(a) The focal length of the lens is 6.86 cm, and the lens is converging.
(b) The height of the image is 6.375 mm, and it is inverted.
To solve this problem, we need to use the lens formula and magnification formula, which are given below:
The lens formula is:
[tex]\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}[/tex]
where:
[tex]f[/tex] is the focal length of the lens[tex]d_o[/tex] is the object distance (distance from the object to the lens)[tex]d_i[/tex] is the image distance (distance from the image to the lens)The magnification formula is:
[tex]m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}[/tex]
where:
[tex]m[/tex] is the magnification[tex]h_i[/tex] is the height of the image[tex]h_o[/tex] is the height of the objectNow we can solve the problem step-by-step.
Part (a): Finding the focal length
Given:
Object distance, [tex]d_o = 16.0[/tex] cm
Image distance, [tex]d_i = 12.0[/tex] cm (since the image is on the same side as the object, [tex]d_i[/tex] is positive)
Using the lens formula:
[tex]\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}[/tex]
Substituting the given values:
[tex]\frac{1}{f} = \frac{1}{16.0} + \frac{1}{12.0}[/tex]
Calculating the right-hand side:
[tex]\frac{1}{f} = \frac{3 + 4}{48} = \frac{7}{48}[/tex]
Therefore, the focal length of the lens, [tex]f[/tex], is:
[tex]f = \frac{48}{7} = 6.86[/tex] cm
Since the focal length is positive, the lens is converging.
Part (b): Finding the height and orientation of the image
Given:
Object height, [tex]h_o = 8.50[/tex] mm
Object distance, [tex]d_o = 16.0[/tex] cm
Image distance, [tex]d_i = 12.0[/tex] cm
Using the magnification formula:
[tex]m = -\frac{d_i}{d_o}[/tex]
Substituting the given values:
[tex]m = -\frac{12.0}{16.0} = -0.75[/tex]
Using the magnification formula for height:
[tex]m = \frac{h_i}{h_o}[/tex]
Therefore:
[tex]-0.75 = \frac{h_i}{8.50}[/tex]
Giving:
[tex]h_i = -0.75 \times 8.50 = -6.375[/tex] mm
Since the height of the image, [tex]h_i[/tex], is negative, the image is inverted.
A 89.1 g piece of copper (specific heat 0.38 J/g･°C) is heated and then placed into 400.0 g of water initially at 20.7°C. The water increases in temperature to 22.2°C. What is the initial temperature of the copper? (The specific heat of water is 4.18 J/g･°C).
Answer:
The initial temperature of copper is 96.3 °C
Explanation:
Specific heat capacity is the energy needed to raise the temperature of one gram of material by one degree celsius. The energy absorbed or released by a material during temperature change can be calculated by the below formula:
[tex]Q=mc(T_{2}-T_{1})[/tex] , where:
Q = energy (J)
m = mass (g)
c = specific heat capacity (J/g·°C)
[tex]T_{1}[/tex] = initial temperature (°C)
[tex]T_{2}[/tex] = final temperature (°C)
In an ideal situation, it can be assumed that all the energy lost by the piece of copper is gained by the water, resulting in its temperature rise and that there is no change of mass/state for either material. Thus, the equation can be written as below:
[tex]+Q_{c}=-Q_{w}[/tex]
[tex]+m_{c}c_{c}(T_{c2}-T_{c1})=-m_{w}c_{w}(T_{w2}-T_{w1})[/tex]
It can also be assumed that the final temperature of both the copper and water are the same. Thus substituting below values in above equation will give:
[tex]m_{w}[/tex] = 400 g
[tex]c_{w}[/tex] = 4.18 J/g·°C
[tex]T_{w1}[/tex] = 20.7 °C
[tex]T_{w2}[/tex] = 22.2 °C
[tex]m_{c}[/tex] = 89.1 g
[tex]c_{c}[/tex] = 0.38 J/g·°C
[tex]T_{c1}[/tex] = ? °C
[tex]T_{c2}[/tex] = 22.2 °C
[tex]+89.1*0.38*(22.2-T_{c1})=-400*4.18(22.2-20.7)[/tex]
Solving for [tex]T_{c1}[/tex] gives:
[tex]T_{c1}[/tex] = 96.3 °C
Here, we are required to determine the initial temperature of the copper.
The initial temperature of the copper is;
T(c1) = 96.27 °C
The energy required to raise the temperature of one gram of a material by one degree celsius is termed the Specific Heat Capacity of that material.
Mathematically, we have;
Q=mc{T(2) -T(1)}
Q=mc{T(2) -T(1)} where:
Q=mc{T(2) -T(1)} where:Q = energy (J)Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:
Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)
Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)
m(c) × c(c) × (T2(c) - T1(c)) = -{m(w) × c(w) × (T2(w) - T1(w))}
However, the final temperature of the copper piece and water can be assumed to be equal
i.e. T2(c) = T1(w)
m(w) = 400gc(w) = 4.18 J/g·°CT (w1) = 20.7 °CT(w2) = 22.2 °Cm(c) = 89.1 gc(c) = 0.38 J/g·°CT(c1) = ? °CT(c2) = 22.2 °C+89.1*0.38*(22.2-T(c1)) = -400*4.18(22.2-20.7)
(22.2−T(c1) ) = −2508/33.858
(22.2−T(c1) ) = -74.07
T(c1) = 74.07 + 22.2
T(c1) = 96.27 °C.
