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Doss [256]
2 years ago
13

Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th

at is carried and the larger the sediment particles. As the river slows down, the heavier particles of the load will settle out. How can you explain the particle movement based on Newton's laws of motion?
A) An object remains at rest until a force acts on it.
B) For every action there is an equal and opposite reaction.
C) The greater the force of the water, the faster the particle accelerates.
D) The water and the particle exert a gravitational attractive force on each other.
Physics
2 answers:
Alekssandra [29.7K]2 years ago
4 0

Explanation :

Newton's laws of motion tells us the work of gravity. In this case, initially the sediments are at rest. As the faster water flows, the more sediments carries out with larger sediment particles and when the flow of water is less, the heavier particle will settle out. This implies that the greater the force of water, the faster the particle accelerates.

katovenus [111]2 years ago
3 0

Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

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Using the formula for kinetic energy of a moving particle k=12mv2, find the kinetic energy ka of particle a and the kinetic ener
Flauer [41]
<span>Answer: KE = (11/2)mω²r², particle B must have mass of 2m, while A has mass m. Then the moment of inertia of the system is I = Σ md² = m*(3r)² + 2m*r² = 11mr² and then KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2 So I'll proceed under that assumption. For particle A, translational KEa = ½mv² but v = ω*d = ω*3r, so KEa = ½m(3ωr)² = (9/2)mω²r² For particld B, translational KEb = ½(2m)v² but v = ω*r, so KEb = ½(2m)ω²r² so total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2 which is equal to our rotational KE.</span>
5 0
2 years ago
f a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance D40 compared to the stopping distance
Oksana_A [137]

Answer:

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

Explanation:

given data

speed = 40 miles / hour

distance = D40

speed limit = 25 miles / hour

distance = D25

to find out

express number a multiple of stopping distance @ 25 mph

solution

we know here stopping distance is directly proportional to (speed)²

so here speed ratio is

initial speed = \frac{40}{25}

so initial speed = 1.6

so

stopping distance increase = (1.6)²

\frac{D40}{D25} = (1.6)²

\frac{D40}{D25} = 2.56

so here

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

5 0
2 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
2 years ago
g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb
Law Incorporation [45]

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

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total buoyant force = vg + .885 g

For floating

vg + .885 g  = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115  / 33.5

= .36  

= 36 %

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2 years ago
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Sergeeva-Olga [200]
By v^2 = u^2 + 2gh 
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