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madam [21]
3 years ago
5

The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (Th

is arrangement is called a double pendulum.) The counterclockwise couple C0, having moment 150????⋅m, is applied to B2beginning at t=0. Find the angular accelerations of B1and B2upon application of the couple, and the force exerted on B2at P.
Physics
1 answer:
Bezzdna [24]3 years ago
3 0

Answer:

hello the diagram relating to this question is attached below

a) angular accelerations : B1 = 180 rad/sec,  B2 = 1080 rad/sec

b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>

at point P ; Co = I* ∝B2'

                150  = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

at point 0 ; Co = Inet * ∝B1

                  150 = [ (2*0.5^2) / 3  + (2*0.5^2) / 3  + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

hence angular acceleration of B1 =  180 rad/sec

<u>b) Determine the force exerted on B2 at P</u>

T2 = mB1g + T1  -------- ( 1 )

where ; T1 = mB2g  ( at point p )

                 = 2 * 9.81 = 19.6 N

back to equation 1

T2 = (2 * 9.8 ) + 19.6 = 39.2 N

<u />

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8 0
2 years ago
Convert 123,453 to a scientific notation
Natalija [7]

Answer:

1.23453*10^5 is scientific way

3 0
2 years ago
A gray kangaroo can bound across level ground with each jump carrying it 9.6 m from the takeoff point. Typically the kangaroo le
xz_007 [3.2K]

Answer:

(A) 11 m/s

(B) 1.3 m

Explanation:

Horizontal range, R = 9.6 m

Angle of projection, theta = 28 degree

(A)

Use the formula of horizontal range

R = u^2 Sin 2 theta / g

u^2 = R g / Sin 2 theta

u^2 = 9.6 × 9.8 / Sin ( 2 × 28)

u = 10.65 m/s

u = 11 m/s

(B)

Use the formula for maximum height

H = u^2 Sin ^2 theta / 2g

H =

10.65 × 10.65 × Sin^2 (28) / ( 2 × 9.8)

H = 1.275 m

H = 1 .3 m

4 0
2 years ago
A bowling ball weighing 71.2 N is attached to the ceiling by a 3.30 m rope. The ball is pulled to one side and released; it then
wolverine [178]

Answer:

a. The acceleration of the bowling ball is 9.5 m/s² toward the center

b.  The tension in the rope is 140.24 N

Explanation:

given information:

ball weight,  W = 71.2 N

the length of rope, R = 3.30 m

ball speed, v = 5.60 m/s

a. The acceleration of the bowling ball, α

α = \frac{v^{2} }{R}

where

α = the acceleration

v = the speed

R = radius

thus

α = \frac{v^{2} }{R}

  = \frac{5.60^{2} }{3.3}

  = 9.5 m/s² toward the center

b. The tension in the rope?

according to the Newton's second law

ΣF = m a

where

F = force

m = mass

a = acceleration

so,

ΣF = m a

T - W = m a

T = m a + W

  = (W a/g) + W

  = (71.2 x 9.5/9.8) + 71.2

  = 140.24 N

4 0
3 years ago
Where do magnetic fields occur?
Tresset [83]

Answer:

<h2>The Magnetosphere and Magnets</h2>

Explanation:

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According to the <em>National Geographic, </em><em>"Earth’s magnetic field dominates a region called the magnetosphere, which wraps around the planet and its atmosphere. Solar wind, charged particles from the sun, presses the magnetosphere against the Earth on the side facing the sun and stretches it into a teardrop shape on the shadow side. The magnetosphere protects the Earth from most of the particles, but some leak through it and become trapped. When particles from the solar wind hit atoms of gas in the upper atmosphere around the geomagnetic poles, they produce light displays called auroras. These auroras appear over places like Alaska, Canada and Scandinavia, where they are sometimes called “Northern Lights.” The “Southern Lights” can be seen in Antarctica and New Zealand. The magnetic field is the area around a magnet that has magnetic force. All magnets have north and south poles."</em>

--------------------------------------------

<em>Hope this helps! <3</em>

--------------------------------------------

5 0
3 years ago
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