Answer:
49.3 N
Explanation:
Given that Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in the rope?
The weight of the bucket of water = mg.
Weight = 4.25 × 9.8
Weight = 41.65 N
The tension and the weight will be opposite in direction.
Total force = ma
T - mg = ma
Make tension T the subject of formula
T = ma + mg
T = m ( a + g )
Substitutes all the parameters into the formula
T = 4.25 ( 1.8 + 9.8 )
T = 4.25 ( 11.6 )
T = 49.3 N
Therefore, the tension in the rope is 49.3 N approximately.
Answer:
Average force = 67 mn
Explanation:
Given:
Initial velocity u = 0 m/s
Final velocity v = 67 m/s
Time t = 1 ms = 0.001 sec.
Computation:
Using Momentum theory
Change in momentum = F × Δt
(v-u)/t = F × Δt
F × 0.001 = (67 - 0)/0.001
F= 67,000,000
Average force = 67 mn
Answer:
Explanation:
Given that
g=9.8m/s²
The spring constant is
k=50N/m
The length of the bungee cord is
Lo=32m
Height of bridge which one end of the bungee is tied is 91m
A steel ball of mass 92kg is attached to the other end of the bungee.
The potential energy(Us) of the steel ball before dropped from the bridge is given as
P.E= mgh
P.E= 92×9.8×91
P.E= 82045.6 J
Us= 82045.6 J
Potential energy)(Uc) of the cord is given as
Uc= ½ke²
Where 'e' is the extension
Then the extension is final height extended by cord minus height of cord
e=hf - hi
e=hf - 32
Uc= ½×50×(hf-32)²
Uc=25(hf-32)²
Using conservation of energy,
Then,
The potential energy of free fall equals the potential energy in string
Uc=Us
25(hf-32)²=82045.6
(hf-32)² = 82045.6/25
(hf-32)²=3281.825
Take square root of both sides
√(hf-32)²=√(3281.825)
hf-32=57.29
hf=57.29+32
hf=89.29m
We neglect the negative sign of the root because the string cannot compressed
Answer:
1)Observe a phenomenon
2)Ask a question/ start inferring
3)Form a hypothesis
4)Create an experiment
5)Collect data
6)Compare results
7)Analyze
8)Report findings
9)Compare with other experiments
