Answer:
a) 
= 928 J
, b)U = -62.7 J
, c) K = 0
, d) Y = 11.0367 m, e) v = 15.23 m / s
Explanation:
To solve this exercise we will use the concepts of mechanical energy.
a) The elastic potential energy is
= ½ k x²
= ½ 2900 0.80²
= 928 J
b) place the origin at the point of the uncompressed spring, the spider's potential energy
U = m h and
U = 8 9.8 (-0.80)
U = -62.7 J
c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also
K = ½ m v²
K = 0
d) write the energy at two points, maximum compression and maximum height
Em₀ = ke = ½ m x²
= mg y
Emo =
½ k x² = m g y
y = ½ k x² / m g
y = ½ 2900 0.8² / (8 9.8)
y = 11.8367 m
As zero was placed for the spring without stretching the height from that reference is
Y = y- 0.80
Y = 11.8367 -0.80
Y = 11.0367 m
Bonus
Energy for maximum compression and uncompressed spring
Emo = ½ k x² = 928 J
= ½ m v²
Emo =
Emo = ½ m v²
v =√ 2Emo / m
v = √ (2 928/8)
v = 15.23 m / s
1 nanowatt = 1 nanojoule/sec
1 watt = 1 joule/sec
10 watts = 10 joules/sec
100 watts = 100 joules/sec
742.914 watts = 742.914 joules/sec
1,000 watts = 1,000 joules/sec
10,000 watts = 10,000 joules/sec
100,000 watts = 100,000 joules/sec
1 megawatt = 1 megajoule/sec
1 gigawatt = 1 gigajoule/sec
1 petawatt = 1 petajoule/sec
We don't care what frequency the transmission is using,
or who their morning DJ is.
Answer:
See Explanation
Explanation:
The relationship between angle of an incline and the acceleration of an object moving down the incline.
As the angle of an incline increases, so does the acceleration of the body moving down the incline increases, resolving the force acting on an inclined object
Parallel force = mgsin, perpendicular = mgcosΘ
With th weigh component 'mg' of the parallel force accounting for the acceleration of the body down the incline.
mgsinΘ = ma
Fnet = ma
B.) From Fnet = ma
Fnet = ma
a = Fnet / m
Where Fnet = Net force = mgsinΘ, a = acceleration
1 mole of arsenic 75g
238 moles of arsenic x g
x = 17850g of arsenic
Answer:
Explanation:
A )
When empty , H₀ length of barge is inside water .
volume of barge inside water = A x H₀
Weight of displaced water = AH₀ x ρ x g
Buoyant force = weight of displaced water = AH₀ ρg
B)
It should balance the weight of barge
Weight = buoyant force
Weight = AH₀ ρg
mass of barge = weight / g
weight / g = AH₀ ρ
= 550 x .55 x 1000
= 302500 kg
