Max height occurs when v = 0.
v(t) = ds(t)/dt
v(t) = 80 - 32t
0 = 80 - 32t
t = 5/2
s(5/2) = 80(5/2) - 16(5/2)^2
s(5/2) = 100
Answer: 100 ft
96 = 80t - 16t²
t = 3, 2
(80 ± √256) / 32 using the quadratic equation.
v(2) = 16
v(3) = -16
<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:
a)
b) north edge will rise up
Explanation:
torque on the coil is given as
where N is number of loop = 9 loops
i is current = 7.80 A
-B -earth magnetic field = 
A- area of circular coil
A =0.015 m2
PUTITNG ALL VALUE TO GET TORQUE
b) north edge will rise up
Answer:
h = 48.077 ft
Explanation:
given,
distance between two observer = 300 ft
angle of elevation to top pole = 16° and 20°
height of the flagpole = ?
now,
Let h be the height of the flagpole
Let x be the distance of the pole
now,
again applying
h = 48.077 ft
Gas or fog in the air due to evaporation.
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