Answer:
Explanation:
Let v be the velocity acquired by electron in electric field
V q = 1/2 m v²
V is potential difference applied on charge q , m is mass of charge , v is velocity acquired
2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²
v² = 844 x 10¹²
v = 29.05 x 10⁶ m /s
Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .
Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron
= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶
= 79.02 x 10⁻¹³ N .
Minimum force will be zero when electron moves along the direction of magnetic field .
Answer:
(a) Wavelength is 0.436 m
(b) Length is 0.872 m
(c) 11.518 m/s
Solution:
As per the question:
The eqn of the displacement is given by:
(1)
n = 4
Now,
We know the standard eqn is given by:
(2)
Now, on comparing eqn (1) and (2):
A = 1.22 cm
K = 

where
A = Amplitude
K = Propagation constant
= angular velocity
Now, to calculate the string's wavelength,
(a) 
where
K = propagation vector


(b) The length of the string is given by:


(c) Now, we first find the frequency of the wave:



Now,
Speed of the wave is given by:


Answer:
10m/s
Explanation:
Using the law of conservation of momentums
M1u1+m2u2 = (m1+m2)v
Substitute.
4000(10)+1500(10) = (4000+1500)v
40,000+15,000 = 5,500v
55000 = 5500v
v = 55000/5500
v= 10m/s
Hence the velocity of the truck after Collision is 10m/s
Differentiate the components of position to get the corresponding components of velocity :


At <em>t</em> = 5.0 s, the particle has velocity


The speed at this time is the magnitude of the velocity :

The direction of motion at this time is the angle
that the velocity vector makes with the positive <em>x</em>-axis, such that
