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2 years ago

Do banked curves help car make turns more softly

Physics

2 answers:

solong [7]2 years ago

4 0

Bank turned help drivers maintain speed they keep the car from skidding

KonstantinChe [14]2 years ago

3 0

Answer:

yes

Explanation:

banked curve :

is a road that looks like the top part of a race track

its looks elevated

often seen in bicycle race track (velodrome)

Banking the curve can help keep cars from skidding. When the curve is banked, the centripetal force can be supplied by the horizontal component of the normal force.

sfuca

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What is the thinnest film of a coating with n = 1.39 on glass (n=1.52) for which destructive interference of the red component (

Illusion [34]

In optics, you have to create a layer of coating that is approximately 1/4 of the light's wavelength. The working equation for this problem is:

d = λ₀/4n,
where
λ₀ is the wavelength of the incident light
n is the refractive index of the coating

Substituting the values,

d = (650 nm)/4(1.39)
<em>d = 116.9 nm</em>

8 0

4 years ago

You are traveling in a car that is moving at a velocity of 30 m/s. Suddenly, a car 15 meters in front of you slams on its brakes

Artemon [7]

Answer:

The acceleration of car is 6.67 m/s².

Explanation:

Given that,

Initial velocity = 30 m/s

Distance = 15 m

Final velocity = 10 m/s

Time = 3 sec

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v_{f}-v_{i}}{t}$

$a=\dfrac{10-30}{3}$

$a=\dfrac{-20}{3}$

$a=-6.67\ m/s^2$

Negative sign shows the car is slowing down.

Hence, The acceleration of car is 6.67 m/s².

7 0

4 years ago

At one instant a bicyclist is 21.0 m due east of a park's flagpole, going due south with a speed of 13.0 m/s. Then 21.0 s later,

andre [41]

Answer:

Part a)

$d = 21\sqrt2 = 29.7 m$

Part b)

Direction is 45 degree North of West

Part c)

$v_{avg} = 1.41 m/s$

Part d)

direction of velocity will be 45 Degree North of West

Part e)

$a = 0.875 m/s^2$

Part f)

$\theta = 45 degree$ North of East

Explanation:

Initial position of the cyclist is given as

$r_1 = 21.0 m$ due East

final position of the cyclist after t = 21.0 s

$r_2 = 21.0 m$ due North

Part a)

for displacement we can find the change in the position of the cyclist

so we have

$d = r_2 - r_1$

$d = 21\hat j - 21\hat i$

so magnitude of the displacement is given as

$d = 21\sqrt2 = 29.7 m$

Part b)

direction of the displacement is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{21}{-21}$

so it is 45 degree North of West

Part c)

For average velocity we know that it is defined as the ratio of displacement and time

so here the magnitude of average velocity is defined as

$v_{avg} = \frac{\Delta x}{t}$

$v_{avg} = \frac{29.7}{21}$

$v_{avg} = 1.41 m/s$

Part d)

As we know that average velocity direction is always same as that of average displacement direction

so here direction of displacement will be 45 Degree North of West

Part e)

Here we also know that initial velocity of the cyclist is 13 m/s due South while after t = 21 s its velocity is 13 m/s due East

So we have

change in velocity of the cyclist is given as

$\Delta v = v_f - v_i$

$\Delta v = 13\hat i - (-13\hat j)$

now average acceleration is given as

$a = \frac{\Delta v}{\Delta t}$

$a = \frac{13\hat i + 13\hat j}{21}$

so the magnitude of acceleration is given as

$a = \frac{13\sqrt2}{21} = 0.875 m/s^2$

Part f)

direction of acceleration is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{13}{13}$

$\theta = 45 degree$ North of East

4 0

3 years ago

Find time when boy catches the girl or when they are at their closest separation..

soldier1979 [14.2K]

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\implies{Time=10\:sec\:and\:30\:sec}}} \\\\$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}} \\ \\$

$\green{\underline{\bold{Given :}}} \\ \\ \:\:\:\: \bullet\:\:\tt\red{ Velocity \: of \: boy = 50 \: m/s} \\ \\ \:\:\:\: \bullet\:\: \tt\orange{Velocity \: of \: girl = 30 \: m/s }\\ \\ \:\:\:\: \bullet\:\:\tt\green{ Acceleration \: of \: boy = {1 \: m/s}^{2}} \\ \\ \:\:\:\:\bullet\:\: \tt\blue{Acceleration \: of \:girl= {2\: m/s}^{2} }\\ \\ \:\:\:\: \bullet \:\:\tt\purple{Sepration \: between \: them = 150 \: m }\\ \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \\ \:\:\:\: \bullet\:\: \tt\blue{Time \: taken \: to \: catch \: the \: girl }\\$

<u>According to given question</u> :

$\\ \green{\star} \: \text{Using \: relative \: motion \: method} \\ \\ \green{ \circ} \: \tt Net \: velocity = 50 - 30 = 20 \: m/s \\ \\ \green{ \circ} \: \tt Net \: acceleration = 1 - 2 = - 1\: m /{s}^{2} \\\\ \\ \star\:\bold\red{\underline{\:As \: we \: know \: that\:}} \\\\ \tt\purple{: \implies s = ut + \frac{1}{2} {at}^{2}} \\ \\ \tt\green{: \implies 150 = 20 \times t + \frac{1}{2} \times -1 \times {t}^{2}} \\ \\ \tt\purple{: \implies 300 = 40t - {t}^{2}} \\ \\ \tt\green{: \implies {t}^{2} - 40t + 300 = 0} \\ \\ \tt\purple{: \implies t = \frac{ - ( - 40) \pm\sqrt{ { (- 40)}^{2} - 4 \times 1 \times 300} }{2 \times 1} }\\ \\ \tt\green{: \implies t = \frac{40 \pm \sqrt{1600 - 1200} }{2} }\\ \\ \tt\purple{: \implies t = \frac{40 \pm 20}{2} }\\ \\ \green{\tt: \implies t = 10 \: sec \: and \: 30 \: sec}$

7 0

3 years ago

The displacement ‘x’of a particle is dependent on time ‘t’according to the relation X= 3 -5t +<img src="https://tex.z-dn.net/?f=

rusak2 [61]

Answer:

Explanation:

8 0

3 years ago

Do banked curves help car make turns more softly​

Do banked curves help car make turns more softly