<span>A = area of styrofoam
M = mass of stryofoam = A*h*rho_s
m = mass of swimmer
Total mass = m + M = m + A*h*rho_s
Downward force = g*(total mass) = g*[m + A*h*rho_s]
The slab is completely submerged.
Buoyant force = g*(mass of water displaced) = g*[A*h*rho_w]
Equate these
g*[m + A*h*rho_s] = g*[A*h*rho_w]
m + A*h*rho_s = A*h*rho_w
A*h*[rho_w - rho_s] = m
A = m/[h*(rho_w - rho_s)]</span>
Complete question is;
A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².
What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦
Answer:
14.08°
Explanation:
The time covered will be given by the formula;
t = (2V_x•tan θ)/g
t = (2 × 24 × tan 59)/9.8
t = 8.152 s
Now, the slope of the flight path at the point of impact will be given by the formula;
tan α = V_y/V_x
We are given V_x = 24 m/s
V_y will be gotten from the formula;
v = gt
Thus;
V_y = gt
V_y = 9.8 × (8.152) = 78.89 m/s
Thus;
tan α = 78.89/24
tan α = 3.2871
α = tan^(-1) 3.2871
α = 73.08°
Thus ;
Relative angle φ = α - θ = 73.08 - 59 = 14.08°
To solve this question, we need to use the component method and split our displacements into their x and y vectors. We will assign north and east as the positive directions.
The first movement of 25m west is already split. x = -25m, y = 0m.
The second movement of 45m [E60N] needs to be split using trig.
x = 45cos60 = 22.5m
y = 45sin60 = 39.0m
Then, we add the two x and two y displacements to get the total displacement in each direction.
x = -25m + 22.5m = -2.5m
y = 0m + 39.0m
We can use Pythagorean theorem to find the total displacement.
d² = x² + y²
d = √(-2.5² + 39²)
d = 39.08m
And then we can use tan to find the angle.
inversetan(y/x) = angle
inversetan(39/2.5) = 86.3
Therefore, the total displacement is 39.08m [W86.3N]
Answer:
B = 15μT
Explanation:
In order to calculate the magnitude of the magnetic field generated by the coaxial cable you use the Ampere's law, which is given by:
(1)
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
I: current
r: distance from the wire to the point in which B is calculated
In this case you have two currents with opposite directions, which also generates magnetic opposite magnetic fields. Then, you have (but only for r > radius of the cylindrical conductor) the following equation:
(2)
I1: current of the central wire = 2.00A
I2: current of the cylindrical conductor = 3.50A
r: distance = 2.00 cm = 0.02 m
You replace the values of all parameters in the equation (2), and you use the absolute value because you need the magnitude of B, not its direction.
The agnitude of the magnetic field outside the coaxial cable, at a distance of 2.00cm to the center of the cable is 15μT
