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natulia [17]
3 years ago
12

When a jet lands on an aircraft carrier, a hook on the tail of the plane grabs a wire that quickly brings the plane to a halt be

fore it overshoots the deck. in a typical landing, a jet touching down at 240 km/h is stopped in a distance of 95 m .\?
Physics
1 answer:
Veronika [31]3 years ago
8 0
The problem seems to be incomplete because there is no question. However, from the problem description, the logical question is to find he acceleration needed by the jet to land on the airplane carrier. The working equation would be:

2ad = v₂² - v₁²
Since the jet stops, v₂ = 0. Substituting the values:
2(a)(95 m) = 0² - [(240 km/h)(1000 m/1 km)(1h/3600 s)]²
Solving for a,
<em>a = -23.39 m/s² (the negative sign indicates that the jet is decelerating)</em>
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Please can someone solve this physics question with a good explenation.
zimovet [89]

Answer:

The coefficient of dynamic friction is 0.025.

Explanation:

Given:

Initial speed after the push is 'v' as seen in the graph.

Final speed of the stone is 0 m/s as it comes to rest.

Total distance traveled is, D=29.8\ m

Total time taken is, t_{total}=17.5\ s

Time interval for deceleration is 3.5 to 17.5 s which is for 14 s.

Now, average speed of the stone is given as:

v_{avg}=\frac{D}{t_{total}}=\frac{29.8}{17.5}=1.703\ m/s

Now, we know that, average speed can also be expressed as:

v_{avg}=\frac{v_i+v_f}{2}\\1.703=\frac{v+0}{2}\\v=2\times 1.703=3.41\ m/s

Now, from the graph, the vertical height of the triangles is, v=3.41\ m/s

The deceleration is given as the slope of the line from time 3.5 s to 17.5 s.

Therefore, deceleration is:

a=\frac{\textrm{Vertical height}}{\textrm{Time interval}}\\a=\frac{v-0}{17.5-3.5}\\a=\frac{3.41}{14}=0.244\ m/s^2

Frictional force is the net force acting on the stone. Frictional force is given as:

f=\mu_dN\\Where, \mu_d\rightarrow \textrm{coefficient of dynamic friction}\\N\rightarrow \textrm{Normal force}\\N=mg\\\therefore f=\mu_dmg

Now, from Newton's second law, net force is equal to the product of mass and acceleration.

Therefore,

\mu_dmg=ma\\\mu_d=\frac{a}{g}

Plug in 0.244 for 'a' and 9.8 for 'g'. This gives,

\mu_d=\frac{a}{g}=\frac{0.244}{9.8}=0.025

Therefore, the coefficient of dynamic friction is 0.025.

3 0
3 years ago
At what point on the position-time graph shown is the object's instantaneous velocity greatest?
ioda

I'm probably going to have to say C. E as it seems the steepest right around there. If I'm wrong on that, it has to be B. B

7 0
3 years ago
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Can someone help me please
makvit [3.9K]

Answer:

Time, I believe. Pretty sure it's time lol

3 0
2 years ago
What do ocean waves and sound waves have in common?
Anettt [7]
Since they are both examples of moving waves, they both transmit energy.
7 0
3 years ago
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If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o
tangare [24]
The correct answer is letter A. 6 millimeters. <span>If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front of the lens, the height of the image is 6 millimeters.
</span>
Solution:
18 / x = 12 / 4 
12x = 72
x = 6mm
7 0
2 years ago
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