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gogolik [260]
2 years ago
14

J. Ram travels 50 m on a straight road in 30 s. Hari covers the same distance

Physics
1 answer:
ira [324]2 years ago
6 0

Explanation:

Explanation:

In the given question, there are two conditions.

Ram traveled 50 m in 30 seconds in a straight road.

Hari covers the same distance in the road; that means 50 m in 45 seconds

The formula for work w=\frac{d}{t}w=td

Ram's work w=\frac{50}{30}=1.66666w=3050=1.66666

Hari's work w=\frac{50}{45}=1.111w=4550=1.111

Ram is slower than Hari and the velocity of Ram is slower than Hari.

Ram has done less work than Hari.

Hence, Hari has done more work and he has more power than Ram.

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A thin rod of length 0.75 m and mass 0.42 kg is suspended
MrRissso [65]

Answer:

a)  K = 0.63 J, b)  h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

         w² = \frac{m g d}{I}

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

              d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

              I = ⅓ m L²

we substitute

            w = \sqrt{\frac{mgL}{2}  \ \frac{1}{\frac{1}{3} mL^2} }

            w = \sqrt{\frac{3}{2}  \ \frac{g}{L} }

            w = \sqrt{ \frac{3}{2} \ \frac{9.8}{0.75}  }

            w = 4.427 rad / s

an oscillatory system is described by the expression

              θ = θ₀ cos (wt + Φ)

the angular velocity is

             w = dθ /dt

             w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

           w = 4.0 rad / s

the kinetic energy is

           K = ½ I w²

           K = ½ (⅓ m L²) w²

           K = 1/6 m L² w²

           K = 1/6 0.42 0.75² 4.0²

           K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

             Em₀ = K = ½ I w²

final point. Highest point

             Em_f = U = m g h

energy is conserved

             Em₀ = Em_f

             ½ I w² = m g h

             ½ (⅓ m L²) w² = m g h

             h = 1/6 L² w² / g

             h = 1/6 0.75² 4.0² / 9.8

             h = 0.153 m

5 0
2 years ago
In three sentences, please describe SIMPLE HARMONIC motion, and give two examples. Thank you! :-)
11Alexandr11 [23.1K]
A system that repeats to and from its mean or rest point. that executes harmonic motion. a few examples I've heard of are since the springtime a mass-spring system,a swing, simple pendulum, one more example is a steel ball rolling in a curved is this what you need or do you need three more sentences dish. to get S.H.M a body just displaced away from the resting position and of course then is released. the human body oscillates due to the reinforce that pulls it back do you need anything else answered on this and I'll answer it
3 0
3 years ago
How much potintial energy is in a closed system where the height of the object is 2 m, and the mass is 10 kg?
Lorico [155]

Answer:

D. 196 J

Explanation:

PE=mgh=10×9.8×2=196J

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The volume would be 287cm³. Multiply all the 3 numbers by each other
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