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3 years ago

12

3. What is the power of a derby race car that uses a 50 N force to travel 20 meters in 10 seconds?

1 answer:

makkiz [27]3 years ago

7 0

I think it is B. 100 W

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1. If the strength of the magnetic field at B is 3 units, the strength of the magnetic field at A is _____.

Ymorist [56]

<h3><u>Answer;</u></h3>

C. 12 units

<h3><u>Explanation;</u></h3>

If the strength of the magnetic field at B is 3 units, the strength of the magnetic field at A is 12 units
Magnetic field strength is one of two ways that the intensity of a magnetic field can be expressed.
<em><u>The strength of the field is inversely proportional to the square of the distance from the source. This means that If the distance between two points in magnetic filed is doubled the magnetic force between them will fall to a quarter of the initial value. </u></em>
<em><u>On the other hand, if the distance between two magnets is halved the magnetic force between them will increase to four times the initial value.</u></em>

6 0

3 years ago

Which actions are examples of conserving resources? Check all that apply.

uysha [10]

Answer:

1st, 2nd, and 4th

Explanation:

1st conserves gasoline/petroleum

2nd conserves electricity

4th conserves paper

6 0

3 years ago

Read 2 more answers

Which of the following is NOT a characteristic of noble gases?

Rudik [331]

Solid at room temperature

8 0

3 years ago

Read 2 more answers

A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part

Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

<u>Polar moment of Inertia</u>

$(I_p)s = \frac{\pi(0.55)4}2$

= 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

$T_{max} = T_{allow} \frac{I_p}{r}$

= $(14 \times 10^3) \times (\frac{0.14374}{0.55})$

= 3658.836 lb.in

= $\frac{3658.836}{12}$ lb.ft

= 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

$T_{max} = T_{allow}( \frac{Ip}{r})$

= $(14 \times10^3) \times ( \frac{0.13101}{0.55})$

= 3334.8 lb.in

= $(\frac{3334.8}{12} )$ lb.ft

= 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

$P_{max} = 2 \pi f_T$

= $2\pi(2.1) \times 304.9$

= 4023.061 lb. ft/s

= $(\frac{4023.061}{550})$ hp

= 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

$P _{max,t} = 2\pi f_T$

= $2\pi(2.1) \times 277.9$

= 3666.804 lb.ft /s

= $(\frac{3666.804}{550})$ hp

= 6.667 hp

7 0

3 years ago

NEVERMIND ANSWER FOR SOME POINTS

saw5 [17]

Answer:

Answer what?

Explanation:

5 0

3 years ago