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Aliun [14]
3 years ago
15

While in a park, you walk west for 52 m, then you walk 33.9° north of west for 44 m, and finally you walk due north for 25 m. Fi

nd the components of your final displacement, from your initial to final point, along the north and west directions. (a) displacement component due north (b) displacement component due west
Physics
1 answer:
vodka [1.7K]3 years ago
4 0

Answer:

(a)  Total north component = (26.89 + 25) = 51.89 metres

(b) Total west component = (-34.83 + 52) = 17.17 metres.

Explanation:

Do northwest components first.

North component = (sin 33.9 x 44) =26.89  metres.

West component = (cos 33.9 x 44) = -34.83metres.

Total west component = (-34.83 + 52) = 17.17 metres.

Total north component = (26.89 + 25) = 51.89 metres

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A 3.1 kg ball is dropped from the top of a 38 m tall building. What is the speed of the ball when it is halfway from the buildin
Archy [21]

Answer:

19.3m/s

Explanation:

Use third equation of motion

v^2-u^2=2gh

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity

insert values to get answer

v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s

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2 years ago
Which has the same meaning as air pressure?
SVETLANKA909090 [29]
B) atmosphere pressure, i believe

7 0
3 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
2 years ago
When preparing to strip insulation in order to splice conductors, a person should strip the insulation back far enough so that t
Fiesta28 [93]

Answer:

Closely fits into the connector.

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It's one of the steps used for the splicing of aluminium conductors in the underground connections. Where we do the strip insulation to splice the conductors by using compression type connectors.

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