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salantis [7]
3 years ago
14

A jet with mass m = 5 × 104 kg jet accelerates down the runway for takeoff at 1.9 m/s2. What is the net vertical force on the ai

rplane as it levels off?
Physics
1 answer:
OLEGan [10]3 years ago
3 0
The first sentence got me all psyched up to answer the question "What
horizontal force do the engines generate in order to accelerate it ?". 

But the actual question, in the second sentence, turned out to be
a completely different one.

When the plane levels off and continues on at a constant altitude, it's
not accelerating up or down, so the net vertical force on it is zero. 
The lift generated by the wings is exactly balancing the downward
force of gravity on the airplane.
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Zinaida [17]

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7 0
3 years ago
You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f
Alborosie

Answer:

3.25 m

Explanation:

t = Time taken = 0.166 seconds

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s²

s = 1 because meter stick is 1 meter in length

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1-\frac{1}{2}\times 9.81\times 0.166^2}{0.166}\\\Rightarrow u=5.21\ m/s

Here, the initial velocity of point B is calculated from the time which is given. This velocity will be the final velocity of the acorn which falling from point A.

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.21^2-0^2}{2\times 9.81}\\\Rightarrow s=1.38\ m

The distance of the acorn from the ground is 1.87+1.38 = 3.25 m

4 0
3 years ago
A man exerts a constant force to pull a 51-kg box across a floor at constant speed. He exerts this force by attaching a rope to
Vitek1552 [10]

Answer:

W=561.41 J

Explanation:

Given that

m = 51 kg

μk = 0.12

θ = 36.9∘

Lets F is the force applied by man

Given that block is moving at constant speed it mans that acceleration is zero.

Horizontal force = F cos θ

Vertical force = F sinθ

Friction force Fr=  μk N

N +  F sinθ = m g

N = m g -  F sinθ

Fr =  μk (m g -  F sinθ)

For equilibrium

F cos θ =  μk (m g -  F sinθ)

F ( cos θ +μk  sinθ)  = μk (m g

Now by putting the values

F ( cos 36.9∘ + 0.12 x sin36.9∘)=0.12 x 51 x 10

F= 70.2 N

We know that Work

W= F cos θ .d

W= 70.2 x cos 36.9∘ x 10

W=561.41 J

3 0
3 years ago
The graph below represents the motion of four cars, labeled W, X, Y, and Z.
ycow [4]

Answer:

Answer is A

Explanation:

As we know , for constant velocity we get a straight line.

The formula for this problem is s = vt which is similar to a straight line formula like y = mx + c.

If we put here c = 0 we get the formula for distance and velocity.

So the answer is A.

6 0
3 years ago
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