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Firlakuza [10]
2 years ago
6

Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.

Engineering
1 answer:
Liono4ka [1.6K]2 years ago
7 0

Answer: Eye injury

Explanation: small material such as dust, dirt, and metal shards can harm your eyes with potential blindness or infection.

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If a particle moving in a circular path of radius 5 m has a velocity function v = 4t2 m/s, what is the magnitude of its total ac
rjkz [21]

Answer:

8.62m/s²

Explanation:

the particle is experiencing both translational and circular motion

v=4t²

a_{t}=\frac{dv}{dt}=8t

at t=1s, \frac{dv}{dt}=8(1)=8m/s²

a_{c} = \frac{v^{2} }{r}

at t=1, v= 4(1)² = 4m/s

a_{c}=4²/5

a_{c}=3.2m/s²

∴ magnitude of total acceleration, a

a=\sqrt{a_{t} ^{2} + a_{c} {2} }

a=\sqrt{8^{2} +3.2^{2}  }

a=\sqrt{64+10.24}

a=\sqrt{74.24}

a=8.62m/s²

5 0
3 years ago
Read 2 more answers
There is a proposal in Brooklyn to construct a new mid-rise apartment building on a vacant lot at the intersection of Avenue A a
Soloha48 [4]

Answer:

a. Park should be adjacent to 48th Street, b. Difference in noise level = 707dBa, c. Yes

Explanation:

Data given for Avenue A

Cars = 496, Medium Truck = 52, Heavy Truck = 19, Buses = 10

Data given for 48th Street

Cars = 822, Medium Truck = 22, Heavy Truck = 8, Buses = 3

Consider the PCEs to be Cars = 1, Medium Truck = 13, Heavy Truck = 47, Buses = 18

a. Noise Level = Number of vehicles x PCE

For Avenue A

Noise level = (496 x 1) + (52 x 13) + (19 x 47) + (10 x 18) = 2245dBa

For 48th Street

Noise level = (822 x 1) + (22 x 13  + (8 x 47) + (3 x 18) = 1538dBa

The park should adjacent to 48th street as it is quieter than Avenue A

b. Let the Setback be 50ft. We know that the reduction of noise for 100ft = 5-8 dBa, hence

For Avenue A Noise Reduction due to 50 ft = (8/100) x 50 = 4dBa

Noise at Setback distance = 2245 - 4 = 2241dBa

Considering the same setback the noise at 48th street would be = 1538 - 4 = 1534 dBa

The difference is noise level between the two sides would be = 2241 - 1534 = 707 dBa

c. Yes the developer concerns are valid because there is a clear difference in noise levels of the two sites. This can be seen even after setting the same Setback. Locating the park next to Avenue A will cause serious noise problems.

8 0
4 years ago
The collar A, having a mass of 0.75 kg is attached to a spring having a stiffness of k = 200 N/m . When rod BC rotates about the
gladu [14]

Answer:

Speed=1.633 m/s

Force= 20 N

Explanation:

Ideally, v^{2}=\frac {ks^{2}}{m} hence v=s\sqrt {\frac {k}{m}} where v is the speed of collar, m is the mass of collar, k is spring constant and s is the displacement.

In this case, s=100-0=100mm=0.1m since 1 m is equivalent to 1000mm

k is given as 200 N/m and mass is 0.75 Kg

Substituting the given values

v=0.1 m\sqrt \frac {200 N/m}{0.75 Kg}=1.632993162 m/s\approx 1.633 m/s

Therefore, <u>the speed is 1.633 m/s</u>

The sum of vertical forces is given by mg where g is acceleration due to gravity and it's value taken as 9.81 m/s^{2}

Therefore, F_y=0.75\times 9.81=7.3575 N\approx 7.36 N

The sum of forces in normal direction is given by Ma_n=Ks therefore

Ma_n=200*0.1=20 N

Therefore, <u>normal force on the rod is 20 N</u>

5 0
3 years ago
Describe the make-up of an internal combustion engine.<br> Pls answer quickly.
Whitepunk [10]

Answer:

The engine consists of a fixed cylinder and a moving piston. The expanding combustion gases push the piston, which in turn rotates the crankshaft. Ultimately, through a system of gears in the power-train, this motion drives the vehicle's wheels.

Explanation:

8 0
2 years ago
The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular veloc
madreJ [45]

Answer:

Option B

116 ft/s^{2}

Explanation:

\theta=2 rev=2(2\pi)=4\pi

\alpha \theta=0.5(\omega_f^{2}-\omega_i^{2})

\alpha (4\pi)= 0.5(\omega_f^{2}-\omega_i^{2})

\alpha (8\pi)= (\omega_f^{2}-\omega_i^{2})

(2) (8\pi)= (\omega_f^{2}-\omega_i^{2})

(2) (8\pi)= (\omega_f^{2}-4^{2})

\omega_f=8.14 rads/s

v=r\omega=1.75*8.14=14.245 ft/s

Centripetal acceleration =\omega_f^{2} r=8.14^{2}*1.75=115.95 ft/s^{2}

Tangential component=dr=2*1.75=3.5

Resultant=\sqrt{3.5^{2}+115.95^{2}}\approx 116 ft/s^{2}

5 0
3 years ago
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