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madreJ [45]
3 years ago
8

Riley is doing an experiment working with friction. He wants to find two ways to modify an object that could increase or decreas

e the same type of friction
Physics
2 answers:
Dmitrij [34]3 years ago
8 0
If the type of friction is sliding friction, it may be increased by increasing the roughness of the surface and may be decreased by smoothing the surface or applying a lubricant.
elena-14-01-66 [18.8K]3 years ago
3 0

Answer:

As we know that friction force is a contact force which opposes the relative motion between two surfaces in contact.

So here we can change the force between two surface when they move relative to each other.

Here when Lubrication is used between two surface then the relative motion of two planes becomes easy. So by lubrication of the surface either by oil or by smoothing the plane we can say that friction force will be reduced.

Also friction force depends on the effective contact points between two surfaces. So as we increase the contact points between two surfaces then the effective frictional force will increase.

Also we know that friction force is given by

F_f = \mu F_n

now as we increase the normal force between two contact planes either by increasing the weight or by applying normal external force we can say that friction force will increase.

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N = 9 × 10−6 C / 1.6 × 10−19 C
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3 years ago
(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind
Vilka [71]
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
F_f = \mu N
where \mu is the coefficient of friction, while N is the normal force. So we have:
F=\mu N
since we know that F=300 N and \mu=0.3, we can find N, the magnitude of the normal force:
N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is \mu=0.03 due to the presence of the oil. Therefore, we have:
N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N
8 0
3 years ago
Veronica claims that she can throw a dart at a dartboard from a distance of 2 m and hit the 5 cm wide bullseye if she throws the
wariber [46]
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t = 2 m/ 15 m/s = 0.133 s

Then, find the height the dart has fallen from its initial point using this equation: h = 0.5gt²

h = 0.5(9.81 m/s²)(0.133 s)² = 0.0872 m or 8.72 cm

Since the diameter of the bull's eye is only 5 cm, and you started at the same level of the top of the bull's eye, that means the maximum allowance would only be 5 cm. Since it exceeded to 8.72 cm, it means that <em>Veronica will not hit the bull's eye.</em>
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Answer:

A. A login vty mode subcommand

Explanation:

since we are protecting co-workers from connecting to the switches from their desktop PCs, we would need a Telnet line which is used to connect to devices remotely from other network devices on the same network segment as the device we want to connect to. A login local vty subcommand configures a local username for login access but since our design constraint is to configure without usernames, option A is the correct answer.

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Answer:

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Explanation:

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