Answer:
t = 3.516 s
Explanation:
The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:
Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0
If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.
This can be done knowing that
1 km = 1000 m
1 h = 3600 s
Therefore
1 km/h = (1000/3600) m/s = 0.2777... m/s
100 km/h = 27.777... m/s
Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:
27.777 m/s = 7.9 m/s^2 t
Solving for t
t = (27.7777 / 7.9) s = 3.516 s
<h3><u>Answer;</u></h3>
the north end to the south end.
<h3><u>Explanation;</u></h3>
- Magnetic field lines from a bar magnet form lines that are closed. The direction of magnetic field is taken to be outward from the North pole of the magnet and in to the South pole of the magnet.
- A magnetic field refers to the area surrounding a magnet where a force is exerted on certain objects. These lines are spread out of the north end of the magnet.
- The magnetic field lines resemble a bubble.
Answer:
F = 7.68 10¹¹ N, θ = 45º
Explanation:
In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges
The net force is
F_ {net} = F₂₁ + F₂₃ + F₂₄
bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.
let's use trigonometry
cos 45 = F₂₄ₓ / F₂₄
sin 45 = F_{24y) / F₂₄
F₂₄ₓ = F₂₄ cos 45
F_{24y} = F₂₄ sin 45
let's do the sum on each axis
X axis
Fₓ = -F₂₁ + F₂₄ₓ
Fₓ = -F₂₁₁ + F₂₄ cos 45
Y axis
F_y = - F₂₃ + F_{24y}
F_y = -F₂₃ + F₂₄ sin 45
They indicate that the magnitude of all charges is the same, therefore
F₂₁ = F₂₃
Let's use Coulomb's law
F₂₁ = k q₁ q₂ / r₁₂²
the distance between the two charges is
r = a
F₂₁ = k q² / a²
we calculate F₂₄
F₂₄ = k q₂ q₄ / r₂₄²
the distance is
r² = a² + a²
r² = 2 a²
we substitute
F₂₄ = k q² / 2 a²
we substitute in the components of the forces
Fx = 
Fx = 
( -1 + ½ cos 45)
F_y = k \frac{q^2}{a^2} ( -1 + ½ sin 45)
We calculate
F₀ = 9 10⁹ 4.25² / 0.440²
F₀ = 8.40 10¹¹ N
Fₓ = 8.40 10¹¹ (½ 0.707 - 1)
Fₓ = -5.43 10¹¹ N
remember cos 45 = sin 45
F_y = - 5.43 10¹¹ N
We can give the resultant force in two ways
a) F = Fₓ î + F_y ^j
F = -5.43 10¹¹ (i + j) N
b) In the form of module and angle.
For the module we use the Pythagorean theorem
F = 
F = 5.43 10¹¹ √2
F = 7.68 10¹¹ N
in angle is
θ = 45º
Answer:
57,42 KJ
Explanation:
By a isobaric proces, the expresion for the works in the jpg adjunt. Then:
W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)
By the ideal gases law: PV=RTn
Then, in (1): (remember Pa = Pb)
W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)
Since we have 1 Kg air: How much is this in moles?
From bibliography: 28.96 g/mol
Then, in 1 Kg (1000 g) there are:
n = 34,53 mol
Finally, in (2):
W = (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ
