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postnew [5]
3 years ago
12

A uniform beam of arbitrary, unsymmetrical cross-section and length 2???? is built-in at one end and simply supported in the ver

tical direction at a point halfway along its length. This support, however, allows the beam to deflect freely in the horizontal x direction. For a vertical load W applied at the free end of the beam, calculate bending moments for the sections at A and B.
Physics
1 answer:
leonid [27]3 years ago
5 0

Answer:

 the bending moment will be W from either sides

Explanation:

bending moment= force (load) * perpendicular distance, if I understand the question the distance will be 1/2 of the length

=> f x 1/2(l) =W*1/2(2) =W

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A scuba diver and her gear displace a volume of 68.5 L and have a total mass of 71.8 kg . Part A What is the buoyant force on th
Luba_88 [7]

Answer:

A) Fb = 671.3 N

B) The diver will sink.

Explanation:

A)

The buoyant force applied on an object by a fluid is given by the following formula:

Fb = Vρg

where,

Fb = Buoyant Force = ?

V = Volume of the water displaced by the object = 68.5 L = 0.0685 m³

ρ = Density of Water = 1000 kg/m³

g = 9.8 m/s²

Therefore,

Fb = (0.0685 m³)(1000 kg/m³)(9.8 m/s²)

<u>Fb = 671.3 N</u>

B)

Now, in order to find out whether the diver sinks or float, we need to find weight of the diver with gear.

W = mg = (71.8 kg)(9.8 m/s²)

W = 703.64 N

Since, W > Fb. Therefore, the downward force of weight will make the diver sink.

<u>The diver will sink.</u>

8 0
3 years ago
Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?.
natita [175]

For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s   is mathematically given as

F= 618.9 N

<h3>What is the centripetal force?</h3>

Generally, the equation for the angular speed  is mathematically given as

w = v/R

Therefore

w= 4.7/1.8

w= 2.611 rad/s

Where total momentum

Tm= 642.96 + 272.32

Tm= 915.28

and total inertia

Ti= 184 + 246.24

Ti= 430.24

In conclusion, centripetal force

F= mrw^2

F = m*R*w2^2

F = 76*1.8*2.127^2

F= 618.9 N

Read more about mass

brainly.com/question/15959704

CQ

Flag

a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?

3 0
3 years ago
PLEASE HELP ME! IM TIMED
viktelen [127]
The last answer I think
3 0
3 years ago
A specimen of aluminum having a rectangular cross section 9.5 mm × 12.9 mm (0.3740 in. × 0.5079 in.) is pulled in tension with 3
ryzh [129]

Answer:

The resultant strain in the aluminum specimen is 4.14 \times {10^{ - 3}}

Explanation:

Given that,

Dimension of specimen of aluminium, 9.5 mm × 12.9 mm

Area of cross section of aluminium specimen,

A=9.5\times 12.9=123.84\times 10^{-6}\ mm^2A=122.55\times 10^{-6}\ m^2

Tension acting on object, T = 35000 N

The elastic modulus for aluminum is,E=69\ GPa=69\times 10^9\ Pa

The stress acting on material is proportional to the strain. Its formula is given by :

\epsilon=\dfrac{\sigma}{E}

\sigma is the stress

\epsilon=\dfrac{F}{EA}\\\epsilon=\dfrac{35000}{69\times 10^9\times 122.5\times 10^{-6}}\\\epsilon=4.14\times 10^{-3}

Thus, The resultant strain in the aluminum specimen is 4.14 \times {10^{ - 3}}

5 0
4 years ago
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An archer shoots an arrow into a target using a bow. If the "action" force is the force of the bowstring against the arrow, what
aliya0001 [1]
Weight of an arrow is 20 plus 3
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3 years ago
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