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lana [24]
2 years ago
14

Newton laws escape room

Physics
1 answer:
ivann1987 [24]2 years ago
7 0
I can barely see the image you put
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Nasa’s goal is to help industry reduce emissions from aircraft by how much by 2050 compared to 2005?
lys-0071 [83]

Answer:

50%

Explanation:

Commercial aviation is responsible for 2% of global carbon emission. In the year 2009, the members of IATA (International Air Transport Association) had drawn a pledge to:

1. Halve the carbon emission due to aircraft by 2050, relative to emission of 2005

2. To make growth of industry carbon neutral by 2020

3. To cut CO₂ emission by 1.5% per year till 2020

To achieve these targets, a four pillar action plan was created.

8 0
2 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s

v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s

<u>Time taken to reach 1180 m is 11.29 seconds</u>

v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0
3 years ago
Problem 4: a long wire carries current towards east. a positive charge moves westward and just north from the wire. what is the
Alex787 [66]
The direction of the force experienced by the positive charge is upward.

We can use the right-hand rule to understand the direction of the Lorentz force acting on the charge: let's put the thumb in the same direction of the current in the wire (eastward), while the other fingers "wrap themselves" around the wire. These other fingers give the direction of the Lorentz force in every point of the space around the wire. Since the charge is located north of the wire, in that point the fingers are directed upward, so the positive charge experiences a force directed upward.
(if it was a negative charge, we should have taken the opposite direction)
4 0
2 years ago
The summer camps had a field trip from the campus to Fragrance Hill. They traveled at an average speed of 65 km/h in the first 2
ludmilkaskok [199]

Answer:

Explanation:

They traveled this distance in 2 parts, essentially. Part 1 had an average speed for a certain number of hours, part 2 had an average speed for a certain number of hours, and those 2 parts taken together took them a distance of 364 km. In equation form, that looks like this:

km/hr part 1 + km/hr part 2 = 364 km

Now we need to find each part on the left side of that equation. Part 1 first:

We traveled 65 km/hr for 2 hours, so that took us

65\frac{km}{hr}*2hr and canceling out the hour label, we have that in part 1 we got

65(2) = 130 km. Good. Now onto the second part, where our unknown is.

We traveled 78 km/hr the second part for x hours, so that took us

78\frac{km}{hr}*xhr and canceling out the hour label, we have that in part 2 we got

78x km. Now we can fill in the main equation (the one in bold print)

130 km + 78x km = 364 km and subtracting 130 km from both sides:

78x km = 234 km and dividing by 78 km:

x = 3 hours. Part 2 took 3 hours. Part 1 took 2 hours, so the whole trip took 5 hours.

7 0
2 years ago
How to calculate maximum height, when i know only V<br><br> Pleeeaaassseee as fast as you can
Dahasolnce [82]

Answer: was it this problem?

Explanation:

8 0
2 years ago
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