The correct answer is :
Unit vectors I and j along the x-axis and y-axis, respectively, define the Cartesian coordinate system. The radial unit vector r, which indicates the direction from the origin, and the unit vector t, which is orthogonal (perpendicular) to the radial direction, together create the polar coordinate system.
We can obtain the horizontal component by applying the trigonometric identity of Cos(Ф), and if we obtain the component on the x axle, such as 22000 (m)×Cos(51°) = x, we may determine that x = 13845.05 metres. We need to obtain the vector components because we already know the distance and the angle.
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Answer:
3.1216 m/s.
Explanation:
Given:
M1 = 0.153 kg
v1 = 0.7 m/s
M2 = 0.308 kg
v2 = -2.16 m/s
M1v1 + M2v2 = M1V1 + M2V2
0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2
= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2
0.153V1 + 0.308V2 = -0.55818. i
For the velocities,
v1 - v2 = -(V1 - V2)
0.7 - (-2.16) = -(V1 - V2)
-(V1 - V2) = 2.86
V2 - V1 = 2.86. ii
Solving equation i and ii simultaneously,
V1 = 3.1216 m/s
V2 = 0.2616 m/s
Ok i will answer for real this time. Please give me brainliest.
<span>The Answerr is:
5.12*10^15. Since e=h*f, f=e/h. 3.4*10^(-18)/h.
</span>i am so sorry i was doing a challenge and i needed answers to get 100 pts.
Hope I Helped
~TeenOlafLover <3
Answer:
0.191 s
Explanation:
The distance from the center of the cube to the upper corner is r = d/√2.
When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.
Sum of the torques:
∑τ = Iα
Fr cos θ = Iα
(k r sin θ) r cos θ = Iα
kr² sin θ cos θ = Iα
k (d²/2) sin θ cos θ = Iα
For a cube rotating about its center, I = ⅙ md².
k (d²/2) sin θ cos θ = ⅙ md² α
3k sin θ cos θ = mα
3/2 k sin(2θ) = mα
For small values of θ, sin θ ≈ θ.
3/2 k (2θ) = mα
α = (3k/m) θ
d²θ/dt² = (3k/m) θ
For this differential equation, the coefficient is the square of the angular frequency, ω².
ω² = 3k/m
ω = √(3k/m)
The period is:
T = 2π / ω
T = 2π √(m/(3k))
Given m = 2.50 kg and k = 900 N/m:
T = 2π √(2.50 kg / (3 × 900 N/m))
T = 0.191 s
The period is 0.191 seconds.
They were going at a velocity 4.07m/s
<u>Explanation:</u>
Distance s =5 m
initial velocity u= 0.8 m/s
Acceleration a =1.6m/s2
We have to calculate the velocity with which they were going afterwards i.e final velocity.
Use the equation of motion
They were going with a velocity 4.07 m/s afterwards.
