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4 years ago

A runner traveled 15 kilometers north then backtracked 11 kilometers south before stopping. His resultant displacement was

Physics

2 answers:

Vanyuwa [196]4 years ago

5 0

displacement is 4 km in north direction.

Explanation:

we are given

A runner traveled 15 kilometers north first

then backtracked 11 kilometers south

we have to find displacement=d=?

assume that east side represents +ve x- axis(horizontal)

and

north represent +ve y - axis (vertical)

we can take x-axis (horizontal) as i

and y-axis (vertical) as j.

Displacement = 15 j -11 j = 4 km north

Serhud [2]4 years ago

4 0

Answer:

Resultant displacement is 4 km north.

Explanation:

Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

A runner traveled 15 kilometers north then backtracked 11 kilometers south before stopping,

So initial displacement, 15 kilometers north = 15 j km

Second displacement, 11 kilometers south = -11 j km

Total displacement = 15 j -11 j = 4 j km

Total displacement is 4 km north.

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kipiarov [429]

Answer:

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2 years ago

wo parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude

melisa1 [442]

Answer:

-18896.49 V/m

Explanation:

Distance between the two plates = 10 cm = 10 x $10^{-2}$ m = 0.1 m

Also, one of the plates is taken as zero volt.

a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V

b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V

Potential field strength = -dV/dx

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dx is the difference in distance between these points

For the first case above,

potential field strength = -393/0.0705 = -5574.46 V/m

For the second case ,

potential field strength = -393/0.0295 = -13322.03 V/m

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4 years ago

Give two examples of spatial interference which can be easily observed.

kifflom [539]

Answer : space and kite

3 0

3 years ago

How do I find the tension? its supposed to be in the mid 40s

babunello [35]

Answer:

a = 2.3 m/s²

T = 45 N

Explanation:

Draw a free body diagram for each mass.

For the mass on the incline, there are four forces:

Weight force mg pulling down.

Normal force N perpendicular to the incline.

Friction force Nμ pushing down the incline.

Tension force T pulling up the incline.

For the hanging mass, there are two forces:

Weight force Mg pulling down.

Tension force T pulling up.

Sum of the forces on the hanging mass in the -y direction:

∑F = ma

Mg − T = Ma

T = Mg − Ma

Sum of the forces on the sliding mass in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces on the sliding mass in the parallel direction:

∑F = ma

T − mg sin θ − Nμ = ma

Substitute:

Mg − Ma − mg sin θ − mgμ cos θ = ma

Mg − mg (sin θ + μ cos θ) = ma + Ma

Mg − mg (sin θ + μ cos θ) = (m + M) a

a = [ Mg − mg (sin θ + μ cos θ) ] / (m + M)

Plug in values:

a = [ 6.0×9.8 − 5.0×9.8 (sin 30° + 0.20 cos 30°) ] / (5.0 + 6.0)

a = 2.3 m/s²

Now find tension:

T = Mg − Ma

T = 6.0×9.8 − 6.0×2.3

T = 45 N

3 0

4 years ago

Help I need an answer!

Umnica [9.8K]

You apparently haven't noticed yet . . .

The moon rises in the East, moves across the sky, and sets
in the West, just like the sun and everything else in the sky.

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49 minutes later. So each day, the moon rises about 49 minutes
later than it rose on the previous day.

In order to describe the direction and height of the moon in the sky,
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from which you observed it. And wherever the moon was in the sky
at that time, it was in a noticeably different place 30 minutes later.

Just like the sun.

5 0

3 years ago