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gulaghasi [49]
2 years ago
5

A man walked 50 m to the south, turned, walked 40 m to the west.

Physics
1 answer:
MaRussiya [10]2 years ago
3 0

Answer:

Direction is N 51.3° E

Explanation:

» From trigonometric ratios:

{ \tt{ \red{ \tan( \theta) =  \frac{opposite \: { \blue{(south)}}}{adjacent \: { \blue{(west)}}}  }}} \\  \\ { \tt{ \tan( \theta)  =  \frac{50}{40} =  \frac{5}{4}  }} \\  \\ { \tt{ \theta =  { \tan }^{ - 1} ( \frac{5}{4}) }} \\  \\ { \tt{ \theta = 51.3 \degree}} \\  \\ { \boxed{ \tt{ \: direction : \:{ \green{N }}\: 51.3 \degree \: { \green{ E }}}}}

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Because heat is a path function or the energy in transit.

Explanation:

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What is the dimensional formula of pressure ??​
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A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0
Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). The time is 0.493\times10^{-7}\ sec.

Explanation:

Given that,

Electric field E=-6.00\times10^{5}i\ N/C

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

F=F_{e}

ma=qE

a=\dfrac{Eq}{m}

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}

a=-5.74\times10^{13}\ m/s62

The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). We need to calculate the initial peed

Using equation of motion

v^2-u^2=2as

Where, s = distance

Put the value into the formula

0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}

u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}

u=2834783.94

u=2.83\times10^{6}\ m/s

The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

t=\dfrac{u}{a}

Where, u = initial velocity

a = acceleration

Put the value into the formula

t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}

t=0.493\times10^{-7}\ sec

Hence, (a). The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). The time is 0.493\times10^{-7}\ sec.

5 0
2 years ago
A 120.00 resistor, a 60.00 resistor, and a 40.00 resistor are connected in parallel and placed across a 12.0-V battery. What
Svet_ta [14]

Answer:

C

Explanation:

3 0
2 years ago
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