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Kay [80]
3 years ago
11

The distance between the centers of the wheels of a motorcycle is 146 cm. The center of mass ofthe motorcycle, including the rid

er, is 78.5 cm above the ground and halfway between the wheels.Assume the mass of each wheel is small compared to the body of the motorcycle. The enginedrives the rear wheel only. What horizontal acceleration of the motorcycle will make the frontwheel rise off the ground
Physics
1 answer:
Vitek1552 [10]3 years ago
8 0

Answer:

9.12267515924 m/s²

Explanation:

Here the moment created by the wheels and the moment created by the center of gravity will balance each other.

h = Height of the center of mass = 78.5 cm

d =  Distance from back wheel to the center of mass = \dfrac{146\times 10^{-2}}{2}\ m

g = Acceleration due to gravity = 9.81 m/s²

a = Horizontal acceleration

The equation is of the form

mgd=Fh\\\Rightarrow mgd=mah\\\Rightarrow a=\dfrac{gd}{h}\\\Rightarrow a=\dfrac{9.81\times \dfrac{146\times 10^{-2}}{2}}{78.5\times 10^{-2}}\\\Rightarrow a=9.12267515924\ m/s^2

The horizontal acceleration of the motorcycle that will make the front wheel rise off the ground is 9.12267515924 m/s²

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Stop lamps must be visible within____ to the rear
Charra [1.4K]
The answer is 300 feet. The stop lamp or lamps on the rear of a vehicle must show a red light that is set in motion upon application of the service or foot brake and, in a vehicle manufactured or assembled on or after January 1, 1964, must be visible from a distance of not less than 300 feet to the rear in normal sunlight. Take note, if the vehicle is manufactured or assembled January 1, 1964, the stop lamp or lamps must be visible from a distance of not less than 100 feet. Also, the stop lamp may be combined with one or more other rear lamps. 
3 0
4 years ago
Read 2 more answers
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
Which of the following biomes might be found at a latitude of 33° South?
cupoosta [38]
The answer is going be desert. 
5 0
3 years ago
A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like prot
Rzqust [24]

Answer:

595391.482946 m/s

3.21875\times 10^{6}

Explanation:

E = Energy = 1.85 keV

I = Current = 5.15 mA

e = Charge of electron = 1.6\times 10^{-19}\ C

t = Time taken = 1 second

m = Mass of proton = 1.67\times 10^{-27}\ kg

Velocity of proton is given by

v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s

The speed of the proton is 595391.482946 m/s

Current is given by

I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C

Number of protons is

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons

The number of protons is 3.21875\times 10^{6}

3 0
3 years ago
Betty (mass 40 kg), standing on slippery ice, catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Show that the
MA_775_DIABLO [31]

Answer:

v = 2.18m/s

Explanation:

In order to calculate the speed of Betty and her dog you take into account the law of momentum conservation. The total momentum before Betty catches her dog must be equal to the total momentum after.

Then you have:

Mv_{1o}+mv_{2o}=(M+m)v        (1)

M: mass Betty = 40kg

m: mass of the dog = 15kg

v1o: initial speed of Betty = 3.0m/s

v2o: initial speed of the dog = 0 m/s

v: speed of both Betty and her dog = ?

You solve the equation (1) for v:

v=\frac{Mv_{1o}+mv_{2o}}{M+m}=\frac{(40kg)(3.0m/s)+(15kg)(0m/s)}{40kg+15kg}\\\\v=2.18m/s

The speed fo both Betty and her dog is 2.18m/s

7 0
3 years ago
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