Answer:
a) Magnitude of the average velocity from t = 0 to t = 2 s = 2.24 m/s
b) Magnitude of the average velocity from t = 0 to t = 5 s = 3.16 m/s
Explanation:
a) Velocity is rate of change of position.
A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0
A particle's position coordinates (x, y) are (5.0 m, 5.0 m) at t = 2.0 s
Displacement = (5-1)i + (5-3)j = 4i + 2j
Change in time = 2s
Velocity
[tex]v=\frac{4i+2j}{2}=2i+j[/tex]
Magnitude of velocity
[tex]v=\sqrt{2^2+1^2}=2.24m/s[/tex]
b) Velocity is rate of change of position.
A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0
A particle's position coordinates (x, y) are (14.0 m, 12.0 m) at t = 5.0 s
Displacement = (14-1)i + (12-3)j = 13i + 9j
Change in time = 5s
Velocity
[tex]v=\frac{13i+9j}{5}=2.6i+1.8j[/tex]
Magnitude of velocity
[tex]v=\sqrt{2.6^2+1.8^2}=3.16m/s[/tex]
If 5.0 x 10^21 electrons pass through a 20-Ω resistor in 10 min, what is the potential difference across the resistor?
Answer:
26.67 Volt
Explanation:
number of electrons, n = 5 x 10^21
R = 20 ohm
t = 10 min = 10 x 60 = 600 sec
V = ?
q = n x charge of one electron
q = n x e
q = 5 x 10^21 x 1.6 x 10^-19 = 800 C
i = q / t = 800 / 600 = 4 / 3 A
According to Ohm's law
V = i x R
V = 4 x 20 / 3 = 26.67 Volt
The voltage or potential difference across the resistor is 26.67 Volt.
Given the data in the question;
Number of electrons; [tex]n = 5*10^{21}[/tex]Resistor; [tex]R = 20ohms[/tex]Time elapsed; [tex]t = 10min = 600s[/tex]Ohm's Law and ResistanceOhm’s law states that the potential difference between two points is directly proportional to the current flowing through the resistance.
Hence
[tex]V = IR[/tex]
Where V is the voltage or potential difference, potential difference, I is the current and R is the resistance.
First we determine the total charge.
[tex]Q = n * e[/tex]
Where Q is the charge, n is the number of electrons and e is the charge on one electron. ( [tex]e = 1.6 * 10^{-19}C[/tex] )
Hence
[tex]Q = (5*10^{21}) * (1.6 * 10^{-19C)}\\\\Q = 800C[/tex]
Since current is the electric charge transferred per unit time.
[tex]I = \frac{Q}{t}[/tex]
Hence
[tex]I = \frac{800C}{600s}\\ \\I = 1.33A[/tex]
Now, we input our values into the above expression
[tex]V = I * R\\\\V = 1.33A * 20Ohms\\\\V = 26.67Volt[/tex]
Therefore, the voltage or potential difference across the resistor is 26.67 Volt.
Learn more about potential difference: https://brainly.com/question/2364325
An electric field of 7.50×105 V/m is desired between two parallel plates, each of area 45.0 cm2 and separated by 2.45 mm of air. What charge must be on each plate?
Answer:
Charge, [tex]q=2.98\times 10^{-8}\ C[/tex]
Explanation:
It is given that,
Value of electric field, [tex]E=7.5\times 10^5\ V/m[/tex]
Area of parallel plates, [tex]A=45\ cm^2=0.0045\ m^2[/tex]
Distance between two parallel plates, d = 2.45 mm = 0.00245 m
For a parallel plate capacitor, the capacitance is given by :
[tex]C=\dfrac{\epsilon_oA}{d}[/tex].......(1)
Since, [tex]E=\dfrac{V}{d}[/tex]
V = E . d ............(2)
And [tex]C=\dfrac{q}{V}[/tex].....(3)
From equation (1), (2) and (3) we get :
[tex]\dfrac{q}{V}=\dfrac{\epsilon_oA}{d}[/tex]
[tex]q=\epsilon_o EA[/tex]
[tex]q=8.85\times 10^{-12}\ F/m\times 7.5\times 10^5\ V/m\times 0.0045\ m^2[/tex]
[tex]q=2.98\times 10^{-8}\ C[/tex]
So, the charge on the each plate is [tex]2.98\times 10^{-8}\ C[/tex]. Hence, this is the required solution.
To determine the charge on each plate for a desired electric field in a capacitor with parallel plates, use the formula Q = E x A x ε_0. Convert the area to m^2, insert all values including the permittivity of free space, and solve for Q.
Explanation:To find the charge required on each of the parallel plates to create a specific electric field, you can use the relationship between the electric field (E), the charge (Q) on the plates, and the permittivity of free space (ε0). The electric field between two parallel plates is given by the equation:
E = Q / (ε0 × A)
where E is the electric field, Q is the charge on each plate, A is the area of the plates, and ε0 is the permittivity of free space (ε0 = 8.854 x 10−12 C2/N·m2). Rearranging this for Q gives:
Q = E × A × ε0
First, convert the area from cm2 to m2 by multiplying by (10−4)2. Then, plug in the values:
Q = (7.50 × 105 V/m) × (0.45 × 10−4 m2) × (8.854 × 10−12 C2/N·m2)
Calculate Q to find the charge required on each plate.
A small 1.0 kg steel ball rolls west at 3.0 m/s collides with a large 3.0 kg ball at rest. After the collision, the small ball moves south at 2.0 m/s. What is the direction of the momentum (with respect to east) of the large ball after the collision
Answer:
The direction of the momentum of the large ball after the collision with respect to east is 146.58°.
