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EastWind [94]
3 years ago
9

A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hang

ing vertically, the mass is pulled aside a small distance of 7.6 cm and released from rest. While the mass is swinging the cord exerts an almost-constant force on it. For this problem, assume the force is constant as the mass swings. How much work in J does the cord do to the mass as the mass swings a distance of 8.0 cm
Physics
1 answer:
valina [46]3 years ago
6 0

Answer:

work done is -2.8  × 10⁻⁶ J

Explanation:

Given the data in the question;

mass of the pendulum m = 6 kg

Length of core = 1.7 m

Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad

In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.

Now, the required work done will be;

W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta

W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta

W = -mgl  \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta

W = -mgl[ -cosθ ]^{0.047}_{0.045 }

W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]

W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]

W = -2.8  × 10⁻⁶ J

Therefore, work done is -2.8  × 10⁻⁶ J

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A square piece of tin has 12 inches on a side. An open box is formed by cutting out equal square pieces at the corners and bendi
Citrus2011 [14]

Answer:

Explanation:

Given a square Piece whose side is 12 inches

Now square pieces are cut from each corner to make it a open box

Suppose x is the length of square piece at each corner

then

base square has a length of 12-2x

Dimension of new box is (12-2x)\times (12-2x)\times x

Volume V=(12-2x)\times (12-2x)\times x

V=\left ( 12-2x\right )^2\cdot x

For maximum volume differentiate with respect to x we get

\Rightarrow\frac{\mathrm{d} V}{\mathrm{d} x}=2\times \left ( 12-2x\right )\times \left ( -2\right )\cdot x+\left ( 12-2x\right )^2=0

we get x=6 and 4 but at x=6 volume becomes zero therefore x=4 is valid

V=\left ( 12-2\cdot 4\right )^2\cdot 4

V=4^3

V=64\ in.^3

6 0
3 years ago
A bird flies 3.7 meters in 46 seconds, what is its speed?
victus00 [196]

Answer:

Speed is 0.08 m/s.

Explanation:

Given the distance that the bird flies = 3.7 meters

The time is taken by the bird to fly the 3.7 meters = 46 seconds  

We have given distance and time. Now we have to find the speed at which the bird flies. So, to calculate the speed of the bird we have to divide the distance by the time.  

Below is the formula to find the speed.

Speed = Distance / Time

Now insert the given value in the formula.

Speed = 3.7 / 46 = 0.08 m/s

8 0
3 years ago
What is the impulse of a 3 kg object that starts from rest and moves to 20 m/s?
IrinaK [193]

Answer:

The impulse on the object is 60Ns.

Explanation:

Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.

F = m a

F = m(\frac{v_{2}  - v_{1} }{t})

⇒     Ft = m(v_{2} - v_{1})

where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object, v_{1} is its initialvelocity and v_{2} is the final velocity of the object.

Therefore,

impulse = Ft = m(v_{2} - v_{1})

From the question, m = 3kg, v_{1} = 0m/s and v_{2} = 20m/s.

So that,

Impulse = 3 (20 - 0)

             = 3(20)

             = 60Ns

The impulse on the object is 60Ns.

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3 years ago
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RSB [31]

Answer:

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Explanation:

7 0
3 years ago
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