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a_sh-v [17]
1 year ago
15

Arm abcd is pinned at b and undergoes reciprocating motion such that θ=(0.3 sin 4t) rad, where t is measured in seconds and the

argument for the sine is in radiaus. determine the largest speed of point a during the motion and the magnitude of the acceleration of point d at this instant.
Physics
1 answer:
storchak [24]1 year ago
8 0
<span>θ=0.3sin(4t)
w=0.3cost(4t)(4)=1.2cost(4t)
a=-4.8sin(4t)

cos4t max will always be 1 (refer to cos graph), for same reason, sin4t will always be 0

therefore, wmax=1.2rad/s
 
vAmax=r*w=250*1.2=300mm/s
(may be different if your picture/radius is from a different picture)

adt=a*r=200*-4.8sin(4t)=0 (sin(4t)=0)

adn=r*w^2=200*1.2^2=288

ad= square root of adt^2+adn^2 = 288mm/s^2</span>
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A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t
Reil [10]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

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Consider the static equilibrium diagram here. What is the angle F1 must make with the horizontal?
Nataly [62]

ANSWER


\theta=35\degree


EXPLANATION


Since the body is in equilibrium, total upward forces must equal total downward force.


Also the net horizontal forces acting on the body must be zero.



We need to resolve F_1 into vertical and horizontal components.



The horizontal component is


x=F_1\cos\theta.


The vertical component is


y=F_1\sin\theta.



Equating the up force to the downward forces gives,


F_1\sin\theta + 20N=60N.


This implies that,



F_1\sin\theta =60N-20N.




F_1\sin\theta=40N...eqn1




Also the horizontal forces must be equal.


F_1\cos\theta=57N...eqn2.



Dividing equation (1) by equation (2) gives,


\frac{F_1\sin\theta}{F_1\cos\theta}=\frac{40}{57}.


\Rightarrow \tan\theta=0.70175





\Rightarrow \theta=tan^{-1}(0.70175)


\Rightarrow \theta=35.0594.




Therefore the given angle that F_1 must make with the horizontal is approximately 35° to the nearest degree.

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