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Lyrx [107]
3 years ago
15

The __________ has a center support bearing that carries the load between the inner and the outer flywheel and is fitted with da

mper springs to absorb shocks.
Physics
1 answer:
alexgriva [62]3 years ago
6 0

Answer;

The dual mass flywheel

The dual mass flywheel has a center support bearing that carries the load between the inner and the outer flywheel and is fitted with damper springs to absorb shocks.

Explanation;

The dual mass flywheel eliminates excessive transmission gear rattle, reduces gear change/shift effort, and increases fuel economy. There are two basic types of Dual Mass Flywheels. The first is made up of a primary and secondary flywheel with a series of torsion springs and cushions.

-There is a friction ring located between the inner and outer flywheel that allows the inner and outer flywheel to slip. This feature is designed to alleviate any damage to the transmission when torque loads exceed the vehicle rating of the transmission. The friction ring is the weak spot in the system and can wear out if excessive engine torque loads are applied through it.

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A police car waits in hiding slightly off the highway. A speeding car is spotted by the police car traveling at speed s = 27.6 m
andriy [413]

Answer:

t=\frac{2s}{a}=25.56s

Explanation:

The distance traveled by the speeding car will be d_s=st, where s=27.6m/s.

The distance traveled by the police car will be given by the formula:

d=v_{0}t+\frac{at^2}{2}, where a=2.16m/s^2 and v_0=0m/s since it departs from rest, thus having d=\frac{at^2}{2}.

The police car will reach the speeding car when those distances are the same, or s=d, so we will have:

st=\frac{at^2}{2}

Which means:

t=\frac{2s}{a}

This is the expression asked, but we can use our values:

t=\frac{2s}{a}=\frac{2(27.6m/s)}{(2.16m/s^2)}=25.56s

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4 years ago
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Georgia [21]
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4 0
3 years ago
What is the inverse square law and how does it relate to gravity?
Nesterboy [21]

Answer:

Inverse Square Law Newton proposed the Inverse Square Law. The effect of gravity (and also on forces such as sunlight) works like this. If say we have a half-mass Earth, it would produce a gravity of not half but a quarter (the square of 2).

7 0
4 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

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