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cupoosta [38]
2 years ago
9

A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it

looks like this group varied the amount of mass sitting on the block with each trial - this is not recommended). Nonetheless, what is their average coefficient of static friction?
Trial Mass of block (g) Hanging mass (kg)
1 105 0.053
2 165 0.081
3 220 0.118
4 280 0.149
5 315 0.180
6 385 0.198
Physics
1 answer:
anzhelika [568]2 years ago
8 0

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = \frac{sum of coefficient of static friction}{number of trials}

                                              = \frac{0.779}{6}

                                              = 0.12983

The average coefficient of static friction is 0.130

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OverLord2011 [107]

Answer:

best explanation of this is sentence B

Explanation:

The radiation emission of the bodies is given by the expression

     P = σ A e T⁴

Where P is the power emitted in watts, σ is the Stefan-Boltzmann constant, A is the surface area of ​​the body, e is the emissivity for black body e = 1 and T is the absolute body temperature in degrees Kelvin.

When the values ​​are substituted the power is quite high 2.5 KW, but the medium surrounding the box also emits radiation

   T box ≈ T room

    P box ≈ P room

As the two powers are similar and the box can absorbed, since it has the ability to emit and absorb radiation, as the medium is also close of the temperature of the box, the amount emitted is very similar to that absorbed, so the net change in energy is very small.

   In the case that the box is much hotter or colder than the surrounding medium if there is a significant net transfer.

Consequently, the best explanation of this is sentence B

5 0
2 years ago
The CBR method of flexible pavement design gives an idea about the:
marta [7]

Answer:

A. quality of road-making material

Explanation:

The CBR method of flexible pavement design gives an idea about the:

A. quality of road-making material

B. traffic intensities

C. characteristics of soil

D. All of the above

The California Bearing Ratio (CBR) test is a penetration test used to evaluate the subgrade strength of roads and pavements. The results of these tests are used with the curves to determine the thickness of pavement and its component layers. it is the measure of the resistance of a material against the penetration of a standard plunger.

This can be used to determine the quality of road making material. CBR is expressed as the percentage of the actual load to the standard load.

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2 years ago
D=? V=100mL M=1.5kg=___g
Nana76 [90]
If you go to a website known as 'Wolfram' and type in the question they should be able to answer it.  It's basically a very smart calculator.  I'd add a link but I don't want to seem like I'm advertising, and it's against Brainly rules.
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2 years ago
3. A 75kg man sits at one end of a uniform seesaw pivoted at its center, and his 24kg son sits at the
bulgar [2K]

Answer:

The wife have to sit at 0.46 L from the middle point of the seesaw.

Explanation:

We need to make a sketch of the seesaw and the loads acting over it.

And by the studying of the Newton's law we can find the equation useful to find the distance of the mother sitting on the seesaw with respect to the center ot the pivot point.

A logical intuition will give us the idea that the mother will be on the side of her son to make the balance.

The maximum momentum with respect to the pivot point (0) will be:

M=75 *\frac{L}{2}

Where L/2 is the half of the distance of the seesaw

Therefore the other loads ( mom + son) must be create a momentum equal to the maximum momentum.

7 0
2 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.240 rev/s. The magnitude
bazaltina [42]

Explanation:

Given that,

Angular velocity = 0.240 rev/s

Angular acceleration = 0.917 rev/s²

Diameter = 0.720 m

(a). We need to calculate the angular velocity after time 0.203 s

Using equation of angular motion

\omega_{f}=\omega_{i}+\alpha t

Put the value in the equation

\omega_{f}=0.240+0.917\times0.203

\omega_{f}=0.426\ rev/s

The angular velocity is 0.426 rev/s.

(b). We need to calculate the tangential speed of the blade

Using formula of  tangential speed

v= r\omega

Put the value into the formula

v = \dfrac{0.720 }{2}\times0.426\times2\pi

v=0.963\ m/s

The tangential speed of the blade is 0.963 m/s.

(c). We need to calculate the magnitude at of the tangential acceleration

Using formula of tangential acceleration

a_{t}=r\alpha

Put the value into the formula

a_{t}=0.36\times0.917\times2\pi

a_{c}=2.074\ m/s^2

The tangential acceleration is 2.074 m/s².

Hence, This is required solution.

4 0
2 years ago
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