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const2013 [10]
3 years ago
13

A 1.4-kg block slides freely across a rough surface such that the block slows down with an acceleration of â1.25 m/s2. what is t

he coefficient of kinetic friction between the block and the surface?
Physics
1 answer:
Marianna [84]3 years ago
6 0

Mass of the block = 1.4 kg

Weight of the block = mg = 1.4 × 9.8 = 13.72 N

Normal force from the surface (N) = 13.72 N

Acceleration = 1.25 m/s^2

Let the coefficient of kinetic friction be μ

Friction force = μN

F(net) = ma

μmg = ma

μg = a

μ = \frac{a}{g}

μ = \frac{1.25}{9.8}

μ = 0.1275

Hence, the coefficient of kinetic friction is: μ = 0.1275

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gladu [14]

Answer:

Explanation:

Credit card companies make money in two ways. One is the fees they charge retailers, restaurants, and other sellers of goods and services when you use your card to buy something. The other is the interest and fees they charge you. Here is how credit card interest works—and how you can pay less of it.

8 0
1 year ago
An ideal step-down transformer is needed to reduce a primary voltage of 120 V to 6.0 V. What must be the ratio of the number of
Julli [10]

Answer:

N_s :  N_p = 20 : 1

Explanation:

From the question we are told that

    The primary voltage is  V_p  =  120 \  V

     The secondary voltage is  V_s  =  6 \  V    

     

Generally from the transformer equation we have that

        \frac{V_p}{V_s}  =  \frac{N_p}{N_s}

So

       \frac{120}{6}  =  \frac{N_p}{N_s}

=>      \frac{N_p}{N_s} = 20

Therefore the ratio of the number of turns in the secondary to the number of turns in the primary is  

       N_s :  N_p = 20 : 1

7 0
3 years ago
What type of cells don't have cell walls.
iren [92.7K]

Answer:

Animal Cells

Explanation:

they dont have cell walls because they dont need them. Wall cells are found in plants, which animals dont need.

6 0
3 years ago
Read 2 more answers
The sound intensity level from one solo flute is 70 dB. If 10 flutists standing close together play in unison, what will the sou
Masja [62]

Answer:

80 dB

Explanation:

I_0 = Threshold intensity = 10^{-12}\ W/m^2

I = Intensity of sound

\beta = Intensity level of sound = 70 dB

Intensity level of sound is given by

\beta=10log\dfrac{I}{I_0}\\\Rightarrow 70=10log\dfrac{I}{I_0}\\\Rightarrow \dfrac{70}{10}=log\dfrac{I}{10^{-12}}\\\Rightarrow 10^{\dfrac{70}{10}}=\dfrac{I}{10^{-12}}\\\Rightarrow \dfrac{I}{10^{-12}}=10^{\dfrac{70}{10}}\\\Rightarrow I=10^{7}\times 10^{-12}\\\Rightarrow I=10^{-5}\ W/m^2

If there are 10 flutes I=10\times 10^{-5}\ W/m^2

\beta=10log\dfrac{10\times 10^{-5}}{10^{-12}}\\\Rightarrow \beta=10log10^8\\\Rightarrow \beta=10\times 8\\\Rightarrow \beta=80\ dB

The sound intensity level is 80 dB

5 0
3 years ago
the atomic number of uranium-235 is 92, its half-life is 704 million years, and the radioactive decay of 1 kg of 235u releases 6
wel

The overall efficiency of the conversion is 6.93%.

<h3>What is the overall efficiency?</h3>

We know that generation of electricity from radioactive fuels is one of the major ways by which we can generate electricity in many countries. Now the efficiency of each of the steps have been shown in the question;

  • Conversion efficiency = 35%
  • Transmission efficiency = 90%
  • Light  efficiency = 22%

Overall efficiency = (0.35 * 0.90 * 0.22) * 100 = 6.93%

The energy is obtained from;

6.7X10^13 J * 6.93/100

= 4.6 * 10^12 J

Learn more about nuclear energy:brainly.com/question/12793906

#SPJ4

Missing parts;

The atomic number of uranium-235 is 92, its half-life is 704 million years and the radioactive decay of l kg of 235U releases 6.7X10^13 J. Generally, radioactive material is considered safe after 10 half-lives.

Assume that a nuclear powerplant can convert energy from 235U into electricity with an efficiency of 35%, the electrical transmission lines operate at 90% efficiency, and flourescent lights operate at 22% efficiency.

1.) What is the overall efficiency of converting the energy of 235U into flourescent light?

2.) How much energy from 1 kg of 235U is converted into flourescent light?

7 0
1 year ago
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