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Marizza181 [45]
3 years ago
13

3) For the graph on the right, the graphical representation between speed and mass could best be described as: A. The speed of f

all is directly proportional to the mass. B. The speed of the fall is proportional to the square root of the mass. C. The speed of fall is inversely proportional to the mass. D. The speed of the fall is proportional to the square of the mass. E. There is no relationship between the speed of fall and mass. speed of fall mass
​
Physics
1 answer:
Vitek1552 [10]3 years ago
8 0

Answer:

B

Explanation:

KE=0.5mv^2

The speed is inversely proportional to the square root of mass

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An adiabatic closed system is accelerated from 0 m/s to 34 m/s. Determine the specific energy change of this system, in kJ/kg.
prohojiy [21]

Answer:

Δe=0.578 kJ/kg

Explanation:

Given data

Velocity v₁=0 m/s

Velocity v₂=34 m/s

to find

Specific energy change Δe

Solution

The specific energy change is simply determined from change in velocity

Δe=(v₂²-v₁²)/2

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=(\frac{(34)^{2} *10^{-3} }{2} )\\=0.578kJ/kg

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A 28.0 kg child plays on a swing having support ropes that are 2.30 m long. A friend pulls her back until the ropes are 45.0 ∘ f
Dimas [21]

Answer

A)184.9J

B)=3.63m/s

C) Zero

Explanation:

A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

U=Mgh

Where m=28kg

g= 9.8m/s

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Her Potential Energy will now

= 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

where

m = 28.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

V=√2k/m

V=√(2×184.9/28

=3.63m/s

C)we can find work done by the tension in the rope is given using expresion below

W= Tdcosx

where W= work done

T is the tension

d = displacement of the child

x= angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.

6 0
3 years ago
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