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3 years ago

d is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?

2 answers:

5 0

<span>D is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
</span>Answer: The sled's potential energy is 882 Joules

Lesechka [4]3 years ago

4 0

Answer:

882 J

Explanation:

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The Young’s modulus of nickel is Y = 2 × 1011 N/m2 . Its molar mass is Mmolar = 0.059 kg and its density is rho = 8900 kg/m3 . G

Charra [1.4K]

Answer:

Atomic Size and Mass:

convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s

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3 years ago

Assume you need to design a hydronic system that can deliver 80,000 Btu/hr. What flow rate of water is required if the temperatu

PolarNik [594]

Answer:

At 10°F change in temperature

Mass flowrate = 1.01 kg/s = 2.227 lbm/s

Volumetric flowrate = 1010 m³/s = 35667.8 ft³/s

At 20°F change in temperature

Mass flowrate = 0.505 kg/s = 1.113 lbm/s

Volumetric flowrate = 505 m³/s = 17833.9 ft³/s

Explanation:

80000 btu/hr = 23445.7 W

P = ṁc(ΔT)

ṁ = MASS flowrate

c = specific heat capacity of water = 4182 J/kg.K,

ΔT = change in temperature = 10°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 10°F = 10×10/18 = 5.556°C = 5.556K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 5.556)

ṁ = 23445.7/(4182 × 5.556)

ṁ = 1.01 kg/s = 2.227 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 1.01 = 1010 m³/s = 35667.8 ft³/s

For a change of 20°F,

ΔT = change in temperature = 20°F

To convert, a change of 18°F is equal to a change of 10°C

A change of 20°F = 20×10/18 = 11.1111°C = 11.111K

P = ṁc(ΔT)

23445.7 = ṁ(4182 × 11.111)

ṁ = 23445.7/(4182 × 11.111)

ṁ = 0.505 kg/s = 1.113 lbm/s

In volumetric flow rate, Q = density × mass flowrate = 1000 × 0.505 = 505 m³/s = 17833.9 ft³/s

Hope this Helps!!!

4 0

3 years ago

A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location

amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

$f=-16.0 cm$ (the focal length is negative for a diverging lens)

$p=10.0 cm$ is the distance of the object from the lens

Solvign the equation for q, we find

$\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}$

$q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm$

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

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3 years ago

The invisible line that passes through the North Pole and the South<br> Pole is called Earth's what

REY [17]

Answer:

The Prime Meridian

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3 years ago

When a perfume bottle is opened, some liquid changes to gas and the fragrance spreads around the room. Which sentence explains t

Katyanochek1 [597]

When a perfume bottle is opened, some liquid changes to gas and the fragrance spreads around the room because gases do not have a definite volume.

Hope this helps!

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3 years ago