PLEASE HELP NEED TO KNOW IF CORRECT!! BRAINLIEST ANSWER WILL BE REWARDED.

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

7 0

3 years ago

A body is travelling with a velocity 30 m/s².what will be its velocity after 4s?

Ipatiy [6.2K]

Answer:

70m/s²

Explanation:

we will use the first equation of Dalton to find it

3 0

3 years ago

The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:

Furkat [3]

To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is s=r(B), that has its <span>r=radius,B=angle in rad velocity v=ds/dt= w(r)
Do not forget about </span> w = angular speed in rad/s and $w1 = 1 revolution/sec = 2Pi (rad/s)$
Now we can go to proportion
$v1=v2$
$w1*r1 = w2r2$ $w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)$
$w2 = w3 (which is the angular velocity of the rear wheel) 
$
SOLVING FOR A : $v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec$
$v3 = 14.66 ft/sec(1 mile/5280 ft)( 3600 sec/h)$ $= 9.99$ or something about <span>10 mph --- SOLVING FOR B.
</span>I'm sure it helps!

7 0

3 years ago

A child drops a ball from a window.The ball strikes the ground in 3.0 seconds.What is the velocity of the ball the instant befor

lord [1]

It might be 4.0 or 2.22344 seconds as velocity speed

4 0

3 years ago

Which mineral is commonly pinkish and has light-colored, wavy lines? A) quartz B) potassium feldspar C) pyroxene D) amphibole D)

Nina [5.8K]

Answer:

option B

Explanation:

The correct answer is option B

the mineral which is pinkish and light in color with wavy lines is potassium feldspar .

pyroxene is a mineral is typically dark in color generally it's color is dark green.

amphibole is also dark in color an is in green color.

olivine is a mineral which is in olive green color.

Quartz is a mineral which is light colored and it is also pink in color but it does not have wavy line.

4 0

3 years ago