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3 years ago

HELP WILL MEDAL !!! if you give a inpproper statement you will be repoted

1 answer:

8 0

1: Weather has a somewhat unpredictable pattern.
2: They can misinterpret the pattern of weather that looks like snow.
3: If they predict it early then the weather can change making there interpretation wrong.

Hope that helps a little bit. Please don’t report me I tried.

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An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and

Svetradugi [14.3K]

Answer:

Total contraction on the Bar = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm

Diameter for aluminum bar = 40 mm

Hole diameter = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180 × 10³ N

Calculate the total contraction on the bar = ???

The relation used in calculating the contraction on the bar is:

$\delta L = \dfrac{P *L }{A*E}$

The relation used in calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Total contraction on the Bar = $\dfrac{180 *10^3 \N }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]$

Total contraction on the Bar = 2.117( 0.398 + 0.182)

Total contraction on the Bar = 2.117*(0.58)

Total contraction on the Bar = 1.22786 mm

5 0

4 years ago

Problem 1: Problem 1.58 in Young & Freedman A plane leaves the airport in Galisteo and flies 170 km at 68.0° east of north;

Maurinko [17]

Answer:

Explanation:

We shall convert the displacement in vector form using unit vector i and j

consider east as x axis and north as y axis

170 km at 68.0° east of north

D₁ = 170 sin68 i + 170cos 68 j

= 157 i + 63.68 j

230 km at 36.0° south of east

D₂ = 230 cos36 i - 230 sin 36 j

= 186 i - 135.2 j

Resultant Displacement = D₁ +D₂

= 157 i + 63.68 j + 186 i - 135.2 j

= 343 i - 71.52 j

Resultant magnitude

= √ ( 343² + 71.52²

= 350 km

Angle

= tan⁻¹ ( - 71.52 / 343 )

12⁰ south of east .

4 0

3 years ago

Problem 5 You are playing a game where you drop a coin into a water tank and try to land it on a target. You often find this gam

Dvinal [7]

Answer:

Option D: 1.5in in front of the target

Explanation:

The object distance is $y= 6in$ .

Because the surface is flat, the radius of curvature is infinity .

The incident index is $n_i=\frac{4}{3}$ and the transmitted index is $n_t= 1$ .

The single interface equation is $\frac{n_i}{y}+\frac{n_t}{y^i}=\frac{n_t-n_i}{r}$

Substituting the quantities given in the problem,

$\frac{\frac{4}{3}}{6in}+\frac{1}{y^i}= 0$

The image distance is then $y^i=-\frac{18}{4}in =-4.5in$

Therefore, the coin falls $1.5in$ in front of the target

6 0

3 years ago

A block with a mass of 31.8 kg is pushed on a frictionless

OlgaM077 [116]

Answer: Total work done on the block is 3670.5 Joules.

Step by step:

Work done:

$W = F.d.\cos\theta$

With F the force, d the displacement, and theta the angle of action (which is 0 since the block is pushed along the direction of displacement, and cos 0 = 1)

$W = F.d$

Given:

F = 75 N

m = 31.8 kg

Final velocity $v_f = 15.3 \frac{m}{s}$

In order to calculate the Work we need to determine the displacement, or distance the block travels. We can use the information about F and m to first figure out the acceleration:

$F = ma\\\implies a=\frac{F}{m}=\frac{75 N }{31.8 kg}\approx 2.36 \frac{m}{s^2}\\$

Now we can determine the displacement from the following formula:

$d = \frac{1}{2}a^2+v_0t+d_0$

Here, the initial displacement is 0 and initial velocity is also 0 (at rest):

$d = \frac{1}{2}at^2\\$

Now we still have "t" as unknown. But we are given one more bit of information from which this can be determined:

$v_f = a\cdot t_f\\\implies t_f = \frac{v_f}{a} = \frac{15.2 \frac{m}{s}}{2.36 \frac{m}{s^2}}\approx 6.44 s$

(using vf as final velocity, and tf as final time)

So it takes about 6.44 seconds for the block to move. This allows us to finally calculate the displacement:

$d = \frac{1}{2}at^2=\frac{1}{2}2.36 \frac{m}{s^2}\cdot 6.44^2 s^2 \approx 48.94 m$

and the corresponding work:

$W = F\cdot d=75 N\cdot 48.94 m =3670.5J$

7 0

3 years ago

Why do solar eclipses happen only during a new moon?

mixas84 [53]

Answer:

For a solar eclipse to occur the moon must be between the earth and the sun (the light of the sun is blocked)'

By definition this is a new moon (which occurs when the moon cannot be seen because it is located between the sun and the earth).

6 0

3 years ago