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
The given statement "If you see fire or smoke coming from your vehicle's hood, you should pull over and get away from the vehicle as soon as possible." is True.
What to do when there is smoking engine in your car?
You should take rapid action if the smoke seems to be coming from a fire under your bonnet.
- Don't try to prop the bonnet open; instead, pull the bonnet release lever.
- Remove everyone from the vehicle and maintain your distance.
- Ask for the fire service by dialing 999.
- If it is safe to do so, warn oncoming traffic.
Learn more about the safety precaution with the help of the given link:
brainly.com/question/2347255
#SPJ4
Explanation:
7) Given:
v₀ = 2.0 m/s
v = 0 m/s
t = 3.00 s
Find: Δx
Acceleration isn't included in the problem, so use a kinematic equation that doesn't involve a.
Δx = ½ (v + v₀) t
Δx = ½ (0 m/s + 2.0 m/s) (3.00 s)
Δx = 3.0 m
8) Given:
v₀ = 0 m/s
v = 5 m/s
t = 4 s
Find: a
Displacement isn't included in the problem, so use a kinematic equation that doesn't involve Δx.
v = at + v₀
5 m/s = a (4 s) + 0 m/s
a = 1.25 m/s²
9) Given:
v_avg = Δx / t
0.5 m/s = 8 m / t
t = 16 s
Answer:
L_max= 0
Explanation:
The formula for magnitude of maximum orbital angular momentum is given by
l= orbital quantum number
l= n-1
n= shell number or principal quantum number
for n=1 , l=0
therefore,
L_max= 0