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Lady bird [3.3K]
2 years ago
8

What force must be applied to the end of a rod along the x-axis of length 2.25 m in order to produce a torque on the rod about t

he origin of 11 z-hat N·m? Choose among the available options below. x,y,z,-z
Physics
1 answer:
4vir4ik [10]2 years ago
7 0

Answer:

<h2>4.9N approx.</h2>

Explanation:

Step one:

given data

length of rod r= 2.25m

Required torque = 11 Nm

Required

The force needed to produce a torque of 11Nm

Step two:

<em>"By definition, Torque is the twisting force that tends to cause rotation. </em>

<em>The point where the object rotates is known as the axis of rotation."</em>

Mathematically,

\tau = rF\sin\theta

\tau = torque

r = radius

F = force

\theta = angle between F and the lever arm

in this case is zero

\tau = rF\\\\F=\tau/r

substituting we have

F=11/2.25

F=4.88N

F=4.9N approx.

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After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 49.0 cm . The explorer finds that
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3 years ago
Potassium is a crucial element for the healthy operation of the human
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A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air res
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Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

mass of golf ball m = 46.8 g = 0.0468 kg

initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

so

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef  = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

v₂ = √( 3457.44 - 325.164 )  

v₂ = √( 3132.276 )

v₂ = 55.97 m/s

Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

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