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storchak [24]
3 years ago
7

A car goes round a curve of radius 48m, the road is banked at an angle of 15 with the horizontal,at what maximum speed may the c

ar travel if there is to be low tendency to skid even on very slippery pavement​
Physics
1 answer:
Marizza181 [45]3 years ago
3 0

Answer:

11 m/s

Explanation:

Draw a free body diagram.  There are two forces acting on the car:

Weigh force mg pulling down

Normal force N pushing perpendicular to the incline

Sum the forces in the +y direction:

∑F = ma

N cos θ − mg = 0

N = mg / cos θ

Sum the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Substitute and solve for v:

(mg / cos θ) sin θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Plug in values:

v = √(9.8 m/s² × 48 m × tan 15°)

v = 11.2 m/s

Rounded to 2 significant figures, the maximum speed is 11 m/s.

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1.33 m^3 of fluid flows out of a pipe in 24.5 s. the fluid leaves the pipe at 3.55 m/s. what is the area of the pipe. (unit=m^2)
Margaret [11]

Answer:

0.015meter^{2}

Explanation:

Total volume of water coming out = 1.33meter^{3}

Also volume = Cross sectional area*Length covered

Length covered = Velocity *time

                           =24.5*3.55

                           =86.97 meter

Let the cross sectional area be A.

1.33 = 86.97*A

A =0.015meter^{2}

3 0
4 years ago
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Valentin [98]
Light energy is not kinetic energy
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3 years ago
Process of science identifying stellar corpses mastering astronomy
Roman55 [17]
One of the procedures in identifying the stellar corpses would involve "measuring <span>Doppler shifts in the spectrum of the main-sequence star so that you can determine the mass of the compact object." In addition, the use of advanced measuring devices like a space telescope is most likely used.</span>
4 0
3 years ago
In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 5.00�1015kg and a rad
irakobra [83]

Answer:

a).v_{1}=13.49 \frac{m}{s}

b).v_{2}=17.54\frac{m}{s}

Explanation:

Using the conservation of energy and the kinetic energy of the satellite around the asteroid can model the motion in

\frac{1}{2}*m*v^2=\frac{G*M*m}{a_{o}}

v^2=\frac{2*G*M*m}{a_{o}}

G=6.673x10^{-11}\frac{m^3}{kg*s^2}

M=1.20x10^{16}kg

a).

a_{o}=8.8km*\frac{1000m}{1km}=8800m

v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{8800m}}

v_{1}=13.49 \frac{m}{s}

b).

a_{f}=5.2km*\frac{1000m}{1km}=5200m

v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{5200m}}

v_{2}=17.54\frac{m}{s}

3 0
3 years ago
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A guitar string produces 3 beats/s when sounded with a 352-hz tuning fork and 8 beats/s when sounded with a 357-hz tuning fork.
MA_775_DIABLO [31]

Explanation:

The fluctuating sound heard when two objects vibrate with different frequencies is called beats. It is given that guitar string produces 3 beats/s when sounded with a 352 Hz tuning fork and 8 beats/s when sounded with a 357 Hz tuning fork.

It is assumed to find the vibrational frequency of the string.

For 3 beats/s, beat frequency can be :

352 - 3 or 352 + 3 = 349 Hz or 355 Hz

For 8 beats/s, beat frequency can be :

357 - 8 or 357 + 8 = 349 Hz or 365 Hz

It means that the vibrational frequency is 349 Hz.

5 0
3 years ago
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