Explanation:
Given that,
Mass of large ball = 3.0 kg
Mass of steel ball = 1.0 kg
Velocity = 3.0 kg
After collision,
Velocity = 2.0 m/s
Using conservation of momentum
[tex]m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}[/tex]
[tex]3.0\times0+1.0\times(3.0)(-i)=1.0\times2(-j)+3.0\times v_{2}[/tex]
[tex]-3i+2j=3.0\times v_{2}[/tex]
[tex]v_{2}=-i+0.66j[/tex]
The direction of the momentum
[tex]tan\theta=\dfrac{0.66}{-1}[/tex]
[tex]\theta=tan^{-1}\dfrac{0.66}{-1}[/tex]
[tex]\theta=-33.42^{\circ}[/tex]
The direction of the momentum with respect to east
[tex]\theta=180-33.42=146.58^{\circ}[/tex]
Hence, The direction of the momentum of the large ball after the collision with respect to east is 146.58°.
A woman with a mass of 60 kg runs at a speed of 10 m/s and jumps onto a giant 30 kg skateboard initially at rest. What is the combined speed of the woman and the skateboard?
Answer:
The combined speed of the woman and the skateboard is 6.67 m/s.
Explanation:
It is given that,
Mass of woman, m₁ = 60 kg
Speed of woman, v = 10 m/s
The woman jumps onto a giant skateboard of mass, m₂ = 30 kg
Let V is the combined speed of the woman and the skateboard. It can be calculated using the conservation of momentum as :
[tex]60\ kg\times 10\ m/s+30(0)=(60\ kg+30\ kg)V[/tex]
On solving the above equation we get V = 6.67 m/s
So, the combined speed of the woman and the skateboard is 6.67 m/s. Hence, this is the required solution.
Each plate of a parallel-plate air-filled capacitor has an area of 2×10−3 m2, and the separation of the plates is 5×10−2 mm. An electric field of 8.5 ×106 V/m is present between the plates. What is the surface charge density on the plates? (ε 0 = 8.85 × 10-12 C2/N · m2)
Answer:
The surface charge density on the plate, [tex]\sigma=7.5\times 10^{-5}\ C/m^2[/tex]
Explanation:
It is given that,
Area of parallel plate capacitor, [tex]A=2\times 10^{-3}\ m^2[/tex]
Separation between the plates, [tex]d=5\times 10^{-2}\ mm[/tex]
Electric field between the plates, [tex]E=8.5\times 10^{6}\ V/m[/tex]
We need to find the surface charge density on the plates. The formula for electric field is given by :
[tex]E=\dfrac{\sigma}{\epsilon}[/tex]
Where
[tex]\sigma[/tex] = surface charge density
[tex]\sigma=E\times \epsilon[/tex]
[tex]\sigma=8.5\times 10^{6}\ V/m\times 8.85\times 10^{-12}\ C^2/Nm^2[/tex]
[tex]\sigma=0.000075\ C/m^2[/tex]
[tex]\sigma=7.5\times 10^{-5}\ C/m^2[/tex]
Hence, this is the required solution.
Final answer:
The surface charge density on the plates of a parallel-plate air-filled capacitor 7.5225 x [tex]10^{-5}[/tex] [tex]C/m^2[/tex].
Explanation:
To determine the surface charge density on the plates of the parallel-plate air-filled capacitor, we can use the relationship between the electric field (E), the permittivity of free space
[tex](\sigma\))[/tex]. The electric field is defined as
[tex](E = \frac{\sigma}{\varepsilon_0}).[/tex]
Given that the electric field (E) is
[tex](8.5 \times 10^6 V/m)[/tex] and the permittivity of free space [tex]((\varepsilon_0))[/tex] is
[tex](8.85 \times 10^{-12} C^2/N \cdot m^2)[/tex], we can rearrange the formula to solve for the surface charge density [tex]((\sigma)):[/tex]
[tex](\sigma = E \cdot \varepsilon_0 = (8.5 \times 10^6 V/m) \times (8.85 \times 10^{-12} C^2/N \cdot m^2) = 7.5225 \times 10^{-5} C/m^2).[/tex]
Thus, the surface charge density on the plates is
[tex](7.5225 \times 10^{-5} C/m^2).[/tex]
Two 3.00 cm × 3.00 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC . Part A What is the electric field strength inside the capacitor if the spacing between the plates is 1.30 mm ?
Answer:
[tex]8.89\cdot 10^4 V/m[/tex]
Explanation:
The electric field strength between the plates of a parallel plate capacitor is given by
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where
[tex]\sigma[/tex] is the surface charge density
[tex]\epsilon_0[/tex] is the vacuum permittivity
Here we have
[tex]A=3.00 \cdot 3.00 = 9.00 cm^2 = 9.0\cdot 10^{-4} m^2[/tex] is the area of the plates
[tex]Q=0.708 nC = 0.708 \cdot 10^{-9} C[/tex] is the charge on each plate
So the surface charge density is
[tex]\sigma=\frac{Q}{A}=\frac{0.708\cdot 10^{-9}}{9.0\cdot 10^{-4} m^2}=7.87\cdot 10^{-7} C/m^2[/tex]
And now we can find the electric field strength
[tex]E=\frac{\sigma}{\epsilon_0}=\frac{7.87\cdot 10^{-7}}{8.85\cdot 10^{-12}}=8.89\cdot 10^4 V/m[/tex]
The electric field strength inside a parallel-plate capacitor with given dimensions and charge is approximately 88.89 kV/m. This is calculated using the capacitance and charge to find the voltage, which then helps determine the electric field strength.
