Answer:
the time interval that an earth observer measures is 4 seconds
Explanation:
Given the data in the question;
speed of the spacecraft as it moves past the is 0.6 times the speed of light
we know that speed of light c = 3 × 10⁸ m/s
so speed of spacecraft v = 0.6 × c = 0.6c
time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds
Now, from time dilation;
t = Δt₀ / √( 1 - ( v² / c² ) )
t = Δt₀ / √( 1 - ( v/c )² )
we substitute
t = 3.2 / √( 1 - ( 0.6c / c )² )
t = 3.2 / √( 1 - ( 0.6 )² )
t = 3.2 / √( 1 - 0.36 )
t = 3.2 / √0.64
t = 3.2 / 0.8
t = 4 seconds
Therefore, the time interval that an earth observer measures is 4 seconds
Answer:
a) F = 64.30 N, b) θ = 121.4º
Explanation:
Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components
let's use trigonometry
Force F1
sin 170 = F_{1y} / F₁
cos 170 = F₁ₓ / F₁
F_{1y} = F₁ sin 170
F₁ₓ = F₁ cos 170
F_{1y} = 100 sin 170 = 17.36 N
F₁ₓ = 100 cos 170 = -98.48 N
Force F2
sin 30 = F_{2y} / F₂
cos 30 = F₂ₓ / F₂
F_{2y} = F₂ sin 30
F₂ₓ = F₂ cos 30
F_{2y} = 75 sin 30 = 37.5 N
F₂ₓ = 75 cos 30 = 64.95 N
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = -98.48 +64.95
Fₓ = -33.53 N
Y axis
F_y = F_{1y} + F_{2y}
F_y = 17.36 + 37.5
F_y = 54.86 N
a) the magnitude of the resultant vector
let's use Pythagoras' theorem
F = Ra Fx ^ 2 + Fy²
F = Ra 33.53² + 54.86²
F = 64.30 N
b) the direction of the resultant
let's use trigonometry
tan θ’= F_y / Fₓ
θ'= 
θ'= tan⁻¹ (54.86 / (33.53)
θ’= 58.6º
this angle is in the second quadrant
The angle measured from the positive side of the x-axis is
θ = 180 -θ'
θ = 180- 58.6
θ = 121.4º
Physical activity is movement that is carried out by the skeletal muscles that requires energy. In other words, any movement one does is actually physical activity. Exercise, however, is planned, structured, repetitive and intentional movement intended to improve or maintain physical fitness.
:<span> </span><span>The gradient of the curve 1/x at x=2 is m = -¼
We may choose any length of line to represent the direction of the slope (direction vector) at that point. We could choose a line for which x = 2 and then y would have to be -½ so that the gradient is still = -½/2 = -¼. It is simply convenient to choose a unit length for x, making y = -¼ The length of the resultant of x and y is √(1²+¼²) = √(17/16) = √(17)/4 which is a direction vector. If we had taken the direction vector to be (2, ½) then we would have a resultant direction vector of √17/2. It doesn't really matter what length the direction vector is - it's job is only to show the direction. So their choice of 1 is quite arbitrary but convenient, since it is easy to work with units – that's why we use units!
Now, we know that the magnitude of the velocity vector must be 5 and the magnitude of our direction vector at the moment is √(17)/4. We therefore need to multiply this direction vector by 20/√(17) to get 5 – just try it : √(17)/4 × 20/√(17) = 5.
We could equally well have done this with (2, ½) and would have got 2½ for lambda.</span>
