Well, the figure seems to report that velocity is measured in m/s²... That label should say m/s. (Unless of course this is the graph of acceleration over time, but then the answer would probably be more complicated than the given choices.)
If the graph indeed shows velocity, and the unit is just a typo, then the displacement from A to D is equal to the area under the curve.
From A to B, the area is of a triangle with height 4 m/s and base 1 s, hence the area is 1/2 • (4 m/s) • (1 s) = 2 m.
From B to C, it's a rectangle with length 3 s and height 4 m/s, hence with area (3 s) • (4 m/s) = 12 m.
From C to D, it's a trapezoid with "height" 2 s and bases 4 m/s and 2 m/s, hence with area 1/2 • (4 m/s + 2 m/s) • (2 s) = 6 m.
The total displacement is then 2 m + 12 m + 6m = 20 m.
Frequency = (speed) divided by (wavelength)
= (3 x 10⁸ m/s) / (700 x 10⁻⁹ m)
= (3 / 700) x 10¹⁷ ( per sec )
= 4.286 x 10¹⁴ per sec = 4.286 x 10¹⁴ Hz.
= 42,860 GHz .
Answer:
a, 1.775s
b, 17.04μC
c, 1.28s
Explanation:
Given
R = 1.25MΩ
C = 1.42µF
ε = 12.0 V
q = 8.78 µC
Time constant, τ = RC
τ = (1.25*10^6) * ( 1.42*10^-6)
τ = 1.775s
q• = εC
q• = 12 * 1.42*10^-6
q• = 17.04*10^-6C
q• = 17.04μC
Time t =
q = q• [1 - e^(t/τ)]
t = τIn[q•/(q•-q)]
t = 1.775In[17.04μC/(17.04μC-8.78μC)]
t = 1.775In(2.06)
t = 1.775*0.723
t = 1.28s
This question is incomplete, the complete question is;
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).
Answer:
the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
Explanation:
Given the data in the question and image below and as illustrated in the second image;
distance S = 40 m
V
= 54 km/hr
V
= 72 km/hr
α = 100 m
now, angular velocity of Bxy will be;
ω = V / α
so, we substitute
ω = ( 54 × 1000/3600) / 100
ω = 15 / 100
ω = 0.15 rad/s
Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
Answer:
Explanation:
As we know that the tension in two strings are
now we have
so we can say
also we have
now divide two equations
