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vova2212 [387]
4 years ago
14

Why does the equilibrium position of the spring change when a mass is added to the spring? Will the mass oscillate around the ne

w equilibrium position of the spring or the previous position without a mass attached to the spring? If the equilibrium position of the spring changes by 20 cm (assuming no initial mass) when a mass is added to the spring with constant 4.9 kg/s^2, what is the mass of the object attached to the spring?
Physics
1 answer:
abruzzese [7]4 years ago
7 0
-it is because force is acted against gravity by the spring when we keep the mass on it
-it depends upon the spring constant
-i will answer it later!!
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A box is sliding down an incline tilted at a 16.0 angle above horizontal. The box is initially sliding down the incline at a spe
blsea [12.9K]

Answer:

d) 1.44m

Explanation:

According to Newton's second law:

\sum F=m.a\\

analyzing the horizontal components of the force:

\sum F_x=mg*sin(\theta)-\µ*m.g*cos(\theta)\\\sum F_x=m*9.8*(sin(16^o)-0.420*cos(16^o))\\\sum F_x=-1.25m

applying the second law

-1.25m=m.a\\a=-1.25m/s^2

given the acceleration we can calculate the distances traveled before stopping:

(v_f)^2=(v_o)^2+2.a.\Delta x\\0=(1.90m/s)^2-2(1.25)*\Delta x\\\Delta x=1.44m

8 0
4 years ago
Does current flow through or across a resistor?
Mamont248 [21]

Answer:

Current flows across a resistor.

Explanation:

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3 0
4 years ago
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Power is measured in Joules per second, which is also the<br> _____
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Answer:

Watt

Explanation:

Power is measured in Watts. J/s is the base unit of measurement, but we usually measure power in Watts (W).

3 0
3 years ago
Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
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Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

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 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

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Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

3 0
3 years ago
Explain how land breeze and sea breezes occur?
marshall27 [118]

Explanation:

during the daytime the land gets hot and the hot air blows up to the sky and to feel the space the cool you are from the Sea blows to the land this phenomenon is called sea breeze similarly at night the hot air of sea goes up n cool air from land blows to sea which is called land breexe

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