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Paha777 [63]
2 years ago
11

Two identical blocks, block A and block B, are placed on different horizontal surfaces. Block A is initially at rest on a smooth

surface, while block B is initially at rest on a rough surface. A constant horizontal force of magnitude F0 is exerted on each block. After the force has been applied for a time Δt, the speeds of blocks A and B are vA and vB, respectively. Which of the following claims indicates the correct relation between vA and vB and provides the best justification for the relation? (A) Va < Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block A.(B) Va < Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater, on average for block B.(C) Va > Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block A (D) Va > Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block B.
Physics
1 answer:
Nostrana [21]2 years ago
3 0

Answer:

(D) Va > Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block B.

Explanation:

Given that, Two identical blocks, block A and block B, are placed on different horizontal surfaces. Block A is initially at rest on a smooth surface, while block B is initially at rest on a rough surface.

And now, a constant force F₀ is exerted on each block for a time Δt.

Now, speeds of blocks A and B are vA and vB respectively.

In case of block A the surface is smooth, so there is no opposing force.

Whereas as, in case of block B, the surface is rough which means friction opposes the motion

So, some of the applied force is used to overcome this friction which makes the speed of block B vB to be less than that of block A vA.

⇒ Va > Vb.

And completely eliminating friction is not possible.

so, even smooth surface has friction which is very very little.

so, The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block B.

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Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the
Hoochie [10]

You've told us:

-- 130°x  =  212°F

and

-- 10°x  =  32°F

Thank you.  Those are two points on a graph of °x vs °F .  With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are  (130, 212)  and  (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation.  I'll use the second point  (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are !  Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad.  Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for  x, and the solution will be the same temperature in  F  too.

or

We can write [ F = 1.5F + 17 ], solve it for  F, and the solution will be the same temperature in  x  too.

F = 1.5F + 17

Subtract  F  from each side:  0.5F + 17 = 0

Subtract 17 from each side:   0.5F = -17

Multiply each side by 2 :  F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for  x:  

F = 1.5(-34) + 17

F = -51 + 17

<em>F = -34</em>

It works !  -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

<em>yay !</em>

6 0
2 years ago
A mad scientist wants to collect massive amounts of charge on basketball sized aluminum balls. The scientist wants to place 6 C
vitfil [10]

The distance between two basket ball sized aluminium balls is 9714 m.

Explanation:

Coulomb's law, or Coulomb's inverse-square law, is an experimental law of physics that quantifies the amount of force between two stationary, electrically charged particles. The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force .

Coulomb's law formula => F = (k * Qb1 * Qb2)/r²

Given data :-

charge on ball 1 Qb1 = 6C

charge on ball 2 Qb2 = 14C

Force exerted F = 8000 N

k =  8.988 x 10^9 Nm²C−²(coulomb's constant).

substituting given values in the coulomb's formula

8000 = (( 8.988 x 10^9)*6*14)/r²

shifting r and 8000 to other sides

r² = (756 * 10^9)/8000

r = 9714 m.

Therefore the distance between two balls is r = 9714 m.

4 0
3 years ago
A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa
Nataly_w [17]

Answer:

V_p = 267.258 m/s

V_n = 38.375 m/s      

Explanation:

using the law of the conservation of the linear momentum:

P_i = P_f

where P_i is the inicial momemtum and P_f is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

P_i = m(270 m/s)

P_f = mV_P + M_nV_n

where m is the mass of the proton and V_p is the velocity of the proton after the collision, M_n is the mass of the nucleus and V_n is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = mV_p + 14mV_n

then, m is cancelated and we have:

270 = V_p + 14V_n

This is a elastic collision, so the kinetic energy K is conservated. Then:

K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2

and

Kf = \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

then,

\frac{1}{2}m(270)^2 =  \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

here we can cancel the m and get:

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

now, we have two equations and two incognites:

270 = V_p + 14V_n  (eq. 1)

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

in the second equation, we have:

36450 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2  (eq. 2)

from this last equation we solve for V_n as:

V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

and replace in the other equation as:

270 = V_p + 14\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

so,

V_p = -267.258 m/s

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

V_n = \frac{270+267.258}{14}

V_n = 38.375 m/s  

3 0
2 years ago
Find the mass of a person who weighs 608 N
Anvisha [2.4K]

Answer:

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7 0
2 years ago
Read 2 more answers
Which accounts for an increase in the temperature of a gas that is kept a constant volume
oksian1 [2.3K]

Answer:

An increase in pressure

Explanation:

The ideal gas law states that:

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V is the volume

n is the number of moles

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T is the temperature of the gas

in the equation, n and R are constant. For a gas kept at constant volume, V is constant as well. Therefore, from the formula we see that if the temperature (T) is increase, the pressure (p) must increase as well.

7 0
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