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Step2247 [10]
3 years ago
8

our lab partner wears a new pair of sneakers to lab and, rather than performing the required experiments, you decide to measure

the coefficient of static friction they make with the lab floor. Your partner reports her mass to be 59.0 kg. Hooking up an industrial force sensor to your interface you give your partner a piece of rope attached to the sensor and begin to pull. When the sensor reads 318.3 N your partner starts to slip. Determine the coefficient of static friction between your partner and the floor.
Physics
1 answer:
Dafna1 [17]3 years ago
4 0

Answer:

The coefficient of static friction between your partner and the floor is 0.55

Explanation:

Given:

Mass m = 59 Kg

Frictional force F_{s}  = 318.3 N

From the formula of frictional force,

 F_{s} = \mu_{s} mg

Where \mu _{s} = coefficient of static friction, g = 9.8 \frac{m}{s^{2} }

Put the above values and find the coefficient of static friction.

318.3 = \mu_{s} \times 59 \times 9.8

\mu_{s} = 0.55

Therefore, the coefficient of static friction between your partner and the floor is 0.55

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aliya0001 [1]

Answer:

the mass of the bullet is 10.5 g

Explanation:

Given;

initial velocity, u₁ = 280 m/s

final velocity of the bullet, v₁ = 70 m/s

final velocity of the block, v₂ = 0.2 m/s

mass of the block, m₂ = 11 kg

initial velocity of the block, u₂ = 0

let the mass of the bullet = m₁

Apply the principle of conservation of linear momentum for elastic collision to calculate the mass of the bullet.

m₁u₁ + m₂u₂ = m₁v₁  + m₂v₂

280m₁  +  11(0)  = 70m₁  +  11 x 0.2

280m₁ = 70m₁  + 2.2

280m₁ - 70m₁  = 2.2

210m₁ = 2.2

m₁ = 2.2/210

m₁ = 0.0105 kg

m₁ = 10.5 g

Therefore, the mass of the bullet is 10.5 g

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3 years ago
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3 years ago
Compare the current in the 8-ohm resistors to the current in the 4-ohm resistors.
Gemiola [76]

Answer:

a)   i₈ = 0.5 i₄,  b)   i₁₀ = 0.3 i₃,    i₁₀ = 0.8 i₈

Explanation:

For this exercise we use ohm's law

       V = i R

        i = V / R

we assume that the applied voltage is the same in all cases

let's find the current for each resistance

         

R = 4 Ω

         i₄ = V / 4

R = 8 Ω

         i₈ = V / 8

we look for the relationship between these two currents

         i₈ /i₄ = 4/8 = ½

         i₈ = 0.5 i₄

R = 3 Ω

        i₃ = V3

R = 10 Ω

         

        i₁₀ = V / 10

   

we look for relationships

       i₁₀ / 1₃ = 3/10

       i₁₀ = 0.3 i₃

       i₁₀ / 1₈ = 8/10

       i₁₀ = 0.8 i₈

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3 years ago
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3 years ago
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Electronics and inhabitants of the International Space Station generate a significant amount of thermal energy that the station
katrin2010 [14]

Answer:

4462.0927 W

Explanation:

\epsilon = Emissivity of the panel = 1

\sigma = Stefan-Boltzmann constant = 5.67\times 10^{-8}\ W/m^2K^4

T = Temperature = (273.15+6)

Area of the panel is given by

A=2\times 1.8\times 3.6\\\Rightarrow A=12.96\ m^2

The power radiated is given by

P=\epsilon \sigma AT^4\\\Rightarrow P=1\times 5.67\times 10^{-8}\times 12.96\times (273.15+6)^4\\\Rightarrow P=4462.0927\ W

The power radiated from each panel is 4462.0927 W

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4 years ago
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