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3 years ago

Which of the following is an example of distance equaling displacement?

Physics

1 answer:

AleksandrR [38]3 years ago

3 0

Answer:

Explanation:

The student had displaced their in the class when she left. The phone is what's displaced and student leaving equals distance.

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Use Hooke's Law to determine the work done by the variable force in the spring problem. Nine joules of work is required to stret

natima [27]

Answer:

29.16 J

Explanation:

From Hook's law,

W = 1/2(ke²)..................... Equation 1

Where W = work done, k = Spring constant, e = extension.

Given: W = 9 J, e = 0.5 m.

Substitute into equation 1

9 = 1/2(k×0.5²)

Solve for k

k = 18/0.5²

k = 72 N/m.

The work done required to stretch the spring by additional 0.4 m is

W = 1/2(72)(0.4+0.5)²

W = 36(0.9²)

W = 29.16 J.

6 0

3 years ago

Electromagnetic waves differ from other types of waves because they are

nikklg [1K]

I think it is D because u think of space and Electromagnetic Waves Travel Through Space Moving All Kinds of Particles From The Sun With It Which Creates The Northern and Southern Lights In The Poles :-)

8 0

3 years ago

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

${v}^{2} = {z}^{2} + 2ax$

$v = z + at$

First let's find the final velocity the plane will have at the end of the runway using the first equation:

${v}^{2} = {0}^{2} + 2(5)(1800)$

$v = 60 \sqrt{5}$

Now we can plug this into the second equation to find t:

$60 \sqrt{5} = 0 + 5t$

$t = 12 \sqrt{5}$

Then using 3 significant figures we round to 26.8 seconds

3 0

3 years ago

A car slows down from 27.7 m/s to 10.9 m/s in 2.37 s. What is its acceleration?

Sunny_sXe [5.5K]

Answer:

- 7.088 m/s²

Explanation:

As we know that,

★ Acceleration = Change in velocity/Time

→ a = (v - u)/t

Here,

Initial velocity (u) = 27.7 m/s
Final velocity (v) = 10.9 m/s

→ a = (10.9 m/s - 27.7 m/s)/2.37 s

→ a = -16.8/2.37 m/s²

→ a = -7.088 m/s² [Answer]

Negative sign denotes that the velocity is decreasing.

4 0

3 years ago

Read 2 more answers

A flat sheet of ice has a thickness of 1.4 cm. It is on top of a flat sheet of crown glass that has a thickness of 3.0 cm. Light

MAXImum [283]

Answer:

t = 2.13 10-10 s , d = 6.39 cm

Explanation:

For this exercise we use the definition of refractive index

n = c / v

Where n is the refraction index, c the speed of light and v the speed in the material medium.

The refractive indices of ice and crown glass are 1.13 and 1.52, respectively, therefore the speed of the beam in the material medium is

v = c / n

As the beam strikes perpendicularly, the beam path is equal to the distance of the leaves, there is no refraction, so we can use the uniform motion relationships

v = d / t

t = d / v

t = d n / c

Let's look for the times on each sheet

Ice

t₁ = 1.4 10⁻² 1.31 / 3 10⁸

t₁ = 0.6113 10⁻¹⁰ s

Crown glass (BK7)

t₂ = 3.0 10⁻² 1.52 / 3.0 10⁸

t₂ = 1.52 10⁻¹⁰ s

Time is a scalar therefore it is additive

t = t₁ + t₂

t = (0.6113 + 1.52) 10⁻¹⁰

t = 2.13 10-10 s

The distance traveled by this time in a vacuum would be

d = c t

d = 3 10⁸ 2.13 10⁻¹⁰

d = 6.39 10⁻² m

d = 6.39 cm

3 0

3 years ago