All categories

2 years ago

What is first and second conditions of equilibrium?

1 answer:

7 0

I think the answer is the first condition sum of forces acting on a body is zero ( ∑ F =0 ) and the second condition sum of torque acting on a body is ( ∑ τ = 0 )

Could you put me as brainliest?

You might be interested in

A book is to be produced with pages of thickness 0.125mm.If the book is to be exactly 1cm thick what is the maximum number of pa

miskamm [114]

Answer:

1cm=10mm

10cm/0.125mm

80pages

5 0

2 years ago

PLS ANSWER DUE LATER TODAY !!!

Assoli18 [71]

Magna Carta. I’m pretty sure it’s the correct answer.

4 0

2 years ago

Read 2 more answers

Pls help me quickly ......

marishachu [46]

Answer:

the last one: weight force

3 0

3 years ago

Please help me with this question guys.

katen-ka-za [31]

Answer:

<em>The average speed is 22.2 km/h</em>

Explanation:

<u>Average Speed</u>

Given an object travels a total distance d and took a total time t, then the average speed is:

$\displaystyle \bar v=\frac{d}{t}$

The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:

$\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h$

Then he drives d2=7 km at v2=43 km/h taking a time of:

$\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h$

The total time is

t=0.467 h + 0.163 h = 0.63 h

The total distance is

d = 7 km + 7 km = 14 km

The average speed is:

$\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h$

The average speed is 22.2 km/h

7 0

2 years ago

On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs

user100 [1]

Answer:

A) $\Delta P = 14512.5 Pa = 14.512 kPa$

B) F = 1632.65 N

Explanation:

Given details

outside air speed is given as $v_2 = 150 m/s$

since inside air is atmospheric , $v_1 = 0 m/s$

a) By using bernoulli equation between outside and inside of flight

$P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh$

$\Delta P = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$

$\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]$

$\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]$

$\Delta P = 14512.5 Pa = 14.512 kPa$

b) force exerted on window

Area of window $= 25\times 45 = 1125 cm^2 = 0.1125 m^2$

We know that force is given as

$F = P\times A$

$F = 14512.5 \times 0.1125$

F = 1632.65 N

5 0

3 years ago