Answer:
a = 9.86 m/s²
Explanation:
given,
distance between the centers of wheel = 156 cm
center of mass of motorcycle including rider = 77.5 cm
horizontal acceleration of motor cycle = ?
now,
The moment created by the wheels must equal the moment created by gravity.
take moment about wheel as it touches the ground, here we will take horizontal distance between them.
then, take the moment around the center of mass. Since the force on the ground from the wheels is horizontal, we need the vertical distance.
now equating both the moment
m g d = F h
d is the horizontal distance
h is the vertical distance
m g d = m a h
term of mass get eliminated
g d = a h
so,


a = 9.86 m/s²
<span>this is a dynamometer</span>
Answer:
8F_i = 3F_f
Explanation:
When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.
Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.
Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.
The electrostatic force, Fi, in the initial configuration can be calculated as follows.
![F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f](https://tex.z-dn.net/?f=F_i%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7Bq_Aq_B%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7BQ%5E2%7D%7Br%5E2%7D%5B%2Ftex%7D%3C%2Fp%3E%3Cp%3EThe%20electrostatic%20force%2C%20Ff%2C%20in%20the%20final%20configuration%20is%20%3C%2Fp%3E%3Cp%3E%5Btex%5DF_f%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7Bq_Aq_B%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B3Q%5E2%2F8%7D%7Br%5E2%7D%5B%2Ftex%7D%3C%2Fp%3E%3Cp%3ETherefore%2C%20the%20relation%20between%20Fi%20and%20Ff%20is%20as%20follows%3C%2Fp%3E%3Cp%3E%5Btex%5DF_i%20%3D%20F_f%5Cfrac%7B3%7D%7B8%7D%5C%5C8F_i%20%3D%203F_f)
Answer:
False
Explanation:
Let's consider the definition of the angular momentum,

where
is the moment of inertia for a rigid body. Now, this moment of inertia could change if we change the axis of rotation, because "r" is defined as the distance between the puntual mass and the nearest point on the axis of rotation, but still it's going to have some value. On the other hand,
so
unless
║
.
In conclusion, a rigid body could rotate about certain axis, generating an angular momentum, but if you choose another axis, there could be some parts of the rigid body rotating around the new axis, especially if there is a projection of the old axis in the new one.