Answer:
The sum of the kinetic energies of the alpha particle and the new nucleus = (1.359098 × 10⁻¹²) J
Explanation:
We will use the conservation of energy theorem for extremely small particles,
Total energy before split = total energy after split
That is,
Total energy of the original nucleus = (total energy of the new nucleus) + (total energy of the alpha particle)
Total energy of these subatomic particles is given as equal to (rest energy) + (kinetic energy)
Rest energy = mc² (Einstein)
Let Kinetic energy be k
Kinetic energy of original nucleus = k₀ = 0 J
Kinetic energy of new nucleus = kₙ
Kinetic energy of alpha particle = kₐ
Mass of original nucleus = m₀ = (3.969554 × 10⁻²⁵) kg
Mass of new nucleus = mₙ = (3.902996 × 10⁻²⁵) kg
Mass of alpha particle = mₐ = (6.640678 × 10⁻²⁷) kg
Speed of light = (2.9979246 × 10⁸) m/s
Total energy of the original nucleus = m₀c² (kinetic energy = 0, since it was originally at rest)
Total energy of new nucleus = (mₙc²) + kₙ
Total energy of the alpha particle = (mₐc²) + kₐ
(m₀c²) = (mₙc²) + kₙ + (mₐc²) + kₐ
kₙ + kₐ = (m₀c²) - [(mₙc²) + (mₐc²)
(kₙ + kₐ) = c² (m₀ - mₙ - mₐ)
(kₙ + kₐ) = (2.9979246 × 10⁸)² [(3.969554 × 10⁻²⁵) - (3.902996 × 10⁻²⁵) - (6.640678 × 10⁻²⁷)]
(kₙ + kₐ) = (8.98755191 × 10¹⁶)(1.5122 × 10⁻²⁹) = (1.35909760 × 10⁻¹²) J
A circular coil 17.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this location has magnitude 5.50×10−5T and points into the Earth at an angle of 56.0∘ below a line pointing due north. A 7.80-A clockwise current passes through the coil.
Determine the torque on the coil.
Answer:
The torque in the coil is 4.9 × 10⁻⁵ N.m
Explanation:
T = NIABsinθ
Where;
T is the torque on the coil
N is the number of loops = 9
I is the current = 7.8 A
A is the area of the circular coil = ?
B is the Earth's magnetic field = 5.5 × 10⁻⁵ T
θ is the angle of inclination = 90 - 56 = 34°
Area of the circular coil is calculated as follows;
[tex]A = \frac{\pi d^2}{4} \\\\A = \frac{\pi 0.17^2}{4} =0.0227 m^2[/tex]
T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°
T = 4.9 × 10⁻⁵ N.m
Therefore, the torque in the coil is 4.9 × 10⁻⁵ N.m
One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q is placed concentric with the first and in the same plane. The radius of this ring is a/2. If a = 1m and Q = 3µC, what force is exerted on an electron 5m to the right of these along their common axis?
Answer:
The force exerted on an electron is [tex]7.2\times10^{-18}\ N[/tex]
Explanation:
Given that,
Charge = 3 μC
Radius a=1 m
Distance = 5 m
We need to calculate the electric field at any point on the axis of a charged ring
Using formula of electric field
[tex]E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]
[tex]E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]
Put the value into the formula
[tex]E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}[/tex]
[tex]E_{1}=1.0183\times10^{3}\ N/C[/tex]
Using formula of electric field again
[tex]E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]
Put the value into the formula
[tex]E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}[/tex]
[tex]E_{2}=-1.064\times10^{3}\ N/C[/tex]
We need to calculate the resultant electric field
Using formula of electric field
[tex]E=E_{1}+E_{2}[/tex]
Put the value into the formula
[tex]E=1.0183\times10^{3}-1.064\times10^{3}[/tex]
[tex]E=-0.045\times10^{3}\ N/C[/tex]
We need to calculate the force exerted on an electron
Using formula of electric field
[tex]E = \dfrac{F}{q}[/tex]
[tex]F=E\times q[/tex]
Put the value into the formula
[tex]F=-0.045\times10^{3}\times(-1.6\times10^{-19})[/tex]
[tex]F=7.2\times10^{-18}\ N[/tex]
Hence, The force exerted on an electron is [tex]7.2\times10^{-18}\ N[/tex]
A circular coil of 185 turns has a radius of 1.60 cm. (a) Calculate the current that results in a magnetic dipole moment of magnitude 1.79 A·m2. (b) Find the maximum magnitude of the torque that the coil, carrying this current, can experience in a uniform 47.7 mT magnetic field.
