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ASHA 777 [7]
3 years ago
5

FROM THE _____ WHOLE WATER CYCLE STARTS ALL OVER AGAIN

Physics
2 answers:
soldi70 [24.7K]3 years ago
6 0
Clouds? I am not sure of your options!
Natalka [10]3 years ago
6 0

From the <u>w</u><u>a</u><u>t</u><u>e</u><u>r</u> whole water cycle starts again.

  • Most possibly water should be the answer.
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7. What is the acceleration of the box?<br> a. 2.5 m/s2<br> b. 4 m/s2<br> c. 6 m/s2<br> d. 10 m/s2
faust18 [17]
This is your answer
b) 4m/s2
Here’s my work

6 0
3 years ago
Will give 45 points to who ever solves this
Lubov Fominskaja [6]

Answer:

1.  230 kg...

W = m * g

W= 230 kg  * 9.81 m/s^2

<u>W=  2256,3 N</u>

     m= 230kg ,   W = 2256,3 N , g= 9.81 m/s^2

2.  887 N

W= m * g

887 N = m * 9.81 m/s^2

<u>m= 90,42 kg</u>

     m= 90,42 kg,  W = 887 N ,  g= 9.81 m/s^2

3.  420 kg

W= m * a

w= 420 kg *  9.81 m/s^2

<u>w=</u><u>  </u><u>4120,2 N</u>

     m=  420 kg , W = 4120,2 N , g= 9.81 m/s^2

4. Determine the gravity on Pluto where a 15 kg object weighs 55.5N.

w = m * g

55.5 N = 15 kg * g

<u>a=  3,7 m/s^2</u>

     m = 15 kg , W= 55.5 N , g=  3,7 m/s^2

7 0
2 years ago
Read 2 more answers
A 1.0-in.-diameter hole is drilled on the centerline of a long, flat steel bar that is 1 2 thick and 4 in. wide. The bar is subj
Dominik [7]

Answer:

The answers are

The average stress = 20000 lb/in²

The maximum tensile stress immediately adjacent to the hole

= 31076.92 lb/in²

Explanation:

To solve the question we have

Weight of tensile load = 30,000 lb

Width of steel bar = 4 in

Thickness of steel bar = 1/2 in

Average Stress = Force/Area  

Size of hole drilled = 1.0 in diameter

Available width at cross section where the 1.00 in diameter hole is drilled =

(4 - 1) in = 3 inches

Cross sectional area at the point of reduced cross section due to the drilled hole = Width × Thickness (Since the item is a flat bar)

= 3 in × 1/2 in = 1.5 in²

Therefore Stress = (30000 lb)/(1.5 in²) = 20000 lb/in²

the maximum tensile stress immediately adjacent to the hole.

Bending stress = \sigma_B= \frac{M_y}{I} where I = \frac{(0.5^2 + 4^2)}{12}

0.5*30000/I = 11076.92 lb/in²

Max stress = 31076.92 lb/in²

8 0
3 years ago
Read 2 more answers
an aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. the hole is 30mm in diameter and i
zavuch27 [327]

Answer:

ΔL = 1.011 mm

Explanation:

Let's begin by listing out the given information:

Length (L) = 600 mm = 0.6 m,

Diameter (D) = 40 mm = 0.04 m ⇒ Radius (r) = 20 mm = 0.2 m,

Area (cross sectional) = πr² = 3.14 x .02² = 0.001256 m²,

Modulus of Elasticity (E) = 85 GN/m²,

Compressive load (F) = 180 KN

Using the formula, Stress = Load ÷ Area

Mathematically,

σ = F ÷ A = 180 x 10³ ÷ 0.001256

σ = <u>143312.1 KN/m</u>²

Modulus of elasticity = stress ÷ strain

E = σ ÷ ε

ε = ΔL/L

85 x 10⁹ = 143312.1 x 10³ ÷ (ΔL/L)

ΔL = 143312.1 x 10³ ÷ 85 X 10⁹ = 1686.02 * 10⁻⁶

ΔL = L x 1686.02 * 10⁻⁶

ΔL = 0.6 * 1686.02 * 10⁻⁶ = 1011.61 x 10⁻⁶

ΔL = 1.011 x 10⁻³ m

ΔL = <u>1.011 mm</u>

∴The bar contracts by 1.011 mm

4 0
3 years ago
Is this object showing acceleration for the first 2 seconds? explain your answer.
iVinArrow [24]
Yes. The line is increasing. The flat line at the top of the graph is where there is not acceleration and the decreasing line is deceleration.
3 0
3 years ago
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