Why does a lorry travelling at 30mph have more kinetic energy than a car travelling at the same velocity?
1 answer:
Answer:
Because it has more mass
Explanation:
To understand this, think about the equation of kinetic energy
KE = m
Kinetic energy depends on both the velocity (v) as well as the mass (m).
Because a lorry is bigger and heavier than a car, it will have more mass. With more mass, at the same velocity the lorry with have more kinetic energy.
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Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N
Answer:
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.
Explanation:
Geometrically speaking, the distance between the rocket and the observer (), measured in kilometers, can be represented by a right triangle:
(1)
Where:
- Horizontal distance between the rocket and the observer, measured in kilometers.
- Vertical distance between the rocket and the observer, measured in kilometers.
The angle of elevation of the rocket (), measured in sexagesimal degrees, is defined by the following trigonometric relation:
(2)
If we know that , then the expression is:
And the rate of change of this angle is determined by derivatives:
Where:
- Rate of change of the angle of elevation, measured in sexagesimal degrees.
- Vertical speed of the rocket, measured in kilometers per hour.
If we know that and
, then the rate of change of the angle of elevation is:
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.