When an object is free-falling, no other force is acting upon it but the gravitational force. Because of this, the equations of motion are simplified. We can determine first the initial velocity:
v = √2gy = √2(9.81)(4.9) = 9.805 m/s
Then, we use this to the equation below:
y = vt + 1/2*at²
y = (9.805)(0.5) + 1/2(9.81)(0.5)²
y = 6.13 m
Answer:
a) 578.0 cm²
b) 25.18 km
Explanation:
We're given the density and mass, so first calculate the volume.
D = M / V
V = M / D
V = (6.740 g) / (19.32 g/cm³)
V = 0.3489 cm³
a) The volume of any uniform flat shape (prism) is the area of the base times the thickness.
V = Ah
A = V / h
A = (0.3489 cm³) / (6.036×10⁻⁴ cm)
A = 578.0 cm²
b) The volume of a cylinder is pi times the square of the radius times the length.
V = πr²h
h = V / (πr²)
h = (0.3489 cm³) / (π (2.100×10⁻⁴ cm)²)
h = 2.518×10⁶ cm
h = 25.18 km
Answer:
The maximum height above the point of release is 11.653 m.
Explanation:
Given that,
Mass of block = 0.221 kg
Spring constant k = 5365 N/m
Distance x = 0.097 m
We need to calculate the height
Using stored energy in spring
...(I)
Using gravitational potential energy
....(II)
Using energy of conservation
Where, k = spring constant
m = mass of the block
x = distance
g = acceleration due to gravity
Put the value in the equation
Hence, The maximum height above the point of release is 11.653 m.
Explanation:
Load (l) = 680N
Effort (E) = 500N
Length slope (l) = 12m
Height slope (h) = 8 m
Output = load * height
680 *8 = 5.44 *103 J
The Input = effort * length = 500 *12 = 6000J
the Mechanical advantage (M.A) = load effort= 600500=1.36
the Velocity ratio (V.R) =lh=128 = 1.5
the Efficiency =M.A100%V.R= 90.6%
Answer:
A block device is a computer data storage device that supports reading and (optionally) writing data in fixed-size blocks, sectors, or clusters. These blocks are generally 512 bytes or a multiple thereof in size
