Answer:
Dₓ = -155 sin 23° i + 0 j
Explanation:
The diagram showing the vector has been attached to this response.
As shown in the diagram,
The vector D has an x-component (also called horizontal component) of -D sinθ i. i.e
Dₓ = -D sin θ i [The negative sign shows that D lies in the negative x direction]
Where;
D = magnitude of D = 155m
θ = direction of D = 23°
Therefore;
Dₓ = -155 sin 23° i
Since Dₓ represents the x component, its unit vector, j component has a value of 0.
Therefore, Dₓ can be written in terms of D, θ and the unit vectors i and j as follows;
Dₓ = -155 sin 23° i + 0 j
Answer:
The maximum amount of mechanical energy converted to internal energy during the fall is 26.7 joules
Explanation:
Potential Energy (PE) = weight of baseball × height = 1.47N × 10m = 14.7Nm = 14.7 joules
Kinetic Energy (KE) = 12 joules
Maximum amount of mechanical energy converted to internal energy during the fall = PE + KE = 14.7 joules + 12 joules = 26.7 joules
Answer:
29.4 N/m
0.1
Explanation:
a) From the restoring Force we know that :
F_r = —k*x
the gravitational force :
F_g=mg
Where:
F_r is the restoring force .
F_g is the gravitational force
g is the acceleration of gravity
k is the constant force
xi , x2 are the displacement made by the two masses.
Givens:
<em>m1 = 1.29 kg</em>
<em>m2 = 0.3 kg </em>
<em>x1 = -0.75 m </em>
<em>x2 = -0.2 m </em>
<em>g = 9.8 m/s^2 </em>
Plugging known information to get :
F_r =F_g
-k*x1 + k*x2=m1*g-m2*g
k=29.4 N/m
b) To get the unloaded length 1:
l=x1-(F_1/k)
Givens:
m1 = 1.95kg , x1 = —0.75m
Plugging known infromation to get :
l= x1 — (F_1/k)
= 0.1
Answer :
The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.
Explanation:
Given,
Atomic mass of silver = 107.87 g/mol
Density of silver = 10.35 g/cm^3
Converting to g/m^3,
= 10.35 g/cm^3 × 10^6cm^3/m^3
= 10.35 × 10^6 g/m^3
Avogadro's number = 6.022 × 10^23 atoms/mol
Fraction of lattice sites that are vacant in silver = 1 × 10^-6
Nag = (Na * Da)/Aag
Where,
Nag = Total number of lattice sites in Ag
Na = Avogadro's number
Da = Density of silver
Aag = Atomic weight of silver
= (6.022 × 10^23 × (10.35 × 10^6)/107.87
= 5.778 × 10^28 atoms/m^3
The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6
= 5.778 × 10^22/m^3.
Answer:
The Position of the object L = 0.172 m
Explanation:
The detailed explanation of the question is given in the attach document.