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natta225 [31]
3 years ago
6

Two forces, one of which double the other has resultant of 280N. if the direction of the large force is reversed and the remaini

ng unaltered, the resultant reduces to 160N. determine the magnitude of the force and the angle between the forces.
A body weighing 210N is resting on a horizontal table.A pull of 70 applied at an angle of 21 with the horizontal just causes the body to slide over the table. calculate the normal reaction and the coefficient of friction.​
Engineering
1 answer:
White raven [17]3 years ago
4 0

Answer:

point is 16 N.If their resultant is normal to the smaller force and has a magnitude of 8 N. Then two forces are.

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A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi
raketka [301]

Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation =\frac{N_1 +N_2}{2}

Nm = \frac{500+750}{2} = 625 rpm

w =\frac{2\pi Nm}{60}

  =65.44 rad/sec

F_T = mw^2 e

F_T = mew^2

       = 0.1*(65.44)^2

F_T =428.36 N

Transmission ratio =\frac{300}{428.36} = 0.7

also

transmission ratio = \frac{1}{[\frac{w}{w_n}]^{2} -1}

0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}

SOLVING FOR Wn

Wn = 42 rad/sec

Wn = \sqrt {\frac{k}{m}

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

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3 0
3 years ago
Determine the hydraulic radius for the following rectangular open channel width =23m water depth =3m
Romashka-Z-Leto [24]

Answer:

2.379m

Explanation:

The width = 23m

The depth = 3m

The radius is denoted as R

The wetted area is = A

The perimeter perimeter = P

Hydraulic radius

R = A/P

The area of a rectangular channel

= Width multiplied by Depth

A = 23x3

A = 69m²

Perimeter = (2x3)+23

P = 6+23

P= 29

Hydraulic radius R = 69/29

= 2.379m

This answers the question

Thank you!

8 0
3 years ago
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