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2 years ago

7

A car enters a 300-m radius horizontal curve on a rainy day when the coefficient of static friction between its tires and the ro

ad is 0.600. What is the maximum speed at which the car can travel around this curve without sliding?

1 answer:

Vlada [557]2 years ago

6 0

To solve this problem it is necessary to take into account the concepts related to Centripetal Force and Friction Force.

In the case of the centripetal force, we know that it is defined as

$F_c = \frac{mv^2}{R}$

Where,

m=mass

v= velocity

r= Radius

In the case of the Force of Friction we have to,

$F_f = \mu m*g$

Where,

$\mu =$ Friction Constant

m= mass

g= gravity

According to the information given, the centripetal force must be less than or equal to the friction force to stay on the road, in this way

$\frac{mv^2}{R} \leq \mu m*g$

Re-arrange to find the velocity,

$v \leq \sqrt{\mu gR}$

$v \leq \sqrt{(0.6)(9.8)(300)}$

$v \leq 42m/s$

Therefore la velocidad del carro debe ser igual o menor a 42m/s para mantenerse en el camino

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What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s? The dia

ycow [4]

The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s

The rotational moment of inertia of the puck is
I = (mr²)/2
= 0.5*(0.15 kg)*(0.038 m)²
= 1.083 x 10⁻⁴ kg-m²

The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
= 0.5*(0.15 kg)*(0.5 m/s)²
= 0.0187 j

The rotational KE is
KE₂ = (1/2)*I*ω²
= 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
= 0.0038 J

The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J

Answer: 0.0226 J

4 0

1 year ago

A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f

sasho [114]

We use the kinematic equations,

$v=u+at$ (A)

$S= ut + \frac{1}{2} at^2$ (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given, $u=0$ , $a=0.21 \ m/s^2$ and $s= 280 m$ .

Substituting these values in equation (B), we get

$280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s$ .

Therefore from equation (A),

$v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s$

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s

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1 year ago

On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are unifo

timofeeve [1]

Answer:

a = 0.5 m/s²

Explanation:

Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:

ωf = ω₀ + α*t

⇒ ωf = α*t ⇒ α = $\frac{wf}{t}$ = $\frac{1.4 rad/s}{21 s} = 0.067 rad/s2$

The angular velocity, and the linear speed, are related by the following expression:

v = ω*r

Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:

a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²

3 0

2 years ago

A company designed and sells an ultrasonic receiver, which detects sounds unable to be heard by the human ear. The receiver can

Ksenya-84 [330]

Answer:

28√3 m

Explanation:

A = vertex where receiver is placed

S = focus

Bp = r = radius of the outside edge

Bc = 2r = diameter

The full explanation is shown in the picture attached herewith. Thank you and i hope it helps.

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1 year ago

Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez

Zanzabum

(a) $3.3\cdot 10^{-6} Pa$

The radiation pressure exerted by an electromagnetic wave on a surface that totally absorbs the radiation is given by

$p=\frac{I}{c}$

where

I is the intensity of the wave

c is the speed of light

In this problem,

$I=1000 W/m^2$

and substituting $c=3\cdot 10^8 m/s$ , we find the radiation pressure

$p=\frac{1000 W/m^2}{3\cdot 10^8 m/s}=3.3\cdot 10^{-6}Pa$

(b) $4.4\cdot 10^{-8} m/s^2$

Since we know the cross-sectional area of the laser beam:

$A=6.65\cdot 10^{-29}m^2$

starting from the radiation pressure found at point (a), we can calculate the force exerted on a tritium atom:

$F=pa=(3.3\cdot 10^{-6}Pa)(6.65\cdot 10^{-29} m^2)=2.2\cdot 10^{-34}N$

And then, since we know the mass of the atom

$m=5.01\cdot 10^{-27}kg$

we can find the acceleration, by using Newton's second law:

$a=\frac{F}{m}=\frac{2.2\cdot 10^{-34} N}{5.01\cdot 10^{-27} kg}=4.4\cdot 10^{-8} m/s^2$

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2 years ago