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Ierofanga [76]
3 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]3 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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A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then mo
makkiz [27]

Answer:

(a) The speed of the target proton after the collision is:V_{2f} =433(m/s), and (b) the speed of the projectile proton after the collision is: v_{1f}=250(m/s).

Explanation:

We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}, and y axle:0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}. Now replacing the value given as: v_{1i}=500(m/s), \beta_{1}=+60^{o} for the projectile proton and according to the problem \beta_{1}and\beta_{2} are perpendicular so \beta_{2}=-30^{o}, and assuming that m_{1}=m_{2}, we get for x axle:500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2} and y axle: 0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}, then solving for v_{2f}, we get:v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f} and replacing at the first equation we get:500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}, now solving for v_{1f}, we can find the speed of the projectile proton after the collision as:v_{1f}=250(m/s) and v_{2f}=\sqrt{3}*v_{1f}=433(m/s), that is the speed of the target proton after the collision.

5 0
2 years ago
A 0.75 kilogram apple is thrown upward from the ground. The apple reaches a height of 5.5m, what is the beginning gravitational
MaRussiya [10]

Answer:

U(0.0m) = 0J

U(5.5m) = 40.42 J

Explanation:

The gravitational potential energy is given by the following formula:

U=mgh

h: height

m: mass of the apple = 0.75kg

g: gravitational acceleration = 9.8m/s^2

When the apple is at 5.5m from the ground the gravitational potential energy is:

U=(0.75kg)(9.8m/s^2)(5.5m)=40.42\ J

when the apple is on the ground you have:

U=mg(0m)=0\ J

6 0
3 years ago
The average 8- to 18-year-old spends __________ per day on average in front of a screen doing very little to no physical activit
natta225 [31]

Answer:

D. 3 hours or more

Explanation:

The average 8- to 18-year-old spends at least D. 3 hours every day in front of a screen, performing little to no physical activity. This is because, instead of exercising and socializing with their peers, children and teenagers frequently talk, watch a lot of movies/shows, or play video games on their computers. Unfortunately, this is typically considerably more than three hours every day. Although some children still prefer physical activities over this, the bulk of the population does not.

5 0
2 years ago
A capacitor is to be constructed to have a capacitance of 100uF.The area of the plates is 6.om by 0.030m and the relative permit
lara [203]

Answer:

The answer is below

Explanation:

Given that:

The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m =  0.18 m²

the relative permittivity of dielectric (εr) is 7.0

Permittivity of free space (εo) = 8.854 × 10^(-12)

capacitance of 100uF

potential difference (V) of 12V

d = separation between plate

The capacitance (C) of a capacitor is given by:

C=\frac{\epsilon_o \epsilon_r A }{d}\\ 100*10^{-6}=\frac{8.854*10^{-12}*7*0.18}{d}\\ d=\frac{8.854*10^{-12}*7*0.18}{100*10^{-6}}=1.11*10^{-7}\ m

The electric field between plates is given as:

E = V /d

E = 12 / 1.11*10^{-7}=10.75*10^7\ V/m

3 0
3 years ago
A rock dropped on the moon will increase it's speed from 0 m/s to 8.15 m/s in about 5 seconds what is the acceleration of the ro
Lunna [17]

Using the formula:


a = (Vf - Vi) / t


Our initial velocity is 0 m/s, and our final velocity is 8.15 m/s, with a time period of 5 seconds:


a = (8.15 - 0.0) / 5

a = 1.63 m/s^2


If you know the acceleration due to gravity on the Moon, you can confirm this answer. The recorded gravitational acceleration on the Moon is 1.62 m/s^2.

5 0
3 years ago
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