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3 years ago

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While driving, which of the following principles should be used to keep the appropriate distance between your vehicle and the ve

hicle in front of you?
Golden rule
Two-car length rule
Maximize the distance between the two vehicles
Minimize the distance between the two vehicles

1 answer:

Alex787 [66]3 years ago

5 0

The correct answer for this question is "Two-car length rule." While driving, the principle that you should be used to keep the appropriate distance between your vehicle and the vehicle in front of you is to follow the <span>Two-car length rule. This rule is to be followed for safety.</span>

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the answer is sueist

Explanation:

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2 years ago

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An arrow movirg 48.3 m/s has 5.22<br> kg•m/s of momentum. What is its<br> mass?

spin [16.1K]

Answer:

0.11 kg

Explanation:

Ft = MV

Ft = momentum 5.22kg m/s

M = mass

V = velocity 48.3m/s

Therefore

5.22 = M x 48.3

Divide both sides by 48.3

5.22/48.3 = M x 48.3/48.3

0.11 = M

M = 0.11kg

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3 years ago

When a pendulum is at the position all the way to the left when it is swinging (at the top of the arc), what is true of the kine

JulsSmile [24]

Choice - B is the correct one.

At the top of the arc, at one end of the swing:
-- it's not going to get any higher, so the potential energy is maximum
-- it stops moving for an instant, so the kinetic energy is zero

At the bottom of the arc, in the center of the swing:
-- it's not going to get any lower, so the potential energy is minimum
-- it's not going to move any faster, so the kinetic energy is maximum

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3 years ago

If a car travels 600m west in 25 seconds, what is its velocity?

Elena L [17]

$v = \frac{ \:the covered distance}{time} \\$

$v = \frac{600}{25} = \frac{6 \times 100}{25} = 6 \times 4 = 24 \: \\$

$v = 24 \: \frac{m}{s} \\$

_________________________________

If west means the west of the axis x the velocity equal :

$- 24 \: \frac{m}{s} \\$

4 0

2 years ago

A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

3 years ago