Answer:
(a) the current in the coil is 12.03 A
(b) the maximum magnitude of the torque is 0.0854 N.m
Explanation:
Given;
number of turns, N = 185
radius of the coil, r = 1.6 cm = 0.016 m
magnetic dipole moment, μ = 1.79 A·m²
Part (a) current in the coil
μ = NIA
Where;
I is the current in the coil
A is the of the coil = πr² = π(0.016)² = 0.000804 m²
I = μ / (NA)
I = 1.79 / (185 x 0.000804)
I = 1.79 / 0.14874
I = 12.03 A
Part (b) the maximum magnitude of the torque
τ = μB
Where;
τ is the maximum magnitude of the torque
B is the magnetic field strength = 47.7 mT
τ = 1.79 x 0.0477 = 0.0854 N.m
Given Information:
Number of turns = N = 185 turns
Radius of circular coil = r = 1.60 cm = 0.0160 m
Magnetic field = B = 47.7 mT
Magnetic dipole moment = µ =1.79 A.m
Required Information:
(a) Current = I = ?
(b) Maximum Torque = τ = ?
Answer:
(a) Current = 12.03 A
(b) Maximum Torque = 0.0853 N.m
Explanation:
(a) The magnetic dipole moment µ is given by
µ = NIAsin(θ)
I = µ/NAsin(θ)
Where µ is the magnetic dipole moment, N is the number of turns, I is the current flowing through the circular loop, A is the area of circular loop and is given by
A = πr²
A = π(0.0160)²
A = 0.000804 m²
I = 1.79/185*0.000804*sin(90)
I = 12.03 A
(b) The toque τ is given by
τ = NIABsin(θ)
The maximum torque occurs at θ = 90°
τ = 185*12.03*0.00804*0.0477*sin(90°)
τ = 0.0853 N.m
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is W/m². What is the rms value of the electric field in the electromagnetic wave emitted by the laser?
I think your question should be:
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is
[tex] S = 1.23*10^9 W/m^2 [/tex]
What is the rms value of (a) the electric field and
(b) the magnetic field in the electromagnetic wave emitted by the laser
Answer:
a) [tex] 6.81*10^5 N/c [/tex]
b) [tex] 2.27*10^3 T [/tex]
Explanation:
To find the RMS value of the electric field, let's use the formula:
[tex] E_r_m_s = sqrt*(S / CE_o)[/tex]
Where
[tex] C = 3.00 * 10^-^8 m/s [/tex];
[tex] E_o = 8.85*10^-^1^2 C^2/N.m^2 [/tex];
[tex] S = 1.23*10^9 W/m^2 [/tex]
Therefore
[tex] E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]} [/tex]
[tex] E_r_m_s= 6.81 *10^5N/c [/tex]
b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:
[tex] B_r_m_s = E_r_m_s / C [/tex];
[tex] = 6.81*10^5 N/c / 3*10^8m/s [/tex];
[tex] B_r_m_s = 2.27*10^3 T [/tex]
Complete Question:
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is 1.38 * 10⁹ W/m². What is the rms value of the electric field in the electromagnetic wave emitted by the laser?
Answer:
E = 7.21 * 10⁶ N/C
Explanation:
S = 1.38 * 10⁹ W/m²..............(1)
The formula for the average intensity of light can be given by:
S = c∈₀E²
The speed of light, c = 3 * 10⁸ m/s²
Permittivity of air, ∈₀ = 8.85 * 10⁻¹²m-3 kg⁻¹s⁴ A²
Substituting these parameters into equation (1)
1.38 * 10⁹ = 3 * 10⁸ * 8.85 * 10⁻¹² * E²
E² = (1.38 * 10⁹)/(3 * 10⁸ * 8.85 * 10⁻¹²)
E² = 0.052 * 10¹⁷
E = 7.21 * 10⁶ N/C
For a particular pipe in a pipe-organ, it has been determined that the frequencies 576 Hz and 648 Hz are two adjacent natural frequencies. Using 343 m/s as the speed of sound in air, determine the following.(a) fundamental frequency for this pipe (b) Is the pipe is open at both ends or closed at one end?(c) length of the pipe
Answer:
(A) Fo = 72 Hz
(B) The pipe is open at both ends
(C) The length of the pipe is 2.38m
This problem involves the application of the knowledge of standing waves in pipes.
