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mash [69]
3 years ago
7

Suppose that a particular artillery piece has a range R = 9880 yards . Find its range in miles. Use the facts that 1mile=5280ft

and 3ft=1yard.
Physics
1 answer:
Solnce55 [7]3 years ago
6 0

Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

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Which of the following BEST explains why stability is an important skill for athletes?
Irina-Kira [14]

Answer;

B. Without good stability, athletes are too clumsy to perform well.

Explanation;

Stability refers to the resistance to both linear and angular acceleration, or resistance to disruption of equilibrium.

Developing and enhancing core strength and stability helps athletes to maximize their power output and enhance game day performance. Additionally, Improving core strength and stability can also help athletes reduce their risk of injury.

8 0
3 years ago
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If an object has the same number of positive and negative changes, it’s electrical charge is ?
Marrrta [24]

The electrical charge is neutral. I hope this helps!

8 0
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What is meant by the term 'total internal reflection'? (GCSE Level)
scZoUnD [109]

Explanation:

Total Internal Reflection (TIR) is a phenomenon in optics, by which light experiences complete reflection at an interface between two media.

In the figure i > © experience total internal reflection

6 0
2 years ago
Estimate the inductance L of a coil that is 12 cm long, made of about 235 copper-wire turns and a diameter of about 1.7 cm. Show
ANTONII [103]

Answer:

Inductance as calculated is 13.12 mH

Solution:

As per the question:

Length of the coil, l = 12 cm = 0.12 m

Diameter, d = 1.7 cm = 0.017 m

No. of turns, N = 235

Now,

Area of cross-section of the wire, A = \frac{\pi d^{2}}{4} = \frac{\pi \times 0.017^{2}}{4} = 2.269\times 10^{- 4}\ m^{2}

We know that the inductance of the coil is given by the formula:

L = \frac{mu_{o}AN^{2}}{l} = \frac{4\pi \times 10^{- 7}\times 2.269\times 10^{- 4}\times 235^{2}}{0.12} = 1.312\times 10^{- 4}\ H = 13.12\ mH

4 0
3 years ago
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit
vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

\omega =\sqrt{\dfrac{K}{m}}

The maximum speed v given as

v=\omega A

A=Amplitude

v=\sqrt{\dfrac{K}{m}}\times A

0.32=\sqrt{\dfrac{13.5}{0.75}}\times A

A=0.075 m

A= 0.75 cm

The speed at distance x

v=\omega \sqrt{A^2-x^2}

v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}

v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}

v= 0.21 m/s

5 0
3 years ago
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