To find the electric field strength inside a parallel-plate capacitor, you can use the relationship between the voltage (V), plate separation (d), and electric field (E). The formula is given by:
E = V / d
However, we are not given the voltage directly, but we have the charge (Q) and the plate area (A). The first step is to determine the capacitance (C) using the formula:
C = ε₀ (A / d)
Where ε₀ (epsilon naught) is the permittivity of free space (8.85 × 10⁻¹² F/m), A is the area of one plate, and d is the separation between the plates.
Given:
Plate area,
A = (3.00 cm × 3.00 cm)
= 9.00 cm²
= 9.00 × 10⁻⁴ m²
Plate separation,
d = 1.30 mm
= 1.30 × 10⁻³ m
Now calculate the capacitance:
C = ε₀ (A / d)
= (8.85 × 10⁻¹² F/m) (9.00 × 10⁻⁴ m² / 1.30 × 10⁻³ m)
≈ 6.13 × 10⁻¹⁵ F (Farads)
Next, use the charge (Q) and capacitance (C) to find the voltage (V):
Q = C × V
V = Q / C
Given charge,
Q = 0.708 nC
= 0.708 × 10⁻⁹ C
V = (0.708 × 10⁻⁹ C) / (6.13 × 10⁻¹⁵ F)
≈ 115.56 V
Finally, calculate the electric field strength:
E = V / d
E = 115.56 V / 1.30 × 10⁻³ m
≈ 88.89 kV/m
Thus, the electric field strength inside the capacitor is approximately 88.89 kV/m.
An airplane of mass 39,043.01 flies horizontally at an altitude of 9.2 km with a constant speed of 335 m/s relative to Earth. What is the magnitude of the airplane’s angular momentum relative to a ground observer directly below the plane? Give your answer in scientific notation.
Answer:
1.2 x 10¹¹ kgm²/s
Explanation:
m = mass of the airplane = 39043.01
r = altitude of the airplane = 9.2 km = 9.2 x 1000 m = 9200 m
v = speed of airplane = 335 m/s
L = Angular momentum of airplane
Angular momentum of airplane is given as
L = m v r
Inserting the values
L = (39043.01 ) (335) (9200)
L = (39043.01 ) (3082000)
L = 1.2 x 10¹¹ kgm²/s
A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits in phase, and a fringe pattern is observed on a screen 4.8 m from the slits. If there are 5.0 complete bright fringes per centimeter on the screen near the center of the pattern, what is the wavelength of the monochromatic light?
Answer:
750 nm
Explanation:
[tex]d[/tex] = separation of the slits = 1.8 mm = 0.0018 m
λ = wavelength of monochromatic light
[tex]D[/tex] = screen distance = 4.8 m
[tex]y[/tex] = position of first bright fringe = [tex]\frac{1cm}{5 fringe} = \frac{0.01}{5} = 0.002 m[/tex]
[tex]n[/tex] = order = 1
Position of first bright fringe is given as
[tex]y = \frac{nD\lambda }{d}[/tex]
[tex]0.002 = \frac{(1)(4.8)\lambda }{0.0018}[/tex]
λ = 7.5 x 10⁻⁷ m
λ = 750 nm
An object of inertia 0.5kg is hung from a spring, and causes it to extend 5cm. In an elevator accelerating downward at 2 m/s^2 , how far will the spring extend if the same object is suspended from it? Draw the free body diagrams for both the accelerating and non-accelerating situations.
Answer:3.98cm
Explanation:
given data
mass of object[tex]\left ( m\right )[/tex]=0.5kg
intial extension=5 cm
elevator acceleration=2 m/[tex]s^2[/tex]
From FBD of intial position
Kx=mg
K=[tex]\frac{0.5\times 9.81}{0.05}[/tex]
k=98.1 N/m
From FBD of second situation
mg-k[tex]x_0[/tex]=ma
k[tex]x_0[/tex]=m(g-a)
[tex]x_0[/tex]=[tex]\frac{0.5(9.81-2)}{98.1}[/tex]
[tex]x_0[/tex]=3.98cm
At a distance 30 m from a jet engine, intensity of sound is 10 W/m^2. What is the intensity at a distance 180 m?
Answer:
[tex]I_{2}=0.27 W/m^2[/tex]
Explanation:
Intensity is given by the expresion:
[tex]I_{2}=Io (\frac{r1}{r2} )^{2}[/tex]
where:
Io = inicial intensity
r1= initial distance
r= final distance
[tex]I_{2}=10 W/m^2 (\frac{30m}{180m} )^{2}[/tex]
[tex]I_{2}=0.27 W/m^2[/tex]
in a certain right triangle, the two sides that are perpendicular to each other are 5.9 and 5.1 m long. what is the tangent of the angle for which 5.9 m is the opposite side?
Answer:
1.16
Explanation:
Let the angle is theta.
tan θ = perpendicular / base
tan θ = 5.9 / 5.1 = 1.16
The ability to clearly see objects at a distance but not close up is properly called ________. The ability to clearly see objects at a distance but not close up is properly called ________.
Final answer:
Hyperopia, also known as farsightedness, is the condition where close objects are blurry while distant objects are clearly visible. It is corrected with converging lenses. This is the opposite of myopia, which affects distance vision.
Explanation:
The ability to clearly see objects at a distance but not close up is properly called hyperopia, which is a vision problem where close objects are out of focus but distant vision is unaffected; also known as farsightedness. Farsighted individuals have difficulty seeing close objects clearly because their eyes do not converge light rays from a close object enough to make them meet on the retina. To correct this vision problem, converging lenses are used, which help to increase the power of the eyes, allowing for clear vision of nearby objects. This contrasts with myopia (nearsightedness), where distant objects are out of focus while close vision is typically unaffected.
