The diameter of a sphere is measured to be 5.36 in. Find (a) the radius of the sphere in centimeters, (b) the surface area of th
1 answer:
Answer:
a) r = 6.81 cm : radius of the sphere
b) A = 582.78 cm² : surface area of the sphere
c) V = 1322.91 cm³ : volume of the sphere
Explanation:
Formula to calculate the surface area of the sphere:
A = 4×π×r² Formula (1)
Formula to calculate the volume of the sphere:
V = (4/3)×π×r³ Formula (2)
Problem development
a)
d = 5.36 in
1 in = 2.54 cm
r = d/2
Where:
r: sphere radius (cm)
d: sphere diameter (cm)
r = 13.614/2 = 6.81 cm
b)
We replace in formula (1)
A = 4×π×(6.81)² = 582.78 cm²
c)
We replace in formula (2)
V = (4/3)×π×(6.81)³ = 1322.91 cm³
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Answer:
(a). The speed is 115.28 m/s.
(b). The time is 94 min.
(c). The International Space Station travels 43202.400 km.
(d). Speed is scalar quantity.
Velocity is vector quantity.
Explanation:
Given that,
Top speed = 415 km/h
1 miles = 1609 m
(a). We need to calculate the speed in m/s
Using conversion of km/h to m/s
(b). We need to calculate the distance from London to Edinburgh in km
Using conversion for distance
Distance in meter,
We need to calculate the time
Using formula of time
Put the value into the formula
(c). Speed v' =7.66 km/s
v'=7660 m/s
In the time it takes the car to travel from London to Edinburgh,
We need to calculate the distance
Using formula of distance
Put the value into the formula
(d). Speed :
Speed is equal to the distance divided by time.
It is scalar quantity.
Velocity :
Velocity is equal to the displacement divided by time.
It is vector quantity.
Hence, (a). The speed is 115.28 m/s.
(b). The time is 94 min.
(c). The International Space Station travels 43202.400 km.
(d). Speed is scalar quantity.
Velocity is vector quantity.
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