Therefore, the initial temperature of the copper is; T(c1) = 96.27 °C
Read more:
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Two long, parallel wires are separated by a distance of 2.2 cm. The force per unit length that each wire exerts on the other is 3.6 × 10^-5 N/m, and the wires repel each other. The current in one wire is 0.52 A. What is the magnitude of the current in the second wire? (Give your answer in decimal using "A" (Ampere) as unit)
Answer:
[tex]i_2 = 7.6 A[/tex]
Explanation:
As we know that the force per unit length of two parallel current carrying wires is given as
[tex]F = \frac{\mu_o i_1 i_2}{2\pi d}[/tex]
here we know that
[tex]F = 3.6 \times 10^{-5} N/m[/tex]
[tex]i_1 = 0.52 A[/tex]
d = 2.2 cm
now from above equation we have
[tex]3.6 \times 10^{-5} = \frac{4\pi \times 10^{-7} (0.52)(i_2)}{2\pi (0.022)}[/tex]
[tex]3.6 \times 10^{-5} = 4.73 \times 10^{-6} i_2[/tex]
[tex]i_2 = 7.6 A[/tex]
The peak electric field in an intense IR laser beam is about 25,000 V/m. How much energy does this laser beam deliver in 1 minute if it makes a spot with a diameter of 2mm?
Answer:
156.3 J
Explanation:
E = Amplitude of the electric field = 25000 V/m
t = time interval = 1 minute = 60 sec
d = diameter of the spot = 2 mm = 2 x 10⁻³ m
Area of the spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Energy delivered to the spot is given as
U = (0.5)ε₀ E² c A t
Inserting the values
U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)
U = 156.3 J
To find the energy delivered by the laser beam in 1 minute, we need to calculate the power of the laser beam and then multiply it by the duration.
E = Amplitude of the electric field = 25000 V/m
t = time interval = 1 minute = 60 sec
d = diameter of the spot = 2 mm = 2 x 10⁻³ m
Area of the spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Energy delivered to the spot is given as
U = (0.5)ε₀ E² c A t
Inserting the values
U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)
U = 156.3 J
A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.6 m/s and at an angle of 40.8° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.03 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.73 s, and for the (e) horizontal and (f) vertical components at t = 5.05 s. Assume that the catapult is positioned on a plain horizontal ground.
To answer this question, let us break down the initial velocity into its horizontal and vertical components.
v_{ix} = vcos(θ)
v_{iy} = vsin(θ)
v_{x} is the horizontal component of the initial velocity, and this value will stay constant over time because we assume only gravity acts on the stone, therefore only the vertical component of the stone's velocity will change over time
v_{iy} is the vertical component of the initial velocity, and this will change over time due to gravity.
v is the magnitude of the initial velocity.
θ is the angle the velocity vector is oriented at with respect to the horizontal.
Given values:
v = 19.6m/s
θ = 40.8°
Plug in these values and solve for v_{ix} and v_{iy}:
v_{ix} = 19.6cos(40.8°) = 14.8m/s
v_{iy} = 19.6sin(40.8°) = 12.8m/s
To find both the horizontal and vertical displacement at any time, we will use this kinematics equation:
D = X + Vt + 0.5At²
When solving for the horizontal displacement, the following values are:
t = elapsed time
D = horizontal displacement
X = initial horizontal displacement
V = initial horizontal velocity
A = horizontal acceleration
There is no initial horizontal displacement, so X = 0m
The initial horizontal velocity V = v_{ix} = 14.8m/s
Assuming we don't care about air resistance, no force has a component acting horizontally on the stone, so A = 0m/s²
Therefore the equation for the stone's horizontal displacement is given by:
D = 14.8t
When solving for the vertical displacement, the following values are:
t = elapsed time
D = vertical displacement
X = initial vertical displacement
V = initial vertical velocity
A = vertical acceleration
There is no initial vertical displacement, so X = 0m
The initial vertical velocity V = v_{iy} = 12.8m/s
Gravity acts downward on the stone, therefore A = -9.81m/s²
Therefore the equation for the stone's vertical displacement is given by:
D = 12.8t - 4.905t²
Now we just plug in various values of t...
a) At t = 1.03s, the horizontal displacement is D = 14.8(1.03) = 15.2m
b) At t = 1.03s, the vertical displacement is D = 12.8(1.03)-4.905(1.03)² = 7.98m
c) At t = 1.73s, the horizontal displacement is D = 14.8(1.73) = 25.6m
d) At t = 1.73s, the vertical displacement is D = 12.8(1.73)-4.905(1.73)² = 7.46m
Before you write down the following results, read the following explanation.
e) At t = 5.05s, the horizontal displacement is D = 14.8(5.05) = 74.7m
f) At t = 5.05s, the vertical displacement is D = 12.8(5.05)-4.905(5.05)² = -60.4m
Notice that we have a negative value for the vertical displacement. This isn't possible within the context of the problem, so the vertical displacement at t = 5.05s is actually 0m
The problem hasn't stated whether the ground is frictionless or not. I'm going to assume it is frictionless and therefore the stone will keep moving across the ground after landing, so the horizontal displacement at t = 5.05s is 74.7m
A dense metal sphere is dropped from a 10-meter tower, and at the exact same time an identical metal sphere is thrown horizontally from the same height. If air friction is ignored, which statement below is true?
A. Both spheres land at the same time.
B. The thrown sphere lands before the dropped sphere.
C. The dropped sphere lands before the thrown sphere.
D. The horizontal speed of the thrown sphere will determine whether it will land first or not.
Answer:
A. Both spheres land at the same time.
Explanation:
The horizontal motion doesn't affect the vertical motion. Since the two spheres have the same initial vertical velocity and same initial height, they land at the same time.
Answer:
A. Both spheres land at the same time.
Explanation:
If air resistance and friction are ignored, then both spheres land at the same time. Falling objects fall toward the center of the Earth with the same constant acceleration, independent of their mass, when they are defined to be in free-fall. For an object to be in free-fall, it has to be in a situation in which both air resistance and friction are considered negligible. This is true regardless of the direction of the fall.
