Answer:
a) Fₓ = 23.5 N
b) Net force = Fₓ
Explanation:
An image of the question as described is attached to this solution.
From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)
a) From the diagram, the horizontal component of the force is
Fₓ = 25 cos 20° = 23.49 N = 25 N
b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain
Net force = Fₓ - Frictional force
But frictional force is 0 N
Net force = Fₓ
Hope this Helps!!!
Explanation:
According to Dalton's atomic theory, all the atoms are individual, all the atoms of the same element are identical in properties and mass, the compound is formed from two or more kinds of the atoms, all the matter is made up of small atoms and the chemical reaction is a rearrangement of the atoms.
The discoveries which contradicts the components of Dalton's atomic theory from the given discoveries are:
Nuclear reactions can change an atom of one element into an atom of another element.
Atoms of a given element can have different numbers of neutrons.
Atoms contain smaller particles: protons, neutrons, and electrons.
Answer:
The moment of inertia is 0.7500 kg-m².
Explanation:
Given that,
Mass = 2.2 kg
Distance = 0.49 m
If the length is 1.1 m
We need to calculate the moment of inertia
Using formula of moment of inertia
Where, m = mass of rod
l = length of rod
x = distance from its center
Put the value into the formula
Hence, The moment of inertia is 0.7500 kg-m².
we are given in the problem the following dimensions or specifications
B = 0.000055 T r = 0.25 m constant mu0 = 4*pi*10-7
The formula that is applicable from physics is
B = mu0*I/(2*pi*r) I = 2*B*pi*r/mu0 I = 68.75 Amperes
Answer:
The rate of change of the height is - 4 ft/s
Solution:
As per the question:
Height of the person, y = 5 ft
The rate at which the person walks away, 
Distance of the spotlight from the wall, x = 40 ft
Now,
To calculate the rate of change in the height, 
of the person when, x = 10 m:
From fig 1.
![\Delta ABC[\tex] ≈ [tex]\Delta PQC[\tex]Thus[tex]\frac{BC}{AB} = \frac{PQ}{QC}](https://tex.z-dn.net/?f=%5CDelta%20ABC%5B%5Ctex%5D%20%E2%89%88%20%5Btex%5D%5CDelta%20PQC%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3EThus%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cfrac%7BBC%7D%7BAB%7D%20%3D%20%5Cfrac%7BPQ%7D%7BQC%7D)
xy = 200 (1)
Differentiating the above eqn w.r.t time t:
Thus
(2)
From eqn (1):
When x = 10 ft
10y = 200
y = 20 ft
Using eqn (2):
