At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is
∑ F = ma
n - 430 N = (430 N)/g • a
where m is the child's mass and a is their centripetal acceleration. The child has a linear speed of 3.5 m/s at any point along the path of the wheel whose radius is 17 m, so the centripetal acceleration is
a = (3.5 m/s)² / (17 m) ≈ 0.72 m/s²
and so
n = 430 N + (430 N)/g (0.72 m/s²) ≈ 460 N
Wow ! This one could have some twists and turns in it.
Fasten your seat belt. It's going to be a boompy ride.
-- The buoyant force is precisely the missing <em>30N</em> .
-- In order to calculate the density of the frewium sample, we need to know
its mass and its volume. Then, density = mass/volume .
-- From the weight of the sample in air, we can closely calculate its mass.
Weight = (mass) x (gravity)
185N = (mass) x (9.81 m/s²)
Mass = (185N) / (9.81 m/s²) = <u>18.858 kilograms of frewium</u>
-- For its volume, we need to calculate the volume of the displaced water.
The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³. So the volume of the
displaced water (in cm³) is the same as the number of grams in it.
The weight of the displaced water is 30N, and weight = (mass) (gravity).
30N = (mass of the displaced water) x (9.81 m/s²)
Mass = (30N) / (9.81 m/s²) = 3.058 kilograms
Volume of displaced water = <u>3,058 cm³</u>
Finally, density of the frewium sample = (mass)/(volume)
Density = (18,858 grams) / (3,058 cm³) = <em>6.167 gm/cm³</em> (rounded)
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I'm thinking that this must be the hard way to do it,
because I noticed that
(weight in air) / (buoyant force) = 185N / 30N = <u>6.1666...</u>
So apparently . . .
(density of a sample) / (density of water) =
(weight of the sample in air) / (buoyant force in water) .
I never knew that, but it's a good factoid to keep in my tool-box.
Answer:
a) 1450watts
b) 564watts
c) 1.11
Explanation:
Power consumed = IV
I is the current rating
V is the operating voltage
If a blow-dryer and a vacuum cleaner each operate with a voltage of 120 V and the current rating of the blow-dryer is 12 A, while that of the vacuum cleaner is 4.7 A then their individual power rating is calculated thus;
a) For blow-dryer
Operating voltage = 120V
Its current rating = 12A
Power consumed = IV
= 120×12
= 1440watts
b) For vacuum cleaner:
Operating voltage is the same as that of blow dryer = 120V
Its current rating = 4.7A
Power consumed = IV
= 120×4.7
= 564watts
c) Energy used = Power consumed × time taken
Energy used = Power × time
Energy used by blow dryer = 1440×20×60
= 1,728,000Joules
Energy used up by vacuum cleaner = 564×46×60
= 564×2760
= 1,556,640Joules
Ratio of the energy used by the blow-dryer in 20 minutes to the energy used by the vacuum cleaner in 46 minutes will be 1,728,000/1,556,640 = 1.11
