Answer:
11.37 m/s
Explanation:
We are given;
Mass; m = 0.001 kg
Initial velocity; v1 = 0 m/s (since it was released from rest)
Initial height; h1 = 20.6m
From the attached image of the bead in the loop, we can see that height at A; h2 = twice the radius = 2R = 2 × 7 = 14 m
g = 9.8
To solve for the speed at point, we will use formula for conservation of energy. Thus;
½m(v1)² + mgh1 = ½m(v2)² + mgh2
Divide through by m to get;
½(v1)² + gh1 = ½(v2)² + gh2
Plugging in the relevant values;
½(0²) + (9.8 × 20.6) = ½(v2)² + (9.8 × 14)
Rearranging, we have;
½(v2)² = ½(0²) + (9.8 × 20.6) - (9.8 × 14)
Simplifying, we have;
½(v2)² = 64.68
Multiply both sides by 2 to get;
(v2)² = 2 × 64.68
(v2)² = 129.36
v2 = √129.36
v2 = 11.37 m/s
I believe the answer you are looking for is CMB Radiation
Answer:
6.0 s
98 m/s
Explanation:
The radius of the planet is much bigger than the height of the tower, so we will assume the acceleration is constant. Neglect air resistance.
Acceleration due to gravity on this planet is:
a = GM / r²
a = (6.67×10⁻¹¹ m³/kg/s²) (2.7 × 1.48×10²³ kg) / (1.7 × 750,000 m)²
a = 16.4 m/s²
The height of the tower is:
Δy = 96 × 3.05 m
Δy = 293 m
Given v₀ = 0 m/s, find t and v.
Δy = v₀ t + ½ at²
(293 m) = (0 m/s) t + ½ (16.4 m/s²) t²
t = 6.0 s
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (16.4 m/s²) (293 m)
v = 98 m/s
Answer:
the relative density of the gravel in the field is 0.7365 or 73.65%
Explanation:
Given that;
wet density of gravel 
= 2.3 Mg/m³
field water content w = 16%
density of solids in lab 
= 2.70 Mg/m³
Maximum void ratio 
= 0.59
Minimum void ratio 
= 0.28
first we determine the dry density of gravel 
= / 1+ w
= 2.3 Mg/m³ / 1 + 0.16 = 2.3/1.16 = 1.9827 Mg/m³
we know that; 
= 1000 kg/m³ = 1 g/cm³
Specific Gravity of soil G = / = 2.70 Mg/m³ / 1 = 2.70 Mg/m³
eo = ((G×)/) - 1
eo = (( 2.70 × 1) / 1.9827 ) - 1
eo = 1.3617 - 1
Co = 0.3617
so Relative density 
will be;
= - eo / -
we substitute
= 0.59 - 0.3617 / 0.59 - 0.28
= 0.2283 / 0.31
= 0.7365 or 73.65%
Therefore; the relative density of the gravel in the field is 0.7365 or 73.65%
The efficiency of the steam engine is 78.9%.
<h3>What is the efficiency of a machine?</h3>
The efficiency of a machine is a measure of the useful work done by the machine as against the work input into the machine.
- Efficiency of a machine = work output/work input × 100%
Work output of the steam engine = 225 K
Work input of the steam engine = 285 K
The efficiency of the machine = 225/275 × 100% = 78.9%
Therefore, the efficiency of the steam engine is 78.9%.
Leran more about efficiency of machines at: brainly.com/question/7536036
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