Answer:
when all else remains the same, what effect would decreasing the focal have known a convex lens
Explanation:
It would cause the lens to produce only real images
( Hope This Will Help U Out!!)
Answer:
7.74m/s
Explanation:
Mass = 35.9g = 0.0359kg
A = 39.5cm = 0.395m
K = 18.4N/m
At equilibrium position, there's total conservation of energy.
Total energy = kinetic energy + potential energy
Total Energy = K.E + P.E
½KA² = ½mv² + ½kx²
½KA² = ½(mv² + kx²)
KA² = mv² + kx²
Collect like terms
KA² - Kx² = mv²
K(A² - x²) = mv²
V² = k/m (A² - x²)
V = √(K/m (A² - x²) )
note x = ½A
V = √(k/m (A² - (½A)²)
V = √(k/m (A² - A²/4))
Resolve the fraction between A.
V = √(¾. K/m. A² )
V = √(¾ * (18.4/0.0359)*(0.395)²)
V = √(0.75 * 512.53 * 0.156)
V = √(59.966)
V = 7.74m/s
Answer:
390 J
Explanation:
m = 3 kg
u = 16 i + 2 j
(a) Magnitude of velocity = 
= 16.1245 m/s
KEi = 1/2 m v^2 = 0.5 x 3 x 16.1245 = 390 J
(b) v = 18 i + 14 j
Magnitude of velocity = 
= 22.804 m/s
KEf = 1/2 m v^2 = 0.5 x 3 x 22.804 = 780 J
According to the work energy theorem
Work done = change in KE = KEf - KEi = 780 - 390 = 390 J
<span>Mass of the electron = 9.1 x 10 ^ -28g = 9.1 x 10 ^ -31kg
Velocity of the electron = 1.7 x 10 ^ 4
We have Planck Constant h = 6.626 x 10 ^ -34
Wavelength of the electron w = h/mv
w = 6.626 x 10 ^ -34 / ((9.1 x 10 ^ -31)(1.7 x 10 ^ 4))
= 6.626 x 10 ^ -34 / 15.47 x 10 ^ -27
= 0.428312 x 10 ^ -7
= 4.28 x 10 ^ -8 m</span>
Answer:
3.5434 eV
Explanation:
For a particle with kinetic energy E and mass m , the wavelength associated is given by the following relation,
E = 
Putting the values we get
E = 
=1.063 x 10⁻²¹ J
= .0066 eV.
Energy of¹light in terms of eV
= 1244 / 350 = 3.55 eV.
Work function = 3.55 - 0.0066 = 3.5434 eV.
