You must times the area by the volume, look at it as if the area is just one of 23 layers that makes up the volume.
1960x23=45080
so no it cannot be carried as it is 5080cm^3 over the limit
Displacement (between time 0 and time 25) is the area under the velocity time curve, i.e. ∫ vdt.
Here, v(0)=10, v(25)=34 (approx.)
Therefore
displacement = (1/2)(10+34 m/s)*(25-0) s [ trapezoid area ]
=550 m
Answer:
The answer is "
"
Explanation:
please find the complete question in the attached file.
Given value:




calculating the radiation absorbed per kg:



Answer:
1) Determine the domain of the following functions: d ... 3) If g(x) = x + 3 and f(x)= x² – 2x, find the value of f(g(a)). ... 6) Given the graph of f(x) to the right, determine: ... 8) Given f(x)= x? and g(x)= 2* The inverse of g is a function, but the inverse off is ... -3(x-1)= -5 4 (-3). -3x+ 3 = y. 10) The graph of a function f (x) is given at the.
Explanation:
Answer
given,
radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m
mass of electron, M = 9.11 x 10⁻³¹ Kg
charge of electron, q₁ = 1.6 x 10⁻¹⁹ C
q₂ = 1.6 x 10⁻¹⁹ C
we know, force between two charges


F = 8.20 x 10⁻⁸ N
b) using newton's second law
F = m a
m a = 8.20 x 10⁻⁸

a = 9 x 10²² m/s²
c) speed of the electron


v² = 4.77 x 10¹²
v = 2.18 x 10⁶ m/s
d) the period of the circular motion.



T = 1.53 x 10⁻¹⁶ s