m = 43.2 kg
Explanation:
volume of sphere = (4/3)pi(r)^3
= (4/3)(3.14)(2 m)^3
= 33.5 m^3
density = mass/volume
or solving for mass m,
m = (density)×(volume)
= (1.29 kg/m^3)(33.5 m^3)
= 43.2 kg
Answer:
13.6 cm
Explanation:
From Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
In the air, n₁ = 1, and light from the horizon forms a 90° angle with the vertical, so sin θ₁ = sin 90° = 1.
Given n₂ = 4/3:
1 = 4/3 sin θ
sin θ = 3/4
If x is the radius of the circle, then sin θ is:
sin θ = x / √(x² + 12²)
sin θ = x / √(x² + 144)
Substituting:
3/4 = x / √(x² + 144)
9/16 = x² / (x² + 144)
9/16 x² + 81 = x²
81 = 7/16 x²
x ≈ 13.6
3. Sand has a low specific heat compared to air
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
Answer:
the frequency of this mode of vibration is 138.87 Hz
Explanation:
Given;
length of the copper wire, L = 1 m
mass per unit length of the copper wire, μ = 0.0014 kg/m
tension on the wire, T = 27 N
number of segments, n = 2
The frequency of this mode of vibration is calculated as;
Therefore, the frequency of this mode of vibration is 138.87 Hz
