The answer is letter B. XD
Answer:
The pressure of the air in the tyre is 20 kPa
Explanation:
The parameters for the bicycle pump and tyre are;
The volume of air contained in the bicycle pump, V₁ = 20 cm³
The pressure of the air contained in the bicycle pump, P₁ = 100 kPa
The volume (available) of the tyre, where the air is pumped, V₂ = 100 cm³
Let P₂ represent the pressure in the tyre after the air is pumped
By Boyle's law, we have that at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure;
Mathematically, Boyle's law gives the following equation;
P₁ × V₁ = P₂ × V₂
∴ P₂ = (P₁ × V₁)/V₂
Substituting the known values gives;
P₂ = (100 kPa × 20 cm³)/(100 cm³)
∴ P₂ = 100 kPa × 1/5 = 20 kPa
P₂ = 20 kPa
The pressure of the air in the tyre = P₂ = 20 kPa.
Answer:
(a) m = 1.6 x 10²¹ kg
(b) K.E = 2.536 x 10¹¹ J
(c) v = 7.12 x 10⁵ m/s
Explanation:
(a)
First we find the volume of the continent:
V = L*W*H
where,
V = Volume of Slab = ?
L = Length of Slab = 4450 km = 4.45 x 10⁶ m
W = Width of Slab = 4450 km = 4.45 x 10⁶ m
H = Height of Slab = 31 km = 3.1 x 10⁴ m
Therefore,
V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)
V = 6.138 x 10¹⁷ m³
Now, we find the mass:
m = density*V
m = (2620 kg/m³)(6.138 x 10¹⁷ m³)
<u>m = 1.6 x 10²¹ kg</u>
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(b)
The kinetic energy will be:
K.E = (1/2)mv²
where,
v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)
v = 3.17 x 10⁻¹⁰ m/s
Therefore,
K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²
<u>K.E = 2.536 x 10¹¹ J</u>
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(c)
For the same kinetic energy but mass = 77 kg:
K.E = (1/2)mv²
2.536 x 10¹¹ J = (1/2)(77 kg)v²
v = √(2)(2.536 x 10¹¹ J)
<u>v = 7.12 x 10⁵ m/s</u>
Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m
Answer:
Acosθ
Explanation:
The x-component of a vector is defined as :
Magnitude * cosine of the angle
Maginitude * cosθ
The magnitude is represented as A
Hence, horizontal, x - component of the vector is :
Acosθ
Furthermore,
The y-component is taken as the sin of the of the angle multiplied by the magnitude
Vertical, y component : Asinθ