The total potential energy associated with the jumper at the end of his fall is 90,000 J.
The given parameters;
- <em>mass of the jumper, m = 51 kg</em>
- <em>height of the bridge. h = 321 m</em>
- <em>spring constant of the cord, k = 32 N/m</em>
- <em>extension of the cord, x = 179 m - 104 m = 75 m</em>
The total potential energy associated with the jumper at the end of his fall is calculated as follows;
U = ¹/₂kx² + mgΔh
where;
<em>Δh is the change in height after falling </em>
U = ¹/₂(32)(75)² + (51)(9.8)(0)
U = 90,000 J
Thus, the total potential energy associated with the jumper at the end of his fall is 90,000 J.
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912.
outer ear:
pinna
ear canal
middle ear:
ossicles and ear drum
inner ear:
semcircular canals
cochlea
auditory nerve
13.
frequency = wavespeed ÷ wavelength
14.
if frequency increases you would experience a higher pitch in sound
15.
humans can hear 20Hz to 20kHz
16.
The Doppler effect is the change in frequency or wavelength of a wave for an observer who is moving relative to the wave source. Can be used for machines measuring speed via doppler effect.
17.
Doppler in hospitals can be used for ultrasound to provide images for diagnosis and monitoring.
First let us calculate for the moles of CH3OH formed:
moles CH3OH = 23 g / (32 g / mol) = 0.71875 mol
We see that there are 2 moles of H2 per mole of CH3OH, so:
moles H2 = 0.71875 mol * 2 = 1.4375 mol
Assuming ideal gas behaviour, we use the formula:
PV = nRT
V = nRT / P
V = 1.4375 mol * (62.36367 L mmHg / mol K) * (90 + 237.15
K) / 756 mm Hg
<span>V = 43.06 Liters</span>
The work done on the filled bucket in raising out of the hole is 2, 925 Joules
<h3>How to determine the work done</h3>
Using the formula:
Work done = force * distance
Note that force = mass * acceleration
F = mg + ma
F = 4. 5 * 10 + 28 * 10
F = 45 + 280
F = 325 Newton
Distance = 9m
Substitute into formula
Work done = 325 * 9
Work done = 2, 925 Joules
Therefore, the work done is 2, 925 Joules
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Answer:
P = 10 kPa
Explanation:
Given that,
The mass of a small table, m = 4 kg
The area of each leg = 0.001 m²
We need to find the pressure exerted by the table on the floor. Pressure is equal to the force per unit area. So
So, the required pressure is 10 kPa.
