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babymother [125]
2 years ago
8

A 30.0 kg child, initially at rest, slides down a 2.0 m tall slide. The child reaches the bottom of the slide with a speed of 6

m/s. There is friction between the child and the slide. Write a Law of Conservation of Energy equation to represent the transfer of energy from the top of the slide to the bottom. a. b. Use the equation from part (b) to calculate the energy dissipated by friction between the slide and the child? (g = 9.81 m/s) = 13​
Physics
1 answer:
HACTEHA [7]2 years ago
5 0

Answer:

Explanation:

Total energy is constant

E = mgh + ½mv² + Fd

At the top of the slide, all energy is potential

E = mgh + 0 + 0

At the bottom of the slide, all potential energy has converted to kinetic and work of friction.

mgh = ½mv² + W

W = mgh - ½mv²

W = 30.0[(9.81)(2.0) - ½6²]

W = 48.6 J

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Which of the following are Electromagnetic Waves?
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8 0
3 years ago
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Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli
Masja [62]

Answer:

Not possible

Explanation:

E_{cl} = longitudinal modulus of elasticity = 35 Gpa

E_{ct} = transverse modulus of elasticity = 5.17 Gpa

E_m = Epoxy modulus of elasticity = 3.4 Gpa

V_{\rho l} = Volume fraction of fibre (longitudinal)

V_{\rho t} = Volume fraction of fibre (transvers)

E_f = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764

Transverse modulus of elasticity is given by

E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148

V_{\rho l}\neq V_{\rho t}

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0
3 years ago
A 14.0 gauge copper wire of diameter 1.628 mm carries a current of 12.0 mA . A) What is the potential difference across a 1.80 m
Serga [27]

Answer: a) 139.4 μV; b) 129.6 μV

Explanation: In order to solve this problem we have to use the Ohm law given by:

V=R*I whre R= ρ *L/A  where ρ;L and A are the resistivity, length and cross section of teh wire.

Then we have:

for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω

and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω

Finalle we calculate the potential difference (V) for both wires:

Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V

V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V

8 0
3 years ago
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