The right answer for the question that is being asked and shown above is that: "B.The small car has less momentum because its mass is less than the mass of the truck." A small car and a large truck are both driving south at 40 km/h. The statement that is true is that t<span>he small car has less momentum because its mass is less than the mass of the truck. </span>
You can’t ask a question life this and say “(2 points)”first of all the app pickshow many points it is two that’s just disrespectful
F=ma
Force=mass*acceleration
So plug in the values
Hint: the force would be measured in newtons
Answer:
the light emitting must be of greater wavelength
Explanation:
For this exercise we must use the Planck equation
E = h f
And the speed of light
c = λ f
f = c / λ
We replace
E = h c / λ
The wavelength of the green light is of the order of 500 nm, let's calculate the energy
E = 6.63 10⁻³⁴ 3 10⁸ /λ
E = 1,989 10⁻²⁵ /λ
λ = 500 nm = 500 10⁻⁹ m
E = 1,989 10⁻²⁵ / 500 10⁻⁹
E = 3,978 10⁻¹⁹ J
That is the energy of the transition for a transition is an intermediate state the energy must be less, this implies that the wavelength must increase. For the explicit case of a state with half of this energy
= E / 2
= 3,978 10⁻¹⁹ / 2 = 1,989 10⁻¹⁹
Let's clear and calculate
λ = h c / E
λ = 1,989 10⁻²⁵ / 1,989 10⁻¹⁹
λ = 1 10⁻⁶ m
Let's reduce to nm
λ = 1000 nm
This wavelength is in the infrared region
the light emitting must be of greater wavelength
Answer:
h >5/2r
Explanation:
This problem involves the application of the concepts of force and the work-energy theorem.
The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.
Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)
So from newton's second law of motion,
W – N = mv²/r
N = normal force = 0
W = mg
mg = ma = mv²/r
mg = mv²/r
v²= rg
v = √(rg)
The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop
So
ΔPE = ΔKE = 1/2mv²
The height at the roller coaster starts is usually higher than the top of the loop by design. So
ΔPE =mgh - mg×2r = mg(h – 2r)
2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.
In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.
So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.
ΔPE > ΔKE
mg(h–2r) > 1/2mv²
g(h–2r) > 1/2(√(rg))²
g(h–2r) > 1/2×rg
h–2r > 1/2×r
h > 2r + 1/2r
h > 5/2r