X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{\circ}θ=30 ∘ relative to the incident X-rays, what is the energy of the recoiling electron?
Answer:
37.91594 keV
Explanation:
[tex]E_i[/tex] = Incident energy = 400 keV
θ = 30°
h = Planck's constant = 4.135×10⁻¹⁵ eV s = 6.626×10⁻³⁴ J s
Incident photon wavelength
[tex]\lambda_i=\frac{hc}{E_i}\\\Rightarrow \lambda_i=\frac{4.135\times 10^{-15}\times 3\times 10^8}{400\times 10^3}\\\Rightarrow \lambda_i=3.101\times 10^{-12}\ m[/tex]
Difference in wavelength
[tex]\Delta \lambda=\frac{h}{m_ec}(1-cos\theta)\\\Rightarrow \Delta \lambda=\frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8}(1-cos30)\\\Rightarrow \Delta \lambda=3.248\times 10^{-13}\ m[/tex]
[tex]\lambda_f=\lambda_i+\Delta \lambda\\\Rightarrow \lambda_f=3.101\times 10^{-12}+3.248\times 10^{-13}\\\Rightarrow \lambda_f=3.426\times 10^{-12}[/tex]
Final photon wavelength
[tex]\lambda_f=\frac{hc}{\lambda_f}\\\Rightarrow E_f=\frac{4.135\times 10^{-15}\times 3\times 10^8}{3.426\times 10^{-12}}\\\Rightarrow E_f=362084.06\ eV = 362.08406\ keV[/tex]
Energy of the recoiling electron
[tex]\Delta E=E_i-E_f\\\Rightarrow \Delta E=400-362.08406=37.91594\ keV[/tex]
Energy of the recoiling electron is 37.91594 keV
Answer:
The energy of recoiling electron=192.44 keV
Explanation:
Energy of x-ray[tex]E_o=400 keV[/tex]
Web know that compton shift is
[tex]\Delta \lambda =\dfrac{h}{m_eC(1-cos\theta)}[/tex]
[tex]m_e[/tex] is the mass of electron and C is the velocity of sound.
Given that θ=30°
Now by putting the values
[tex]\Delta \lambda =\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8(1-cos30)}[/tex]
[tex]\Delta \lambda =2.8\times 10^{-3}nm[/tex]
[tex]\Delta \lambda_o=\dfrac{hc}{E_o}[/tex]
By putting the values
[tex]\Delta \lambda_o=\dfrac{6.63\times 10^{-34}3\times 10^8}{400\times 1.602\times 10^{-16}}[/tex]
[tex]\Delta \lambda_o=3.31times 10^{-3}nm[/tex]
[tex]\lambda =\Delta \lambda +\lambda _o[/tex]
[tex]\lambda=2.8\times 10^{-3}+3.31\times 10^{-3}nm[/tex]
[tex]\lambda=6.11`\times 10^{-3}nm[/tex]
Energy [tex]E=\dfrac{hC}{\lambda }[/tex]
So [tex]E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{6.11\times 10^{-12}}[/tex]
E=207.55 keV
The energy of recoiling electron=400-207.55 keV
The energy of recoiling electron=192.44 keV
A communication satellite is orbiting at an altitude of 3 earth radii above the ground. what is the speed?
Answer:
4557.88 m/s
Explanation:
M = mass of earth = 5.98 x 10²⁴ kg
R = radius of earth = 6.4 x 10⁶ m
r = orbital radius of satellite = 3R = 3 x 6.4 x 10⁶ = 19.2 x 10⁶ m
[tex]v[/tex] = speed of the satellite
speed of the satellite is given as
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
[tex]v=\sqrt{\frac{(6.67\times 10^{-11})(5.98\times 10^{24})}{19.2\times 10^{6}}}[/tex]
[tex]v[/tex] = 4557.88 m/s
A0.12 kg yo-yo is whirled around by its string in a circle of radius 0.44 m at a constant speed of 6 m/s. What is the required centripetal force in Newtons to keep this yo -yo whirling in a circle?
Answer:
Centripetal force, F = 9.81 N
Explanation:
It is given that,
Mass of the object (yo-yo), m = 0.12 kg
It is whirled around by its string in a circle
Radius of circle, r = 0.44 m
Speed of object, v = 6 m/s
We need to find the required centripetal force in Newtons to keep this yo -yo whirling in a circle. The centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]F=\dfrac{0.12\ kg\times (6\ m/s)^2}{0.44\ m}[/tex]
F = 9.81 Newton
So, the centripetal force required to keep this yo -yo whirling in a circle is 9.81 N. Hence, this is the required solution.
Since force is dp/dt, the force due to radiation pressure reflected off of a solar sail can be calculated as 2 times the radiative momentum striking the sail per second. In the vicinity of Earth's orbit around the Sun, the energy intensity of sunlight is about 1300 W/m2. What is the approximate magnitude of the pressure on the sail? (For comparison, atmospheric pressure is about 105 N/m2.)
