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Contact [7]
3 years ago
15

A person desires to reach a point that is 2.17 km from her present location and in a direction that is 29.6° north of east. Howe

ver, she must travel along streets that are oriented either north-south or east-west. What is the minimum distance she could travel to reach her destination?
Physics
1 answer:
mestny [16]3 years ago
7 0

She needs to move x km in the east-west direction and y km in the north-south direction so that

\sqrt{x^2+y^2}=2.17

and

\tan29.6^\circ=\dfrac yx

Solve the system to get

x=1.89\,\mathrm{km}

y=1.07\,\mathrm{km}

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Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction
lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Cu^{2+}/Cu]}=0.34V

E^o_{[Ag^{+}/Ag]}=0.80V

E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}

E^o=0.80V-(0.34V)=0.46V

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}

E_{cell}=0.434V

Therefore, the cell potential for this cell 0.434 V

8 0
3 years ago
While a roofer is working on a roof that slants at 39.0 degrees above the horizontal, he accidentally nudges his 88.0 N toolbox,
Ostrovityanka [42]

Answer:

V= 6.974 m/s

Explanation:

Component( box) weight acting parallel and down roof 88(sin39.0°)=55.4 N

Force of kinetic friction acting parallel and up roof = 18.0 N

Fnet force acting on tool box acting parallel and down roof

Fnet= 55.4 - 18.0

Fnet=37.4 N

acceleration of tool box down roof

a = 37.4(9.81)/88.0

a= 4.169 m/s²

d = 4.90 m

t = √2d/a

t= √2(4.90)/4.169

t= 1.662 s

V = at

V= 4.169(1.662)

V= 6.974 m/s

5 0
3 years ago
PLEASE HELP! LAST DAY TO TURN IN AND IM SO BEHIND!!!
kolbaska11 [484]
<h3>Question 1</h3>

Answer

option C) velocity

Explanation

acceleration =  Δv ÷ Δt

<h3>Question 2</h3>

Answer

option C) m/s²

Explanation

Δv ÷ Δt

= m/s ÷ s

= m/s x 1/s

= m/s²

<h3>Question 3</h3>

Answer

option B) velocity has both direction and speed.

That is why velocity can be negative but speed can not and velocity is rate of change of displacement where as speed is rate of change of distance.

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3 years ago
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alexandr402 [8]

Answer:

Im only 12 and i need the points so ima try my best.

Explanation:

574.780616 m6 kg3 s-6 K-3 mol-3

6 0
3 years ago
A boy and his younger sister are at the zoo on a hot day. They each buy a cold lemonade. The boy buys a large lemonade, and his
FromTheMoon [43]

Answer: A and C (i took the test)

Explanation: Hope this helps:)

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3 years ago
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