We can use our results for head-on elastic collisions to analyze the recoil of the Earth when a ball bounces off a wall embedded in the Earth. Suppose a professional baseball pitcher hurls a baseball ( 155 grams) with a speed of 99 miles per hour ( 43.6 m/s) at a wall, and the ball bounces back with little loss of kinetic energy. What is the recoil speed of the Earth ( 6 × 1024 kg)?
Answer:
So recoil speed of the Earth will be
[tex]v = 2.25 \times 10^{-24} m/s[/tex]
Explanation:
Here if we assume that during collision if ball will lose very small amount of energy and rebound with same speed
then the impulse given by the ball is
[tex]Impulse = m(v_f - v_i)[/tex]
[tex]Impulse = (0.155)(43.6 - (-43.6))[/tex]
[tex]Impulse = 13.52 Ns[/tex]
so impulse received by the Earth is same as the impulse given by the ball
so here we will have
[tex]Impulse = mv[/tex]
[tex]13.52 = (6 \times 10^{24})v[/tex]
[tex]v = 2.25 \times 10^{-24} m/s[/tex]
The recoil speed of the Earth is [tex]2.25*10^-24 m/s.[/tex]
What is collision?Collision can be regarded as the forceful coming together of bodies.
The impulse of the ball can be calculated as;
[tex]Impulse= MV[/tex]
where V=( V1 -Vo)
V1= final velocity=(43m/s)
V0= initial velocity= (-43.6m/s)
m= mass of baseball
Hence,[tex]V= [43.6-(-43.6)]= 87.2m/s[/tex]
Then Impulse given by the ball=[tex]( 0.155*87.2)= 13.53Ns.[/tex]
We can now calculate the recoil speed of the earth as;
[tex]Impluse= MVV= Impulse/ mass = 13.52/(6*10^24)[/tex]
=2.25*10^-24 m/s
Therefore, the recoil speed of the Earth is 2.25*10^-24 m/s
Learn more about collision at:
https://brainly.com/question/7538238
One kg of air contained in a piston-cylinder assembly undergoes a process from an initial state whereT1=300K,v1=0.8m3/kg, to a final state whereT2=420K,v2=0.2m3/kg. Can this process occur adiabatically? If yes, determine the work, in kJ, for an adiabatic process between these states. If not, determine the direction of the heat transfer. Assume the ideal gas model withcv=0.72kJ/kg·Kfor the air.
Answer:
1. Yes, it can occur adiabatically.
2. The work required is: 86.4kJ
Explanation:
1. The internal energy of a gas is just function of its temperature, and the temperature changes between the states, so, the internal energy must change, but how could it be possible without heat transfer? This process may occur adiabatically due to the energy balance:
[tex]U_{2}-U_{1}=W[/tex]
This balance tell us that the internal energy changes may occur due to work that, in this case, si done over the system.
2. An internal energy change of a gas may be calculated as:
[tex]du=C_{v}dT[/tex]
Assuming [tex]C_{v}[/tex] constant,
[tex]U_{2}-U_{1}=W=m*C_{v}(T_{2}-T_{1})[/tex]
[tex]W=0.72*1*(420-300)=86.4kJ[/tex]
A gas consists of 1024 molecules, each with mass 3 × 10-26 kg. It is heated to a temperature of 300 K, while the volume is held constant. 1) If the gas is confined to a vertical tube 5 × 103 m high, what is the ratio of the pressure at the top to the pressure at the bottom?
Answer:
The ratio of the pressure at the top to the pressure at the bottom is [tex]\dfrac{701}{1000}[/tex]
Explanation:
Given that,
Number of molecules [tex]n= 10^24[/tex]
Mass [tex]m= 3\times10^{-26}\ kg[/tex]
Temperature = 300 K
Height [tex]h = 5\times10^{3}[/tex]
We need to calculate the ratio of the pressure at the top to the pressure at the bottom
Using barometric formula
[tex]P_{h}=P_{0}e^{\dfrac{-mgh}{kT}}[/tex]
[tex]\dfrac{P_{h}}{P_{0}}=e^{\dfrac{-mgh}{kT}}[/tex]
Where, m = mass
g = acceleration due to gravity
h = height
k = Boltzmann constant
T = temperature
Put the value in to the formula
[tex]\dfrac{P_{h}}{P_{0}}=e^{\dfrac{-3\times10^{-26}\times9.8\times5\times10^{3}}{1.3807\times10^{-23}\times300}}[/tex]
[tex]\dfrac{P_{h}}{P_{0}}=\dfrac{701}{1000}[/tex]
Hence, The ratio of the pressure at the top to the pressure at the bottom is [tex]\dfrac{701}{1000}[/tex]
Answer:
Top pressure : Bottom pressure = 701 : 1000
Explanation:
Number of molecules = n = 10^24
Height = h = 5 × 10^3 m
Mass = m = 3 × 10^-26 kg
Boltzman’s Constant = K = 1.38 × 10^-23 J/K
Temperature = T = 300K
The formula for barometer pressure is given Below:
Ph = P0 e^-(mgh/KT)
Ph/P0 = e^-(3 × 10^-26 × 9.81 × 5 × 10^3)/(1.38 × 10^-23)(300)
Ph/P0 = e^-0.355
Ph/P0 = 1/e^0.355
Ph/p0 =0.7008 = 700.8/1000 = 701/1000
Hence,
Top pressure : Bottom pressure = 701 : 1000
Two converging lenses with focal lengths of 40 cm and 20 cm are 16 cm apart. A 2.0 cm -tall object is 14 cm in front of the 40 cm -focal-length lens.
