The answer to the question is C
Answer:
Vy = 0
y(max) = 137,76 m
x(max) = 318 m
Explanation:
We are dealing with a projectile movement
And from problem statement we know:
θ = 60⁰ then sin θ = sin 60⁰ = √3/2 and cos 60⁰ = 1/2 and g = 9.8 m/sec²
V₀ = 60 m/sec
V₀x = V₀ *cosθ Vx = V₀x Vx = Constant then Vx = V₀ *cosθ
Then Vx = 60* 1/2 Vx = 30 m/sec
V₀y = V₀ *sinθ ⇒ V₀y =60*√3/2 ⇒ V₀y = = 30*√3 m/sec
Vy = V₀y * sin θ - g*t
When Vy = 0 (maximum height point) we are half of the way for the ball to hit the ground, then
Vy = 0 ⇒ V₀y - g*t = 0 ⇒ 30*√3 (m /sec) = 9,8 (m/sec²)* t
t = 30*√3/9.8 t = 5.30 sec
y(max) = y₀ + V₀y*t - 1/2 * g*t²
By substitution:
y (max) = 0 + 30*√3 * 5.30 - 0,5* 9.8* (5.3)²
y(max) = 275,40 - 137,64
y(max) = 137,76 m
And finally x(max)
x(max) = Vx *t = 30* 2*5,3
x(max) = 318 m
Answer:
E = k λ₀ / x₀, the field is in thenegative direction of the x axis (-x)
Explanation:
In this problem the electric field of a line of charge is requested, the expression for the electric field is
E = k ∫ dq / r²
where k is the Coulomb constant that you are worth 9 10⁹ N m²/C², that the charge and r the distance to the point of interest, in this case it is the origin (x = 0)
let's use the definite linear density
λ₀ = dq / dx
dq = λ₀ dx
we replace and integrate
E = k λ₀ ∫ dx / x²
E = k λ₀ ( -1 / x)
we evaluate the integral from the lower limit of load x = x₀ to the upper limit x = ∞
E = - k λ₀ (1 /∞ - 1 / x₀)
E = k λ₀ / x₀
as the field is positive the direction is away from the charges, so it is in the negative direction of the x axis (-x)
Answer:
Acceleration = 0.0282 m/s^2
Distance = 13.98 * 10^12 m
Explanation:
we will apply the energy theorem
work done = ΔK.E ( change in Kinetic energy ) ---- ( 1 )
<em>where :</em>
work done = p * t
= 15 * 10^6 watts * ( 1 year ) = 473040000 * 10^6 J
( note : convert 1 year to seconds )
and ΔK.E = 1/2 mVf^2 given ; m = 1200 kg and initial V = 0
<u>back to equation 1 </u>
473040000 * 10^6 = 1/2 mv^2
Vf^2 = 2(473040000 * 10^6 ) / 1200
∴ Vf = 887918.92 m/s
<u>i) Determine how fast the rocket is ( acceleration of the rocket )</u>
a = Vf / t
= 887918.92 / ( 1 year )
= 0.0282 m/s^2
<u>ii) determine distance travelled by rocket </u>
Vf^2 - Vi^2 = 2as
Vi = 0
hence ; Vf^2 = 2as
s ( distance ) = Vf^2 / ( 2a )
= ( 887918.92 )^2 / ( 2 * 0.0282 )
= 13.98 * 10^12 m