Seconds squared is the time unit of acceleration. It represents the change in distance units per second per second. For example, 3 m/sec² means a distance covering 3 meters in the first second, then 9 meters in the 2nd second, and 37 meters in the third second. (3^1, 3^2, 3^3).
Acceleration is part of Newton's 2nd law: force = mass x acceleration. Units of work: joule = kg·m²/s², and power: watts = kg·m²/s³ all contain accelerations.
Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants ( = 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
= - gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
The final velocity of the train at the end of the given distance is 7.81 m/s.
The given parameters;
- initial velocity of the train, u = 6.4 m/s
- acceleration of the train, a = 0.1 m/s²
- distance traveled, s = 100 m
The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;
v² = u² + 2as
v² = (6.4)² + (2 x 0.1 x 100)
v² = 60.96
v = √60.96
v = 7.81 m/s
Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.
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Answer:
The answer is B. hope this helps
Explanation:
Answer:
Therefore,
The magnitude of the force per unit length that one wire exerts on the other is
Explanation:
Given:
Two long, parallel wires separated by a distance,
d = 3.50 cm = 0.035 meter
Currents,
To Find:
Magnitude of the force per unit length that one wire exerts on the other,
Solution:
Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,
where,
Substituting the values we get
Therefore,
The magnitude of the force per unit length that one wire exerts on the other is