Answer:
The final velocity of the vehicle is 10.39 m/s.
Explanation:
Given;
acceleration of the vehicle, a = 2.7 m/s²
distance moved by the vehicle, d = 20 m
The final velocity of the vehicle is calculated using the following kinematic equation;
v² = u² + 2ah
v² = 0 + 2 x 2.7 x 20
v² = 108
v = √108
v = 10.39 m/s
Therefore, the final velocity of the vehicle is 10.39 m/s.
Answer:
potential energy PE = M g h
KE at bottom = 1/2 M V^2
Regardless of the slope of the slide the change in energy is the same
1/2 V^2 = g h
V = (2 g h)^1/2 = (2 * 9.8 m/s^2 * 10 m)^1/2 = 14 m / s
Perhaps the question says that h = 55 * .1 = 5.5 m
Then V = (2 * 9.8 * 5.5) = 10.4 m/s
Answer:
a) 3.0×10⁸ m
b) 0 m
Explanation:
Displacement is the distance from the starting position to the final position.
a) In half a year, the Earth travels from one point on the circle to the point on the exact opposite side of the circle (from 0° to 180°). The distance between the points is the diameter of the circle.
x = 2r
x = 2 (1.5×10⁸ m)
x = 3.0×10⁸ m
b) In a full year, the Earth travels one full revolution, so it ends up back where it started. The displacement is therefore 0 m.
Answer:
a = 2.275 10⁻⁴ m
Explanation:
This is a diffraction problem that is described by the equation
a sin θ = m λ
The first dark minimum occurs for m = 1
a = λ / sin θ
The angle can be found by trigonometry,
tan θ = y / x
θ = tan⁻¹ y / x
Let's reduce the magnitudes to the SI system
y = 8.24 mm = 8.24 10⁻³ m
λ = 625 nm = 625 10⁻⁹ m
θ = tan⁻¹ 8.24 10⁻³ / 3.00
θ = 0.002747 rad
We calculate
a = 625 10⁻⁹ / sin 0.002747
a = 2.275 10⁻⁴ m
Answer:
(a) a = (2i + 4.5j) m/s^2
(b) r = ro + vot + (1/2)at^2
Explanation:
(a) The acceleration of the particle is given by:
vo: initial velocity = (3.00i -2.00j) m/s
v: final velocity = (9.00i + 7.00j) m/s
t = 3s
by replacing the values of the vectors and time you obtain:
![\vec{a}=\frac{1}{3s}[(9.00-3.00)\hat{i}+(7.00-(-2.00))\hat{j}]\\\\\vec{a}=(2\hat{i}+4.5\hat{j})m/s^2](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D%3D%5Cfrac%7B1%7D%7B3s%7D%5B%289.00-3.00%29%5Chat%7Bi%7D%2B%287.00-%28-2.00%29%29%5Chat%7Bj%7D%5D%5C%5C%5C%5C%5Cvec%7Ba%7D%3D%282%5Chat%7Bi%7D%2B4.5%5Chat%7Bj%7D%29m%2Fs%5E2)
(b) The position vector is given by:
where vo = (3.00i -2.00j) m/s and a = (2.00i + 4.50j)m/s^2
