Complete Question
the maximum force a pilot can stand is about seven times his weight. what is the minimum radius of curvature that a jet plane's pilot, pulling out of a vertical dive, can tolerate at a speed of 250m/s?
Answer:
The value is 
Explanation:
From the question we are told that
The weight of the pilot is 
The maximum force a pilot can withstand is 
The speed is 
Generally the centripetal force acting on the pilot is equal to the net force acting on the pilot i.e
Here N is the normal force acting on the pilot
Now
So
=> 
=>
=> 
Answer:
Explanation:
Given,
Mass of ball 
kg
Speed of ball 
m/s
Mass of persona 
kg
(a) The person is at rest initially. So, by the conservation of momentum
(b) Final speed of ball 
Answer:
41.576 N
Explanation:
Resolving force in the inclined plane in the x and y directions gives us;
F_x = mg sin θ
F_y = mg cos θ
Thus;
F_x = 4.8 × 9.81 × sin 28
F_x = 22.106 N
F_y = 4.8 × 9.81 × cos 28
F_y = 41.576 N
We can use Newton's second law to find the normal force.. Thus;
N - F_y = 0.
Where N is normal force;
N = 41.576 N
Answer:
a)11.6m
b)45.55s
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
a)
for this problem
Vo=0
Vf=319m/min=5.3m/s
a=1.2m/s^2
we can use the ecuation number 1 to calculate the time
t=(Vf-Vo)/a
t=(5.3-0)/1.2=4.4s
then we use the ecuation number 3 to calculate the distance
X=0.5at^2
X=0.5x1.2x4.4^2=11.6m
b)second part
We know that when the elevator starts to accelerate and decelerate, it takes a distance of 11.6m and a time of 4.4s, which means that if the distance is subtracted 2 times this distance (once for acceleration and once for deceleration)
we will have the distance traveled in with constant speed.
With this information we will find the time, and then we will add it with the time it takes for the elevator to accelerate and decelerate
X=218-11.6x2=194.8m
X=VT
T=X/v
t=194.8/5.3=36.75s
Total time=36.75+2x4.4=45.55s
Answer:
-4500 N
Source: Brainly
The police officer must be angry 0_0
