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3 years ago

What is displacement and distance

Physics

1 answer:

iogann1982 [59]3 years ago

7 0

Answer:

Displacement is how far you are between your initial and finishing position.

Distance is how far you went.

Explanation:

For example, using a baseball field... if the sides are each 100 m..

A batter runs to third base... what is his displacement and distance?

Since the batter ran to 3rd base, it means he ran 300 m to get to 3rd base but his displacement is 100 m becuase that's how far they are from home plate. (which was their starting position)

So the answer to my example above would be:

Distance: 300 m

Displacement: 100 m

I hope this helped you.

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A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr

Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of $34.9^{\circ}$

Explanation:

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

$d sin \theta = m \lambda$

where

d is spacing between the lines in the grating

$\theta$ is the angle of the maximum

m is the order of diffraction

$\lambda$ is the wavelength of the light

For laser 1,

$d sin \theta = m_1 \lambda_1$

For laser 2,

$d sin \theta = m_2 \lambda_2$

where

$\lambda_1 = 420 nm\\\lambda_2 = 630 nm$

Since the position of the maxima in the two cases overlaps, then the term $d sin \theta$ on the left is the same for the two cases, therefore we can write:

$m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}$

Therefore:

$m_1 = 3$

$m_2 = 2$

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

$d sin \theta = m_1 \lambda_1$

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

$d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m$

$m_1 = 3$ is the order of the maximum

$\lambda_1 = 420 nm = 420\cdot 10^{-9} m$ is the wavelength of the laser light

Solving for $\theta$ , we find the angle of the maximum:

$sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572$

So the angle is

$\theta=sin^{-1}(0.572)=34.9^{\circ}$

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4 years ago

What do you notice about earths axis during the winter solstice

MrRa [10]

The Earth's axis is tilted "away" from the Sun in the hemisphere. where it's winter. The daytime is very short. That's because the Earth's axis is tilted "away" from the Sun in the hemisphere.

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4 years ago

a 55.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 130 kg football players at a height of 0.800 m . the diamet

suter [353]

The diameter of the football player's piston is 0.55 m

Given that the mass of the cheerleader(m) is 55 kg, mass of football player to be hold (M) is 130 kg, height of the players (h) is 1.30 m, radius of the piston corresponding to the diameter (r) is 0.09 m, Diameter of football player's piston (R), P1 is Pressure on the cheerleader's side, P2 is Pressure on the football player's side

Using Pascal's law,

This law states that if there is a change at a point of a body immersed in a fluid then that change will spread thoroughly to each and every point of the body.

The formula of hydraulic system is,

P1= P2

F1/A1 = F2/A2

mg/πr^2 = 4Mg/πR^2

m/r^2=4M/R^2

R^2=4M×r^2/m

By plugging the values, we get.,

R^2=(4×130×0.09^2)/55

R^2=4.21/55=0.076

R=√0.076 = 0.275 m

Hence, diameter of football player's piston is 0.55 m

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1 year ago

Faraday discovered that a current could be induced in a solenoid (a coil of wire) when _____.

igomit [66]

Faraday discovered that a current could be induced in a solenoid (a coil of wire) when "a magnetized rod is being moved through the coil"

Hope this helps!

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4 years ago

Read 2 more answers

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an

sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The work is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the two tugboats "pull the tanker a distance 0.83km toward the north", that is the displacement, and you have to calculate the net force toward the north.

3. Tugboat #1.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N

4. Tugboat #2:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle: = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N =

5. Total net force, Fn:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. Work, W:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

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3 years ago