Explanation:
The full solution can be found in the attachment below.
For pipes open at both ends the frequency of the pipe is given by
F = nFo = nv/2L where n = 1, 2, 3, 4.....
For pipes closed at one end the frequency of the pipe is given by
F = nFo = nv/4L where n = 1, 3, 5, 7...
The full solution can be found in the attachment below.
suppose we have two masses m1=2000 g and m2=4000g, where m1 is moving with initial velocity v1,i=24m/s and m2 is at rest at t=0s. they collide with each other at some time t=t1 after the collision they stick together and move as one. what is the final velocity, v1f, (magnitude and direction)
Answer:
The final velocity is 8 m/s and its direction is along the positive x-axis
Explanation:
Given :
Mass, m₁ = 2000 g = 2 kg
Mass, m₂ = 4000 g = 4 kg
Initial velocity of mass m₁, v₁ = 24i m/s
Initial velocity of mass m₂, v₂ = 0
According to the problem, after collision the two masses are stick together and moving with same velocity, that is, [tex]v_{1f}[/tex].
Applying conservation of momentum,
Momentum before collision = Momentum after collision
[tex]m_{1} v_{1} +m_{2} v_{2} =(m_{1}+m_{2}) v_{1f}[/tex]
Substitute the suitable values in the above equation.
[tex]2\times24 +4\times0 =(2+4}) v_{1f}[/tex]
[tex]v_{1f}=8i\ m/s[/tex]
A woman walks due west on the deck of a ship at 4 mi/h. The ship is moving north at a speed of 20 mi/h. Find the speed and direction of the woman relative to the surface of the water. (Round your answers to one decimal place.)
Answer:
The speed of woman is 20.39 mi/h at and angle of 11.3 degrees wrt the surface of the water.
Explanation:
Given that,
Speed of women due west, [tex]v_w=4\ mi/h[/tex]
Speed of women due north, [tex]v_n=20\ mi/h[/tex]
We need to find the speed and direction of the woman relative to the surface of the water. The resultant speed is given by :
[tex]v=\sqrt{v_w^2+v_n^2}[/tex]
[tex]v=\sqrt{4^2+20^2}[/tex]
[tex]v=20.39\ mi/h[/tex]
Let [tex]\theta[/tex] is the direction of speed. It is given by :
[tex]\tan\theta=\dfrac{4}{20}[/tex]
[tex]\theta=11.3^{\circ}[/tex]
So, the speed of woman is 20.39 mi/h at and angle of 11.3 degrees wrt the surface of the water.
The speed of the woman relative to the surface of the water is approximately 20.4 mi/h, and she is moving approximately 11.3° south of west.
Explanation:To find the speed and direction of the woman relative to the surface of the water, we can use vector addition. Since the woman is walking due west, her velocity vector is 4 mi/h west. The ship's velocity vector is 20 mi/h north. By adding these two vectors using vector addition, we can find the resultant velocity vector. The magnitude of the resultant velocity represents the speed of the woman relative to the surface of the water, while the direction represents the direction she is moving relative to the surface of the water.
Using the Pythagorean theorem, we can find the magnitude of the resultant velocity:
Resultant Velocity: √(4^2 + 20^2) = √(16 + 400) = √416 ≈ 20.4 mi/h
Using trigonometry, we can find the direction of the resultant velocity:
Direction: tan^(-1)(4/20) ≈ 11.3° south of west
Learn more about Relative Velocity here:https://brainly.com/question/34025828
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The mass luminosity relation L M 3.5 describes the mathematical relationship between luminosity and mass for main sequence stars. It describes how a star with a mass of 2 M⊙ would have a luminosity of _____________ L⊙ while a star with luminosity of 3,160 L⊙ would have an approximate mass of ________________ M⊙.
