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vlabodo [156]
3 years ago
7

The velocity of a point mass that moves along the s-axis is given by s' = 40 - 3t^2 m/s, where t is in seconds. Find displacemen

t of the particle from t = 2 s to t 6 s.
Engineering
1 answer:
strojnjashka [21]3 years ago
6 0

Answer:

The displacement is -48m.

Explanation:

Velocity is the rate of change of displacement. So, the instantaneous displacement is calculated by the integration of velocity function with respect to time. Displacement in range of time is calculated by integrating the velocity function with respect to time with in time range.  

Given:

Velocity along the s-axis is

s{}'=40-3t^{2}

time range is t=2s to t=6s.

Calculation:

Step1

Displacement in the time range t=2s to t=6s is calculated as follows:

\frac{\mathrm{d}s}{\mathrm{d}t}=40-3t^{2}

ds=(40-3t^{2})dt

Step2

Integrate the above equation with respect to time with the lower limit as 2 and upper limit as 6 as follows:

\int ds=\int_{2}^{6}(40-3t^{2})dt

s=(40\times6-40\times2)-3(\frac{1}{3})(6^{3}-2^{3})

s=160-208

s=-48m

Thus, the displacement is -48m.

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Explanation:

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How high a building could fire hoses effectively spray from the ground? Fire hose pressures are around 1 MPa. (It is also said t
Mrac [35]

Answer:

z_{2} = 91.640\,m

Explanation:

The phenomenon can be modelled after the Bernoulli's Principle, in which the sum of heads related to pressure and kinetic energy on ground level is equal to the head related to gravity.

\frac{P_{1}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}= z_{2}+\frac{P_{2}}{\rho\cdot g}

The velocity of water delivered by the fire hose is:

v_{1} = \frac{(300\,\frac{gal}{min} )\cdot(\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot (0.3\,m)^{2}}

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The maximum height is cleared in the Bernoulli's equation:

z_{2}= \frac{P_{1}-P_{2}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}

z_{2}= \frac{1\times 10^{6}\,Pa-101.325\times 10^{3}\,Pa}{(1000\,\frac{kg}{m^{3}} )\cdot(9.807\,\frac{m}{s^{2}} )} + \frac{(0.267\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}

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A homogeneous 800kg bar AB is supported at either end by a cable asshown in the figure
aleksandr82 [10.1K]

The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².

<h3>What is normal stress?</h3>

If the direction of deformation force is perpendicular to the cross-sectional area of ​​the body, the stress is called normal stress. Changes in wire length and body volume will be normal.

σ = P/A

Where, σ = Normal stress

P = Pressure

A = Area

1 Kg = 9.81 N

800 kg = 7848 N

Since the rod is half bronze and half steel

800 kg = 7848/2

= 3924 N

Pₙ = Fₙ = 3924 N                       [n = Bronze]

Pₓ =  3924 N                             [x = steel]

Given,

σₙ = 90MPa

σₓ = 120MPa

Aₙ = ?

Aₓ = ?

Aₙ = Pₙ/σₙ

Aₙ = 3924/90

Aₙ = 43.6 mm²

Aₓ = Pₓ/σₓ

Aₓ = 3924/120

Aₓ = 32.7 mm²

To know more about normal stress, visit:

brainly.com/question/28012990

#SPJ9

4 0
1 year ago
A water tank filled with solar-heated water at 40°C is to be used for showers in a field using gravity-driven flow.
Alenkasestr [34]

Answer:

yes sir

Explanation:

4 0
2 years ago
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