The approximate magnitude of the pressure on the sail due to radiation pressure is [tex]\( 2600 \, \text{N/m}^2 \).[/tex]
To calculate the magnitude of the pressure on the sail due to radiation pressure, we can use the formula for pressure:
[tex]\[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} \][/tex]
The force due to radiation pressure is given by:
[tex]\[ \text{Force} = 2 \times \text{radiative momentum} \][/tex]
The radiative momentum can be calculated using the energy intensity of sunlight:
[tex]\[ \text{Radiative momentum} = \text{Energy intensity} \times \text{Time} \][/tex]
Given:
- Energy intensity of sunlight [tex](\( I \))[/tex] is approximately [tex]\( 1300 \, \text{W/m}^2 \),[/tex]
- Time [tex](\( t \)) is \( 1 \, \text{s} \)[/tex] (per second),
- Atmospheric pressure is about [tex]\( 105 \, \text{N/m}^2 \).[/tex]
Let's calculate the radiative momentum:
[tex]\[ \text{Radiative momentum} = 1300 \times 1 \, \text{N/s} = 1300 \, \text{N/s} \][/tex]
Now, let's calculate the force:
[tex]\[ \text{Force} = 2 \times 1300 \, \text{N/s} = 2600 \, \text{N/s} \][/tex]
Finally, let's calculate the pressure on the sail:
[tex]\[ \text{Pressure} = \frac{2600 \, \text{N/s}}{1 \, \text{m}^2} = 2600 \, \text{N/m}^2 \][/tex]
So, the approximate magnitude of the pressure on the sail due to radiation pressure is [tex]\( 2600 \, \text{N/m}^2 \).[/tex]
At a certain temperature, the ????p for the decomposition of H2S is 0.841. H2S(g)↽−−⇀H2(g)+S(g) Initially, only H2S is present at a pressure of 0.259 atm in a closed container. What is the total pressure in the container at equilibrium?
Answer : The total pressure in the container at equilibrium is, 0.4667 atm
Solution : Given,
Initial pressure of [tex]H_2S[/tex] = 0.259 atm
Equilibrium constant, [tex]K_p[/tex] = 0.841
The given equilibrium reaction is,
[tex]H_2S(g)\rightleftharpoons H_2(g)+S(g)[/tex]
Initially 0.259 0 0
At equilibrium (0.259 - x) x x
Let the partial pressure of [tex]H_2[/tex] and [tex]S[/tex] will be, 'x'
The expression of [tex]K_p[/tex] will be,
[tex]K_p=\frac{(p_{H_2})(p_{S})}{p_{H_2S}}[/tex]
Now put all the values of partial pressure, we get
[tex]0.841=\frac{(x)\times (x)}{(0.259-x)}[/tex]
By solving the term x, we get
[tex]x=0.2077atm[/tex]
The partial pressure of [tex]H_2[/tex] and [tex]S[/tex] = x = 0.2077 atm
Total pressure in the container at equilibrium = [tex]0.259-x+x+x=0.259+x=0.259+0.2077=0.4667atm[/tex]
Therefore, the total pressure in the container at equilibrium is, 0.4667 atm
While entering a freeway, a car accelerates from rest at a rate of 2.1 m/s^2 for 14.5 s. How far does the car travel in those 14.5 s, in meters?
What is the car's final velocity, in meters per second?
Answer:
220.8 m
30.5 m/s
Explanation:
a = acceleration of the car = 2.1 m/s²
t = time of travel = 14.5 s
x = displacement of the car
v₀ = initial velocity of the car = 0 m/s
Displacement of the car is given as
x = v₀ t + (0.5) a t²
Inserting the values
x = (0) (14.5) + (0.5) (2.1) (14.5)²
x = 220.8 m
v = final velocity of the car
Final velocity of the car is given as
v = v₀ + at
v = 0 + (2.1) (14.5)
v = 30.5 m/s
A 55-kg box is being pushed a distance of 7.0 m across the floor by a force whose magnitude is 160 N. The force is parallel to the displacement of the box. The coefficient of kinetic friction is 0.25. Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.
Answer:
work done by applied force = 1120J
work done by frictional force = 943.25J
work done by the gravitational force = work done by the normal force = 0J
Explanation:
Given:
Mass of the box = 55kg
Displacement of the box = 7.0m across the floor
Applied force = 160N
Coefficient of kinetic friction ([tex]\mu_k[/tex]) = 0.25
Now the work done (W) is given as:
W = F×d
Where,
F = Applied force on the body
d = Displacement of the body in the direction of the applied force
Forces acting on the box are:
1) Applied force
2) Frictional Force
3) Gravitational force
4) Normal force due to the weight
Now the work done by the respective forces:
1) By the applied force
W = 160 kg × 7 m = 1120 J
2) By the frictional force
[tex]W_f =-\mu mg\times displacement[/tex]
Where
g = acceleration due to the gravity
and the negative sign here depicts that the frictional force is acting the against the applied force and the direction of displacement.
thus,
[tex]W_f =-0.25\times 55\times 9.8\times 7[/tex]
or
[tex]W_f =-943.25J[/tex]
The work done by the normal force and the gravitational force will be zero as there is no displacement in the direction of the application of both the forces
The work done by both the gravitational force and normal force is equal to zero (0).
Given the following data:
Mass = 55 kg.Distance = 7 meters.Magnitude of force = 160 Newton.Coefficient of kinetic friction = 0.25.The forces acting on the box.In this scenario, there are four forces acting on the box and these include:
An applied forceA frictional ForceA gravitational forceA normal force due to weight.For the applied force:
Mathematically, the work done due to an applied force is given by this formula:
[tex]Work\;done = force \times distance\\\\Work\;done = 160 \times 7[/tex]
Work done = 1,120 Nm.
For the frictional force:
Mathematically, the work done due to a frictional force is given by this formula:
[tex]Work\;done = -\mu mg \times distance\\\\Work\;done = -0.25 \times 55 \times 9.8 \times 7[/tex]
Work done = -943.25.
However, the work done by both the gravitational force and normal force is equal to zero (0) because there isn't any displacement or distance covered in the direction of the application of these forces.
Read more on work done here: brainly.com/question/22599382
A force vector has a magnitude of 720 N and a direction of 38 degree north of east. Determine the magnitude and direction of the components of the force that point along the north-south line and along east-west line.