Answer:
(a) 11.3 L
(b) 10 M
Explanation:
The mass-luminosity relationship states that:
Luminosity ∝ Mass^3.5
Luminosity = (Constant)(Mass)^3.5
So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:
Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)
where,
L = Luminosity of Sun
M = mass of Sun
(a)
It is given that:
Mass of Star = 2M
Therefore, eqn (1) implies that:
Luminosity of star/L = (2M/M)^3.5
Luminosity of Star = (2)^3.5 L
Luminosity of Star = 11.3 L
(b)
It is given that:
Luminosity of star = 3160 L
Therefore, eqn (1) implies that:
3160L/L = (Mass of Star/M)^3.5
taking ln on both sides:
ln (3160) = 3.5 ln(Mass of Star/M)
8.0583/3.5 = ln(Mass of Star/M)
Mass of Star/M = e^2.302
Mass of Star = 10 M
Answer:
1. 11
2. 10
Explanation:
1. Using the formula L=M^3.5, and the fact that we were given the Mass,
L=2^3.5
L=11.3137085
L=11
Thus, the luminosity of the star would be 11 solar lumen
2. If we invert the formula, M=L^(2/7)
M=3160^(2/7)
M=9.99794159
M=10
Thus, the mass of the starwould be 10 solar masses.
0.3-L glass of water at 20C is to be cooled with ice to 5C. Determine how much ice needs to be added to the water, in grams, if the ice is at (a) 0C and (b) –20C. Also (c) determine how much water would be needed if the cooling is to be done with cold water at 0C. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0C and 333.7 kJ/kg, respectively, and the density of water is 1 kg/L.
Answer:
[tex]a. m_i_c_e=54.6g\\b. m_i_c_e=48.7g\\m_c_o_l_d_w_a_t_e_r=900g[/tex]
Explanation:
First we need to state our assumptions:
Thermal properties of ice and water are constant, heat transfer to the glass is negligible, Heat of ice [tex]h_i_f=333.7KJkg[/tex]
Mass of water,[tex]m_w=\rho V =1\times0.3=0.3Kg[/tex].
Energy balance for the ice-water system is defined as
[tex]E_i_n-E_o_u_t=\bigtriangleup E_s_y_s\\0=\bigtriangleup U=\bigtriangleup U_i_c_e+\bigtriangleup U_w[/tex]
a.The mass of ice at [tex]0\textdegree C[/tex] is defined as:
[tex][mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[0+333.7+418\times(5-0)]+0.3\times4.18\times(5-20)=0\\m_i_c_e=0.0546Kg=54.6g[/tex]
b.Mass of ice at [tex]20\textdegree C[/tex] is defined as:
[tex][mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[2.11\times(0-(20))+333.7+4.18\times(5-0)]+0.3\times4.18\times(5-20)=0\\\\m_i_c_e=0.0487Kg=48.7g[/tex]
c.Mass of cooled water at [tex]T_c_w=0\textdegree C[/tex]
[tex]\bigtriangleup U_c_w+\bigtriangleup U_w=0[/tex]
[tex][mc(T_2-T_1)]_c_w+[mc(T_2-T_1)]_w=0\\m_c_w\times4.18\times(5-0)+0.3\times4.18\times(5-20)\\m_c_w=0.9kg=900g[/tex]
(a) The mass of ice needed if the ice is at 0°C is [tex]\( 56.4 \, \text{g} \).[/tex]
(b) The mass of ice needed if the ice is at -20°C is [tex]\( 50.1 \, \text{g} \).[/tex]
(c) The mass of cold water needed if the cooling is done with water at 0°C is [tex]\( 900 \, \text{g} \).[/tex]
To determine how much ice needs to be added to the water to cool it from 20°C to 5°C, we need to consider the heat transfer involved. We'll use the principles of heat balance, assuming no heat is lost to the surroundings.
