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RUDIKE [14]
4 years ago
13

How can a driver best be prepare to enter sharp curves

Physics
1 answer:
Eduardwww [97]4 years ago
7 0
Slow down before entering the curve. Stay close to the outside. Use the gas pedal as you enter the middle of the curve.
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Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 10 m/s.
abruzzese [7]

Hi there!

Kinetic energy can be calculated using the following:

\large\boxed{KE = \frac{1}{2}mv^2}}

Where:

KE = Kinetic energy (J)

m = mass (kg)

v = velocity (m/s)

Plug in the given values:

KE = \frac{1}{2}(625)(10^2) = \boxed{31250J}

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2 years ago
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What happens to the kinetic energy of a snowball as it rolls across the lawn and gains mass?
Delvig [45]
The kinetic energy will increase
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3 years ago
Materials with resistivities between those of good conductors and those of insulators are called:
bogdanovich [222]

semiconductors ... silicon, germanium etc ...

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3 years ago
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A glacier can travel over the surface of earth and reshape it. the chart describes three things a glacier can do.which terms bes
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The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




7 0
3 years ago
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