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3 years ago

The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is _____________.

Physics

2 answers:

zhuklara [117]3 years ago

5 0

Answer:

Explanation:

The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is _____________.

A. on the half of the road with traffic going in the opposite direction

B. on the half of the road that the vehicle is traveling

C. anywhere on a two-way street

According to the law regrading pedestrian, the pedestrian has the right of way immediately she is crossing on the crosswalk and is half of the road that the vehicle is travelling.

These rules are there for the safety of the driver, pedestrians and other road users

RSB [31]3 years ago

4 0

Answer: the correct option is B ( on the half of the road that the vehicle is traveling).

Explanation: according to Georgia Code About Pedestrians; The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is

on the half of the road that the vehicle is traveling.

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15. If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the

goldfiish [28.3K]

The final combined velocity after the collision is 20.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two car and of the truck must be conserved before and after the collision.

This means that we can write the following equation:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 800 kg$ is the mass of the sport car

$u_1 = 13.0 m/s$ is the initial velocity of the car (taking its direction as positive direction)

$m_2 = 1200 kg$ is the mass of the truck

$u_2 = 25.0 m/s$ is the initial velocity of the truck

$v$ is the final combined velocity of the car and the truck, after the collision

Re-arranging the equation and substituting the values, we find the velocity after the collision:

$v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(800)(13)+(1200)(25)}{800+1200}=20.2 m/s$

And the positive sign indicates their final direction is the same as the initial direction of the two vehicles.

Learn more about momentum here:

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#LearnwithBrainly

5 0

3 years ago

The strength of a magnetic field follows _____.

ycow [4]

I think its Coulomb's law<span>
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8 0

3 years ago

Read 2 more answers

Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r

ella [17]

Answer:

$\frac{dR(t)}{dt}=0.06\Omega$

Explanation:

Since $R(t)=\frac{V}{I(t)}$ , we calculate the resistance rate by deriving this formula with respect to time:

$\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})$

Deriving what is left (remember that $(\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)$ ):

$\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}$

So we have:

$\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}$

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

$\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega$

8 0

3 years ago

Car A starts in Sacramento at 11am. It travels along 400 mile route to Los Angeles at 60 mph. Car B starts from Los Angeles at n

damaskus [11]

Answer:

38.89 miles

Explanation:

from the question we have the following:

distance between Sacramento and los angles = 400 miles

speed of car A = 60 mph

start time of car A = 11 am

speed of car B = 75 mph

start time of car B = 12 pm

distance of Fresno from Los Angeles = 150 miles

To start off let's allow car A to travel for one hour (from 11 am to 12 pm), during which it would have covered a distance of 60 miles.
Now the time would be 12 pm and the distance between the two cars would be 400 - 60 (distance traveled by car A within 11 am to 12 pm) = 340 miles
From 12 pm to the time both cars will meet, the distance covered by car A + distance covered by car B would be equal to 340 miles. Therefore
Distance covered by car A = speed x time(t) = 60 x t = 60t
Distance covered by car B = speed x time(t) = 75 x t = 75t
60t + 75t = 340 miles
135t = 340
t = 2.51 hours
Recall that at their meeting point, the distance covered by car B = 75t = 75 x 2.62 = 188.89 miles
Since Fresno is 150 miles from Los Angeles, car B which is 188.89 miles from Los Angeles at their meeting point would be 188.89 - 150 = 38.89 miles from Fresno
38.89 miles would also be the distance of car A from Fresno since that is their meeting point.

4 0

3 years ago

5. If one object has a greater speed than a second object. does the first necessarily have a greater acceleration? Explain, usin

Sveta_85 [38]

Answer:

Explanation:

5. not necessarily so that the first object could have left with initial velocity and the second not, so even if the second has a greater acceleration its velocity is less than that of the first

6. The acceleration of the motorcycle is

SI System Reductions

Vo = 80 km / h (1000m / 1km) (1h / 3600s) = 22.2 m / s

Vf = 90 km / h (1000m / 1km) (1h / 3600s) = 25 m / s

Vf = Vo + at at = Vf-Vo

am = (Vf-Vo) / t

am = (25 -22.2) / t = 2.8 / t

am= 2.8/t

For the bike we have

Vf = 10 km / h (1000m / 1km) (1h / 3600s) = 2.78 m / s

Vo = 0

ab = (Vf -Vo) / t

ab = (2.78 -0) / t

ab = 2.8/t

Since time is the same for both of us, if we round to Significant figures the two accelerations are equal

7. If when an object is slowing or slowing down.

For example, a car goes north and must stop at the traffic light, the acceleration of the brakes goes south

8. Yes, since an object can go to the left and the acceleration to the right, but the object will lose speed over time

9. in the launch of projectiles the acceleration is negative and the speed after half the path is also negative

10. Car B must be moving to car A, because if they leave together B has more acceleration, bone that travels the distance at the same time

11. When we have friction, the velocity of an object increases by an external force, but the friction also increases the acceleration, but since it is positive, the velocity increases until the acceleration is zero and hence the velocity remains constant.

8 0

3 years ago