if an object is moving with constant speed then its distance time graph will be a straight line. a) along time axis b) along dis
1 answer:
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Answer:
a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º
This is 8º north west
Explanation:
This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately
To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.
First vector A = 2 to 20º north west
Measured from the positive x axis is θ = 180 -20 = 160º
We use trigonometry to find the components
Cos 20 = Aₓ / A
sin 20 = / A
Aₓ = A cos 160 = 2 cos 160
= A sin160 = 2 sin160
Aₓ = -1,879 mi
= 0.684 mi
Second vector B = 4 mi 10º west of the south
Angle θ = 270 - 10 = 260º
cos 2600 = Bₓ / B
sin 260 = / B
Bₓ = B cos 260
= B sin 260
Bₓ = 4 cos 260
= 4 sin 260
Bₓ = -0.6946mi
= - 3,939 mi
Third vector C = 3 mi to 15 north east
cos 15 = Cₓ / C
sin15 = / C
Cₓ = C cos 15
= C sin15
Cₓ = 3 cos 15
= 3 sin 15
Cₓ = 2,898 mi
= 0.7765 mi
Now we can find the final position of the person
X = Aₓ + Bₓ + Cₓ
X = -1.879 -0.6949 + 2.898
X = 0.3241 mi
Y = +
+
Y = 0.684 - 3.939 +0.7765
Y = -2.4785 mi
a) We use Pythagoras' theorem
R = √ (x2 + y2)
R = √ (0.3241 2 + (-2.4785) 2)
R = 2.4996 mi
R = 2.5 mi
b) let's use trigonometry
Tan θ = y / x
Tanθ = -2.4785 / 0.3241
θ = tan⁻¹ (-7,647)
θ = -82
Measured from the positive side of the x axis is Te = 360 - 82 = 278º
(90-82) south east
To return to your case you must walk in the opposite direction or Te = 98º
This is 8º north west
Answer:
the magnitude of the velocity of one particle relative to the other is 0.9988c
Explanation:
Given the data in the question;
Velocities of the two particles = 0.9520c
Using Lorentz transformation
Let relative velocity be W, so
v = ( u + v ) / ( 1 + ( uv / c²) )
since each particle travels with the same speed,
u = v
so
v = ( u + u ) / ( 1 + ( u×u / c²) )
v = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )
we substitute
v = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )
v = 1.904c / ( 1 + 0.906304 )
v = 1.904c / 1.906304
v = 0.9988c
Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c
Answer:
Generally given that the electric field is negative it mean that its direction is opposite to that of the force
Explanation:
From the question we are told that
The charge on the small object is
The force is
Generally the magnitude of the electric field is mathematically represented as
=>
=>
Generally given that the electric field is negative it mean that its direction is opposite to that of the force