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LekaFEV [45]
3 years ago
11

If it requires 7.0 JJ of work to stretch a particular spring by 1.7 cmcm from its equilibrium length, how much more work will be

required to stretch it an additional 3.6 cm ?
Physics
1 answer:
Lynna [10]3 years ago
3 0

Answer:

Explanation:

First of all, well calculate the spring constant k

K = 2Ei/x^2

Where Ei = initial work required

x = initial stretch length

k = 2×7/0.017^2 = 48443J/m^2

Now work done in stretching it to 5.3cm (1.7 + 3.6) or 0.053m

EF = kx^2/2

48443 × 0.053^2/2 = 68J

Work done in stretching additional 3.6cm is equal to

68J-7J = 61J

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If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?
ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: (m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})

\frac{8.6}{2.5}  = \log_{10}(b_{1}/b_{2})

\log_{10}(b_{1}/b_{2}) = 3.44

Thus,

(b_{1}/b_{2}) = 2754.22

It means star A is 2754.22 time brighter than Star B.

3 0
3 years ago
When practicing an oral presentation what can you do to prepare for the presentation
Igoryamba
When practicing an oral presentation, you can prepare by writing a draft and practice reading aloud what you are going to say before your oral presentation.
8 0
3 years ago
Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024
myrzilka [38]
Given: Mass of earth Me = 5.98 x 10²⁴ Kg

           Radius of earth r = 6.37 x 10⁶ m

            G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma;  F = GMeMsat/r²     Centripetal acceleration ac = V²/r

               but V = 2πr/T

equate T from all equation.

F = ma

GMeMsat/r² = Msat4π²/rT²    

GMe = 4π²r³/T²

T² = 4π²r³/GMe  

T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds    or around   1.4 Hour

3 0
3 years ago
A child slides down a hill on a toboggan with an acceleration of 1.8 m/s^2. If she starts at rest, how far has she traveled in :
DENIUS [597]

Explanation:

It is given that,

The acceleration of the toboggan, a=1.8\ m/s^2

Initial speed of the toboggan, u = 0

We need to find the distance covered by the toboggan. Using the second equation of motion as :

s=ut+\dfrac{1}{2}at^2

At t = 1 s

s=\dfrac{1}{2}\times 1.8\times 1^2

s_1=0.9\ m

At t = 2 s

s=\dfrac{1}{2}\times 1.8\times 2^2

s_2=3.6\ m

At t = 3 s

s=\dfrac{1}{2}\times 1.8\times 3^2

s_3=8.1\ m

Hence, this is the required solution.

4 0
3 years ago
A repelling force occurs between two charged objects when the charges are of Equal magnitude. Equal magnitude. a Unequal magnitu
Ahat [919]

Answer:

c Like/same signs

Explanation:

A repelling force occurs between two or more charged objects with the charges are of like or same sign.

  • According to Coulombs law, like charges repel on another, unlike charges attracts on another.
  • If a positive charge comes into the vicinity of another positive charge, there will be repulsion.
  • When oppositely charge species are brought near each other, there is an attraction.

Therefore, repulsion occurs when like charges are brought close to each other.

3 0
3 years ago
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