Calculate the image position.
Calculate the image height.
The final image position is 4.34 cm to the right of the second lens, and the final image height is 2.4 cm.
First, we will consider the object in front of the first lens (with a focal length of 40 cm). The object is placed 14 cm in front of this lens.
For the first lens, the lens equation is given by:
[tex]\[\frac{1}{f_1} = \frac{1}{d_o} + \frac{1}{d_i}\][/tex]
where [tex]\(f_1\)[/tex] is the focal length of the first lens, [tex]\(d_o\)[/tex] is the object distance from the first lens, and [tex]\(d_i\)[/tex] is the image distance from the first lens.
Rearranging the equation to solve for [tex]\(d_i\)[/tex], we get:
[tex]\[d_i = \frac{f_1 \cdot d_o}{d_o - f_1}\][/tex]
Plugging in the values:
[tex]\[d_i = \frac{40 \cdot 14}{14 - 40}\] \[d_i = \frac{560}{-26}\] \[d_i = -21.54 \text{ cm}\][/tex]
This means the image formed by the first lens is 21.54 cm to the left of the first lens (virtual image).
Next, we calculate the height of this image using the magnification equation:
[tex]\[m_1 = -\frac{d_i}{d_o}\] \[m_1 = -\frac{-21.54}{14}\] \[m_1 = 1.54\][/tex]
The height of the image formed by the first lens is:
[tex]\[h_i = m_1 \cdot h_o\][/tex]
where [tex]\(h_o\)[/tex] is the object height (2.0 cm).
So, [tex]\(h_i = 1.54 \cdot 2.0\)[/tex]
[tex]\[h_i = 3.08 \text{ cm}\][/tex]
Now, this image becomes the object for the second lens (with a focal length of 20 cm). The distance between the two lenses is 16 cm, so the object distance for the second lens is the image distance from the first lens plus 16 cm.
For the second lens, the object distance [tex]\(d_o'\)[/tex] is:
[tex]\[d_o' = -21.54 + 16\] \[d_o' = -5.54 \text{ cm}\][/tex]
Using the lens equation for the second lens:
[tex]\[\frac{1}{f_2} = \frac{1}{d_o'} + \frac{1}{d_i'}\][/tex]
where [tex]\(f_2\)[/tex] is the focal length of the second lens, [tex]\(d_o'\)[/tex] is the object distance from the second lens, and [tex]\(d_i'\)[/tex] is the image distance from the second lens.
Rearranging the equation to solve for [tex]\(d_i'\)[/tex], we get:
[tex]\[d_i' = \frac{f_2 \cdot d_o'}{d_o' - f_2}\][/tex]
Plugging in the values:
[tex]\[d_i' = \frac{20 \cdot (-5.54)}{-5.54 - 20}\] \[d_i' = \frac{-110.8}{-25.54}\] \[d_i' = 4.34 \text{ cm}\][/tex]
This means the final image is 4.34 cm to the right of the second lens (real image).
To find the height of the final image, we use the magnification equation for the second lens:
[tex]\[m_2 = -\frac{d_i'}{d_o'}\] \[m_2 = -\frac{4.34}{-5.54}\] \[m_2 = 0.78\][/tex]
The height of the final image is:
[tex]\[h_i' = m_2 \cdot h_i\] \[h_i' = 0.78 \cdot 3.08\] \[h_i' = 2.4 \text{ cm}\][/tex]
A capacitor is storing energy of 3 Joules with a voltage of 50 volts across its terminals. A second identical capacitor of the same value is storing energy of 1 Joule. What is the voltage across the terminals of the second capacitor?
Answer:
28.87 volt
Explanation:
Let the capacitance of the capacitor is C.
Energy, U = 3 J, V = 50 volt
U = 1/2 C V^2
3 = 0.5 x C x 50 x 50
C = 2.4 x 10^-3 F
Now U' = 1 J
U' = 1/2 C x V'^2
1 = 0.5 x 2.4 x 10^-3 x V'^2
V' = 28.87 volt
In a hydraulic lift, the diameter of the input piston is 10.0 cm and the diameter of the output piston is 50.0 cm. (a) How much force must be applied to the input piston so that the output piston can lift a 250N object? (b) If the object is lifted a distance of 0.3 m, then how far is the input piston moved?
Answer:
a) Force must be applied to the input piston = 10N
b) The input piston moved by 7.5 m
Explanation:
a) For a hydraulic lift we have equation
[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]
F₁ = ?
d₁ = 10 cm = 0.1 m
F₂ = 250 N
d₂ = 50 cm = 0.5 m
Substituting
[tex]\frac{F_1}{\frac{\pi \times 0.1^2}{4}}=\frac{250}{\frac{\pi \times 0.5^2}{4}}\\\\F_1=10N[/tex]
Force must be applied to the input piston = 10N
b) We have volume of air compressed is same in both input and output.
That is A₁x₁ = A₂x₂
A is area and x is the distance moved
x₂ = 0.3 m
Substituting
[tex]\frac{\pi \times 0.1^2}{4}\times x_1=\frac{\pi \times 0.5^2}{4}\times 0.3\\\\x_1=7.5m[/tex]
The input piston moved by 7.5 m
In an experiment, you determined the density of the wood to be 0.45g/cc, whereas the standard value was 0.47g/cc. Determine the percentage difference. [Hint: Look at the procedure section of Part Al[1 Point] 5. 6. How do you determine y-intercept from a graph? [1 Point]
Answer:
Percentage difference is 4.25 %.