Answer:
443.3 N
567.4 N
Explanation:
Consider the triangle ABC
AC = hypotenuse = magnitude of force vector = F = 720 N
AB = adjacent = Component of force along east-west line = [tex]F_{x}[/tex]
BC = Opposite = Component of force along north-south line = [tex]F_{y}[/tex]
θ = Angle = 38 deg
In triangle ABC
[tex]Sin38 = \frac{BC}{AC}[/tex]
[tex]Sin38 = \frac{F_{y}}{F}[/tex]
[tex]0.616 = \frac{F_{y}}{720}[/tex]
[tex]F_{y}[/tex] = 443.3 N
Also, In triangle ABC
[tex]Cos38 = \frac{AB}{AC}[/tex]
[tex]Cos38 = \frac{F_{x}}{F}[/tex]
[tex]0.788 = \frac{F_{x}}{720}[/tex]
[tex]F_{x}[/tex] = 567.4 N
The coil of a generator has a radius of 0.25 m. When this coil is unwound, the wire from which it is made has a length of 11.0 m. The magnetic field of the generator is 0.17 T, and the coil rotates at a frequency of 13 Hz. What is the peak emf of this generator?
Answer:
19.06 Volt
Explanation:
r =0.25 m, l = 11 m, B = 0.17 T, f = 13 Hz
length for one turn = 2 x 3.14 x r = 2 x 3.14 x 0.25 = 1.57 m
Total number of turns in 11 m
N = 11 / 1.57 = 7
Peak emf, e0 = N x B x A x 2 x 3.14 x f
e0 = 7 x 0.17 x 3.14 x 0.25 x 0.25 x 2 x 3.14 x 13 = 19.06 Volt
A 4m^3 container is being filled at a rate of 0.051m^3/s. What is the time required to completely fill this container?
Answer:
Time required to completely fill this container = 78.43 seconds
Explanation:
Volume of container = 4 m³
Rate of filling of container = 0.051 m³/s
We have the equation
[tex]\texttt{Time required}=\frac{\texttt{Volume of container}}{\texttt{Rate of filling of container}}[/tex]
Substituting
[tex]\texttt{Time required}=\frac{4}{0.051}=\frac{4000}{51}=78.43 s[/tex]
Time required to completely fill this container = 78.43 seconds
A student is skateboarding down a ramp that is 8.51 m long and inclined at 23.7° with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is 4.97 m/s. Neglect friction and find the speed at the bottom of the ramp.
To find the speed at the bottom of the ramp, we can use the principles of physics. By equating the potential energy at the top to the kinetic energy at the bottom, we can solve for the velocity at the bottom of the ramp. Using the given values and trigonometry, we can calculate the height of the ramp and then determine the velocity.
Explanation:To find the speed of the skateboarder at the bottom of the ramp, we can use the principles of physics. Since we are neglecting friction, we can assume that the mechanical energy of the skateboarder is conserved. The mechanical energy consists of both kinetic energy and potential energy. At the top of the ramp, the skateboarder has only potential energy, which is given by the formula: potential energy = mass * gravity * height. At the bottom of the ramp, the skateboarder has only kinetic energy, which is given by the formula: kinetic energy = 1/2 * mass * velocity^2. By equating the potential energy at the top to the kinetic energy at the bottom, we can solve for the velocity at the bottom of the ramp.
In this case, the mass of the skateboarder is not provided, but it cancels out when equating the potential and kinetic energies. So we can assume any mass value as long as it is consistent throughout the calculation. Let's assume the mass of the skateboarder is 1 kg for simplicity. The height of the ramp is not provided, but we can calculate it using trigonometry. The height is the vertical component of the ramp's length, which is given by: height = length * sin(angle). Plug in the values and calculate the height.
Next, we calculate the potential energy at the top of the ramp using the calculated height. The potential energy is then equated to the kinetic energy at the bottom of the ramp and solve for the velocity. Plug in the values and calculate the velocity at the bottom of the ramp.
An FM radio consists of a series RLC circuit with a 6.57 pF (peco – 10-12) capacitor. If a person is dialed into a station broadcasting at 103.9 MHz, what is the inductance needed?
Answer:
3.6 x 10⁻⁷ H
Explanation:
C = capacitance of the capacitor = 6.57 x 10⁻¹² F
L = inductance of the inductor = ?
f = frequency of broadcasting system = 103.9 MHz = 103.9 x 10⁶ Hz
frequency is given as
[tex]f = \frac{1}{2\pi \sqrt{LC}}[/tex]
[tex]103.6\times 10^{6} = \frac{1}{2(3.14)\sqrt{(6.57\times 10^{-12})L}}[/tex]
Inserting the values
L = 3.6 x 10⁻⁷ H
A long, straight wire carrying a current is placed along the y-axis. If the direction of the current is in the +y direction, what is the direction of the magnetic field due to this wire?
Final answer:
The magnetic field at a point on the +x-axis near a wire carrying current in the +y-direction will be directed into the page, as determined by the right-hand rule. To cancel the magnetic field at a specific point on the +x-axis, a second wire carrying equal but opposite current must be arranged parallel to the first.
Explanation:
The question concerns the determination of the magnetic field direction around a long, straight wire carrying a current. According to the right-hand rule, if you point your right thumb in the direction of the current, then the direction in which your fingers curl will show the direction of the magnetic field. In this case, the current is running in the +y direction, so if one were to stand at a point on the +x-axis near the wire and apply the right-hand rule, the fingers would curl into the page, indicating that the magnetic field at that point is directed into the page.
An experimenter wanting to make the total magnetic field at the coordinate x = 1.0 m zero must arrange a second wire parallel to the first with current in the opposite direction. This second wire would also need to be carrying the same magnitude of current but in the -y direction to cancel out the magnetic field from the first wire at the point of interest.
