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Alex17521 [72]
3 years ago
7

If a person runs a distance of 0.7 km in 3 min, what is his average speed in kilometres/hour ​

Engineering
1 answer:
SVEN [57.7K]3 years ago
8 0

Answer:

14 km/hour

Explanation:

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The bulk modulus of a material is 3.5 ✕ 1011 N/m2. What percent fractional change in volume does a piece of this material underg
kiruha [24]

Answer:

percentage change in volume  = 0.00285 %

Explanation:

given data

bulk modulus = 3.5 × 10^{11}  N/m²

bulk stress = 10^{7}  N/m²

solution

we will apply here bulk modulus formula that is

bulk modulus = \frac{bulk\ stress}{bulk\ strain}   ...............1

put here value and we get

3.5 × 10^{11} = \frac{10^7}{bulk\ strain}  

solve it we get

bulk strain = 2.8571 × 10^{-5}

and

bulk strain = \frac{change\ volume}{original\ volume}  

so that percentage change in volume is = 2.8571 × 10^{-5}  × 100

percentage change in volume  = 0.00285 %

6 0
4 years ago
Two players find themselves in a legal battle over a patent. The patent is worth 20 for each player, so the winner would receive
Alborosie

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

5 0
3 years ago
A coil with an average diameter of 5 inch will have an area of ""blank"" square meters
nadezda [96]

Answer:

19.64 square inches

Explanation:

Area will be (¶d^2)/4

= (3.142 x 5^2)/4

= 19.64 square inches

8 0
3 years ago
The end of a large tubular workpart is to be faced on a NC vertical boring mill. The part has an outside diameter of 38.0 in and
nata0808 [166]

Answer:

(a) the cutting time to complete the facing operation = 11.667mins

b) the cutting speeds and metal removal rates at the beginning= 12.89in³/min and end of the cut. = 8.143in³/min

Explanation:

check attached files below for answer.

5 0
3 years ago
A small lake with volume of 160,000 m^3 receives agricultural drainage waters that contain 150 mg / L total dissolved solids (TD
Stels [109]

Answer:

Explanation:

Given that : -

The desirable limit is 500 mg / l , but

allowable upto 2000 mg / l.

The take volume is V = 160.000 m3

V = 160 , 000 x 103 l

The crainage gives 150 mg / l and lake has initialy 100 mg / l

Code of tpr frpm drawn = 150 x 60, 000 x 1000

Ci = 9000 kg / gr

Cl = 100 x 160,000 x 1000

Cl = 16, 000 kg

Since allowable limit = 2000 mg / l

Cn = ( 2000 x 160, 00 x 1000 )

= 320, 000 kg

so, each year the rate increases, by 9000 kg / yr

Read level = ( 320, 000 - 16,000 )

Li = 304, 000 kg

Tr=<u>304,000</u>

      900

=33.77

5 0
3 years ago
Read 2 more answers
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