Answer:
Explanation:
noun
a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical reaction.
The molecular mass, in atomic mass unit, of aspartame would be 294 amu while the mass, in kg, of an aspartame molecule would be 0.294 kg
Aspartame has the chemical formula C14H18N2O5
C = 12, H = 1, N = 14, O = 16
(a) Molecular weight = (12x14) + (1x18) + (14x2) + (16x5)
= 168 + 18 + 28 + 80
= 294 amu
(b) Mass of 1 molecule of aspartame = mole x molar mass
= 1 x 294
= 294 g
Converting 294 g to kg = 294/1000
= 0.294 kg
More on mole and molar mass can be found here: brainly.com/question/6613610
If a mass of a neutron is 1 the electron mass is 0.00054386734 and it's charge is negative. Hope this helps! ;)
Answer:
Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction: Dry friction is a force that opposes the relative lateral motion of two solid surfaces in contact.
Coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.
Answer:
a) # buses = 7
Explanation:
For this exercise we use the kinematic equations, let's find the time it takes to reach the same height
y =
t - ½ g t²
Let's decompose the speed, with trigonometry
v₀ₓ = v₀ cos θ
= v₀ sin θ
v₀ₓ = 40 cos 32
v₀ₓ = 33.9 m / s
= 40 sin32
= 21.2 m / s
When it arrives it is at the same initial height y = 0
0 = ( - ½ gt) t
That has two solutions
t = 0 when it comes out
t = 2 / g when it arrives
t = 2 21.2 /9.8
t = 4,326 s
We use the horizontal displacement equation
x = vox t
x = 33.9 4.326
x = 146.7 m
To find the number of buses we can use a direct proportions rule
# buses = 146.7 / 20
# buses = 7.3
# buses = 7
The distance of the seven buses is
L = 20 * 7 = 140 m
b) let's look for the scope for this jump
R = vo2 sin2T / g
R = 40 2 without 2 32 /9.8
R = 146.7 m
As we can see the range and distance needed to pass the seven (7) buses is different there is a margin of error of 6.7 m in favor of the jumper (security)