Discuss Joule-Thompson effect with relevant examples and formulae.
Answer:
[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]
Explanation:
Joule -Thompson effect
Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.
Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.
Now lets take Steady flow process
Let
[tex]P_1,T_1 [/tex] Pressure and temperature at inlet and
[tex]P_2,T_2 [/tex] Pressure and temperature at exit
We know that Joule -Thompson coefficient given as
[tex]\mu _j=\left(\frac{\partial T}{\partial p}\right)_h[/tex]
Now from T-ds equation
dh=Tds=vdp
So
[tex]Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp[/tex]
⇒[tex]dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]
So Joule -Thompson coefficient
[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]
This is Joule -Thompson coefficient for all gas (real or ideal gas)
We know that for Ideal gas Pv=mRT
[tex]\dfrac{\partial v}{\partial T}=\dfrac{v}{T}[/tex]
So by putting the values in
[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]
[tex]\mu _j=0[/tex] For ideal gas.
An electron and a proton are separated by a distance of 1.0 m. What happens to the force between them if the electron moves 0.5 m away from the proton?
Explanation:
It is given that, an electron and a proton are separated by a distance of 1.0 m i.e d = 1 m . At this position, F is the force between them
[tex]F=k\dfrac{q_1q_2}{1^2}[/tex]...............(1)
We need to find effect on force between them if the electron moves 0.5 m away from the proton. Let the force is F'.
[tex]F'=k\dfrac{q_1q_2}{0.5^2}[/tex]...............(2)
On dividing equation (1) and (2) we get :
[tex]\dfrac{F}{F'}=\dfrac{k\dfrac{q_1q_2}{1^2}}{k\dfrac{q_1q_2}{0.5^2}}[/tex]
[tex]\dfrac{F}{F'}=0.25[/tex]
F' = 4 F
So, the distance between the electron and the proton is 0.5 m, the new force will be 4 times of previous force.
Final answer:
The electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m can be calculated using Coulomb's Law. The force is inversely proportional to the square of the distance, so if the distance is increased, the force decreases. In this case, if the electron moves 0.5 m away from the proton, the new force can be calculated using Coulomb's Law.
Explanation:
The electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m can be calculated using Coulomb's Law. Coulomb's Law states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
So, if the distance between the charges is increased to 1 m, the force between them will decrease. The force is inversely proportional to the square of the distance, which means that if the distance is doubled, the force will decrease by a factor of four.
In this case, if the electron moves 0.5 m away from the proton, the new distance becomes 1.5 m. Using Coulomb's Law, we can calculate the new force:
F = k * (q1 * q2) / r^2
Where F is the force, k is the Coulomb's constant (approximately 9 x 10^9 Nm^2/C^2), q1 and q2 are the charges (in this case, both 1 C), and r is the distance (1.5 m). Plugging in these values, we get:
F = (9 x 10^9 Nm^2/C^2) * (1 C * 1 C) / (1.5 m)^2
F ≈ 4 x 10^9 N
The fixed hydraulic cylinder C imparts a constant upward velocity v = 2.2 m/s to the collar B, which slips freely on rod OA. Determine the resulting angular velocity ωOA when point B has a displacement s = 220 mm, and the fixed distance d = 510 mm. The angular velocity is positive if counterclockwise, negative if clockwise.
We can determine the angular velocity using the formula ω = v/r, where v is the tangential speed (2.2 m/s) and r is the radius of the circle (510 mm). After converting the radius to meters, we can then plug these values into the formula and solve for ω, yielding an angular velocity of 3.9636 rad/s.
To determine the resulting angular velocity ωOA of rod OA when point B has a displacement s = 220 mm, we can use the following formula:
ωOA = v / r
Where:
ωOA = Angular velocity of rod OA
v = Linear velocity of point B
r = Radius or distance from point O to point B
Given:
v = 2.2 m/s
s = 220 mm = 0.22 m
d = 510 mm = 0.51 m
First, we calculate the radius (r) using the Pythagorean theorem because the motion is along a right-angled triangle:
r² = d² + s²
r² = (0.51 m)² + (0.22 m)²
r² = 0.2601 m² + 0.0484 m²
r² = 0.3085 m²
Now, we calculate the square root of r² to get r:
r = √0.3085 m
r ≈ 0.555 m
Now, we can calculate the angular velocity ωOA:
ωOA = v / r
ωOA = 2.2 m/s / 0.555 m
ωOA ≈ 3.9636 rad/s
The angular velocity ωOA is approximately 3.9636 rad/s. Since the motion is counterclockwise (as indicated by the positive linear velocity of point B), the angular velocity is also positive.
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An object of mass 5 kg and velocity 10 m/s has a linear momentum of: (A) 5 kg-m/s, (B) 10 kg-m/s, (C) 50 kg-m/s, (D) 250 kg-m/s^2, (E) none of the above.
Answer:
Linear momentum, p = 50 kg-m/s
Explanation:
It is given that,
Mass of an object, m = 5 kg
Velocity, v = 10 m/s
We need to find the linear momentum of that object. It is given by :
[tex]p=m\times v[/tex]
[tex]p=5\ kg\times 10\ m/s[/tex]
p = 50 kg-m/s
So, the linear momentum of that object is 50 kg-m/s. Hence, this is the required solution.
The linear momentum of an object with a mass of 5 kg and velocity of 10 m/s is calculated using the formula P = mv, which results in 50 kg*m/s.
The student has asked to calculate the linear momentum of an object that has a mass of 5 kg and a velocity of 10 m/s. To find the linear momentum, we use the formula:
P = mv
Where P is the momentum, m is the mass, and v is the velocity. By plugging in the numbers:
P = 5 kg \\* 10 m/s
P = 50 kg\\*m/s
Therefore, the linear momentum of the object is 50 kg\\*m/s, which corresponds to option (C).
