Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
= - gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
Answer:
15.19°, 31.61°, 51.84°
Explanation:
We need to fin the angle for m=1,2,3
We know that the expression for wavelenght is,
Substituting,
Once we have the wavelenght we can find the angle by the equation of the single slit difraction,
Where,
W is the width
m is the integer
the wavelenght
Re-arrange the expression,
For m=1,
For m=2,
For m=3,
<em>The angle of diffraction is directly proportional to the size of the wavelength.</em>
A futuristic design for a car is to have a large solid disk-shaped flywheel within the car storing kinetic energy. The uniform flywheel has mass 370 kg with a radius of 0.500 m and can rotate up to 320 rev/s. Assuming all of this stored kinetic energy could be transferred to the linear velocity of the 3500-kg car, find the maximum attainable speed of the car.
Correct answer is:
<h2>The maximum number of orbits in an atom is <u>Seven.</u></h2><h3>Explanation:</h3>
Every energy level has a limited one orbital including two electrons. The orbits are settled in the sub-levels and there can be further than 1 sub-level as the number of energy levels rises. On energy level 1, there is 1 sub-level and 1 orbital. Energy level 2 can possess 2 sub-levels and 2 orbitals. These remain to develop as you progress from the nucleus of the atom, closing up with an infinite potential number of levels and orbits.
<span>118 C
The Clausius-Clapeyron equation is useful in calculating the boiling point of a liquid at various pressures. It is:
Tb = 1/(1/T0 - R ln(P/P0)/Hvap)
where
Tb = Temperature boiling
R = Ideal Gas Constant (8.3144598 J/(K*mol) )
P = Pressure of interest
Hvap = Heat of vaporization of the liquid
T0, P0 = Temperature and pressure at a known point.
The temperatures are absolute temperatures.
We know that water boils at 100C at 14.7 psi. Yes, it's ugly to be mixing metric and imperial units like that. But since we're only interested in relative pressure differences, it's safe enough. So
P0 = 14.7
P = 14.7 + 12.3 = 27
T0 = 100 + 273.15 = 373.15
And for water, the heat of vaporization per mole is 40660 J/mol
Let's substitute the known values and calculate.
Tb = 1/(1/T0 - R ln(P/P0)/Hvap)
Tb = 1/(1/373.15 K - 8.3144598 J/(K*mol) ln(27/14.7)/40660 J/mol)
Tb = 1/(0.002679887 1/K - 8.3144598 1/K ln(1.836734694)/40660)
Tb = 1/(0.002679887 1/K - 8.3144598 1/K 0.607989372/40660)
Tb = 1/(0.002679887 1/K - 5.055103194 1/K /40660)
Tb = 1/(0.002679887 1/K - 0.000124326 1/K)
Tb = 1/(0.002555561 1/K)
Tb = 391.3034763 K
Tb = 391.3034763 K - 273.15
Tb = 118.1534763 C
Rounding to 3 significant figures gives 118 C</span>
