Answer:
The gravitational potential energy equals the work needed to lift the object.
Explanation:
here we know that

work done is given as

Potential energy is given as

force due to gravity is given as

now here if we plug in the value of distance and force in the formula of work done then we will have

so here we got

so we can concluded that
The gravitational potential energy equals the work needed to lift the object.
<span>E = h x f </span>
<span>. . . then : </span>
<span>f = E / h </span>
<span>f = 4,41•10^-19 / 6,62•10^-34 </span>
<span>f = 6,66•10^14 Hz (s^-1) </span>
<span>b/ What is the wavelength of this light ? </span>
<span>- - - - - - - - - - - - - - - - - - - - - - - - - - - - </span>
<span>λ = c / f </span>
<span>λ = 3•10^8 / 6,66•10^14 </span>
<span>λ = 4,50•10^-7 m </span>
Answer:
The equilibrium temperature is
21.97°c
Explanation:
This problem bothers on the heat capacity of materials
Given data
specific heat capacities
copper is Cc =390 J/kg⋅C∘,
aluminun Ca = 900 J/kg⋅C∘,
water Cw = 4186 J/kg⋅C∘.
Mass of substances
Copper Mc = 235g
Aluminum Ma = 135g
Water Mw = 825g
Temperatures
Copper θc = 255°c
Water and aluminum calorimeter θ1= 16°c
Equilibrium temperature θf =?
Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water
McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)
Substituting our data into the expression we have
235*390(255-θf)=
(135*900+825*4186)(θf-16)
91650(255-θf)=(3574950)(θf-16)
23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6
Collecting like terms and rearranging
23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf
8.2*10^6=3.66*10^6θf
θf=80.5*10^6/3.6*10^6
θf =21.97°c
Answer:
3.62 m and - 1.4 m
Explanation:
Consider a location towards the positive side of x-axis beyond the location of charge Q₂
x = distance of the location from charge Q₂
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location


x = 1.62 m
So location is 2 + 1.62 = 3.62 m
Consider a location towards the negative side of x-axis beyond the location of charge Q₁
x = distance of the location from charge Q₁
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location


x = - 1.4 m
Answer:
1.36
Explanation:
= Index of refraction of air = 1
= Index of refraction of plastic = ?
i = angle of incidence in air = 32.0° deg
r = angle of refraction in plastic = 23.0° deg
Using Snell's law
Sini =
Sinr
(1) SIn32.0 =
Sin23.0
= 1.36