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3 years ago

If I walk 30 m north and turn around and walk 10 m to the south, what is the magnitude

Physics

2 answers:

Karo-lina-s [1.5K]3 years ago

8 0

Answer:

40m

Explanation:

Your total displacement would be 40m! Let me explain.

Going 30m north adds 30 to 0... so your total right now is 30m.

if you turn around and walk 10 m you have to add that 10 to the 30.

so your total displacement is 40m.

miss Akunina [59]3 years ago

5 0

Answer:

Hey

it would be 20 m

If you walk away 30 m then come back 10 you are now 20 m away from your starting point.

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3 years ago

Two long, straight parallel wires 8.2 cm apart carry currents of equal magnitude I. The parallel wires repel each other with a f

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Answer:

36.22 mA

Explanation:

i1 = I , i2 = I, d = 8.2 cm = 0.082 m

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4 years ago

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Answer:

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2 years ago

Read 2 more answers

Before a rifle is fired, the linear momentum of the bullet-rifle system is zero.

marusya05 [52]

Hi there!

II. Linear momentum of the system is zero.

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3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

4 years ago