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3 years ago

You arrive at an intersection with traffic lights that are not working because of a power outage. What do you do?

Physics

2 answers:

d1i1m1o1n [39]3 years ago

8 0

You act as it is a four way stop

solmaris [256]3 years ago

5 0

I'm not sure if this is a school question or a question that you asked because you want to know (Quora is better for this), but I would:

Look around and see what everyone else is doing. Are any of them on phones? How many? If a lot of them are, they're probably calling police and I'd wait for police to arrive. If there was a lot if distress I'd get out of my car and call in my loudest voice to wait for police to come regulate traffic.

If no one else is calling police, I'd call. Tell them to send someone over as fast as they can to regulate traffic, and tell them what I think the importance level is, based on how many cars are backed up.

If people are still trying to drive, I'd try to get everyone to stop and wait for police to come regulate traffic. Stand on my car, call out, use a sign, make a sign, whatever I could think of. Use a whistle if I had one.

When the police came, I'd do what they told me to.

If I was on my way to work, I'd call my boss and say, "you probably know by now, but the traffic lights aren't working. I don't think I can be there in time."

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One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at

Zolol [24]

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature $(T_i)=300 K$

$P_i=10 atm$

$P_f=2 atm$

Work done in iso-thermal process $=P_iV_iln\frac{P_i}{P_f}$

$P_i$ =initial pressure

$P_f$ =Final Pressure

$W=10\times 2.463\times ln\frac{10}{2}=39.64 J$

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is $=P\Delta V$

$V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L$

$V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L$

$\Delta W=1\times (12.315-2.463)=9.852 J$

$\Delta q=\Delta W=9.852 J$

$\Delta U=0$

8 0

3 years ago

A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin

BARSIC [14]

Answer:

$U_k = 0.113$

Explanation:

using the law of the conservation of energy:

$E_i -E_f=W_f$

$\frac{1}{2}Kx^2=NU_kd$

where K is the spring constant, x is the spring compression, N is the normal force of the block, $U_k$ is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

$\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)$

solving for $U_k$ :

$U_k = 0.113$

8 0

3 years ago

A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m toward

Anni [7]

Answer:

$m=18000kg$

Explanation:

From the question we are told that:

Crane Length $l=20m$

Crane Mass $m_a=3000kg$

Arm extension at lifting end $l_l=15m$

Arm extension at counter weight end $l_c=5m$

Load $M_l=5000kg$

Generally the equation for Torque Balance is mathematically given by

$T_1 *l_c-(m_a*g) *l_c-(T_2)*l_l=0$

$mg*5 *-(3000*9.8) *5-(5000*9.8)*15=0$

$m=18000kg$

7 0

2 years ago

A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction ti

vagabundo [1.1K]

Answer:

t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

$d = v. t + \dfrac{v^2}{2a}$

$t = \dfrac{d}{v} -\dfrac{v}{2a}$

t is the reaction time

$t = \dfrac{65}{15.57} -\dfrac{15.57}{2\times 3}$

t = 1.58 s

hence, the reaction time of the ranger is equal to 1.58 s.

3 0

2 years ago

a projectile is lunched with an initial speed of 60.0mm/s at an angle of 30.0° above the horizontal.The projectile lands on a hi

alexandr402 [8]

Answer:

52 mm/s (approximately)

Explanation:

Given:

Initial speed of the projectile is, $u=60.0\ mm/s$

Angle of projection is, $\theta=30.0\°$

Time taken to land on the hill is, $t=4\ s$

In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.

So, the velocity in the horizontal direction always remains the same.

The horizontal component of initial velocity is given as:

$u_x=u\cos\theta\\u_x=60\times \cos(30)\\u_x=30\sqrt3\approx52\ mm/s$

Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.

Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.

6 0

3 years ago