Answer:
t=67.7s
Explanation:
From this question we know that:
Vo = 6m/s
a = 1.8 m/s2
D = 1500m
And we also know that:
Replacing the known values:
Solving for t we get 2 possible answers:
t1 = -44.3s and t2 = 67.7s Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:
t = 67.7s
Circumference C=2πr
<span>C=2π(1.5x10^8)=9.42x10^8 </span>
<span>In 365 Days there are 8760hr </span>
<span>V=distance/time </span>
<span>V=(9.42x10^8)/8760=107534.2km/hr </span>
<span>Answer:
Assuming that I understand the geometry correctly, the combine package-rocket will move off the cliff with only a horizontal velocity component. The package will then fall under gravity traversing the height of the cliff (h) in a time T given by
h = 0.5*g*T^2
However, the speed of the package-rocket system must be sufficient to cross the river in that time
v2 = L/T
Conservation of momentum says that
m1*v1 = (m1 + m2)*v2
where m1 is the mass of the rocket, v1 is the speed of the rocket, m2 is the mass of the package, and v2 is the speed of the package-rocket system.
Expressing v2 in terms of v1
v2 = m1*v1/(m1 + m2)
and then expressing the time in terms of v1
T = (m1 + m2)*L/(m1*v1)
substituting T in the first expression
h = 0.5*g*(m1 + m2)^2*L^2/(m1*v1)^2
solving for v1, the speed before impact is given by
v1 = sqrt(0.5*g/h)*(m1 + m2)*L/m1</span>
The momentum of the car is 4.4x10^3 kg•m/sec
<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>
