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beks73 [17]
3 years ago
7

Define the types of friction and give FOUR examples of each

Physics
1 answer:
Verdich [7]3 years ago
4 0

Answer:

Static Friction - acts on objects when they are resting on a surface

Sliding Friction -  friction that acts on objects when they are sliding over a surface

Rolling Friction - friction that acts on objects when they are rolling over a surface

Fluid Friction - friction that acts on objects that are moving through a fluid

Explanation:

Examples of static include papers on a tabletop, towel hanging on a rack, bookmark in a book , car parked on a hill.

Example of sliding include sledding, pushing an object across a surface, rubbing one's hands together, a car sliding on ice.

Examples of rolling include truck tires, ball bearings, bike wheels, and car tires.

Examples of fluid include water pushing against a swimmer's body as they move through it , the movement of your coffee as you stir it with a spoon,  sucking water through a straw, submarine moving through water.

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93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures
lubasha [3.4K]

Answer:

828 kg/m³ or 0.828 g/cm³

Explanation:

Applying,

D = m/V............. Equation 1

Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.

From the question,

Given: m = 77 g , V = 93 cm³

Substitute these values into equation 1

D = 77/93

D = 0.828 g/cm³

Converting to kg/m³

D = 828 kg/m³

6 0
3 years ago
What do ocean waves and sound waves have in common?
Anettt [7]
Since they are both examples of moving waves, they both transmit energy.
7 0
3 years ago
Read 2 more answers
I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t
dezoksy [38]
Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }
where L is the length of the string, T the tension and m its mass. If  we plug the data of the problem into the equation, we find
f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz

The wavelength of the standing wave is instead twice the length of the string:
\lambda=2 L= 2 \cdot 24 m=48 m

So the speed of the wave is
v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s
7 0
3 years ago
A spring gun is made by compressing a spring in a tube and then latching the spring at
inn [45]

Answer:

a)v=13.2171\,m.s^{-1}

b)H=8.9605\,m

Explanation:

Given:

mass of bullet, m=4.97\times 10^{-3}\,kg

compression of the spring, \Delta x=0.0476\,m

force required for the given compression, F=9.12 \,N

(a)

We know

F=m.a

where:

a= acceleration

9.12=4.97\times 10^{-3}\times a

a\approx 1835\,m.s^{-2}\\

we have:

initial velocity,u=0\,m.s^{-1}

Using the eq. of motion:

v^2=u^2+2a.\Delta x

where:

v= final velocity after the separation of spring with the bullet.

v^2= 0^2+2\times 1835\times 0.0476

v=13.2171\,m.s^{-1}

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

u=13.2171\,m.s^{-1}

∵At maximum height the final velocity will be zero

v=0\,m.s^{-1}

Using the equation of motion:

v^2=u^2-2g.h

where:

h= height

g= acceleration due to gravity

0^2=13.2171^2-2\times 9.8\times h

h=8.9129\,m

is the height from the release position of the spring.

So, the height from the latched position be:

H=h+\Delta x

H=8.9129+0.0476

H=8.9605\,m

4 0
3 years ago
"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two firs
Fantom [35]

Answer:

A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if

the dispersion is great

6 0
3 years ago
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