A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun is 300.0 m/s, how far vertically will the bullet have dropped by the time it hits the target? Ignore the effects of air resistance.
Answer:
The bullet will have been dropped vertically h= 0.54 meters by the time it hits the target.
Explanation:
d= 100m
V= 300 m/s
g= 9.8 m/s²
d= V*t
t= d/V
t= 0.33 s
h= g*t²/2
h=0.54 m
Consider water flowing through a cylindrical pipe with a variable cross-section. The velocity is v at a point where the pipe diameter is 1.0 cm. At a point where the pipe diameter is three times larger, the velocity is:
nine times larger than the initial velocity
one ninth the initial velocity
three times larger than the initial velocity
the same as the initial velocity
one third the initial velocity
Answer:
one ninth
Explanation:
d = 1 cm , v = v
D = 3d, V = ?
By the equation of continuity,
A V = a v
3.14 x D^2 / 4 x V = 3.14 x d^2 / 4 x v
9d^2 x V = d^2 x v
V = v / 9
Thus, the velocity becomes one ninth the initial velocity
The uncertainty in a proton's position is 0.015 nm. a) What is the minimum uncertainty Δp in its momentum? b) What is the kinetic energy of a proton whose momentum is equal to this uncertainty (Δp=p)? (Unit: meV)
Answer:
[tex]\Delta P_{x}=3.51 *10^{-24}[/tex] kg m/s
[tex]E = 3.68 * 10^{-21}[/tex] J
Explanation:
given data
uncertainty in proton position [tex]\Delta x[/tex] = 0.015 nm
according to Heisenberg's principle of uncertainty
[tex]\Delta x \Delta P_{x}= \frac{\frac{h}{2\pi }}{2}[/tex]
Where h is plank constant = [tex]6.6260 * 10^{-34}[/tex] j-s
[tex]\Delta P_{x}= \frac{\frac{h}{2\pi }}{2\Delta x}[/tex]
[tex]\Delta P_{x}= \frac{\frac{6.6260 * 10^{-34}}{2\pi }}{2*0.015*10^{-9}}[/tex]
[tex]\Delta P_{x}=3.51 *10^{-24}[/tex] kg m/s
b) kinetic energy of proton whose momentum
[tex]P =\Delta p[/tex]
[tex]E =\frac{\Delta p^{2}}{2m}[/tex]
where m is mass is proton
[tex]E =\frac{(3.51*10^{-24})^{2}}{2*1.67*10^{-27}}[/tex]
[tex]E = 3.68 * 10^{-21}[/tex] J
When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner, which cools the entire house to 20°C in 38 min. If the COP of the air-conditioning system is 2.8, determine the power drawn by the air conditioner. Assume the entire mass within the house is equivalent to 800 kg of air for which cv = 0.72 kJ/kg·°C and cp = 1.0 kJ/kg·°C
The power drawn by the air conditioner to cool the house from 35°C to 20°C in 38 minutes, given a COP of 2.8, is calculated to be 1.35 kW.
Explanation:To calculate the power drawn by the air conditioner, we first determine the energy needed to cool the house from 35°C to 20°C. The energy required (Q) can be calculated using the formula Q = m⋅cv⋅ΔT, where m is the mass of the air, cv is the specific heat capacity at constant volume, and ΔT is the change in temperature.
Plugging the given values into the formula: Q = 800 kg ⋅ 0.72 kJ/kg°C ⋅ (35°C - 20°C) = 8,640 kJ.
Next, we use the Coefficient of Performance (COP) of the air conditioner to find the work input (W) required for this cooling process. The formula relating these quantities is COP = Q/W, rearranging for W gives W = Q/COP.
Therefore, W = 8,640 kJ / 2.8 = 3,085.71 kJ. To find the power drawn, we need the work in kilowatts (kW), knowing that 1 kW = 1 kJ/s, and that the total duration is 38 min or 2,280 seconds. Thus, the power drawn is Power = W / time = 3,085.71 kJ / 2,280 s = 1.35 kW.
Power drawn by the air conditioner is approximately 6.76 kW, calculated from COP, heat transfer, and time taken for cooling.
To solve this problem, we can use the energy balance equation for the air within the house:
[tex]\[ Q = mc\Delta T \][/tex]
where:
- [tex]\( Q \)[/tex] is the heat transfer,
- [tex]\( m \)[/tex] is the mass of air,
- [tex]\( c \)[/tex] is the specific heat capacity of air,
- [tex]\( \Delta T \)[/tex] is the change in temperature.
We also know that the Coefficient of Performance (COP) is defined as:
COP = Q_cooling/W_input
where:
- [tex]\( Q_{\text{cooling}} \)[/tex] is the heat removed from the house (in this case),
- W_input is the work input (power consumed) by the air conditioner.
We are given that the COP [tex](\( \text{COP} = 2.8 \))[/tex] and the initial and final temperatures [tex](\( T_{\text{initial}} = 35^\circ C \)[/tex], [tex]\( T_{\text{final}} = 20^\circ C \))[/tex].
