A motorcycle and its driver has a mass of 145 kg. Answer the following questions about it, using correct units.
2 answers:
Answer:
(A) 45312.5 J
(B) 5075 J
(C) 62640 J
Explanation:
m = 145 kg
(a) v = 25 m/s
The energy possessed by the body by virtue of its motion is called kinetic energy. The formula for kinetic energy is given by
K = 45312.5 Joule
(b) h = 3.5 m
The energy possessed by the body by virtue of its position is called potential energy.
P = mgh
P = 145 x 10 x 3.5
P = 5075 Joule
(c) v = 28 m/s, h = 4 m
The sum of the kinetic energy and potential energy is called mechanical energy.
M.E. = 1/2 mv^2 + mgh
M.E. = 0.5 x 145 x 28 x 28 + 145 x 10 x 4
M.E. = 56840 + 5800
M.E. = 62640 Joule
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The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
Speed of the proton = 5.02 × 10 ⁶ m /a
Angel of between the velocity and the magnetic force = 60 °
The magnitude of magnetic field B = 0.180 T
The magnitude of the magnetic force on the proton is,
Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
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Answer:
T = 0.003 s
(Period is written as T)
Explanation:
Period = time it takes for one wave to pass (measured in seconds)
frequency = number of cycles that occur in 1 second
(measured in Hz / hertz / 1 second)
Period : T
frequency : f
So, if we know that the frequency of a wave is 300 Hz, we can find the period of the wave from the relation between frequency and period
T = f =
to find the period (T) of this wave, we need to plug in the frequency (f) of 300
T =
T = 0.00333333333
So, the period of a wave that has a frequency of 300 Hz is 0.003 s
[the period/T of this wave is 0.003 s]
1) Analyze the equation
Vₓ = α + β t²
Vₓ = 3.00 + 0.100 t²
That is a quadratic equation, so the graph is a parabola.
2) Therefore, take into account that the main points to draw a parabola are:
i) vertex
ii) concavity
iii) y-intercepts
iv) x - intercepts
Also, for all graphs you need the domain and the range.
3) Find the y-intercept (t = 0)
t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00
4) Find the x-intercepts (Vₓ = 0)
Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²
⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.
5) Concavity
Since, the coefficient of t² is positive, the parabola open upwards.
6) Vertex
It is the local minimum of the equation. You can find it by the first derivative
Vₓ' = 2×0.100 t = 0 ⇒ t = 0
Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s
⇒ vertex = (0,3.00)
7) The domain is given t ∈ [0,5.00]
8) You can also build a table with several points in the domain
t =0; Vₓ = 3.00 + 0.100 (0)² = 0
t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10
t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40
t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90
t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60
t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50
9) Range: Vₓ ∈ [ 3.00, 5.50]
10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.
