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3 years ago

The force that one surface exerts on another when the two rub against each other is called

1 answer:

7 0

1. Friction (Example: when you rub your hands together and you start to feel heat.)

2. Attract each other. (Example: Every object has a gravitational pull of some sort. The larger the object, the stronger the gravitational pull. This is why the sun is able to keep all eight planets in orbit around it.)

3. Hope this helped :)

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A 0.50-kg toy is attached to the end of a 1.0-m very light string. The toy is whirled in a horizontal circular path on a frictio

Deffense [45]

Answer:

26.5 m/s

Explanation:

The tension in the string provides the centripetal force that keeps the toy in circular motion. So we can write:

$T=m\frac{v^2}{r}$

where:

T is the tension in the spring

m = 0.50 kg is the mass of the toy

r = 1.0 m is the radius of the circle (the length of the string)

v is the speed of the toy

The maximum tension in the string is

T = 350 N

If we substitute this value into the equation, we find the maximum speed that the mass can have before the string breaks:

$v=\sqrt{\frac{Tr}{m}}=\sqrt{\frac{(350)(1.0)}{0.50}}=26.5 m/s$

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3 years ago

A bicyclist accelerates from a stop to a speed of 12m/s in 3 seconds.what is her acceleration

AVprozaik [17]

Answer:

4m/s2

Explanation:

acceleration=v-u/t...v=12m/s,u=0m/s,t=3sec

a=12-0/3

=4m/s2

8 0

3 years ago

Answer this

arlik [135]

Answer:

a) Ink X is likely to be pure because it only contain 1 spot.

b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.

c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.

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3 years ago

What are two processes involved<br> in making proteins from the<br> DNA code

Svetach [21]

Answer:

Transcription and Translation. I hope this helps!

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4 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:

$d=vt$