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svetlana [45]
3 years ago
14

A small sphere attached to a light rigid rod rotates about an axis perpendicular to and fixed to the other end of the rod. relat

ive to the positive direction of the axis of rotation, the angular positions of the sphere are positive, its angular velocity is positive, and its angular acceleration is negative. the sphere is
a.rotating clockwise and speeding up.
b.rotating clockwise and slowing down
c.rotating counterclockwise and speeding up
d.rotating counterclockwise and slowing down.
e.first rotating clockwise and then counterclockwise. ? a solid ylinder shown rotates about a fixed axis perpendicular to
Physics
1 answer:
Crank3 years ago
7 0
<span>d.rotating counterclockwise and slowing down This is a matter of understanding the notation and conventions of angular rotations. Positive rotations are counter clockwise and negative rotations are clockwise. An easy way to remember this is the "right hand rule". Make a closed fist with your right hand and have the thumb sticking outwards. If you orient your thumb such that it's pointing in the direction of the positive value along the axis, your fingers will be curled in the positive rotational direction. So in the described scenario, the sphere is rotating in the positive direction (counter clockwise) and decelerating due to the negative angular acceleration. That immediately indicates that options "a", "b", and "e" are wrong since they mention the sphere going clockwise at the beginning. Of the two remaining options "c" and "d", we can discard option "c" since it has the rotation speeding up, and that leaves us with option "d" where the sphere is rotating counter clockwise and slowing down.</span>
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3 years ago
A rowboat passenger uses an oar to push the boat off the dock by exerting a force of 40N for 3.0s. What impulse acts on the boat
jarptica [38.1K]

Answer:

120 Ns

Explanation:

The impulse exerted on an object is given by:

I=Ft

where

F is the force applied

t is the time taken

In this problem, we have:

F = 40 N

t = 3.0 s

So, the impulse acting on the boat is

I=Ft=(40 N)(3.0 s)=120 Ns

8 0
3 years ago
Question 2.
boyakko [2]

The tensile stress of the wire supporting 2 kg mass is determined as 6.1 x 10⁷ N/m².

<h3>Tensile stress of the wire</h3>

The tensile stress of the wire is calculated as follows;

σ = F/A

where;

  • A is area of the wire

A = πr² = πD²/4

where;

  • D is diameter = 0.64 mm

A = π x (0.64 x 10⁻³)²/4

A = 3.22 x 10⁻⁷ m²

σ = F/A = (mg)/A = (2 x 9.8)/( 3.22 x 10⁻⁷)

σ = 6.1 x 10⁷ N/m²

Learn  more about tensile stress here: brainly.com/question/25748369

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3 0
2 years ago
A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t
Jlenok [28]

Answer:

14.36 N

Explanation:

T_{1} = Tension in string 1

T_{2} = Tension in string 2

m_b = mass of the bar = 2.7 kg

W_b = weight of the bar

weight of the bar is given as

W_b = m_{b} g = (2.7) (9.8) = 26.46N

m_m = mass of the bar = 1.35 kg

W_m = weight of the monkey

weight of the monkey is given as

W_m = m_{m} g = (1.35) (9.8) = 13.23N

Using equilibrium of torque about left end

W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33 N

Using equilibrium of force in vertical direction

T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36 N

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3 years ago
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