Explanation:
Standard value of the density of wood, [tex]\rho_s=0.47\ g/cc[/tex]
Experimental value of the density of wood, [tex]\rho_e=0.45\ g/cc[/tex]
We need to find the percentage difference on the density of wood. It is given by :
[tex]\%=|\dfrac{\rho_s-\rho_e}{\rho_s}|\times 100[/tex]
[tex]\%=|\dfrac{0.47-0.45}{0.47}|\times 100[/tex]
Percentage difference in the density of the wood is 4.25 %. Hence, this is the required solution.
It has been suggested that a heat engine could be developed that made use of the fact that the temperature several hundred meters beneath the surface of the ocean is several degrees colder than the temperature at the surface. In the tropics, the temperature may be 6 degrees C and 22 degrees C, respectively. What is the maximum efficiency (in %) such an engine could have?
Answer:
efficiency = 5.4%
Explanation:
Efficiency of heat engine is given as
[tex]\eta = \frac{W}{Q_{in}}[/tex]
now we will have
[tex]W = Q_1 - Q_2[/tex]
so we will have
[tex]\eta = 1 - \frac{Q_2}{Q_1}[/tex]
now we know that
[tex]\frac{Q_2}{Q_1} = \frac{T_2}{T_1}[/tex]
so we have
[tex]\eta = 1 - \frac{T_2}{T_1}[/tex]
[tex]\eta = 1 - \frac{273+6}{273+22}[/tex]
[tex]\eta = 0.054[/tex]
so efficiency is 5.4%
A 280-g mass is mounted on a spring of constant k = 3.3 N/m. The damping constant for this damped system is b = 8.4 x 10^-3 kg/s. How many oscillations will the system undergo during the time the amplitude decays to 1/e of its original value?
Answer:
The number of oscillation is 36.
Explanation:
Given that
Mass = 280 g
Spring constant = 3.3 N/m
Damping constant [tex]b=8.4\times10^{-3}\ Kg/s[/tex]
We need to check the system is under-damped, critical damped and over damped by comparing b with [tex]2m\omega_{0}[/tex]
[tex]2m\omega_{0}=2m\sqrt{\dfrac{k}{m}}[/tex]
[tex]2\sqrt{km}=2\times\sqrt{3.3\times280\times10^{-3}}=1.92kg/s[/tex]
Here, b<<[tex]2m\omega_{0}[/tex]
So, the motion is under-damped and will oscillate
[tex]\omega=\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}[/tex]
The number of oscillation before the amplitude decays to [tex]\dfrac{1}{e}[/tex] of its original value
[tex]A exp(\dfrac{-b}{2m}t)=A exp(-1)[/tex]
[tex]\dfrac{b}{2m}t=1[/tex]
[tex]t = \dfrac{2m}{b}[/tex]
[tex]t=\dfrac{2\times280\times10^{-3}}{8.4\times10^{-3}}[/tex]
[tex]t=66.67\ s[/tex]
We need to calculate the time period of one oscillation
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]T=\dfrac{2\times3.14}{\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}}[/tex]
[tex]T=\dfrac{2\times3.14}{\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}}}[/tex]
[tex]T=\dfrac{2\times3.14}{\sqrt{\dfrac{3.3}{280\times10^{-3}}-\dfrac{(8.4\times10^{-3})^2}{4\times(280\times10^{-3})^2}}}[/tex]
[tex]T=1.83\ sec[/tex]
The number of oscillation is
[tex]n=\dfrac{t}{T}[/tex]
[tex]n=\dfrac{66.67}{1.83}[/tex]
[tex]n=36[/tex]
Hence, The number of oscillation is 36.
At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 3.28 m/s, and an 84.4-kg person feels a 488-N force pressing against his back. What is the radius of a chamber?
The radius of the spinning chamber can be calculated using the centripetal force formula, F = m*v^2/r. The values for mass (m), velocity(v) and force(F) are given in the question, which can be substituted into the rearranged version of the formula for radius (r), r = m*v^2/F.
Explanation:The subject question is based in physics, specifically the 'Circular Motion and Gravitation' topic, and is about finding the radius of a spinning chamber in an amusement park, with the information the force exerted and the speed of the chamber.
To find the radius, we would need to use the formula for the circular force, F = m*v^2/r, where F is the force, m is the mass, v the velocity and r the radius. Rearranging for r, the radius, we get r = m*v^2/F.
Substituting the values from the question, we have m = 84.4 kg (person's weight), v = 3.28 m/s (speed of the outer wall) and F = 488 N. Calculating these, we get r = (84.4 kg * (3.28 m/s)^2)/488 N.
This will give the radius of the chamber.
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The radius of the chamber can be determined using the concept of centripetal force. By substituting the given values into the formula for centripetal force (F = m * v² / r) and rearranging for r, the radius is found to be approximately 2.2 meters.
Explanation:In order to solve the problem, one must understand the concept of centripetal force, which is described as a force that makes a body follow a curved path, with its direction orthogonal to the velocity of the body, towards the fixed point of the instantaneous center of curvature of the path.
In this scenario, the centripetal force is equal to the force pressing against the rider's back, which is 488N. This force can be calculated using the formula: F = m * v² / r, where F is the centripetal force, m is the mass of the rider, v is the speed of the outer wall, and r is the radius of the chamber.
Given that m = 84.4 kg, v = 3.28 m/s, and F = 488 N, by substituting these values into the formula and rearranging it we find that r = m * v² / F. Consequently, r = (84.4 kg * (3.28 m/s)²) / 488 N, which equals approximately 2.2 meters. Therefore, the radius of the chamber is approximately 2.2 meters.
