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3 years ago

When you encounter a yellow light as you approach an intersection what is the safest approach to take?

1 answer:

5 0

This isn't a physics question really, but you should slow down unless you don't believe you can stop in time or don't believe it is safe to stop quickly (say you see someone driving very close behind you and you don't think they're paying attention). In general, if you think that by staying at your current speed your back tires will cross the ending lines of the intersection by the time the light turns red, it is safe to go through the yellow light. However, this is a thing you'll develop a feel for as you're driving, when in doubt, just slow down, just watch slamming on your breaks if there is someone behind you, sometimes people will see the yellow light when they're behind you and they'll speed up behind you so they'll "make it" before the light turns yellow. While this isn't illegal, since they're not technically running the red light, you should never speed up going up to a yellow light, if you need to speed up to make it before it turns red, you shouldn't make the light, just stop, this is especially bad if someone is directly in front of you and is likely to stop at the yellow light while the person behind them speeds up. This causes a lot of accidents.

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Hitungkan pecutan bagi blok di bawah: / Cal<br>(a)<br>m= 2 kg<br>F= 8.0 N

ioda

Answer:

Acceleration = 4 m/s²

Explanation:

Given the following data;

Force = 8 N

Mass = 2 kg

To find the acceleration of the block;

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

$Acceleration = \frac {Net \; force}{mass}$

Substituting into the formula, we have;

$Acceleration = \frac {8}{2}$

Acceleration = 4 m/s²

4 0

3 years ago

The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increa

Nostrana [21]

Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

<h3>What is simple harmonic motion?</h3>

A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
We have the equation of time period of a SHM as,

$T=2\pi \sqrt{\frac{m}{k} }$

Where, m is the mass of the body and k is the spring constant.

<h3>How to solve the problem?</h3>

Given that,

$T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s$

We have to find the value of m,

$T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\$

$m=\frac{5}{8} =0.625kg$

Thus, we can conclude that, the mass m will be 0.625kg.

Learn more about simple harmonic motion here:

brainly.com/question/28045110

#SPJ4

3 0

2 years ago

On isolated ground receptacles, the metal yoke ____ integrally bonded to the equipment grounding terminal of the receptacle.

Ede4ka [16]

On isolated ground receptacles, the metal yoke is not allowed to be integrally bonded to the equipment grounding terminal of the receptacle.

Any device with two distinct switches or receptacles is a duplex device. It can be shaped to fit a Decora opening or a typical duplex plate opening. It should be noted that they can be combination devices with a switch/outlet, switch/pilot light, etc.

Because of grounding connection removal and receptacle, it is utterly undesirable to connect the two bare equipment grounding conductors together directly.

The equipment grounding conductor associated with those circuits must be connected to the box when circuit conductors are terminated on equipment inside a metal box to prevent unneeded current discharge.

Learn more about grounding conductors here brainly.com/question/14886979

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4 0

2 years ago

A spaceship is traveling through deep space towards a space station and needs to make a course correction to go around a nebula.

expeople1 [14]

Answer:

Magnitude = 3.64 × $10^6$

စ = 43.9°

Explanation:

given data

ship to travel = 1.7 × $10^6$ kilometers

turn = 70°

travel an additional = 2.7 × $10^6$ kilometers

solution

we will consider here

Px = 1.7 × $10^6$

Py = 0

Qx =2.7 × $10^6$ cos(70)

Qy= 2.7 × $10^6$ sin(70)

so that

Hx = Px + Qx ............1

Hx = 2.62 × $10^6$

and

Hy = Py + Qy ..........2

Hy = 2.53 × $10^6$

so Magnitude = $\sqrt{((2.62 \times 10^6 )^2+(2.53 \times 10^6)^2)}$

Magnitude = 3.64 ×

so direction will be

tan စ = Hy ÷ Hx ......................3

tan စ = $\frac{2.53}{2.62}$

tan စ = 0.9656

စ = 43.9°

3 0

3 years ago

Suppose Car 1 collides with Car 2; Car 2 was not moving. In an ideal situation with no friction, according to the Law of Conserv

insens350 [35]

Your answer is C it is slightly reduced

4 0

3 years ago

Read 2 more answers