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alex41 [277]
3 years ago
9

The two-second rule applies to any speed __________ A. in all weather conditions. B. in ideal weather and road conditions. C. in

the worst weather and road conditions. D. none of the above
Physics
2 answers:
kumpel [21]3 years ago
5 0

Answer:

The answer is A

Explanation:

A large risk of tailgating is the collision avoidance time being much less than the driver reaction time. Driving instructors advocate that drivers always use the "two-second rule" regardless of speed or the type of road. During adverse weather, downhill slopes, or hazardous conditions such as black ice, it is important to maintain an even greater distance.

Flura [38]3 years ago
5 0

Answer:

in ideal weather and road conditions.

Explanation:

Minimum Safe Following Distances

Leave plenty of space between you and the vehicle ahead, including bicycles. If it stops quickly, you will need time to see the danger and stop.

Using the Two-Second Rule

At any speed, you can use the two-second rule to see if you are far enough behind the car in front of you:

Watch the vehicle ahead pass some fixed point - an overpass, sign, fence corner, or other marker.

Count off the seconds it takes you to reach the same spot in the road ("one thousand and one, one thousand and two...").

If you reach the mark before you finish counting, you are following too closely. Slow down and check your following distance again.  

The two-second rule applies to any speed in ideal weather and road conditions. If road or weather conditions are not good, double your following distance. You should also double your following distance when driving a motor home or towing a trailer.

Following Distance For Trucks

A truck or any vehicle towing another vehicle may not follow within 300 feet of another truck or vehicle towing a vehicle. This law does not apply to overtaking and passing, and it does not apply within cities or towns.

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Which property makes radio waves most useful for communication? They have low frequencies. They have long wavelengths. They tran
Mila [183]
I think it would be that they have long wavelengths
6 0
4 years ago
Read 2 more answers
A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole poin
Llana [10]

Answer:

<em>There is no induced current on the coil.</em>

Explanation:

Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.

7 0
3 years ago
24 POINTS
vesna_86 [32]
80 meters should be it
3 0
3 years ago
A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex
irina1246 [14]

a)

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

F_{B_x}=0

F_{B_y}=0

b)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=0

F_{B_y}=3.21\cdot 10^{-3}N (+z axis)

c)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=3.21\cdot 10^{-3} N (+y axis)

F_{B_y}=3.21\cdot 10^{-3}N (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

F=qE

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:

q=+4.9\mu C=+4.9\cdot 10^{-6}C is the charge

E_x=+242 N/C is the electric field, along the x-direction

So the electric force (along the x-direction) is:

F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N

towards positive x-direction.

The magnetic force instead is given by

F=qvB sin \theta

where

q is the charge

v is the velocity of the charge

B is the magnetic field

\theta is the angle between the directions of v and B

Here the charge is stationary: this means v=0, 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,

F_{E_x}=1.19\cdot 10^{-3} N (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- B_x: this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=0^{\circ}, so the force due to this field is zero.

- B_y: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=90^{\circ}. Therefore, \theta=90^{\circ}, so the force due to this field is:

F_{B_y}=qvB_y

where:

q=+4.9\cdot 10^{-6}C is the charge

v=345 m/s is the velocity

B_y = +1.9 T is the magnetic field

Substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

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:

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

For the field B_x, the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

F_{B_x}=qvB_x

And by substituting,

F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

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 B_y, the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

F_{B_y}=qvB_y

And by substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0
4 years ago
8.0kg rocket fired horizontally encounters a force of air resistance of 4.9 N. The force supplied by the rocket's engine is 60.9
Gelneren [198K]
60.9 - 4.9= 56
56 is the net force
using the formula F = ma
56= 8a
a = 7
acceleration is 7ms^-2
6 0
3 years ago
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