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A plane flying with a constant speed of 150 km/h passes over a ground radar station at an altitude of 3 km and climbs at an angle of 30°. At what rate is the distance from the plane to the radar station increasing a minute later? (Round your answer to the nearest whole number.)
After a minute of flight, the plane's altitude changes due to its climb. The speed at which the distance from the plane to the radar station is increasing equals the resultant of its horizontal and vertical speed components. This can be computed as approximately 1.26 km or 1260 meters.
Explanation:The student's question pertains to both kinematics and trigonometry in Physics. In this scenario, the plane is climbing at an angle, while its horizontal speed is constant. The speed at which the distance from the plane to the radar station increases involves understanding the principle of vector addition and the concept of resultant velocity.
We can construct a right-angled triangle where one side is the horizontal speed component (= 150 km/h), the other side is the vertical speed component (altitude change over time, given by climbing speed = altitude/duration), and the hypotenuse is the resultant velocity, i.e., the speed at which the distance from the plane to the radar station is increasing.
After a minute, the altitude gains due to the climb is 3 km + (150 km/h * sin(30°) * 1/60 hr) = 3.0375 km, where sin(30°) represents the vertical ratio of the velocity. Radar station distance change can be calculated using Pythagoras theorem. In one minute, the plane travels horizontally by 150 km/h * 1/60 hr = 2.5 km. Thus, the change in distance is sqrt{(3.0375 km)^2 + (2.5 km)^2 } - 3 km (original altitude), which approximately equals 1.26 km or 1260 meters when rounded to the nearest whole number.
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The distance from the plane to the radar station is increasing at a rate of approximately 5 km/h one minute later.
Let's define the variables:
v = 150 km/h (speed of the plane)
h = 3 km (altitude of the radar station)
θ = 30° (angle of ascent)
We need to find the rate at which the distance from the plane to the radar station is increasing 1 minute (or 1/60 hours) after the plane passes over the station.
We'll use the following steps:
Determine the horizontal and vertical components of the plane's velocity:
Horizontal component,v_x = v × cos(θ) = 150 km/h × cos(30°)
v_x = 150 × (√3 / 2) ≈ 129.9 km/h
Vertical component,v_y = v × sin(θ) = 150 km/h × sin(30°)
v_y = 150 ×0.5 = 75 km/h
Calculate the horizontal distance traveled in 1 minute:d_x = v_x × (1/60) hours
d_x = 129.9 km/h × (1/60) = 2.165 km
Determine the new altitude after 1 minute:New altitude, h_new = h + (v_y × (1/60))
h_new = 3 km + (75 × (1/60)) = 3 + 1.25 km = 4.25 km
Calculate the distance from the plane to the radar station:Using the Pythagorean Theorem: d = sqrt(d_x² + h_new²)
d = square root of (2.165² + 4.25²)
d = square root of (4.687 + 18.06) = √22.75 = 4.77 km
Differentiate the distance with respect to time to find the rate of change:The rate of distance increase is approximately 4.77 km/h rounded to the nearest whole number, which is 5 km/h
How much work in joules must be done to stop a 920-kg car traveling at 90 km/h?
Answer:
Work done, W = −287500 Joules
Explanation:
It is given that,
Mass of the car, m = 920 kg
The car is travelling at a speed of, u = 90 km/h = 25 m/s
We need to find the amount of work must be done to stop this car. The final velocity of the car, v = 0
Work done is also defined as the change in kinetic energy of an object i.e.
[tex]W=\Delta K[/tex]
[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]
[tex]W=\dfrac{1}{2}\times 920\ kg(0-(25\ m/s)^2)[/tex]
W = −287500 Joules
Negative sign shows the work done is done is opposite direction.
Calculate the pressure exerted on the ground when a woman wears high heals. Her mass is 65 kg. and the area of each heal is 1 cm^2.
Answer:
318.5 x 10^4 Pa
Explanation:
weight of woman = m g = 65 x 9.8 = 637 N
Area of both the heels = 1 x 2 = 2 cm^2 = 2 x 10^-4 m^2
Pressure is defined as the thrust acting per unit area.
P = F / A
Where, F is the weight of the woman and A be the area of heels
P = 637 / (2 x 10^-4) = 318.5 x 10^4 Pa
A current-carrying wire passes through a region of space that has a uniform magnetic field of 0.92 T. If the wire has a length of 2.6 m and a mass of 0.60 kg, determine the minimum current needed to levitate the wire. A
Answer:
Current, I = 2.45 T
Explanation:
It is given that,
Magnetic field, B = 0.92 T
Length of wire, l = 2.6 m
Mass, m = 0.6 kg
We need to find the minimum current needed to levitate the wire. It is given by balancing its weight to the magnetic force i.e.
[tex]Ilb=mg[/tex]
[tex]I=\dfrac{mg}{lB}[/tex]
[tex]I=\dfrac{0.6\ kg\times 9.8\ m/s^2}{2.6\ m\times 0.92\ T}[/tex]
I = 2.45 A
So, the minimum current to levitate the wire is 2.45 T. Hence, this is the required solution.
The minimum current needed to levitate the wire in the given magnetic field is approximately 2.58 Amps, determined by setting the magnetic force acting on the wire equal to the gravitational force and solving for current.
