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Ierofanga [76]
2 years ago
13

A jetliner leaves san francisco for new york, 4600 km away. with a strong tailwind, its speed is 1100 km/hr. at the same time, a

second jet leaves new york for san francisco. flying into the wind, it makes only 700 km/hr. when and where do the two planes pass?

Physics
1 answer:
Troyanec [42]2 years ago
8 0
First, make an illustration of the problem. In reality, the direction traveled by the planes is not linear. Strictly speaking, it should curve out following the shape of the sphere. But I suppose, we could do away with that because we are given with the quantity for distances. I'm just showing a guide to see where the planes would meet.

So, we know that the total distance is 4,600 km. It would consequently mean that x and y would equal to 4,600. The distances x and y could be determined by multiplying their respective speeds to the time. Since they meet at the same time, let's denote it as t.

1,100t + 700t = 4,600
Solving for t,
1,800t = 4600
t = 4600/1800
t = 2.556 hours

So, we know that the two planes would meet after 2.56 hours from the time they leave the airports. Next, we have to know where they meet exactly.

1,100(2.556) = 2,812 km
700(2.556) = 1,788 km

Therefore, they meet at approximately 2,812 km from San Francisco airport, or 1,788 km from New York airport after 2.56 hours since they left their respective ports.

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Lina20 [59]

The amount of heat required is B) 150 J

Explanation:

The amount of heat energy required to increase the temperature of a substance is given by the equation:

Q=mC\Delta T

where:

m is the mass of the substance

C is the specific heat capacity of the substance

\Delta T is the change in temperature of the substance

For the sample of copper in this problem, we have:

m = 25 g (mass)

C = 0.39 J/gºC (specific heat capacity of copper)

\Delta T = 15^{\circ}C (change in temperature)

Substituting, we find:

Q=(25)(0.39)(15)=146 J

So, the closest answer is B) 150 J.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0
3 years ago
Galileo _____.
AnnZ [28]

Answer:

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5 0
3 years ago
Read 2 more answers
If f(x)=4/x+2 and g is the inverse of f,then g'(10)=​
ch4aika [34]

Answer:

g'(10) = \frac{-1}{16}

Explanation:

Since g is the inverse of f ,

We can write

g(f(x)) = x    <em> </em><em>(Identity)</em>

Differentiating both sides of the equation we get,

g'(f(x)).f'(x) = 1

g'(10) = \frac{1}{f'(x)}    --equation[1]    Where f(x) = 10

Now, we have to find x when f(x) = 10

Thus 10 = \frac{4}{x} + 2

\frac{4}{x} = 8

x = \frac{1}{2}

Since f(x) = \frac{4}{x} + 2

f'(x) = -\frac{4}{x^{2} }

f'(\frac{1}{2})  =  -4 × 4 = -16            

Putting it in equation 1, we get:

We get g'(10) = -\frac{1}{16}

5 0
2 years ago
A grindstone increases in angular speed from 6.00 rad/s to 12.20 rad/s in 16.00 s. Through what angle does it turn during that t
Akimi4 [234]

Answer:

Explanation:

Given that the grand stone has initial angular velocity of

w(ini)= 6rad/

And it has a final angular velocity of

w(fin)=12.20rad/sec

Time taken is t=16s

Using equation of angular motion

To get angular acceleration (α)

w(fin)=w(ini)+αt

12.20=6+16α

16α=12.20-6

16α=6.2

α=6.2/16

α=0.3875rad/sec²

The angular acceleration is 0.39rad/s²

Angle that he turn using

w(fin)²=w(ini)²+2αθ

12.2²=6²+2×0.3875θ

12.2²-6²=0.775θ

0.775θ=112.84

Then, θ=112.84/0.775

θ=145.6radian

The angular displacement is 145.6rad

6 0
2 years ago
The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the magnitude of the
PIT_PIT [208]

The drag force is directly proportional to the square of the velocity of motion of the object. So as the speed is doubled, the magnitude of drag force will get quadrupled.

<u>Explanation: </u>

Drag force is the opposing or resisting force acting on any object by the medium in which it is moving. So in this case, you are riding a bicycle, thus the medium can be considered as air.

The formula for calculating drag force is as below:

               \text {Drag force }=\frac{1}{2} \rho v^{2} C_{D} A

Here, ρ is the density of the air molecules, v is the velocity or speed of the bicycle, CD is the drag coefficient and A is the area of the bicycle.

In the above equation, only the term velocity will be a varying quantity with respect to time and other quantities will remain constant throughout the single situation of riding of bicycle.

So, the equation can be,

             \text { Drag force }=k v^{2}

Where ,  

     k=\frac{1}{2} \rho C_{D} A (constant for this whole condition)

Now given the speed of bicycle increased from v to 2v, so the initial drag force will be

                   N_{i}=k v^{2}

After increase in speed, the final drag force will be  

                   N_{f}=k(2 v)^{2}

                   N_{f}=4 k v^{2}=4 N_{i}

Thus, if the speed of the bicycle is doubled, the drag force will get increased by four times.

4 0
3 years ago
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