The circuit element whose purpose is to convert electrical energy into another form of energy is the _____.
2 answers:
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3.4m/s
Explanation:
Given parameters:
Distance to school = 14.4km
Time taken by Amy = 49min
Time taken by bill = 20min after Amy = 20+49 = 69min
Unknown parameters:
How much faster is Amy's average speed = ?
Solution:
Average speed is the rate of change of total distance with total time taken.
Average speed =
convert units to meters and seconds
1000m = 1km
60s = 1min
Distance to school = 14.4 x 1000 = 14400m
Time taken by Amy = 49 x 60 = 2940s
Time taken by Bill = 69 x 60 = 4140s
Average speed of Amy = = 4.9m/s
Average speed of Bill = = 1.4m/s
Differences in speed = 4.9 - 1.5 = 3.4m/s
Amy was 3.4m/s faster than Bill
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Average speed brainly.com/question/8893949
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The initial potential energy of the wagon containing gold boxes will enable
it roll down the hill when cut loose.
The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.
Reasons:
Mass wagon and gold = 166 kg
Location of the wagon = 77 meters up the hill
Slope of the hill = 8°
Location of the rangers = 41 meters from the canyon
Mass of Lone Ranger, m₁ = 65 kg
Mass of Tonto m₂ = 66 kg
Solution;
Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m
Potential energy = m·g·h
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912
Potential energy of wagon, P.E. ≈ 17457.0912 J
By energy conservation, P.E. = K.E.
Where;
v = The velocity of the wagon a the bottom of the cliff
Therefore;
Velocity of the wagon, v ≈ 14.5 m/s
Momentum = Mass, m × Velocity, v
Initial momentum of wagon = m·v
Final momentum of wagon and ranger = (m + m₁ + m₂)·v'
By conservation of momentum, we have;
m·v = (m + m₁ + m₂)·v'
Which gives;
The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s
The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the
gold and jump out of the wagon before the wagon heads over the cliff.
Learn more here:
Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s