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Lena [83]
2 years ago
5

An airplane is flying from Dallas, Texas to Pensacola, Florida. Flying at maximum velocity, it encounters strong winds moving at

half the speed of the plane in the opposite direction. How long will it take the plane to arrive in Pensacola, relative to the original arrival time?
F. It will take half the time.
G. It will take twice as long.
H. It will take 1.5 times as long.
I. It will take three times as long.
Physics
1 answer:
jenyasd209 [6]2 years ago
4 0

Answer:

G. It will take twice as long.

Explanation:

Let's call v the original speed of the plane and d the distance between Dallas and Pensacola. The time the plane originally takes to complete the flight is

t=\frac{d}{v}

In this problem, we are told that the plane encounters wind moving at half of its speed: \frac{v}{2}, in the opposite direction. This means that the new speed of the plane is

v'=v-\frac{v}{2}=\frac{v}{2}

And so, the time the plane takes now to complete the flight is

t'=\frac{d}{v/2}=2\frac{d}{v}=2t

So, the plane takes twice the time as before.

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A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a
Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

Fc = f

m×(v^2)/(R) = μ×m×g

    (v^2)/(R) = g×μ

               μ = (v^2)/(R×g)

                  =  ((25)^2)/((100)×(9.8))

                  = 0.64

Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.

4 0
2 years ago
How much force is needed to accelerate a 1,500 kg car at a rate of 8 m/s2?
lesya [120]
Force = mass * acceleration = 1500kg * 8m/s²
3 0
3 years ago
Given the function f(x) = 8x3 − 3x2 − 5x 8, what part of the function indicates that the left and right ends point in opposite d
ale4655 [162]

Answer: The degree of the first term.

Explanation:

The function:

f(x) = 8x^3-3x^2-5x^8

The left and right ends would be indicated when x is changed to -x. When this is substituted, the change is indicated by the first term because only the degree of first term is odd.

Let the left hand side be donated by -x.

Then,

f(-x) =8(-x)^3-3(-x)^2-5(-x)^8\\ \Rightarrow f(-x)=-8x^3-3x^2-5x^8

Hence, the correct option is the degree of the first term indicates the left and right end points of the function.

8 0
3 years ago
Read 2 more answers
Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim
blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

x=x_{0}+V t \\ x=\frac{1}{2}+V t

The distance between the police car and the intersection is,

y=y_{0}+V t

y=\frac{1}{2}-40 t

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })

z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)

The rate of change is,

2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)

2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots

Now finding z when t=0, from (1) we have

z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}

z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071

The officer's radar gun indicates 25 mph pointed at the other car then, \frac{d z}{d t}=25 when t=0, from

From (2) we get

2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)

2(0.7071)(25)=V+2 V^{2}(0)-40

35.36=V-40

V=35.36+40=75.36

Hence the speed of the car is 75.36 mph

7 0
3 years ago
State Pascal's principle of transmission of pressure​
bulgar [2K]

Answer:

Pascal's law (also Pascal's principle or the principle of transmission of fluid-pressure) is a principle in fluid mechanics given by Blaise Pascal that states that a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere.

4 0
3 years ago
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