Ar what minimum speed must a roller coaster be traveling when upside down at the top of a circle?
at the top of the circle, net force acted on passenger is just the force due to gravity Fg (there is normal force). So apply Newton's 2nd law we have: Fg = m*a(centripetal) m*g = m*a(centripetal) ; m will be canceled out. g = a(centripetal) = v^2/ r v^2 = g*r v = sqrt (g*r) = sqrt (9*9.8) = 9.39 m/s
A train 200 meters long and travelling at a speed of 60 km/h in how many seconds did train cross a pole?
Its a simple math. We know the Universal formula ofDistance = Speed * Time (Formula 1)Here, the distance is 200 metres and speed is 60 km/h, which we require in metre per second. So we need to convert our speed into m/s. Just multiply our speed with 5/18 (1000 (metre) / 60 (minute) * 60 (second)).Speed = (60 * 5)/18 = 16.66 metres per second (round off to two decimal places)Distance = 200 metresBack to Formula 1Distance = Speed * Time => Time = Distance / SpeedTime = 200 / 16.66 = 12 seconds approx.
What does it feel like to be in a high-speed car crash? Did you hit a car, or were you hit by another car? Did you drive into a stationary object?
I was driving back from college for Thanksgiving break when I was involved in my serious wreck. It had rained the night before but the roads were not really wet anymore. I was on a highway and as I started to switch lanes I hydroplaned due to a small space between the concrete of the lanes (idk how it was big enough to even make me hydroplane). So one second I was going 75 MPH and the next thing I know I am sideways on the road. I look out my driver-side window and see that I am headed straight for a telephone pole. I quickly thought “Okay, this can only end two ways: (1) I will die from this impact or (2) I will be paralyzed from it.” I took a deep breath and said something that my friends and I would say in high school when trouble was a major possible outcome. So I outloud and calmly said “well I hope this works out.” Luckily my tires were on the shoulder, which was grass, and it was wet enough to allow my car to slide down enough to where I missed the pole by inches. Right after watching my car barely clear the pole I let out some air as relaxation then all of the sudden…my car flipped 4 times and slammed into a concrete drainage ditch. I remember seeing everything too. From the windshield braking, dirt flying around, to all my stuff getting tossed about. Upon impact I got out of the car feeling just fine and with only a surface scrape smaller than the size of a dime on a knuckle. The police and emergency crews were telling me I needed to sit down since I had to have broken something and was just in shock so I could not feel it. However, I didn’t get hurt at all (outside of the surface scrap). They said that the fact that I was so calm and chill during it saved my life as they had never seen anyone walk away much less uninjured from a wreck like that in the 25+ years they had been doing their jobs.
What can cause a steering wheel to shake and vibrate?
As others have said, it’s usually down to wheel alignment or balance and is usually noticeable after a tyre change if the change wasn’t properly handled. Occasionally this can also be caused by one of the weights placed on the wheel by the garage to balance it suddenly detaching (though this is rare).However, if you notice any sudden change, I’d certainly advise you to stop IMMEDIATELY and check what’s wrong.One day, a few days after a tyre change, my steering started to noticeably start wobbling badly as I was overtaking someone at 70Mph on the motorway. Due to the suddenness and violence of the wobble, I assumed one of the lead balances had suddenly come off and I decided to pull in immediately to the hard shoulder but as I was pulling in below 40 Mph, there was an almighty bang and shudder in the car and I found myself watching the left wheel of the car running away into the distance - luckily off the motorway, into a field.I managed to steer onto the hard shoulder with the car running on 3 wheels and the brake disk of the missing wheel - it turns out that the garage hadn’t tightened the nuts well enough and they came loose and out mid drive.Motorway recovery eventually retrieved the wheel for me and as the tyre garage was within a mile of the motorway (and I was quite close to that junction) they drove me there with my car on their flatbed truck. The manager of the garage went ashen when I explained what had happened but as the only damage was an almost unnoticeable large but gentle dent in the bodywork where the wheel had hit as it went off on its bid for freedom, they were able to fix it with new nuts and get me back on the road quickly.Extremely scary experience though. Modern tyre garages tend to do a final tighten with a wrench for safety if they’ve used the compressed air spanner - something that hadn’t happened in my case. I now always ensure this is done before I leave the premises.
What is the maximum speed, at which a car turns around a curve of radius 30m on a level road, if the coefficient of friction between tyres and road is 0.4?
To go around a curve, the car needs to have a centripetal force applied, and friction with the road supplies that. We will then use the centripetal acceleration below.Constants:[math]g = 10\ m/s^2[/math]Givens:[math]r = 30\ m[/math][math]\mu = 0.4[/math]Unknowns:[math]v = ?[/math]Other values to be used:[math]m[/math]It is useful to draw a force diagram on the car, which will have three forces:[math]F_g[/math] down[math]F_{normal}[/math] up[math]F_{friction}[/math] to the left or right.The relevant relationships are[1] [math]m\ \vec{a} = \sum \vec{F}[/math][2] [math]F_{friction} = \mu \ F_{normal}[/math][3] [math]F_{gravity} = m\ g[/math][4] [math]a_{centripetal} = \frac{v^2}{r}[/math]Eq [1] in the vertical direction gives[math]m (0) = F_{normal} - F_{gravity} \rightarrow F_{normal} = F_{gravity} = mg[/math]Eq [1] in the horizontal direction gives[math]m a_{centripetal} = F_{friction} \rightarrow m \frac{v^2}{r} = \mu F_{normal} = \mu m g[/math]So working with[math]m \frac{v^2}{r} = \mu m g[/math][math]\frac{v^2}{r} = \mu g[/math][math]v^2 = \mu g r[/math][math]v = \sqrt{\mu g r}[/math]It is worth explicitly noting that, while we added [math]m[/math] to solve the problem, it dropped out by the end.
A person in a hurry averages 65 MPH on a trip covering a distance off 400 miles. What time was required to travel that distance?
DIstance (length) = Rate (Speed is Distance/time) * Time On the right side, you'll notice time cancels and we're left with Distance. You can multiply or dived both sides of equation to get R=, T= or D= For instance D=RT then divide both sides by R You get D/R = T T= D/R Just remember the one you can remember, I like D=Rate*Time, and you can divide or multiply to get the other versions. In your problem: T= (400Miles)/65mph) = 6.2 hr I rounded off to 2 digits since 65mph has 2 digits. 400 accuracy is not known. You can add subtract multiply divide square or root one side of an equation, as long as you do the same thing to the other side of equal sign. Keep them in balance!
Is driving easy in Germany?
I'm thinking about renting a car there so I can drive to all the tourist attractions. I speak German, so reading the road signs will not be a problem. I have never driven anywhere other than Canada, so I am a little concerned that I will panic.