Explanation:
The given data is as follows.
k = 130 N/m, 
= 17 cm = 0.17 m (as 1 m = 100 cm)
mass (m) = 2.8 kg
When the spring is compressed then energy stored in it is as follows.
Energy = 
Now, spring energy gets converted into kinetic energy when the box is launched.
So, = 
= 
= 1.34
v = 1.15 m/sec
Now,
Frictional force = 
= 
= 4.116 N
Also, Kinetic energy = work done by friction
1.8515 = 
d = 0.449 m
Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.
<span>In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, </span>where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.
<span>ρwAd1g = Mg</span>
ρw<span>Ad2g = (M + m) g</span>
<span>d2∕d1 = (M + m)/g</span>
m = [(d2<span>∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg</span>
This means that Bubba’s mass is 120 kg.
Answer:
This is false becuase different object weigh different
Thank you!
Explanation:
Answer:
V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s
Explanation:
The volume flow rate of the blood in the artery can be given by the following formula:
where,
V = Volume flow rate = ?
A = cross-sectional area of artery = πd²/4 = π(0.004 m)²/4 = 1.26 x 10⁻⁵ m²
v = velcoity = 0.28 m/s
Therefore,
<u>V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s</u>
