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3 years ago

A bicyclist steadily speeds up from rest to 6.00m/s in 7.30s.

Physics

1 answer:

Simora [160]3 years ago

8 0

Answer:

21.899 m

Explanation:

The explanation here is a little differe because we are starting from rest. So, everyone does physics a little differently, but what I would start with is trying to find the accelaration. I've given the formula, now you are able to plug this into the displacement formula where you Velocity Initial = 0m/s (Since you are starting from rest) and V final = 6.00 m/s...

You just plug numbers into the formula after solving for accelaration and do some simple algebra! I hope this helps! I am attaching a new pdf for visuals! If you can't open it let me know! Also there is a star by accelaration because I didn't know if I spelled it right, sorry!

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A skier is moving down a snowy hill with an acceleration of 0.40 m/s2. The angle of the slope is 5.0∘ to the horizontal. What is

kirill115 [55]

Answer:

1.25377 m/s²

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of friction

$\theta$ = Slope

From Newton's second law

$mgsin\theta-f=ma\\\Rightarrow mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow \mu=\frac{gsin\theta-a}{gcos\theta}\\\Rightarrow \mu=\frac{9.81\times sin5-0.4}{9.81\times cos5}\\\Rightarrow \mu=0.04655$

Applying $\mu$ to the above equation and $\theta=10^{\circ}$

$mgsin\theta-\mu mgcos\theta=ma\\\Rightarrow a=gsin\theta-\mu gcos\theta\\\Rightarrow a=9.81\times sin10-0.04655\times 9.81\times cos10\\\Rightarrow a=1.25377\ m/s^2$

The acceleration of the same skier when she is moving down a hill is 1.25377 m/s²

3 0

3 years ago

What was the maximum speed of the car in your experiment?

MA_775_DIABLO [31]

The measurements used in the experiment is the amount of speed over time.

The measurement of speed is indicated along the “y” axis.

Upon viewing the graph, the highest point along the “y” axis shown is 25 m/s. This would be the maximum.

The maximum speed of the car would be 25 m/s.

7 0

3 years ago

A 1,500 kg car’s speed changes from 30 m/s to 15 m/s after the brakes are applied. Calculate the work done onto the car from the

HACTEHA [7]

The work done onto the car is 506,250 J

The work done on a system implies an increase in the internal energy of the system as a result of some forces acting on the system from the outside.

From the parameters given:

The mass of the car = 1500 kg
The initial speed = 30 m/s
The final speed = 15 m/s

The work done onto the car refers to the change in the kinetic energy (i.e. ΔK.E)

$\mathbf{=\dfrac{1}{2} mv_1^2 -\dfrac{1}{2} mv_2^2}$

$\mathbf{=\dfrac{1}{2} m(v_1^2 - v_2^2)}$

$\mathbf{=\dfrac{1}{2} \times 1500 \times (30^2 - 15^2)}$

= 506,250 J

Therefore, we can conclude that the work done on the car is 506,250 J

Learn more about work done here:

brainly.com/question/18762601

7 0

3 years ago

In the desert air, sound travels at 358 meters per second (m/s). In the polar air, sound travels at 330 meters per

Sati [7]

Answer:

The correct option is;

C. The temperature of the air affects the speed of sound

Explanation:

The information given are;

The speed with which sound travels in the desert air = 358 m/s

The speed with which sound travels in the polar air = 330 m/s

The major difference between the desert air and the polar air is that the temperature of the desert air is hotter than the temperature of the air in the polar region. Therefore the speed of sound is affected by the air temperature.

The equation that gives the relationship of the speed of sound in air, $v_{air}$ , to temperature is presented as follows;

$v_{air} = 331\frac{m}{s} \times \sqrt{\dfrac{T_K}{273 K} } = 331\frac{m}{s} \times \sqrt{1 + \dfrac{T_{^{\circ}C}}{273 ^{\circ} C} }$

Which shows that the speed of sound in air, $v_{air}$ , rises as the temperature of the air rises.

Where;

$T_{^{\circ}C}$ = The temperature of the air in degrees Celsius

$T_K$ = The temperature of the air in degrees Kelvin.

7 0

3 years ago

A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What wil

mash [69]

Answer:

The wavelength will be 33.9 cm

Explanation:

Given;

frequency of the wave, F = 1200 Hz

Tension on the wire, T = 800 N

wavelength, λ = 39.1 cm

$F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}$

Where;

F is the frequency of the wave

T is tension on the string

μ is mass per unit length of the string

λ is wavelength

$\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\$

when the tension is decreased to 600 N, that is T₂ = 600 N

$T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm$

Therefore, the wavelength will be 33.9 cm

5 0

3 years ago