First, let's calculate the heat transfer using the energy balance equation:
[tex]\[ Q = mc\Delta T \][/tex]
[tex]\[ Q = (800 \, \text{kg})(1.0 \, \text{kJ/kg}^\circ C)(35 - 20) \][/tex]
[tex]\[ Q = (800 \, \text{kg})(1.0 \, \text{kJ/kg}^\circ C)(15) \][/tex]
Q = 12000kJ
Now, we can find the work input using the COP formula:
COP = Q_cooling/{W_input
W_input = [tex]\frac{Q_{\text{cooling}}}{\text{COP}}[/tex]
W_input = [tex]\frac{12000 \, \text{kJ}}{2.8}[/tex]
W_input ≈ 4285.71 kJ
To find the power drawn by the air conditioner, we need to convert the energy to power. Since the time taken for cooling is 38 minutes, or [tex]\( \frac{38}{60} \)[/tex] hours:
Power = W_input /Time
[tex]\[ \text{Power} = \frac{4285.71 \, \text{kJ}}{\frac{38}{60} \, \text{hours}} \][/tex]
Power ≈ 6762.45 W
So, the power drawn by the air conditioner is approximately [tex]\( 6762.45 \, \text{W} \)[/tex] or [tex]\( 6.76 \, \text{kW} \)[/tex].
A current runs through a wire in the +y direction along the y-axis. What is the direction of the magnetic field at a point (x,y,z) = (0,0,1.045 m)
Answer:
Magnetic field at given point is along + x direction
Explanation:
As we know that position of the wire is along Y axis
current is flowing along +y direction
so here we will have
[tex]\vec B = \frac{\mu_0 i (\vec {dl} \times \hat r)}{4\pi r^2}[/tex]
now the direction of length vector is along +Y direction
also the position vector is given as
[tex]\hat r = \hat k[/tex]
now for the direction of magnetic field we can say
[tex]\vec B = (\hat j \times \hat k)[/tex]
[tex]\vec B = \hat i[/tex]
so magnetic field is along +X direction
Using the right-hand rule for a wire carrying current in the +y direction along the y-axis, the magnetic field at the point (0,0,1.045 m) on the x-axis would be directed in the +x-direction.
When a current runs through a wire in the +y direction along the y-axis, we can use the right-hand rule to determine the direction of the magnetic field at a point near the wire. For a point on the x-axis such as (0,0,1.045 m), you would point your thumb in the direction of the current (upward along the y-axis) and curl your fingers.
Your fingers will curl in the direction of the magnetic field lines, which at the given point on the x-axis would circle around the wire. This means the magnetic field at that point would be directed in the +x-direction.
A uniform thin rod is hung vertically from one end and set into small amplitude oscillation. If the rod has a length of 3.6 m, this rod will have the same period as a simple pendulum of length ____ cm. Round your answer to the nearest whole number.
Answer:
The length of the simple pendulum is 2.4 meters.
Explanation:
Time period of simple pendulum is given by :
[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]
L is the length of pendulum
The time period of the rope is given by :
[tex]T=2\pi\sqrt{\dfrac{2L'}{3g}}[/tex]
L' is the length of the rod, L' = 3.6 m
It is given that, the rod have the same period as a simple pendulum and we need to find the length of simple pendulum i.e.
[tex]2\pi\sqrt{\dfrac{L}{g}}=2\pi\sqrt{\dfrac{2L'}{3g}}[/tex]
On solving the above equation as :
[tex]\dfrac{L}{g}=\dfrac{2L'}{3g}[/tex]
L = 2.4 m
So, the length of the thin rod that is hung vertically from one end and set into small amplitude oscillation 2.4 meters. Hence, this is the required solution.
A person steps off the end of a 3.45 � high diving board and drops to the water below. (a) How long does it take for the person to reach the water? (b) What is person’s speed on entering the water? (c) What is person’s speed on entering the water if they step off a 12.0 � diving
(a) It takes approximately 0.841 seconds to reach the water from a 3.45 m diving board, with a speed of about 8.24 m/s.
(b) If stepping off a 12.0 m diving board, it takes approximately 1.565 seconds to reach the water, with a speed of about 15.34 m/s.
To solve this problem, we can use the kinematic equations of motion to calculate the time taken to reach the water and the speed upon entering the water.
(a) To find the time taken to reach the water, we can use the kinematic equation:
Height = (1/2) * gravity * time^2
Where:
Height = 3.45 m (height of the diving board)
Gravity = 9.8 m/s² (acceleration due to gravity)
Time is the time taken to reach the water.
Rearranging the equation to solve for time:
time = sqrt((2 * height) / gravity)
time = sqrt((2 * 3.45 m) / 9.8 m/s²)
time ≈ sqrt(0.7061)
time ≈ 0.841 s
(b) To find the speed upon entering the water, we can use the equation:
Speed = gravity * time
Where:
Speed is the speed upon entering the water.
Substituting the known values:
Speed = 9.8 m/s² * 0.841 s
Speed ≈ 8.24 m/s
(c) If the person steps off a 12.0 m diving board, we can repeat the above calculations using height = 12.0 m:
time = sqrt((2 * 12.0 m) / 9.8 m/s²)
time ≈ sqrt(2.449)
time ≈ 1.565 s
Speed = 9.8 m/s² * 1.565 s
Speed ≈ 15.34 m/s
If the speed of light in a substance is 2.26 x 10^8 m/s, what is the index of refraction of that substance?
The index of refraction can be calculated using the speed of light in vacuum divided by the speed of light in a specific medium. Given the speed of light in a medium is 2.26 x 10^8 m/s, the index of refraction would be calculated as roughly 1.33.
Explanation:To calculate the index of refraction for a certain medium, you would use the speed of light in vacuum (c) divided by the speed of light in that medium (v). This equation is denoted as n = c/v.
The speed of light in vacuum is approximately 3.00×10^8 m/s. Given that the speed of light in a certain substance is 2.26 x 10^8 m/s, we can substitute these values into the equation to find the material's index of refraction.
So, n = (3.00×10^8 m/s) / (2.26 x 10^8 m/s). Therefore, the index of refraction of that substance is approximately 1.33. This property of a substance tells us how much the path of light is bent, or refracted, when entering the material.
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