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n200080 [17]
3 years ago
7

When fuel and air are compressed in the compression stroke, ...... a. each molecule of fuel is heated to its flash point b. each

molecule of air is heated to its flash point c. the exhaust valve opens to release excessive pressure
Physics
1 answer:
alexira [117]3 years ago
7 0

None of the choices is an appropriate response.

There's no such thing as the temperature of a molecule.  Temperature and
pressure are both outside-world manifestations of the energy the molecules
have.  But on the molecular level, what it is is the kinetic energy with which
they're all scurrying around.

When the fuel/air mixture is compressed during the compression stroke,
the temperature is raised to the flash point of the mixture.  The work done
during the compression pumps energy into the molecules, their kinetic
energy increases, and they begin scurrying around fast enough so that
when they collide, they're able to stick together, form a new molecule,
and release some of their kinetic energy in the form of heat.


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A capacitor is charged until its stored energy is 7.54 J. A second capacitor is then connected to it in parallel. If the charge
Ivan

Answer:

2 J

Explanation:

A charged capacitor of capacitance C_1 with energy of 7.54 J, is connected in parallel with another capacitor C_2 , so the charge is equally distributed between them.

(a) The energy stored in the capacitor before it being connected to the other capacitor is:

U_O=q_0^2/2C_1=7.54 J\\

The energy stored in the electric field is the sum of the energies of the two capacitors:

U=U_1+U_2\\U=q_1^2/2C_1+q_2^2/2C_2

since the charge equally distributed,  q_1 = q_2 = q_o/2. and since they are connected in parallel the potential difference on both of them is the same :

V_1=V_2\\q_1/C_1=q_2/C_2\\q_0/C_1=q_0/C_2\\C_1=C_2=C_3\\

hence,

U=q_0^2/8C+q_0^2/8C\\U=q_0^2/4C\\U=2J

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3 years ago
The coefficient of linear expansion of ordinary glass is three times that of Pyrex glass. An ordinary glass rod and a Pyrex glas
DochEvi [55]

Answer:

b. 3 times

Explanation:

Lets take

Coefficient for ordinary glass = α₁

Coefficient for pyrex glass = α₂

 Given that α₁ = 3 α₂

Initial length of both glasses are equal = L

Change in the temperature is also same .= ΔT

We know that change in the length  given as

ΔL =  L α ΔT

Therefore

\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{L\alpha_1\Delta T}{L\alpha_2\Delta T}

\dfrac{\Delta L_1}{\Delta L_2}=\dfrac{3\alpha_2}{\alpha_2}

ΔL₁ = 3ΔL₂

Therefore change in the length of original glass is three time of pyrex glass.

b. 3 times

6 0
3 years ago
A man wishes to lift a stone weighing 1440 N, using a first-class lever that measures 5 meters. What force should it perform if
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Answer:

3360 N

Explanation:

In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.

The lever is 5 m long.  The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.

The torques are equal:

Fr = Fr

(1440 N) (3.5 m) = F (1.5 m)

F = 3360 N

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3 years ago
Which one of the following statements best describes the just world hypothesis?   A. People deserve what happens to them, whethe
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3 years ago
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A sphere of mass m and radius r is released from rest at the top of a curved track of height H. The sphere travels down the curv
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Explanation:

<em>(a) On the dots below, which represent the sphere, draw and label the forces (not components) that are exerted on the sphere at point A and at point B, respectively.  Each force must be represented by a distinct arrow starting on and pointing away from the dot.</em>

At point A, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing left, and static friction force Fs pushing down.

At point B, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing down, and static friction force Fs pushing right.

<em>(b) i. Derive an expression for the speed of the sphere at point A.</em>

Energy is conserved:

PE = PE + KE + RE

mgH = mgR + ½mv² + ½Iω²

mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²

mgH = mgR + ½mv² + ⅕mv²

gH = gR + ⁷/₁₀ v²

v² = 10g(H−R)/7

v = √(10g(H−R)/7)

<em>ii. Derive an expression for the normal force the track exerts on the sphere at point A.</em>

Sum of forces in the radial (-x) direction:

∑F = ma

N = mv²/R

N = m (10g(H−R)/7) / R

N = 10mg(H−R)/(7R)

<em>(c) Calculate the ratio of the rotational kinetic energy to the translational kinetic energy of the sphere at point A.</em>

RE / KE

= (½Iω²) / (½mv²)

= ½(⅖mr²)(v/r)² / (½mv²)

= (⅕mv²) / (½mv²)

= ⅕ / ½

= ⅖

<em>(d) The minimum release height necessary for the sphere to travel around the loop and not lose contact with the loop at point B is Hmin.  The sphere is replaced with a hoop of the same mass and radius.  Will the value of Hmin increase, decrease, or stay the same?  Justify your answer.</em>

When the sphere or hoop just begins to lose contact with the loop at point B, the normal force is 0.  Sum of forces in the radial (-y) direction:

∑F = ma

mg = mv²/R

gR = v²

Applying conservation of energy:

PE = PE + KE + RE

mgH = mg(2R) + ½mv² + ½Iω²

mgH = 2mgR + ½mv² + ½(kmr²)(v/r)²

mgH = 2mgR + ½mv² + ½kmv²

gH = 2gR + ½v² + ½kv²

gH = 2gR + ½v² (1 + k)

Substituting for v²:

gH = 2gR + ½(gR) (1 + k)

H = 2R + ½R (1 + k)

H = ½R (4 + 1 + k)

H = ½R (5 + k)

For a sphere, k = 2/5.  For a hoop, k = 1.  As k increases, H increases.

<em>(e) The sphere is again released from a known height H and eventually leaves the track at point C, which is a height R above the bottom of the loop, as shown in the figure above.  The track makes an angle of θ above the horizontal at point C.  Express your answer in part (e) in terms of m, r, H, R, θ, and physical constants, as appropriate.  Calculate the maximum height above the bottom of the loop that the sphere will reach.</em>

C is at the same height as A, so we can use our answer from part (b) to write an equation for the initial velocity at C.

v₀ = √(10g(H−R)/7)

The vertical component of this initial velocity is v₀ sin θ.  At the maximum height, the vertical velocity is 0.  During this time, the sphere is in free fall.  The maximum height reached is therefore:

v² = v₀² + 2aΔx

0² = (√(10g(H−R)/7) sin θ)² + 2(-g)(h − R)

0 = 10g(H−R)/7 sin²θ − 2g(h − R)

2g(h − R) = 10g(H−R)/7 sin²θ

h − R = 5(H−R)/7 sin²θ

h = R + ⁵/₇(H−R)sin²θ

4 0
3 years ago
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