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adelina 88 [10]
2 years ago
13

abin is doing work by lifting a bowling ball. Which statement could be made about the energy in this situation?

Physics
2 answers:
PtichkaEL [24]2 years ago
8 0
The statement that could be made about the energy in this situation would be :
It being transferred from his arms muscles to the ball.

The muscle contraction from his arms created a force that could be used to lift the ball up.<span />
bekas [8.4K]2 years ago
3 0

Answer:

b.

Explanation:

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The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, temperature on left side is 1.48 times the temperature on right

6 0
2 years ago
A balloon is at a height of 81m and is ascending upwards with a velocity of 12m/s. A body of 2kg weight is dropped from it. If g
kap26 [50]
I know you are Indian by your question, HC Verma class 9 or 11 !!

if you got any problem, comment !!

7 0
2 years ago
What is the net force required to give an automobile with a mass of 1,600 kg an acceleration of 4.5 m/s2
Paladinen [302]

Answer:

Fnet=7200 N

Explanation:

Fnet=mass x acceleration

mass= 1600kg acceleration=4.5m/s^2

8 0
2 years ago
A slender uniform rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considere
Nataliya [291]

Answer:

The correct answer is D    I_{30} /I_{50} =   1.5

Explanation:

In this exercise the moment of inertia equation should be used

    I = ∫ r² dm

In addition to the parallel axis theorem

    I = I_{cm} + M D²

Where  I_{cm} is the moment of the center of mass, M is the total mass of the body and D the distance from this point to the axis of interest

Let's apply these relationships to our problem, the center of mass of a uniform rod coincides with its geometric center, in this case the rod is 1 m long, so the center of mass is in

    L = 100.00 cm (1m / 100 cm) = 1.0000 m

     x_{cm} = 50 cm = 0.50 m

Let's calculate the moment of inertia for this point, suppose the rod is on the x-axis and use the concept of linear density

    λ = M / L = dm / dx

    dm =  λ dx

Let's replace in the moment of inertia equation

    I = ∫ x² ( λ dx)

We integrate

    I =  λ x³ / 3

We evaluate between the lower limits x = -L/2 to the upper limit x = L/2

    I =  λ/3 [(L/2)³ - (-L/2)³] = lam/3  [L³/8 - (-L³ / 8)]

    I =  λ/3  L³/4

    I = 1/12  λ L³

Let's replace the linear density with its value

    I = 1/12  (M/L)  L³

    I = 1/12  M L²

Let's calculate with the given values

   I = 1/12  M 1²

   I = 1/12 M

This point is the center of mass of the rod

    Icm = I = 1/12 M  = 8.333 10-2 M

Now let's use the parallel axis theorem to calculate the moment of resection of the new axis, which is 0.30 m from one end, in this case the distance is

    D = x_{cm} - x

    D = 0.50 - 0.30

    D = 0.20  m

Let's calculate

   I_{30} =I_{cm} + M D²

   I_{30} = 1/12 M + M 0.202

   I_{30} = M (1/12 + 0.04)

   I_{30} = M 0.123

To find the relationship between the two moments of inertia, divide the quantities

  I_{30} / I_{50} = M 0.123 / (M 8.3 10-2)

   I_{30} /I_{50} = 1.48

The correct answer is d 1.5

6 0
2 years ago
g Suppose Howard is pulling a bucket of bricks up along the side of a building with a rope. The bricks have a mass of 20 kg and
cupoosta [38]

Answer:

= 236N

Explanation:

tension T = mg + ma

Given that,

m = 20kg

g = 9.8 m/s²

a = 2.0 m/s²

T = m(g + a)

T = 20( 9.8 + 2.0)

  = 20(11.8)

  = 236N

4 0
2 years ago
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