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ollegr [7]
3 years ago
15

Please help

Physics
1 answer:
Sergio039 [100]3 years ago
8 0

2) The velocity of the wheel is increasing

3) The slope is the rate of change of velocity

4) The unit of the slope is metres per second squared.

5) Acceleration

6) The symbol used for acceleration is a

7) The relationship between velocity and time is v=gsin \theta t

Explanation:

2)

The graph in the problem represents the velocity of the wheel as a function of the time.

As we can see from the graph, the velocity is increasing as the time passes. This means that the wheel is accelerating (its velocity is changing constantly)

3)

The slope of the graph represents the rate of change of the velocity.

Mathematically, it can be written as:

m=\frac{\Delta v}{\Delta t}

where

m is the slope

\Delta v is the change in velocity

\Delta t is the time interval considered

As we see from the graph, the slope of the line is positive and constant: this means that the velocity is increasing at a constant rate.

4)

The unit of the slope can be determined starting by the units of the two variables involved.

On the y-axis, we have the velocity, which is measured in metre per second (m/s)

On the x-axis, we have the time, which is measured in seconds (s)

The slope is the ratio between these two quantities:

m=\frac{\Delta v}{\Delta t}

Therefore, the units of the slope are

m=\frac{[m/s]}{[s]}=[m/s^2]

So, metres per second squared.

5)

The rate of change of velocity is also known as acceleration.

In fact, acceleration is defined as the ratio between the change in velocity and the change in time:

a=\frac{\Delta v}{\Delta t}

By comparing with the formula of the slope in part 3), we see that the two equations are identical, therefore the acceleration corresponds to the slope of the graph.

6)

The symbol used to represent the acceleration is a:

a=\frac{\Delta v}{\Delta t}

7)

In any uniformly accelerated motion, the relationship between velocity and time is given by the following suvat equation:

v=u+at

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

In this problem, the wheel starts from rest, so

u = 0

Also, for an object rolling down a ramp, the acceleration is given by

a=g sin \theta

where g is the acceleration of gravity and \theta is the angle of the ramp. Substituting, we find the final expression of the velocity:

v=gsin \theta t

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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A beam of light has a wavelength of 650 nm in vacuum. (a) What is the speed of this light in a liquid whose index of refraction
Lady_Fox [76]

Answer:

The speed of this light and wavelength in a liquid are 2.04\times10^{8}\ m/s and 442 nm.

Explanation:

Given that,

Wavelength = 650 nm

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Using formula of speed

n = \dfrac{c}{v}

Where, n = refraction index

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Put the value into the formula

1.47=\dfrac{3\times10^{8}}{v}

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(b). We need to calculate the wavelength

Using formula of wavelength

n=\dfrac{\lambda_{0}}{\lambda}

\lambda=\dfrac{\lambda_{0}}{n}

Where, \lambda_{0} = wavelength in vacuum

\lambda = wavelength in medium

Put the value into the formula

\lambda=\dfrac{650\times10^{-9}}{1.47}

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Hence, The speed of this light and wavelength in a liquid are 2.04\times10^{8}\ m/s and 442 nm.

3 0
3 years ago
) Suppose a particle travels along a straight line with velocity v(t) = t 2 e −3t meters per second after t seconds. How far doe
pshichka [43]

Answer:

x(t=3s) = 0.07 m to the nearest hundredth

Explanation:

v(t) = t² e⁻³ᵗ

Find displacement after t = 3 s.

Recall, velocity, v = (dx/dt)

v = (dx/dt) = t² e⁻³ᵗ

dx = t² e⁻³ᵗ dt

∫ dx = ∫ t² e⁻³ᵗ dt

This integration will be done using the integration by parts method.

Integration by parts is done this way...

∫ u dv = uv - ∫ v du

Comparing ∫ t² e⁻³ᵗ dt to ∫ u dv

u = t²

∫ dv = ∫ e⁻³ᵗ dt

u = t²

(du/dt) = 2t

du = 2t dt

∫ dv = ∫ e⁻³ᵗ dt

v = (-e⁻³ᵗ/3)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ (-e⁻³ᵗ/3) 2t dt

= (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt

But the integral (∫ 2t (-e⁻³ᵗ/3) dt) is another integration by parts problem.

∫ u dv = uv - ∫ v du

u = 2t

∫ dv = ∫ (-e⁻³ᵗ/3) dt

u = 2t

(du/dt) = 2

du = 2 dt

∫ dv = ∫ (-e⁻³ᵗ/3) dt

v = (e⁻³ᵗ/9)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ 2t (-e⁻³ᵗ/3) dt = 2t (e⁻³ᵗ/9) - ∫ 2 (e⁻³ᵗ/9) dt = 2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)

Putting this back into the main integration by parts equation

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt = (-t²e⁻³ᵗ/3) - [2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)]

x(t) = ∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k (k = constant of integration)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k

At t = 0 s, v(0) = 0, hence, x(0) = 0

0 = 0 - 0 - (2/27) + k

k = (2/27)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + (2/27)

At t = 3 s

x(3) = (-9e⁻⁹/3) - (6e⁻⁹/9) - (2e⁻⁹/27) + (2/27)

x(3) = -0.0003702294 - 0.0000822732 - 0.0000091415 + 0.0740740741 = 0.07361243 m = 0.07 m to the nearest hundredth.

7 0
3 years ago
Read 2 more answers
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