (-j) = (m1 + m2) v (-i)

0.3 v1 - 0.2* 4 j = - 0.5* 2 i

v1 = - 3.33 i + 2.667 j

so  for each components of velocity v1 on the x and y axis are shown below in a respective order

x = - 3.33 i

y = 2.667 j

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3 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

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F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

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F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

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What can you say about the magnitudes of the forces that the balloons exert on each other?
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Answer:

F_G=G. \frac{m_1.m_2}{R^2} gravitational force

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Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

F_G=G. \frac{m_1.m_2}{R^2}

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But when  there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.

Given as:

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3 0
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. Light travels at a speed of about 300 000 km/s. a. Express this value in scientific notation. b. Convert this value to meters
olganol [36]

Answer:

The answer to your question is below

Explanation:

Data

light speed = 300 000 km/s

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to do it, we just move the decimal point 5 places to the left

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v = 300 000 km/s

d = ?

t = 1 s

 formula   v = d/t      we clear distance     d = vxt

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Car A moves 20 meters every second:
20 x 40 = 800 meters
8 0
3 years ago
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