(a) Fx = 1.464 N
(b) Fy = 1.952 N
(c) F(x, y) = 1.464 i + 1.952 j
Given
Mass = 1kg
Acceleration = 2.44 m/s2
Angle with positive X axis = 53°
As we know
F = ma
By substituting value
F= 1×2.44 N
F= 2.44 N
(a) Component of force in X direction
Fx = F Cosθ
Fx = 2.44 Cos(53°)
Fx = 2.44 × 0.60 = 1.464 N
(b) Component of force in Y direction
Fy = F Sinθ
Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N
(c) Net force in vector notation
F(x, y) = 1.464 i + 1.952 j
Thus we got net force.
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Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
Answer: Heterogeneous mixture - the parts are not uniformly mixed.
A mixture contains components having distinct chemical properties. There are two types of mixtures: homogeneous and heterogeneous. In a homogeneous mixture there is uniform distribution of components. we cannot distinguish one portion of the mixture from another. for example salt mixed in water. In heterogeneous mixture, the components are not uniformly mixed. hence, we are able to distinguish different parts of a mixture, like the mixture of iron, sand and salt given in this question.
Answer:
λ = 5940 Angstroms
Explanation:
This is an exercise of the relativistic Doppler effect
f’= f √((1- v / c) / (1 + v / c))
Where the speed in between the strr and the observer is positive if they move away
Let's use the relationship
c = λ f
f = c /λ
We replace
c /λ’ = c /λ √ ((1- v / c) / (1 + v / c))
λ = λ’ √ ((1- v / c) / (1 + v / c))
Let's calculate
v = 0.01 c
v = 0.01 3 10⁸
v= 3 10⁶ m / s
λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]
λ = 6000 √ [0.99 / 1.01]
λ = 5940 Angstroms
Answer:
2156 J
Explanation:
From the question,
Work done = Combined mass of the bucket and water×height×gravity.
W = (M+m)hg............................. Equation 1
Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.
Given: M = 20 kg, m = 2 kg, h = 10 m
Constant: g = 9.8 m/s²
Substitute these value into equation 1
W = (20+2)×10×9.8
W = 22×98
W = 2156 J