All categories

2 years ago

How do you know which forces works for free bodies

Engineering

1 answer:

miss Akunina [59]2 years ago

5 0

Answer:

Gravitational force (pulled downward by the Earth)

Normal force (pushed upward by the ground)

Applied force (pushed by the person)

Friction force (pulled opposite the direction of motion by the roughness of the ground)

You might be interested in

How wold you classify the earliest examples of S.T.E.M discoveries provided in this lesson?

Gnoma [55]

Answer:

the answer is C

Explanation:

8 0

3 years ago

Can you solve this question

Alecsey [184]

Answer:

eojcjksjsososisjsiisisiiaodbjspbcpjsphcpjajosjjs ahahhahahahahahahahahahahahahhhahahahaahahhahahahahaahahahahaha

6 0

3 years ago

Read 2 more answers

Six housing subdivisions within a city area are target for emergency service by a centralized fire station. Where should the new

Marina86 [1]

Answer:

Explanation:

Since there are six points, the minimum distance from all points would be the centroid of polygon formed by A,B,C,D,E,F

To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.

$C_{x}=\frac{1}{6A} sum(({x_{i} +x_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))$ where i=1 to N-1 and N=6

$C_{y}=\frac{1}{6A} sum(({y_{i} +y_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))$

A area of the polygon can be found by the following formula $A=\frac{1}{2} sum(x_{i} y_{i+1} -x_{i+1} y_{i})$ where i=1 to N-1

$A=\frac{1}{2}[ (x_{1} y_{2} -x_{2} y_{1})+ (x_{2} y_{3} -x_{3} y_{2})+(x_{3} y_{4} -x_{4} y_{3})+(x_{4} y_{5} -x_{5} y_{4})+(x_{5} y_{6} -x_{6} y_{5})]$

A=0.5[(20×25 -25×15) +(25×32 -13×25)+(13×21 -4×32)+(4×8 -18×21)+(18×14 -25×8)

A=225.5 miles²

Now putting the value of area in Cx and Cy

$C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(x_{2}+x_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(x_{3}+x_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(x_{4}+x_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(x_{5}+x_{6})(x_{5} y_{6} -x_{6} y_{5})]]$

putting the values of x's and y's you will get

$C_{x} =15.36$

For Cy

$C_{y} =\frac{1}{6A}[ [(y_{1}+y_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(y_{2}+y_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(y_{3}+y_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(y_{4}+y_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(y_{5}+y_{6})(x_{5} y_{6} -x_{6} y_{5})]]$

putting the values of x's and y's you will get

$C_{y} =22.55$

So coordinates for the fire station should be (15.36,22.55)

5 0

3 years ago

Air enters the 1 m² inlet of an aircraft engine at 100 kPa and 20° C with a velocity of 180 m/s. Determine: a) The volumetric fl

Shkiper50 [21]

Answer:

a) 180 m³/s

b) 213.4 kg/s

Explanation:

$A_1$ = 1 m²

$P_1$ = 100 kPa

$V_1$ = 180 m/s

Flow rate

$Q=A_1V_1\\\Rightarrow Q=1\times 180\\\Rightarrow Q=180\ m^3/s$

Volumetric flow rate = 180 m³/s

Mass flow rate

$\dot{m}=\rho Q\\\Rightarrow \dot m=\frac{P_1}{RT} Q\\\Rightarrow \dot m=\frac{100000}{287\times 293.15}\times 180\\\Rightarrow \dotm=213.94\ kg/s$

Mass flow rate = 213.4 kg/s

3 0

3 years ago

The viscosity of a fluid is 10 be measured by a viscometer constructed of two 75-cm-long concentric cylinders. The outer diamete

a_sh-v [17]

Answer:

0.023 Pa*s

Explanation:

The surface area of the side of the inner cylinder is:

A = π*d*l

A = π*0.15*0.75 = 0.35 m^2

At 200 rpm the inner cylinder has a tangential speed of:

u = w * r

u = w * d/2

w = 200 rpm * 2π / 60 = 20.9 rad/s

u = 20.9 * 0.15 / 2 = 1.57 m/s

The torque is of 0.8 N*m, this means that the force is:

T = F * r

F = T / r

F = 2*T / d

For Newtoninan fluids with two plates moving respect of each other with a fluid between the viscous friction force would be:

F = μ*A*u / y

Where

μ: viscocity

y: separation between pates

A: surface area of the plates

Then:

2*T / d = μ*A*u/y

Rearranging:

μ = 2*T*y / (d*A*u)

μ = 2*0.8*0.0012 / (0.15*0.35*1.57) = 0.023 Pa*s

5 0

3 years ago

How do you know which forces works for free bodies​

How do you know which forces works for free bodies