Given Data:
Volume of water: [tex]\( V = 0.3 \, \text{L} \)[/tex]Initial temperature of water: [tex]\( T_i = 20^\circ \text{C} \)[/tex]Final temperature of water: [tex]\( T_f = 5^\circ \text{C} \)[/tex]Specific heat capacity of water: [tex]\( c_w = 4.18 \, \text{kJ/kg}^\circ \text{C} \)[/tex]Density of water: [tex]\( \rho_w = 1 \, \text{kg/L} \)[/tex]Heat of fusion of ice: [tex]\( L_f = 333.7 \, \text{kJ/kg} \)[/tex]Specific heat capacity of ice: [tex]\( c_i = 2.1 \, \text{kJ/kg}^\circ \text{C} \)[/tex]Melting temperature of ice: [tex]\( 0^\circ \text{C} \)[/tex]Calculations:
(a) Ice at 0°C:
1. Heat required to cool the water from 20°C to 5°C:
[tex]\[ Q_{\text{water}} = m_w \cdot c_w \cdot (T_i - T_f) \][/tex]
where:
[tex]\( m_w = \rho_w \cdot V \)[/tex][tex]\( \rho_w = 1 \, \text{kg/L} \)[/tex][tex]\( V = 0.3 \, \text{L} \)[/tex]So,
[tex]\[ m_w = 1 \, \text{kg/L} \times 0.3 \, \text{L} = 0.3 \, \text{kg} \][/tex]
[tex]\[ Q_{\text{water}} = 0.3 \, \text{kg} \times 4.18 \, \text{kJ/kg}^\circ \text{C} \times (20^\circ \text{C} - 5^\circ \text{C}) \][/tex]
[tex]\[ Q_{\text{water}} = 0.3 \times 4.18 \times 15 \][/tex]
[tex]\[ Q_{\text{ice}} = m_{\text{ice}} \cdot L_f \][/tex]
2. Heat absorbed by ice to melt at 0°C:
[tex]\[ Q_{\text{ice}} = m_{\text{ice}} \cdot L_f \][/tex]
To find the mass of ice needed:
[tex]\[ m_{\text{ice}} = \frac{Q_{\text{water}}}{L_f} \][/tex]
[tex]\[ m_{\text{ice}} = \frac{18.81 \, \text{kJ}}{333.7 \, \text{kJ/kg}} \][/tex]
[tex]\[ m_{\text{ice}} = 0.0564 \, \text{kg} = 56.4 \, \text{g} \][/tex]
(b) Ice at -20°C:
1. Heat required to cool ice from -20°C to 0°C:
[tex]\[ Q_{\text{cool}} = m_{\text{ice}} \cdot c_i \cdot (0^\circ \text{C} - (-20^\circ \text{C})) \][/tex]
[tex]\[ Q_{\text{cool}} = m_{\text{ice}} \cdot 2.1 \, \text{kJ/kg}^\circ \text{C} \cdot 20 \][/tex]
[tex]\[ Q_{\text{cool}} = m_{\text{ice}} \cdot 42 \, \text{kJ/kg} \][/tex]
2. Heat absorbed by ice to melt at 0°C:
[tex]\[ Q_{\text{melt}} = m_{\text{ice}} \cdot L_f = m_{\text{ice}} \cdot 333.7 \, \text{kJ/kg} \][/tex]
3. Total heat absorbed by the ice:
[tex]\[ Q_{\text{total}} = Q_{\text{cool}} + Q_{\text{melt}} \][/tex]
[tex]\[ Q_{\text{total}} = m_{\text{ice}} \cdot 42 \, \text{kJ/kg} + m_{\text{ice}} \cdot 333.7 \, \text{kJ/kg} \][/tex]
[tex]\[ Q_{\text{total}} = m_{\text{ice}} \cdot (42 + 333.7) \, \text{kJ/kg} \][/tex]
[tex]\[ Q_{\text{total}} = m_{\text{ice}} \cdot 375.7 \, \text{kJ/kg} \][/tex]
Using the heat required to cool the water:
[tex]\[ m_{\text{ice}} = \frac{Q_{\text{water}}}{Q_{\text{total}}} \][/tex]
[tex]\[ m_{\text{ice}} = \frac{18.81 \, \text{kJ}}{375.7 \, \text{kJ/kg}} \][/tex]
[tex]\[ m_{\text{ice}} = 0.0501 \, \text{kg} = 50.1 \, \text{g} \][/tex]
(c) Cooling with Water at 0°C:
1. Heat required to cool the water from 20°C to 5°C:
[tex]\[ Q_{\text{water}} = 18.81 \, \text{kJ} \][/tex]
2. Heat absorbed by the cold water to warm from 0°C to 5°C:
[tex]\[ Q_{\text{cold water}} = m_{\text{cold water}} \cdot c_w \cdot (5^\circ \text{C} - 0^\circ \text{C}) \][/tex]
[tex]\[ 18.81 \, \text{kJ} = m_{\text{cold water}} \cdot 4.18 \, \text{kJ/kg}^\circ \text{C} \cdot 5 \][/tex]
[tex]\[ m_{\text{cold water}} = \frac{18.81 \, \text{kJ}}{4.18 \, \text{kJ/kg}^\circ \text{C} \times 5} \][/tex]
[tex]\[ m_{\text{cold water}} = \frac{18.81}{20.9} \][/tex]
[tex]\[ m_{\text{cold water}} \approx 0.900 \, \text{kg} = 900 \, \text{g} \][/tex]
What must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 670 N/CN/C ?
