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3 years ago

Describe the forces that make a bicycle move.

Physics

1 answer:

sasho [114]3 years ago

3 0

Answer:

this is an example of Newton's Second Law, acceleration happens when a force acts on a mass.

the force that makes the bicycle move is your foot pushing against the pedal

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The speed of light in vinegar is 2.30 x 10^8 m/s. Determine the index of refraction. (2

pantera1 [17]

Answer:

$n _{v} = \frac{c}{v} \\ = \frac{3 \times {10}^{8} }{2.30 \times {10}^{8} } \\ = 1.30$

6 0

3 years ago

n general, which trophic level has the LEAST energy available to it? A) producer B) primary consumer C) secondary consumer D) te

zaharov [31]

D - tertiary consumer

This is because it is the farther up to food chain.

4 0

3 years ago

Someone help me like please thank you

lianna [129]

The car should have less kinetic energy.
They are both going the same speed, but the truck is bigger and heavier. The more mass an object has, the more kinetic energy it has. There is more mass being moved, so it makes more kinetic energy. The car does not have as much mass, so it makes less kinetic energy compared to the truck.

Good luck with the rest of your test or quiz :)

3 0

3 years ago

A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.

marin [14]

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

$\omega = 25 rpm$

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

$\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s$

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

$\omega = 2 \pi f$ (1)

where

$\omega$ is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

$\omega=0.42 rev/s$

Keeping in mind that $1 rev = 2\pi rad$ , the angular speed can be rewritten as

$\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42$

And re-arranginf eq.(1), we can find the frequency:

$f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz$

And the frequency is the number of complete revolutions made per second.

For an object in circular motion, the tangential speed is related to the angular speed by the equation

$v=\omega r$

where

$\omega$ is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

$\omega = 2\pi \cdot 0.42 rad/s$ is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

$v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s$

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

$d=vt$

where

v is the tangential speed

t is the time elapsed

Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

$d=(26.4)(20)=528 cm$

Learn more about circular motion:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

7 0

4 years ago

Read 2 more answers

A star emits electromagnetic radiation which closely resembles the spectrum of a blackbody. The three star system named Albireo

aksik [14]

Answer:

A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm , this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.

we have a fluctuation of the intensity emitted by the stars. Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.

Explanation:

The radiation of a black body is characterized by its temperature, with Wien's law of displacement we can find the maximum wavelength emitted by each star.

λ T = 2,898 10⁻³

therefore the emission the star of A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm

The emission of the premiere is in the ultraviolet light range, as this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.

The burst with A = 4300K has a bad emission maximum = 6.7395 10⁻⁷ m = 673.95 nm, which corresponds to an emission in the visible in the orange range, giving a blackbody spectrum of this range, but since the emission is formed by two stars, we see that when the two are placed one in front of the other the intensity of the emission must increase significantly and when they are placed next to each other the intensity reaches its minimum, consequently we have a fluctuation of the intensity emitted by the stars.

Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.

7 0

3 years ago