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2 answers:

5 0

I believe your answer is C. Dissolving of limestone, as because this is a type of chemical weathering I believe

Please correct me if wrong
Have a Nice Day! =D

Reptile [31]3 years ago

4 0

C is not mechanical weathering. It is chemical weathering.

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Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg⋅m2 and for arms and legs in is 0.70

Aleksandr-060686 [28]

Given:
I₁ = 0.70 kg-m², the moment of inertia with arms and legs in
I₂ = 3.5 kg-m², the moment of inertia with arms and a leg out.
ω₁ = 4.8 rev/s, the angular speed with arms and legs in.
That is,
ω₁ = (4.8 rev/s)*(2π rad/rev) = 30.159 rad/s

Let ω₂ = the angular speed with arms and a leg out.
Because momentum is conserved, therefore
I₂ω₂ = I₁ω₁
ω₂ = (I₁/I₂)ω₁
= (0.7/3.5)*(30.159)
= 6.032 rad/s

ω₂ = (6.032 rad/s)*(1/(2π) rev/rad) = 0.96 rev/s

Answer: 0.96 rev/s

3 0

3 years ago

if a sine wave representing ac voltage has an effective value of 70 vac what would be the waves peak value

Morgarella [4.7K]

Answer:

99 V

Explanation:

The effective voltage of an AC current (also called rms voltage) is given by

$V_{rms} = \frac{V_0}{\sqrt{2}}$

where

Vrms is the rms voltage

V0 is the peak voltage

In this problem, we know the effective voltage:

$V_{rms} = 70 V$

Therefore, we can re-arrange the equation to find the peak voltage:

$V_0 = V_{rms} \sqrt{2} =(70 V)(\sqrt{2})=99 V$

3 0

3 years ago

An infinitely long line of charge has linear charge density 6.00×10−12 C/m . A proton (mass 1.67×10−27 kg,charge +1.60×10−19 C)

Bogdan [553]

Answer:

$A) \,K.E=1.405\times 10^{-20}J$

$B)\,r_f=0.268\,m$

Explanation:

$Charge\,\,density=\lambda=6\times 10^{-12}C/n\\\\Mass\,\, of \,\,proton=m_p=1.67\times 10^{-27}kg\\\\charge\,\, of\,\, proton=q_p=1.609\times 10^{-19}C\\\\r=12\,cm=0.12\, m\\\\v=4.103\times 10^{3}m/s$

A) Initial kinetic energy of proton

$K.E=\frac{1}{2}m_pv^2\\\\K.E=1.405\times 10^{-20}J$

B) How close does the proton get to the line of charge?

Potential energy and kinetic energy are related as:

$K_i+U_i=K_f+U_f\\\\U_f-U_i=K_i-K_f\\\\q(V_f-V_i)=1.40\times 10^{-20}\\\\V_f-V_i=0.087--(1)\\$

Change in voltage is

$V_f-V_i=\frac{\lambda}{2\pi \epsilon_o}ln\frac{r_f}{r_i}\\\\ln|\frac{r_f}{r_i}|=(0.087)(\frac{2\pi \epsilon_o}{\lambda})\\\\ln|\frac{r_f}{r_i}|==0.8059\\\\\frac{r_f}{r_i}=2.24\\\\r_f=(2.24)(.12)\\\\r_f=0.268 m$