Explain why a magnet from your refrigerator could not be used to life something as heavy as a car ?

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

2 years ago

You are driving your motorcycle in a circle of radius 75 m on wet pavement. what is the fastest you can go before you lose tract

stira [4]

On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s

Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.

Maximum velocity of a road with friction is given by the formula,

v = μRg

where, v is the maximum velocity

μ is the coefficient of static friction

R is the radius of the circle road

g is the acceleration due to gravity

Given,

μ = 0.20

R = 75m

g = 9.8m/s²

On substituting the given values in the above formula,

v = 0.20× 75 ×9.8

v = 147m/s

So, the Maximum velocity of the wet road is 147m/s.

Learn more about Velocity here, brainly.com/question/18084516

#SPJ4

3 0

1 year ago

What causes an object to rotate

frez [133]

A spinning force acting upon it

3 0

3 years ago

What is the frequency of blue light that has a wavelength of 446 nm?

Maslowich

When I went through with the math, the answer I came upon was:
6.67 X 10^14 

Here is how I did it: First of all we need to know the equation. 

c=nu X lamda 
(speed of light) = (frequency)(wavelength) 
(3.0 X 10^8 m/s) = (frequency)(450nm) 

We want the answer in meters so we need to convert 450nm to meters. 
450nm= 4.5 X 10^ -7 m 
(3.0 X 10^8 m/s) = (frequency)(4.5 X 10^ -7 m) 

Divide the speed of light by the wavelength. 
(3.0 X 10^8m/s) / (4.5 X 10^ -7m) =6.67 X 10^ 14 per second or s- 

Answer: 6.67 X 10^14 s- hope this helps

7 0

3 years ago

Read 2 more answers

¿Qué significa que la investigación debe ser replicable? Analiza.

Inga [223]

Significa que en su investigación, debe proporcionar suficiente información para que otras personas que lean su investigación puedan hacer la investigación nuevamente.

4 0

3 years ago