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xeze [42]
2 years ago
8

If 5J of work are done on a spring, compressing it by 12cm, what is the spring constant?

Physics
1 answer:
laiz [17]2 years ago
7 0

Answer:

hi

Explanation:

hello

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25. How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?
Lina20 [59]
  • Length=l=4m

\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{l}{g}}

\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{4}{9.8}}

\\ \rm\rightarrowtail T=2\pi(0.63887)

\\ \rm\rightarrowtail T=1.27774\pi

\\ \rm\rightarrowtail T=4.012s

4 0
2 years ago
A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal
PilotLPTM [1.2K]
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline 
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
3 0
3 years ago
(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down
slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

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

Here, u = 0 and a = g

h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0
3 years ago
What conventions are used in SI to indicate units
klemol [59]

Answer:

<u>Conventions used in SI to indicate units are as follows:</u>

  • Only singular form of units are used. for example: use kg and not kgs.
  • Do not use full stop after the abbreviations of any unit. for example: do not use kg. or cm.
  • Use one space between last numeric digit and SI unit. for example: 10 cm, 9 km.
  • Symbols and words should not be mixed. for example: use Kilogram per cubic and not kilogram/m3.
  • While writing numerals, only the symbols of the units should be written. for example: use 10 cm and not Ten cm.
  • Units named after a scientist should be written in small letters. for example: newton, henry.
  • Degree sign should not be used when the kelvin unit is used. for exmaple: use 37° and not 37°k

5 0
3 years ago
A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s
Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

T=m\frac{v^2}{r}

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N

5 0
3 years ago
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