Explanation:
Given:
v₀ = 22 m/s
v = 0 m/s
t = 17.32 s
Find: a
v = at + v₀
(0 m/s) = a (17.32 s) + (22 m/s)
a = -1.270 m/s²
Round as needed.
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
Answer:
4 hoop, disk, sphere
Explanation:
Because
We are given data that
Hoop, disk, sphere have Same mass and radius
So let
And Initial angular velocity, = 0
The Force on each be F
And Time = t
Also let
Radius of each = r
So let's find the inertia shall we!!
I1 = m r² /2
= 0.5 mr² the his is for dis
I2 = m r² for hoop
And
Moment of inertia of sphere wiil be
I3 = (2/5) mr²
= 0.4 mr²
So
ωf = ωi + α t
= 0 + ( τ / I ) t
= ( F r / I ) t
So we can see that
ωf is inversely proportional to moment of inertia.
And so we take the
Order of I ( least to greatest ) :
I3 (sphere) , I1 (disk) , I2 (hoop) , ,
Order of ωf: ( least to greatest)
That of omega xf is the reverse of inertial so
hoop, disk, sphere
Option - 4
Answer:
when the mass of the bottle is 0.125 kg, the average height of the beanbag is 0.35 m.
when the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is 0.91m.
when the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is 1.26m.
when the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is 1.57m.
Explanation:
To answer the following questions for this specific problem:
a. 11.48 secs
b. Vp = a*t*3.6 =
3*11.48*3.6 = 124.0 km/h
<span>c. 9.1 secs. </span>
I am hoping that this answer has satisfied your query about
and it will be able to help you.
