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valentina_108 [34]
3 years ago
6

CAN SOMEONE PLEASE HELP ME WITH MY PHYSICS QUESTIONS? I NEED CORRECT ANSWERS ONLY

Physics
1 answer:
creativ13 [48]3 years ago
5 0
B) moves with a constant speed until hitting the other end.
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A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di
navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

X(van)=5.65t+154

X(driver)=34.4t+\frac{(-2)t^{2} }{2}

or by rearanging the drivers equation.

X(driver)=34.4t+t^{2}

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

X(van)=X(driver)

5.65t+154=34.4t-t^{2}

0=t^{2} -(34.4-5.65)t+1540=t^{2} -28.75t+154

To solve this equation we use the following formulas

t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}

t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}

Where a=1; b=-28.75; c=154

So we get:

t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63st=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

V(driver)=V_{0} +at}

V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer  A).

Best of luck

8 0
3 years ago
A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force
Vlada [557]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

vo = 25 m/sec 
<span>vf = 0 m/sec </span>
<span>Fμ = 7100 N (Force due to friction) </span>
<span>Fg = 14700 N </span>
<span>With the force due to gravity, you can find the mass of the car: </span>
<span>F = ma </span>
<span>14700 N = m (9.8 m/sec²) </span>
<span>m = 1500 kg </span>
<span>Now, we can use the equation again to find the deacceleration due to friction: </span>
<span>F = ma </span>
<span>7100 N = (1500 kg) a </span>
<span>a = 4.73333333333 m/sec² </span>
<span>And now, we can use a velocity formula to find the distance traveled: </span>
<span>vf² = vo² + 2a∆d </span>
<span>0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d </span>
<span>0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d </span>
<span>-625 m²/sec² = (-9.466666666667 m/sec²) ∆d </span>
<span>∆d = 66.0211267605634 m </span>
<span>∆d = 66.02 m</span>
7 0
4 years ago
HELP ASAP!!! PLEASE!!!! A driver traveling on the highway at 100km/hr notices the speed limit changes to 50km/hr as a speed came
Talja [164]

Answer:

90m

Explanation:

Speed * time

100*0.9=90m

hope this helped xx

8 0
4 years ago
you are driving at 18m/s down Lyndale​ avenue. A car backs out a driveway 25 meters in front of you. You continue at that speed
zimovet [89]

Hello!

For this, first let's calculate time of stop:

t = (V - Vi) / a

Replacing:

t = (0 m/s - 18 m/s) / -5,4 m/s^2

Resolving:

t = -18 m/s / -5,4 m/s^2

t = 3,33 s + 0,25 s = 3,58 s

Now lets calculate distance traveled, with formula:

d = Vi*t + (a*t^2)/2

Replacing:

d = 18 m/s * 3,58 s + (-5,4 m/s^2 * (3,58 s)^2) /2

Resolving:

d = 64,44 m + (-34,604 m)

d = 29,83 m

Then, the vehicle will CRASH

7 0
3 years ago
Suppose you sound a 1056-hertz tuning fork at the same time you strike a note on the piano and hear 2 beats/second. You tighten
Sindrei [870]

Answer:

The frequency of the piano string is either 1053 HZ or 1059 HZ.

Explanation:

Here we know that frequency of beats is equal to the difference between the frequencies between two waves .

     Given that frequency of tuning fork is 1056 HZ  .

     Let the frequency of the piano be ' f ' .

   Given that number of beats = 3.

    We know that   | 1056 - f | = 3 ;

                      ⇒       1056- f = ±3,

         Upon solving this we get  

                        f = 1056-3 and 1056 + 3

                   ⇒ f = 1053 or 1059 .

                     

6 0
3 years ago
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