my hypothesis is that If you drop a piece of buttered toast, it will land butter side down.
I tested it by dropping 10 pieces of buttered toast off the table and noted on which side it landed
It could be falsified cause I just made all of this up. In essence, it's like flipping a coin, 50/50 chance so I could say that 5 landed butter up and 5 landed butter down.
<span>
The needle of a compass will always lies along the magnetic
field lines of the earth.
A magnetic declination at a point on the earth’s surface
equal to zero implies that
the horizontal component of the earth’s magnetic field line
at that specific point lies along
the line of the north-south magnetic poles. </span>
The presence of a
current-carrying wire creates an additional <span>
magnetic field that combines with the earth’s magnetic field.
Since magnetic
<span>fields are vector quantities, therefore the magnetic field of
the earth and the magnetic field of the vertical wire must be
combined vectorially. </span></span>
<span>
Where:</span>
B1 = magnetic field of
the earth along the x-axis = 0.45 × 10 ⁻ ⁴ T
B2 = magnetic field due to
the straight vertical wire along the y-axis
We can calculate for B2
using Amperes Law:
B2 = μ₀ i / [ 2 π R ]
B2 = [ 4π × 10 ⁻ ⁷ T • m / A ] ( 36 A ) / [ 2 π (0.21 m ) ] <span>
B2 = 5.97 × 10 ⁻ ⁵ T = 0.60 × 10 ⁻ ⁴ T </span>
The angle can be
calculated using tan function:<span>
tan θ = y / x = B₂ / B₁ = 0.60 × 10 ⁻ ⁴ T / 0.45 × 10 ⁻ ⁴ T <span>
tan θ = 1.326</span></span>
θ = 53°
<span>
<span>The compass needle points along the direction of 53° west of
north.</span></span>
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
