Answer:
![h = 18.75 m](https://tex.z-dn.net/?f=h%20%3D%2018.75%20m)
Now when it will reach at point B then its normal force is just equal to ZERO
![N_B = 0](https://tex.z-dn.net/?f=N_B%20%3D%200)
![F_n = 1.72 \times 10^4](https://tex.z-dn.net/?f=F_n%20%3D%201.72%20%5Ctimes%2010%5E4)
Explanation:
Since we need to cross both the loops so least speed at the bottom must be
![v = \sqrt{5 R g}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B5%20R%20g%7D)
also by energy conservation this is gained by initial potential energy
![mgh = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![v = \sqrt{2gh}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2gh%7D)
so we will have
![\sqrt{2gh} = \sqrt{5Rg}](https://tex.z-dn.net/?f=%5Csqrt%7B2gh%7D%20%3D%20%5Csqrt%7B5Rg%7D)
now we have
![h = \frac{5R}{2}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B5R%7D%7B2%7D)
here we have
R = 7.5 m
so we have
![h = \frac{5(7.5)}{2}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B5%287.5%29%7D%7B2%7D)
![h = 18.75 m](https://tex.z-dn.net/?f=h%20%3D%2018.75%20m)
Now when it will reach at point B then its normal force is just equal to ZERO
![N_B = 0](https://tex.z-dn.net/?f=N_B%20%3D%200)
now when it reach point C then the speed will be
![mgh - mg(2R_c) = \frac{1}{2]mv_c^2](https://tex.z-dn.net/?f=mgh%20-%20mg%282R_c%29%20%3D%20%5Cfrac%7B1%7D%7B2%5Dmv_c%5E2)
![v_c^2 = 2g(h - 2R_c)](https://tex.z-dn.net/?f=v_c%5E2%20%3D%202g%28h%20-%202R_c%29)
![v_c = 13.1 m/s](https://tex.z-dn.net/?f=v_c%20%3D%2013.1%20m%2Fs)
now normal force at point C is given as
![F_n = \frac{mv_c^2}{R_c} - mg](https://tex.z-dn.net/?f=F_n%20%3D%20%5Cfrac%7Bmv_c%5E2%7D%7BR_c%7D%20-%20mg)
![F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)](https://tex.z-dn.net/?f=F_n%20%3D%20%5Cfrac%7B700%5Ctimes%2013.1%5E2%7D%7B5%7D%20-%20%28700%20%5Ctimes%209.8%29)
![F_n = 1.72 \times 10^4](https://tex.z-dn.net/?f=F_n%20%3D%201.72%20%5Ctimes%2010%5E4)
Answer:
the third law (for every action there is an equal and opposite reaction).
Explanation:
The skateboarder pushes backwards on the road (that is he applies a force on the road in a direction opposite the direction of intended motion). By Newton's third law, this action of the skateboarder causes an equal reaction of the road on the skateboarder in the opposite direction. Newton's third law states that action and reaction are equal but opposite in direction. So, the road in response to this backward force pushes the skateboarder in the forward direction causing the skateboarder and the skateboard to move in the forward direction.
Answer:
a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω
Explanation:
For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances
all resistors connected
V = i (R₁ + R₂)
with R₁ connected
V = (i + 0.5) R₁
with R₂ connected
V = (i + 0.25) R₂
We have a system of three equations with three unknowns for which we can solve it
We substitute the last two equations in the first
V = i (
)
1 = i (
)
1 = i (
) =
i² + 0.75 i + 0.125 = 2i² + 0.75 i
i² - 0.125 = 0
i = √0.125
i = 0.35355 A
with the second equation we look for R1
R₁ =
R₁ = 12 /( 0.35355 +0.5)
R₁ = 14.1 Ω
with the third equation we look for R2
R₂ =
R₂ =
R₂ = 19.9 Ω
The answer is that He is gonna struggle
The minimum uncertainty in the electron's momentum is
![Δp = 2.2822 \times 10 {}^{ - 20} kg/ms](https://tex.z-dn.net/?f=%CE%94p%20%3D%202.2822%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2020%7D%20kg%2Fms)
Given:
Uncertainty in position (ΔX)
![= 24 \times 10 {}^{ - 15}m](https://tex.z-dn.net/?f=%20%3D%2024%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2015%7Dm%20)
planck's constant (h)
![= 6.26 \times 10 {}^{ - 34} js](https://tex.z-dn.net/?f=%20%3D%206.26%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2034%7D%20js)
To find:
uncertainty in momentum (Δp)
Δx.Δph/4π
![24 \times 10 {}^{ - 15} .Δp = \frac{6.26 \times 10 {}^{ - 34} }{4 \times \frac{22}{7} }](https://tex.z-dn.net/?f=24%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2015%7D%20.%CE%94p%20%3D%20%20%5Cfrac%7B6.26%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2034%7D%20%7D%7B4%20%5Ctimes%20%20%5Cfrac%7B22%7D%7B7%7D%20%7D%20)
![24 \times 10 {}^{ - 15}. Δp = \frac{6.26 \times 10 {}^{ - 34} \times 7 }{8}](https://tex.z-dn.net/?f=%2024%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2015%7D.%20%CE%94p%20%3D%20%20%5Cfrac%7B6.26%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2034%7D%20%5Ctimes%207%20%7D%7B8%7D%20)
![Δp = \frac{43.82 \times 10 {}^{ - 34} }{8 \times 24 \times 10 {}^{ - 15} }](https://tex.z-dn.net/?f=%CE%94p%20%3D%20%20%5Cfrac%7B43.82%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2034%7D%20%7D%7B8%20%5Ctimes%2024%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2015%7D%20%7D%20)
![Δp = \frac{43.82 \times 10 {}^{ - 34} \times 10 {}^{ 15} }{192}](https://tex.z-dn.net/?f=%CE%94p%20%3D%20%20%5Cfrac%7B43.82%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2034%7D%20%20%5Ctimes%2010%20%7B%7D%5E%7B%2015%7D%20%7D%7B192%7D%20)
![Δp = \frac{4382 \times 10 {}^{ - 21} }{192}](https://tex.z-dn.net/?f=%CE%94p%20%3D%20%20%5Cfrac%7B4382%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2021%7D%20%7D%7B192%7D%20)
![Δp =22.822 \times 10 {}^{ - 21}](https://tex.z-dn.net/?f=%CE%94p%20%3D22.822%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2021%7D%20)
![Δp = 2.2822 \times 10 {}^{ - 20} kg/ms](https://tex.z-dn.net/?f=%CE%94p%20%3D%202.2822%20%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2020%7D%20kg%2Fms)
learn more about electron's momentum from here: brainly.com/question/28203580
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