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Tema [17]
3 years ago
15

A car increases its speed at a constant rate of 1.5 m/s2 as it rounds a curve of radius 200 m. The magnitude of the total accele

ration is 2.5 m/s2 . What is the car’s speed at this instant?

Engineering
1 answer:
vladimir2022 [97]3 years ago
7 0

Answer:

The car's instant speed is 20m/s

Explanation:

a = at + an

The above equation is a vector equation

a is the resultant acceleration

at   is the tangent acceleration

an  is the normal acceleration

See the attached diagram for solution

<h2 />

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Which of the following factors does not promote safety in the shop?
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3 years ago
A motor is mounted on a platform that is observed to vibrate excessively at an operating speed of 6000 rpm producing a 250-N for
vichka [17]

Answer:

The amplitude of the absorbed mass can be found

for ka:

X_{a} =0.002m=\frac{F_{0} }{K_{a} } =\frac{250}{K_{a} } =125000N/m

now

w^2=\frac{K_{a} }{m_{a} } \\m_{a} =\frac{K_{a} }{w^2} =\frac{125000}{[6000*2\pi /60]^2} =0.317kg

4 0
3 years ago
Use the concept that y = c, −[infinity] &lt; x &lt; [infinity], is a constant function if and only if y' = 0 to determine whethe
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attached below

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5 0
3 years ago
A spring-mass-damper instrument is employed for acceleration measurements. The spring constant is 12000 N/m. The mass is 5 g. Th
shepuryov [24]

Answer:

a) 246.56 Hz

b) 203.313 Hz

c) Add more springs

Explanation:

Spring constant = 12000 N/m

mass = 5g = 5 * 10^-3 kg

damping ratio = 0.4

<u>a) Calculate Natural frequency </u>

Wn = √k/m = \sqrt{12000 /  5*10^{-3}  }

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<u>b) Bandwidth of instrument </u>

W / Wn = \sqrt{1-2(0.4)^2}

W / Wn = 0.8246

therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz

C ) To increase the bandwidth we have to add more springs

5 0
3 years ago
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