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3 years ago

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Peregrine falcons are known for their maneuvering ability. In a tight circular turn, a falcon can attain a centripetal accelerat

ion 1.5 times the free-fall acceleration. What is the radius of the turn if the falcon is flying at 17 m/s ?

1 answer:

kati45 [8]3 years ago

8 0

Answer:

19.7 m

Explanation:

The centripetal acceleration of the falcon is given by

$a=\frac{v^2}{r}$

where

a is the acceleration

v is the speed

r is the radius of the circular path

In this problem, we have:

$a=1.5 g = 1.5(9.8)=14.7 m/s^2$ is the acceleration

v = 17 m/s is the speed

Solving for r, we find the radius of the turn:

$r=\frac{v^2}{a}=\frac{17^2}{14.7}=19.7 m$

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Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more m

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Answer:

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

τ = I α

T r = I α

T gR = I α

Alf = T R / I (1)

T = α I / R

Now let's use Newton's second law in the mass that descends

W- T = m a

a = (m g -T) / m

The two accelerations need related

a = R α

α = a / R

a = (m g - α I / R) / m

R α = g - α I /m R

α (R + I / mR) = g

α = g / R (1 + I / mR²)

We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

Let's review the claims

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases

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3 years ago

The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an

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Answer:

$v_1 = 3.5 \ m/s$

Explanation:

Given that :

mass of the SUV is = 2140 kg

moment of inertia about G , i.e $I_G$ = 875 kg.m²

We know from the conservation of angular momentum that:

$H_1= H_2$

$mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2$

$2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2$

$1637.1 v_1$ $= 3841.575 \omega_2$

$\omega_2 = \frac{1637.1 v_1}{3841.575}$

$\omega _2 = 0.4626 \ v_1$

From the conservation of energy as well;we have :

$T_2 +V_{2 \to 3} = T_3 \\ \\ \\ \frac{1}{2} I_A \omega_2^2 - mgh =0$

$[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0$

$706.93 \ v_1^2 - 8657.49 =0$

$706.93 \ v_1^2 = 8657.49$

$v_1^2 = \frac{8657.49}{706.93 }$

$v_1 ^2 = 12.25$

$v_1 = \sqrt{ 12.25$

$v_1 = 3.5 \ m/s$

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3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

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Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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