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2 years ago

Between any object and surface, the coefficient of static friction is _____ the coefficient of kinetic friction.

Physics

1 answer:

GarryVolchara [31]2 years ago

7 0

Answer

Greater than.

Static friction is the force that opposes motion between two objects that are not in motion. For the object to move one have to overcome the static friction. This force is usually greater than the force that exist between two surfaces that are in relative motion. This force is called kinetic friction.

For this reason the coefficient of static friction is greater that coefficient of kinetic friction.

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Elabora una tabla, como la del ejemplo, con los resultados obtenidos en los test que desarrollaste en actividades anteriores. Lu

rodikova [14]

Explanation:

Clear rendering reads;

"Make a table, like the one in the example, with the results obtained in the tests you carried out in previous activities. Then answer: What do these results indicate regarding your physical condition and how do they relate to your health? From your answers, consider the challenge of improving your physical condition by maintaining or improving your exercise routine or permanent practice of physical activity".

So the incomplete text above it seems you've been instructed to perform an experiment and then state your result/analysis.

8 0

2 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )

Also at equilibrium the moment about the point where the string holding the bar (where ( $m_1 +m_2$ ) and $m_3$ are hanged ) is attached is zero

So basically

$(m_1 + m_2 ) * 20 = m_3 * 30$

$(0.42 + 0.47) * 20 = 30 * m_3$

Making $m_3$ subject

$m_3 = \frac{(0.42 + 0.47) * 20 }{30 }$

$m_3 = 0.59 kg$

3 0

2 years ago

A weightlifter can lift 220lbs. (assume 1kg weighs 2.2lbs) If he lifts the weight 2m above the floor, how much work did he do?

Anton [14]

Convert 220 lb to kg.

220/2.2 = 100kg.

W = Fd (In this case, F is the weight)
W = (100)(2)
W = 200J

P = W/t
P = (200)/(1.2)
P = 166.67W

5 0

2 years ago

Can you please help me find the answer

Serggg [28]

Im sorry may you please retake the picture then i will answer

8 0

2 years ago

If a 80kg diver jumps off of a 5 m high dive into a regulation diving pool, how much should the temperature of the pool go up?

Agata [3.3K]

Answer:

The answer cannot be determined.

Explanation:

The energy of the diver when he hits the pool will be equal to its potential energy $mgh$ , and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.

The change in temperature ${\Delta}T$ then will be

${\Delta}T=\frac{{\Delta}Q}{mc} .$

Where m is the mass of water in the pool, c is the specific heat capacity of water, and ${\Delta}Q$ is the added heat which in this case is the energy of the diver.

Since we do not know the mass of the water in the pool, we cannot make this calculation.

7 0

2 years ago