Answer:
150J
Explanation:
work output/work input=100%
so just make work output the subject
Answer: E = 941738.537J
Explanation:
to begin,
given that the mass = 2320 pound = 1052.334 kg
Δh = 110 ft = 33.528 m
given that distance (d) = 1283 ft = 391.058 m
also the speed (v) is 65 mph = 29.058 m/s
force (F) = 87 pounds = 386.995 N
from our knowledge in work energy theory;
E = Fd + 1/2mv² + mgh
E = (386.995 × 391.058) + (1/2×1052.334×29.058²) + (1052.334×9.81×33.528)
E = 151337.491 + 444278.2 + 346122.84
E = 941738.537J
i hope this helps, cheers.
Answer:
The y-axis should be labelled as W in Newtons (kg·m/s²)
Explanation:
The given data is presented here as follows;
Mass (kg)
Newtons (kg·m/s²)
3.2
31.381
4.6
45.1111
6.1
59.821
7.4
72.569
9
89.241
10.4
101.989
10.9
106.892
From the table, it can be seen that there is a nearly linear relationship between the amount of Newtons and the mass, as the slope of the data has a relatively constant slope
Therefore, the data can be said to be a function of Weight in Newtons to the mass in kilograms such that the weight depends on the mass as follows;
W(m) in Newtons = Mass, m in kg × g
Where;
g is the constant of proportionality
Therefore, the y-axis component which is the dependent variable is the function, W(m) = Weight of the body while the x-axis component which is the independent variable is the mass. m
The graph of the data is created with Microsoft Excel give the slope which is the constant of proportionality, g = 9.8379, which is the acceleration due to gravity g ≈ 9.8 m/s²
We therefore label the y-axis as W in Newtons (kg·m/s²)
Answer:
λ = 102.78 nm
This radiation is in the UV range,
Explanation:
Bohr's atomic model for the hydrogen atom states that the energy is
E = - 13.606 / n²
where 13.606 eV is the ground state energy and n is an integer
an atom transition is the jump of an electron from an initial state to a final state of lesser emergy
ΔE = 13.606 (1 /
- 1 / n_{i}^{2})
the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon
DE = 13.606 (1/1 - 1/3²)
DE = 12.094 eV
let's reduce the energy to the SI system
DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J
let's find the wavelength is this energy, let's use Planck's equation to find the frequency
E = h f
f = E / h
f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴
f = 2.9186 10¹⁵ Hz
now we can look up the wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 2.9186 10¹⁵
λ = 1.0278 10⁻⁷ m
let's reduce to nm
λ = 102.78 nm
This radiation is in the UV range, which occurs for wavelengths less than 400 nm.
Answer:
The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.