Answer:
The answer is B. When the magnet is placed on a globe to correctly align with Earth’s magnetic field, it is considered to be suspended freely. The Earth has geographical poles as well with North and South poles. Since unlike poles attract, the South Pole of the magnet will be attracted to the geographical North.
Explanation:
Answer:
Proton: v=0.689 m/s
Neutron: v=0.688 m/s
Electron: v=1265.078 m/s
Alpha particle: v=0.173 m/s
Explanation:
De Broglie equation allows you to calculate the “wavelength” of an electron or any other particle or object of mass m that moves with velocity v:
λ=
h is the Planck constant: 6.626×10⁻³⁴
We know that the wavelength of the particle is 575 nm (575×10⁻⁹m), so we find the velocity v for each particle:
λ=
v=h÷(mλ)
<u>Proton:</u>
m=1.673×10⁻²⁴ g · 
=1.673×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(1.673×10⁻²⁷ kg×575×10⁻⁹m)
v=0.689 m/s
<u>Neutron:</u>
m=1.675×10⁻²⁴ g · =1.675×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(1.675×10⁻²⁷ kg×575×10⁻⁹m)
v=0.688 m/s
<u>Electron:</u>
m= 9.109×10⁻²⁸ g · =9.109×10⁻³¹ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(9.109×10⁻³¹ kg×575×10⁻⁹m)
v=1265.078 m/s
<u>Alpha particle:</u>
m=6.645×10⁻²⁴ g · =6.645×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(6.645×10⁻²⁷ kg×575×10⁻⁹m)
v=0.173 m/s
Answer:
3.62 V
Explanation:
L = 80 cm = 0.8 m
f = 15 rps
B = 60 m T = 0.060 T
ω = 2 x π x f = 2 x 3.14 x 15 = 94.2 rad/s
v = r ω
here, r be the radius of circular path. Here r = length of rod = L
v = 0.80 x 94.2 = 75.36 m/s
The motional emf is given by
e = B v L = 0.060 x 75.36 x 0.8 = 3.62 V
A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π
b. vy = 4π cos (4πt + π/2)
vy = 0
c. m = sin(4πt + π/2) / [<span>πt + cos(4πt + π/2)]
d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f. t = -0.35
g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax
h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax
i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)
h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt
k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.
