Light travels in waves AND in bundles called "photons".
It's hard to imagine something that's a wave and also a bundle.
But it turns out that light behaves like both waves and bundles.
If you design an experiment to detect waves, then it responds to light.
And if you design an experiment to detect 'bundles' or particles, then
that one also responds to light.
False?
science can be a motivation out of curiosity AND societal needs.
For example:battery operated cars, are a curiosity and beneficial to the earths environment
Answer:
78 km/h
Explanation:
If I normally drive a 12 hour trip at an average speed of 100 km/h, my destination has a total distance of:
- 100 km/h · 12 h = 1,200 km
Today, I drive the first 2/3 of the distance at 116 km/h. Let's first calculate what 2/3 of the normal distance is.
I've driven 800 km already. I need to drive 400 km more to reach my final destination. I need to figure out my average speed during this last 1/3 of the distance.
To do this, I first need to calculate how much time I spent driving 116 km/h for the past 800 km.
- 116 km/1 h = 800 km/? h
- 800 = 116 · ?
- ? = 800/116
- ? = 6.89655172
I spent 6.89655172 hours driving during the first 2/3 of the distance.
Now, I need to subtract this value from 12 hours to find the remaining time I have left.
- 12 h - 6.89655172 h = 5.10344828 h
Using this remaining time and my remaining distance, I can calculate my average speed.
- ? km/1 hr = 400 km/5.10344828 h
- 5.10344828 · ? = 400
- ? = 400/5.10344828
- ? = 78.3783783148
My average speed during the last third of the distance is around 78 km/h.
Explanation:
Below is an attachment containing the solution.
Answer: 2.49×10^-3 N/m
Explanation: The force per unit length that two wires exerts on each other is defined by the formula below
F/L = (u×i1×i2) / (2πr)
Where F/L = force per meter
u = permeability of free space = 1.256×10^-6 mkg/s^2A^2
i1 = current on first wire = 57A
i2 = current on second wire = 57 A
r = distance between both wires = 26cm = 0.26m
By substituting the parameters, we have that
Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26
= 4080.744×10^-6/ 1.634
= 4.080×10^-3 / 1.634
= 2.49×10^-3 N/m
