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horrorfan [7]
3 years ago
5

The disk rotates about a fixed axis through point O with a clockwise angular velocity ω0 = 16 rad/s and a counterclockwise angul

ar acceleration α0 = 5.6 rad/s2 at the instant under consideration. The value of r is 280 mm. Pin A is fixed to the disk but slides freely within the slotted member BC. Determine the velocity and acceleration of A relative to slotted member BC and the angular velocity and angular acceleration of BC. The relative velocity and acceleration are positive if they point from B to C and negative if they point from C to B. The angluar velocity and angular accelerations are positive if counterclockwise, negative if clockwise.
Physics
1 answer:
Lapatulllka [165]3 years ago
3 0

Answer:

V = 4.48m/s a = 1.57m/s²

Explanation:

ω₀ = 16rad/s

α₀ = 5.6rad/s²

r = 280mm = 0.28m

a = ?

v = ?

Angular velocity (ω₀) = velocity of acceleration / length of path

ω₀ = v / r

V = ω₀ * r

V = 16 * 0.28

V = 4.48m/s

Acceleration = ?

Angular acceleration α₀ = angular velocity (ω) / time take (t)

α₀ = ω / t .... equation i

But acceleration (a) = velocity (v) / time (t)

a = v / t

t = v / a

Put t = v / a into equation i

α₀ = ω / (v / a)

α₀ = ω * a / v

α * v = ω * a

a = (α * v) / ω

a = (5.6 * 4.48) / 16

a = 1.568m/s²

a = 1.57m/s²

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A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m verticall
lorasvet [3.4K]

Explanation:

Given:

t = 20 seconds

x = 3000 m

y = 450 m

a) To find the vertical component of the initial velocity v_{0y}, we can use the equation

y = v_{0y}t - \frac{1}{2}gt^2

Solving for v_{0y},

v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}

\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}

\:\:\:\:\:\:\:=120.5\:\text{m/s}

b) We can solve for the horizontal component of the velocity v_{0x} as

x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}

or

v_{0x} = 150\:\text{m/s}

4 0
3 years ago
Part D- Isolating a Variable with a Coefficient In some cases, neither of the two equations in the system will contain a variabl
Shtirlitz [24]

Answer:

D)     D = \frac{5}{4}  - \frac{3}{4} \ C, E)  (C, D) = ( \frac{17}{7}, \ \frac{-4}{7}

Explanation:

Part D) two expressions are indicated

          3C + 4D = 5

          2C +5 D = 2

let's simplify each expression

         3C + 4D = 5

         4D = 5 - 3C

we divide by 4

            D = \frac{5}{4}  - \frac{3}{4} \ C

The other expression

       2C +5 D = 2

       2C = 2 - 5D

        C = 1 -  \frac{5}{2} \ D

we can see that the correct result is 1

Part E.

It is asked to solve the problem by the substitution method, we already have

          D =  \frac{5}{4}  - \frac{3}{4} \ C

we substitute in the other equation

            2C +5 D = 2

             2C +5 (5/4 - ¾ C) = 2

we solve

            C (2 - 15/4) + 25/4 = 2

             -7 / 4 C = 2 - 25/4

             -7 / 4 C = -17/4

              7C = 17

               C = \frac{17}{7}

now we calculate D

               D = \frac{5}{4} - \frac{3}{4} \ \frac{17}{7}

               D = 5/4 - 51/28

               D =\frac{35-51}{28}

               D = - 16/28

               D = - \frac{4}{7}

the result is (C, D) = ( \frac{17}{7}, \ \frac{-4}{7} )

8 0
3 years ago
The strength of the electric field 0.5 m from a 6 uC charge is
Triss [41]

Answer:

E= 2.158× 10*5N/C

Explanation:

K= 8.99×10*9, q= 6×10*-6C, d= 0.5m

E= kq/d*2

E= (8.99×10*9× 6×10*-6)/0.5*2

E= 215760

E= 2.158 ×10*5N/C

5 0
3 years ago
An athlete does one push-up period in the process, she moves half of her body weight, 250 N, a distance of 20 cm, this distance
ioda
In the push-up, the athlete's weight  (or force) of 250 N moves through a distance of 20 cm.

By definition,
Work = force * distance.

Therefore, work done in one push-up is
W=(250 \, N)*(20 \, cm)*(10^{-2} \,  \frac{cm}{m}) = 50 \, J

Note that 1 J = 1  N-m.

Answer:  1 J
3 0
4 years ago
What happens when a mechanical wave travels through a medium?
lesantik [10]
Your answer would be Option A. Energy is transferred.

I just took this quiz, and can 100% confirm this is the answer.

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Hope this helps! :) 


7 0
3 years ago
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