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Otrada [13]
3 years ago
8

A 88.6-kg wrecking ball hangs from a uniform heavy-duty chain having a mass of 26.9kg . (Use 9.80m/s2 for the gravitational acce

leration at the earth's surface.)
Part A

Find the maximum tension in the chain.

Tmax = N
Part B

Find the minimum tension in the chain.

Tmin = N
Part C

What is the tension at a point three-fourths of the way up from the bottom of the cha
Physics
1 answer:
Dafna11 [192]3 years ago
6 0

Answer:

Tension maximum =1131.9 N

Tension minimum =868.28 N

Tension at 3/4= 1065.995 N

Explanation:

a)

Given Mass of wrecking ball M1=88.6 Kg

Mass of the chain M2=26.9 Kg

Maximum Tension Tension max=(M1+M2) × (9.8 m/s²)

=(88.6+26.9) × (9.8 m/s²)

=115.5 × 9.8 m/s²

Tension maximum =1131.9 N

b)

Minimum Tension Tension minimum=Mass of the wrecking ball only × 9.8 m/s²

=88.6 × 9.8 m/s²

Tension minimum =868.28 N

c)

Tension at 3/4 from the bottom of the chain =In this part you have to use 75% of the chain so you have to take 3/4 of 26.9

= (3/4 × 26.9)+88.9) × 9.8 m/s²

= (20.175+88.6) × 9.8 m/s²

=(108.775) × 9.8 m/s²

=1065.995 N

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Upthrust given by Water to the body
erica [24]

Answer:

According to Archimedes principle the upthrust on the body is equal to the weight of the water displaced by the body. ... Here, the mass would be the net difference in the weight of the object.

7 0
2 years ago
A 1.31 kg object is attached to a horizontal spring of force constant 2.70 N/cm and is started oscillating by pulling it 6.20 cm
goldfiish [28.3K]

Answer:

The answer is below

Explanation:

a) The change in energy is the difference between the final energy and the initial energy.

ΔE (energy change) = Ef (final energy) - Ei (initial energy)

\Delta E=\frac{1}{2}kA_f^2 -\frac{1}{2}kA_i^2\\\\k=force\ constant=2.7\ N/cm=270\ N/m, A_f=final\ dispalacment= 3.7\ cm=0.037\ m,\\ A_i=initial \ displacement = 6.2\ cm=0.062\ m\\\\Hence:\\\\\Delta E=\frac{1}{2}(270)(0.037)^2 -\frac{1}{2}(270)(0.062)^2\\\\\Delta E=-0.334 \ J

The negative sign shows that energy is lost to the environment. Hence 0.334 J is lost to the environment.

b) According to the law of conservation of energy, energy cannot be created or destroyed but transformed from one form to another.

The oscillating object loses energy due to wind resistance, friction between the spring and the object. Given that the air is frictionless, hence the energy loss is due to friction which is converted to heat.

6 0
3 years ago
The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top
vaieri [72.5K]

Answer:

v = 44,16 m/s

Explanation:

We will fixate our reference in the starting point from where Dan jumped of, at the top of the Casino. Therefore, the displacement made when dan reached the airbag would be of y= -99,4 m viewed from our reference. We describe the motion of dan with the equation:

v_y^2 =v_0^2 +2ay

Dan jumped from the rest, that means that the initial velocity v_0=0, therefore:

 v_y^2 =2ay \rightarrow v_y = \sqrt{2ay}

Since Dan is moving in the negative axis regarding our reference point, we take the negative root of the equation.

v_y=-√(2*(-9,81 m/s^2 )*(-99,4 m) )=44,1613 m/s  v_y =- \sqrt{2*(-9,81 m/s^2)*(-99,4 m)} = 44,1613 m

So, if we don’t take the air resistance into account, Dan would have achieved an velocity of 44,16 m/s when he reached the airbag.

I hope everything was clear with my explanation. If you need anything else, let me know. Have a great day :D

7 0
3 years ago
When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is
padilas [110]

Answer:

P_{C} = 3.2\, atm

Explanation:

Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:

Bulb A (2 L, 2 atm) - Before opening:

P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T

Bulb B (3 L, 4 atm) - Before opening:

P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T

Bulbs A & B (5 L) - After opening:

P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T

After some algebraic manipulation, a formula for final pressure is derived:

P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}

And final pressure is obtained:

P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}

P_{C} = 3.2\, atm

5 0
2 years ago
An ideal solenoid 20 cm long is wound with 5000 turns of very thin wire. What strength magnetic field is produced at the center
KIM [24]
<h2>Answer:</h2>

0.31425 Tesla

<h2>Explanation:</h2>

The magnetic field strength of a solenoid can be found by using the Ampere's law as follows;

BL = μ₀ x N x I      -------------------(i)

Where;

B = magnetic field strength

L = length of the solenoid

μ₀ = magnetic constant = 1.257 x 10⁻⁶H/m

N = number of turns in the coil of the solenoid

I = current flowing through the coil of the solenoid.

<em>From the question, </em>

L = 20cm = 0.2m

N = 5000 turns

I = 10A

<em>Substitute these values into equation (i) as follows;</em>

B x 0.2 = 1.257 x 10⁻⁶ x 5000 x 10

0.2B = 6.285 x 10⁻²

<em>Solve for B;</em>

B = 6.285 x 10⁻² / 0.2

B = 31.425 x 10⁻²

B = 0.31425 T

Therefore, the magnetic field strength is 0.31425 Tesla

7 0
3 years ago
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