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Gre4nikov [31]
3 years ago
5

Manufactured bolts can’t be too long or too short or they might not fit where they need to. Help determine which bolts will work

and which won’t. Show your work for all steps.
1. A machine makes bolts that are 12 mm long. The bolts are considered acceptable if they are within 0.2 mm of 12 mm.

a. What is the longest the bolt can be and still be acceptable?



b. What is the shortest the bolt can be and still be acceptable?
Physics
1 answer:
Sever21 [200]3 years ago
5 0
What is the longest the bolt can be and still be acceptable
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Define force ? ☹❣️️ Degine weight? ​
mylen [45]

Explanation:

force is anything that allows a body to be in motion..

6 0
3 years ago
If an object is accelerating at a rate of 8 m/s^2 what is the mass of the box. <br><br> See attached
Orlov [11]

Answer:

50 kg

Explanation:

fnet=ma

600-200=m8

divide both side by 8 to make m the subject of the formula Thus m=50kg

6 0
2 years ago
Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take
KiRa [710]

Answer:

\frac{V_{e}}{V_{h}}=0.428*10^{2}

Explanation:

From conservation of energy states that

K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}

5 0
3 years ago
Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co
aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m

The radius is 1.195 m

Time period would be given by

T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s

Time period of the motion is 2.8375 s

Angular speed is given by

\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s

The angular speed of the motion is 2.21433 rad/s

4 0
4 years ago
The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for
abruzzese [7]

Answer:

The answer is "\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}"

Explanation:

\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta    - \omega_{0}^{r} \\\\\to \omega^{r} = 2  (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}

5 0
3 years ago
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