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tresset_1 [31]
2 years ago
5

15. When you cannot stop safely at a yellow traffic light before entering an intersection, ______________. A. stop in the inters

ection as soon as the light turns red B. accelerate so you'd cross the intersection before the light turns red C. enter the intersection carefully and continue across
Physics
1 answer:
Elan Coil [88]2 years ago
4 0

Answer:  When you cannot stop safely at a yellow traffic light before entering an intersection, enter the intersection carefully and continue across.

Explanation: To find the correct answer, we need to know more about the traffic signal rules.

<h3>What is the traffic signal rules?</h3>
  • Red light- Indicator for the motorists to stop.
  • Green-Signal for safety and word GO.
  • Yellow- This signal let you know that the red signal is about to be displayed.
  • when it's turned on, you can start slowing down to come to a stop in anticipation of red light.
  • when we cannot stop safely at a yellow traffic light before entering an intersection, enter the intersection carefully and continue across.

Thus, we can conclude that, the option C is correct.

Learn more about the traffic signal rules here:

brainly.com/question/28044804

#SPJ4

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What is the work required for a penguin to push a box 2 meters with a force of 8 newtons?
choli [55]

Work done is given by product of force and displacement due to that force

So here we will have

Work = Force \times displacement

here we know that

Force = 8 N

displacement = 2 m

Now work done is given as

W = 8\times 2

W = 16 J

so it will do 16 J work to move the box

3 0
3 years ago
If both mass and speed are doubled, what happens to its momentum?
puteri [66]

The general formula is:      Momentum = (mass) x (speed)

I never like to just write a bunch of algebra without explaining it.
But in this particular case, there's really not much to say, and
I think the algebra will pretty well explain itself.  I hope so:


Original momentum = (original mass) x (original speed)


New momentum = (2 x original mass) x (2 x original speed)

                           = (2) x (original mass) x (2) x (original speed)

                           = (2) x (2) x (original mass) x (original speed)

                           =  (4) x (original mass) x (original speed)

                           =  (4) x (original momentum).

7 0
3 years ago
The movement of a magnetic pole away from the actual pole
Lunna [17]

The North Magnetic Pole is the point on the surface of Earth's Northern Hemisphere at which the planet's magnetic field points vertically downwards (in other words, if a magnetic compass needle is allowed to rotate about a horizontal axis, it will point straight down). There is only one location where this occurs, near (but distinct from) the Geographic North Pole and the Geomagnetic North Pole.
3 0
3 years ago
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel
Zanzabum
Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m

Answer: ymax = 10084.2m
8 0
3 years ago
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