Explanation:The minimum current necessary to levitate the wire in a uniform magnetic field can be determined by equating the magnetic force acting on the wire to the gravitational force acting on it. The magnetic force exerted on a current-carrying wire in a magnetic field is given by F = IℓBsinθ, where F is the force, I is the current, ℓ is the length of the wire, B is the magnetic field strength, and θ is the angle between the current and the magnetic field. Given the wire is levitating, the angle θ is 90°, meaning sinθ is 1. Additionally, the gravitational force is F = mg, where m is the mass of the wire and g is the acceleration due to gravity. Setting the magnetic force equal to the gravitational force gives IℓB = mg, which we can solve for I to get I = mg/(ℓB). Using the given values, I = (0.60 kg * 9.8 m/s²) / (2.6 m * 0.92 T) = 2.58 A. So, the minimum current needed to levitate the wire is approximately 2.58 Amps.
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A ball is thrown from the edge of a 40.0 m high cliff with a speed of 20.0 m/s at an angle of 30.0° below horizontal. What is the speed of the ball when it hits the ground below the cliff?
Answer:
Velocity is 34.42 m/s at an angle of 56.91° below horizontal
Speed is 34.42 m/s
Explanation:
Velocity = 20.0 m/s at an angle of 30° below horizontal
Vertical velocity = 20 sin 20 = 6.84 m/s downward.
Horizontal velocity = 20 cos 20 = 18.79 m/s towards right.
Let us consider the vertical motion of ball we have equation of motion
v² = u² + 2as
We need to find v, u = 6.84 m/s, a = 9.81 m/s² and s = 40 m
Substituting
v² = 6.84² + 2 x 9.81 x 40 = 831.59
v = 28.84 m/s
So on reaching ground velocity of ball is
Vertical velocity = 28.84 m/s downward.
Horizontal velocity = 18.79 m/s towards right.
Velocity
[tex]v=\sqrt{28.84^2+18.79^2}=34.42m/s[/tex]
[tex]tan\theta =\frac{28.84}{18.79}\\\\\theta =56.91^0[/tex]
So velocity is 34.42 m/s at an angle of 56.91° below horizontal
Speed is 34.42 m/s
A diverging lens has a focal length of -30.0 cm. An object is placed 18.0 cm in front of this lens.
(a) Calculate the image distance.
(b) Calculate the magnification.
Answer:
A) Calculate the distance
To find the image distance for a diverging lens with a focal length of -30.0 cm and an object placed 18.0 cm in front, use the lens formula to calculate di = -77.14 cm, indicating a virtual image. The magnification equation yields a magnification of 4.285, indicating an upright image.
To calculate the image distance and magnification for a diverging lens, we can use the lens formula and the magnification equation. Given that a diverging lens has a focal length of -30.0 cm, and an object is placed 18.0 cm in front of this lens, we first need to use the lens formula:
1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance.
Substituting the values, we get:
1/(-30) = 1/18 + 1/di
Following the steps:
Solving the equation for 1/di gives us 1/di = 1/(-30) - 1/18.
Finding a common denominator and subtracting the fractions we get 1/di = (-2 - 5)/(-540), which simplifies to 1/di = -7/540.
Therefore, di = -540/7 = -77.14 cm.
The negative sign indicates that the image is virtual and located on the same side of the lens as the object.
For magnification (m), use the equation m = - di/do:
m = - (-77.14)/18
m = 77.14/18
m = 4.285
The positive magnification value indicates that the image is upright compared to the object.
An object is moving at a constant speed along a straight line. Which of the following statements is not true? A. There must be a non-zero net force acting on the object.
B. The acceleration of the object is zero.
C. The net force acting on the object must be zero.
D. The velocity of the object is constant.
Answer:
False statement = There must be a non-zero net force acting on the object.
Explanation:
An object is moving at a constant speed along a straight line. If the speed is constant then its velocity must be constant. We know that the rate of change of velocity is called acceleration of the object i.e.
[tex]a=\dfrac{dv}{dt}[/tex]
a = 0
⇒ The acceleration of the object is zero.
The product of force and acceleration gives the magnitude of force acting on the object i.e.
F = m a = 0
⇒ The net force acting on the object must be zero.
So, the option (a) is not true. This is because the force acting on the object is zero. First option contradicts the fact.
A rock is thrown into a still pond. The circular ripples move outward from the point of impact of the rock so that the radius of the circle formed by a ripple increases at the rate of 5 feet per minute. Find the rate at which the area is changing at the instant the radius is 12 feet.
Answer:
376.9911ft²/minute
Explanation:
In the given question the rate of chage of radius in given as
[tex]\frac{\mathrm{d}r }{\mathrm{d} t}[/tex]=5ft per minute
we know ares of circle A=pi r^{2}
differentiating w.r.t. t we get
[tex]\frac{\mathrm{d} A}{\mathrm{d} t}=2\pi r\frac{\mathrm{d}r }{\mathrm{d} t}[/tex]
Now, we have find [tex]\frac{\mathrm{d}A }{\mathrm{d} t} at r=12 feet[/tex]
[tex]\frac{\mathrm{d} A}{\mathrm{d} t}=2\times\pi\times12\times5=120\pi=376.9911ft^{2}/minute[/tex]
The rate at which the area of the ripple is changing when the radius is 12 feet is 120π square feet per minute.
Explanation:The concept in question is related to mathematical calculus and the principle of rates. The area of a circle is represented by the formula A = πr², with 'A' representing the area and 'r' the radius. To determine how the area changes with changes in the radius, we differentiate this function with respect to time, resulting in dA/dt = 2πr (dr/dt). We know the radius is increasing at a rate of 5 feet per minute (dr/dt = 5 ft/min). At the instant when the radius is 12 feet, we simply substitute into our differentiated equation to find that dA/dt = 2π(12ft)(5ft/min) = 120π ft²/min.
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