he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug
1 answer:
Answer:
207.4 N
Explanation:
The torque on a body is
where r is the radius vector from the point of rotation to the point at which force F is applied.
The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.
Therefore, torque is also given by
Where is the angle between r and F.
Use the expression of torque.
Substitute L for r in the equation
Where L is the length of the wrench.
Making F the subject
Force required to pull the wrench is given as,
Substitute for
, 25 cm for L, and 115o for
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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
a)
(+x axis)
b)
(+x axis)
(+z axis)
c)
(+x axis)
(+y axis)
(-x axis)
Explanation:
a)
The electric force exerted on a charged particle is given by
where
q is the charge
E is the electric field
For a positive charge, the direction of the force is the same as the electric field.
In this problem:
is the charge
is the electric field, along the x-direction
So the electric force (along the x-direction) is:
towards positive x-direction.
The magnetic force instead is given by
where
q is the charge
v is the velocity of the charge
B is the magnetic field
is the angle between the directions of v and B
Here the charge is stationary: this means , therefore the magnetic force due to each component of the magnetic field is zero.
b)
In this case, the particle is moving along the +x axis.
The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,
(towards positive x-direction)
Concerning the magnetic force, we have to analyze the two different fields:
- : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore,
, so the force due to this field is zero.
: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore,
. Therefore,
, so the force due to this field is:
where:
is the charge
is the velocity
is the magnetic field
Substituting,
And the direction of this force can be found using the right-hand rule:
- Index finger: direction of the velocity (+x axis)
- Middle finger: direction of the magnetic field (+y axis)
- Thumb: direction of the force (+z axis)
c)
As in part b), the electric force has not change, since it does not depend on the veocity of the particle:
(+x axis)
For the field , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is
And by substituting,
And by using the right-hand rule:
- Index finger: velocity (+z axis)
- Middle finger: magnetic field (+x axis)
- Thumb: force (+y axis)
For the field , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is
And by substituting,
And by using the right-hand rule:
- Index finger: velocity (+z axis)
- Middle finger: magnetic field (+y axis)
- Thumb: force (-y axis)