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sattari [20]
3 years ago
6

Tammy leaves the office, drives 34 km due north, then turns onto a second highway and continues in a direction of 35◦ north of e

ast for 79 km. What is her total displacement from the office? Answer in units of km. \ At what angle is her displacement? (C
Physics
1 answer:
Natasha2012 [34]3 years ago
4 0

Answer:

The total displacement is 102 km 51^o north of east.

Explanation:

We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.

d_n=34km+79km*sin(35^o)=79km

and

d_e=79*cos(35^o)=65km

The total displacement is given by:

D=\sqrt{(79km)^2+(65km)^2}=102km

with an agle of:

\alpha =arctg(\frac{79km}{65km})=51^o

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A uranium and iron atom reside a distance R = 44.10 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize
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Answer:

distance r from the uranium atom is 18.27 nm

Explanation:

given data

uranium and iron atom distance R = 44.10 nm

uranium atom = singly ionized

iron atom = doubly ionized

to find out

distance r from the uranium atom

solution

we consider here that uranium electron at distance = r

and electron between uranium and iron so here

so we can say electron and iron  distance = ( 44.10 - r ) nm

and we know single ionized uranium charge q2= 1.602 × 10^{-19} C

and charge on iron will be q3 = 2 × 1.602 × 10^{-19} C

so charge on electron is q1 =  - 1.602 × 10^{-19} C

and we know F = k\frac{q*q}{r^{2} }  

so now by equilibrium

Fu = Fi

k\frac{q*q}{r^{2} }  =  k\frac{q*q}{r^{2} }

put here k = 9*10^{9} and find r

9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{r^{2} }  =  9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{(44.10-r)^{2} }

\frac{1}{r^{2} } = \frac{2}{(44.10 -r)^2}

r = 18.27 nm

distance r from the uranium atom is 18.27 nm

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3 years ago
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