Answer:
a=0.212 m/s²
Explanation:
Given that
q= 10⁻⁹ C
m = 5 x 10⁻⁹ kg
Magnetic filed ,B= 0.003 T
Speed ,V= 500 m/s
θ= 45°
Lets take acceleration of the mass is a m/s²
The force on the charge due to magnetic filed B
F= q V B sinθ
Also F= m a ( from Newton's law)
By balancing these above two forces
m a= q V B sinθ



a=0.212 m/s²
The work done by
along the given path <em>C</em> from <em>A</em> to <em>B</em> is given by the line integral,

I assume the path itself is a line segment, which can be parameterized by

with 0 ≤ <em>t</em> ≤ 1. Then the work performed by <em>F</em> along <em>C</em> is
![\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1%20%5Cleft%286x%28t%29%5E3%5C%2C%5Cvec%5Cimath-4y%28t%29%5C%2C%5Cvec%5Cjmath%5Cright%29%5Ccdot%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5Bx%28t%29%5C%2C%5Cvec%5Cimath%20%2B%20y%28t%29%5C%2C%5Cvec%5Cjmath%5Cright%5D%5C%2C%5Cmathrm%20dt%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E1%20%28288%283t-1%29%5E3-8%282t%2B5%29%29%20%5C%2C%5Cmathrm%20dt%20%3D%20%5Cboxed%7B312%7D)
Answer:
Therefore, the moment of inertia is:
Explanation:
The period of an oscillation equation of a solid pendulum is given by:
(1)
Where:
- I is the moment of inertia
- M is the mass of the pendulum
- d is the distance from the center of mass to the pivot
- g is the gravity
Let's solve the equation (1) for I


Before find I, we need to remember that
Now, the moment of inertia will be:
Therefore, the moment of inertia is:
I hope it helps you!
Answer: 5000N
Explanation:
The basic principle of a circular orbit is that Fg = m × ac, so as we have the mass and the centripetal acceleration (also called normal acceleration) we just have to operate. Fg = 1000kg × 5m/s² = 5000N
Answer:
power emitted is 1.75 W
Explanation:
given data
length l = 5 cm = 5 ×
m
diameter d = 0.074 cm = 74 ×
m
total filament emissivity = 0.300
temperature = 3068 K
to find out
power emitted
solution
we find first area that is π×d×L
area = π×d×L
area = π×74 ×
×5 ×
area = 1162.3892 ×
m²
so here power emitted is express as
power emitted = E × σ × area × (temperature)^4
put here all value
power emitted = 0.300× 5.67 × 1162.3892 ×
× (3068)^4
power emitted = 1.75 W