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Leni [432]
3 years ago
12

A curve of radius 80 m is banked at 45 deg.

Physics
1 answer:
goldfiish [28.3K]3 years ago
4 0

Answer:

v=28m/s

Explanation:

The vertical component of the normal force must cancel out with the weight of the car taking the curve:

N_y=W

Ncos\theta=mg

(Notice it has to be <em>cos </em>and not <em>sin</em>, because the angle \theta is the slope, for null slope N_y=Ncos(0)=N, as it should be).

The horizontal component of the normal force must be the centripetal force, that is:

N_x=F_{cp}

Nsen\theta=ma_{cp}

(\frac{mg}{cos\theta})sin\theta=m\frac{v^2}{r}

gtan\theta=\frac{v^2}{r}

v=\sqrt{grtan\theta}=\sqrt{(9.8m/s^2)(80m)tan(45^{\circ})}=28m/s

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Why does diamond sparkles stars twinkles?​
Helga [31]

Answer:

Because they want attention

6 0
3 years ago
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
MatroZZZ [7]

Answer:

The wavelength is \lambda_2 =  534 *10^{-9} \ m

Explanation:

From the question we are told that

   The wavelength of the first light is  \lambda _ 1 =  587 \ nm

    The order of the first light that is being considered is  m_1  =  10

     The order of the second light that is being considered is  m_2  =  11

Generally the distance between the fringes for the first light is mathematically represented as

      y_1 =  \frac{ m_1  * \lambda_1 *  D}{d}

 Here  D is the distance from the screen

 and    d  is the distance of separation of the slit.

      For the second light the distance between the fringes is  mathematically represented as

         y_2 =  \frac{ m_2  * \lambda_2 *  D}{d}

Now given that both of the light are passed through the same double slit

       \frac{y_1}{y_2}  =  \frac{\frac{m_1 *  \lambda_1 * D}{d}  }{\frac{m_2 *  \lambda_2 * D}{d}  } = 1

=>    \frac{ m_1 *  \lambda _1  }{ m_2  *  \lambda_2} =  1

=>     \lambda_2 =  \frac{m_1 *  \lambda_1}{m_2}

=>    \lambda_2 =  \frac{10  *   587 *10^{-9}}{11}

=>   \lambda_2 =  534 *10^{-9} \ m

4 0
2 years ago
A 60.00-cm guitar string under a tension of 50.000 N has a mass per unit length of 0.100 00 g/cm chegg
Evgen [1.6K]

Fundamental frequency,

f=v2l=T/μ−−−−√2l

=(50)/0.1×10−3/10−22×0.6−−−−−−−−−−−−−−−−−−−√

=58.96Hz

Let, n th harmonic is the hightest frequency, then

(58.93)n = 20000

∴N=339.38

Hence, 339 is the highest frequency.

∴fmax=(339)(58.93)Hz=19977Hz.

<h3>What is frequency?</h3>

In physics, frequency is the number of waves that pass a given point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration. See also simple harmonic motion and angular velocity.

learn more about frequency refer:

brainly.com/question/254161

#SPJ4

7 0
1 year ago
A 5cm tall object is placed 4cm in front of a converging lens that has a focal length of 8cm. Where is the image located in ____
OverLord2011 [107]

Answer:

a. -8 cm

Explanation:

d_{o} = distance of the object = 4 cm

d_{i} = distance of the image = ?

f = focal length of the converging lens  = 8 cm

using the lens equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}

\frac{1}{4} + \frac{1}{d_{i}} = \frac{1}{8}

d_{i} = - 8 cm

4 0
2 years ago
A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh
Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 rad/sec^2

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius r=\frac{d}{2}=\frac{1.63}{2}=0.815m

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec

(b) We know that linear speed is given by

v=r\omega =0.815\times 8.362=6.815m/sec

(c) We have given final angular velocity \omega _f=675rev/min

And \omega _i=79.9rev/min

Time t = 63 sec

Angular acceleration is given by \alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2

(d) Change in angle is given by

\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev

7 0
2 years ago
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