V = 1/3 Bh v = 1/3 (13 ac)(43560ft^2/ac)(481ft) v = 90793560 ft^3 * 0.3048m/ft * 0.3048m/ft * 0.3048m/ft = 2570987m^3
Answer:
If the temperature of the air in the balloon is less than the temperature of the air surrounding the balloon then the balloon will appear slightly deflated because of the difference in temperature.
As the temperature of the air in the balloon reaches the surrounding air temperature, then the balloon will appear to be fully inflated because the temperature of the air in the balloon is the same as the surrounding air temperature.
The calculated coefficient of kinetic friction is 0.33125.'
The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
given mass of the block=10 kg
spring constant k= 2250 Nm
now according to principal of conservation of energy we observe,
the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.
mgh= μ (mgl) +1/2 kx²
10 x 10 x 3= μ(600) +(1125) (0.09)
μ(600) =300 - 101.25
μ = 198.75÷600
μ =0.33125
The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)
Learn more about kinetic friction here-
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Answer:
The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.
Explanation:
This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S
= 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S = 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S > σ
