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Fynjy0 [20]
3 years ago
5

A block with mass 0.470 kg sits at rest on a light but not long vertical spring that has spring constant 85.0 N/m and one end on

the floor. A second identical block is dropped onto the first from a height of 4.40 m above the first block and sticks to it. What is the maximum elastic potential energy stored in the spring during the motion of the blocks after the collision?
Physics
1 answer:
a_sh-v [17]3 years ago
4 0

Answer: elastic potential energy = 20.27 J

Explanation:

Given that the

Mass M = 0.470 kg

Height h = 4.40 m

Spring constant K = 85 N/m

The maximum elastic potential will be equal to the maximum kinetic energy experienced by the block.

But according to conservative of energy, the maximum kinetic energy is equal to the maximum potential energy experienced by the block of mass M.

That is

K .E = P.E = mgh

Where g = 9.8m/s^2

Substitutes all the parameters into the formula

K.E = 0.470 × 9.8 × 4.4

K.E = 20.27 J

Where K.E = maximum elastic potential energy stored in the spring during the motion of the blocks after the collision which is 20.27J.

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solution:

radius of steel ball(r)=5cm=0.05m

density of ball =8000kgm

terminal velocity(v)=25m/s^2

density of air( d) =1.29 kgm

now

volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3

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or, 8000=m/0.00052

or, m=4.16 kg

weight of the ball (W)= mg=4.16×10=41.6 N

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=6×3.14×eta×0.05×25

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To attain the terminal velocity,

Fiscous force=Weight

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5 0
3 years ago
If stellar parallax can be measured to a precision of about 0.01 arcsec using telescopes on the Earth to observe stars, to what
marin [14]

Answer:

It corresponds to a distance of 100 parsecs away from Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

It is defined in a analytic way as it follows:

       

\tan{p} = \frac{1AU}{d}

Where d is the distance to the star.

p('') = \frac{1}{d} (1)  

Equation (1) can be rewritten in terms of d:

d(pc) = \frac{1}{p('')} (2)

Equation (2) represents the distance in a unit known as parsec (pc).

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (1AU).

For the case of   (p('') = 0.01):

d(pc) = \frac{1}{0.01}

d(pc) = 100

Hence, it corresponds to a distance of 100 parsecs away from Earth.

<em>Summary:</em>

Notice how a small parallax angle means that the object is farther away.

Key terms:

Parsec: Parallax of arc second

7 0
3 years ago
A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,
ICE Princess25 [194]
Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂                (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392        (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

7 0
3 years ago
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