The amount of light that enters the pupil is controlled by the: retina. lens. inis.

An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop-the-loop maneuver. The acceleration

Debora [2.8K]

Answer:

The radius of the loop is 20.66 km

Explanation:

let the radius of the loop be r

mass of airplane is m

At the top, the pilot experiences two radial forces, which are

1) Gravitational force is mg

2) Centrifugal forces mv²/r out of the center

When the pilot experiences no weight,

then, mg = mv²/r

r = v² / g

= 450² / 9.8

= 20.66 x 10³3

= 20.66 km

3 0

3 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

3 years ago

Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?

lesantik [10]

The equilibrium conditions allow to find the results for the balance forces are:

F₁ = 225.4 N

F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

∑ F = 0

∑ τ = 0

Where F are the forces and τ the torques.

The torque is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

F₂ 2 - W₁ 1 - W₂ 1.5 = 0 $\frac{W_1 \ 1 + W_2 \ 1.5}{2}$