Answer:
[tex]q = 2.067 \times 10^{-5}\ C[/tex]
Explanation:
Given,
mass = 1.41 g = 0.00141 Kg
Electric field,E = 670 N/C.
We know,
Force in charge due to Electric field.
F = E q
And also we know
F = m g
Equating both the equation of motion
m g = E q
[tex]q =\dfrac{mg}{E}[/tex]
[tex]q =\dfrac{0.00141 \times 9.81}{670}[/tex]
[tex]q = 2.067 \times 10^{-5}\ C[/tex]
Charge of the particle is equal to [tex]q = 2.067 \times 10^{-5}\ C[/tex]
P7.36 A ship is 125 m long and has a wetted area of 3500 m2. Its propellers can deliver a maximum power of 1.1 MW to seawater at 20C. If all drag is due to friction, estimate the maximum ship speed, in kn.
The solution is in the attachment
A 1050 kg sports car is moving westbound at 13.0 m/s on a level road when it collides with a 6320 kg truck driving east on the same road at 14.0 m/s . The two vehicles remain locked together after the collision.
a. What is the velocity (magnitude and direction) of the two vehicles just after the collision?
b. At what speed should the truck have been moving so that both it and the car are stopped in the collision?
c. Find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b). For which situation is the change in kinetic energy greater in magnitude?
Answer:
(A) V = 9.89m/s
(B) U = -2.50m/s
(C) ΔK.E = –377047J
(D) ΔK.E = –257750J
Explanation:
The full solution can be found in the attachment below. The east has been chosen as the direction for positivity.
This problem involves the principle of momentum conservation. This principle states that the total momentum before collision is equal to the total momentum after collision. This problem is an inelastic kind of collision for which the momentum is conserved but the kinetic energy is not. The kinetic energy after collision is always lesser than that before collision. The balance is converted into heat by friction, and also sound energy.
See attachment below for full solution.
Whats the temperature -15°F in degrees Celsius?
Answer:
-26.1111
Explanation:
Hello!
The formula for converting Fahrenheit to Celsius can be written as follows:
C = [tex]\frac{(F-32)(5)}{9}[/tex]
Now let’s insert the value provided for Fahrenheit:
C = [tex]\frac{((-15)-32)(5)}{9}[/tex]
Now simplify the numerator:
C = [tex]\frac{(-235)}{9}[/tex]
Now divide:
C = -26.111
We have now proven that -15 degrees Fahrenheit is equal to -26.111 degrees Celsius.
I hope this helps!
Two satellites, one in geosynchronous orbit (T = 24 hrs) and one with a period of 12 hrs, are orbiting Earth. How many times larger than the radius of Earth is the distance between the orbits of the two satellites.
The distance between the orbits of two satellites is 7.97 m.
Explanation:
Johannes Kepler was the first to propose three laws for the planetary motion. According to him, the orbits in which planets are rotating are elliptical in nature and Sun is at the focus of the ellipse. Also the area of sweeping is same.
So based on these three assumptions, Kepler postulated three laws. One among them is Kepler's third law of planetary motion. According to the third law, the square of the time taken by a planet to cover a specified region is directly proportional to the cube of the major elliptical axis or the radius of the ellipse.
So, [tex]T^{2} = r^{3}[/tex]
Thus, for the geosynchornous satellite, as the time taken is 24 hours, then the radius or the major axis of this satellite is
[tex](24)^{2}= r^{3} \\(2*2*2*3)^{2} = r^{3}\\r = \sqrt[\frac{2}{3} ]{2*2*2*3} =(2)^{2} * (6)^{\frac{2}{3} } =13.21 m[/tex]
Similarly, for the another satellite orbiting in time period of 12 hours, the major axis of this satellite is
[tex](12)^{2}= r^{3} \\(2*2*3)^{2} = r^{3}\\r = \sqrt[\frac{2}{3} ]{12} =5.24 m[/tex]
So, the difference between the two radius will give the distance between the two orbits, 13.21-5.24 = 7.97 m.
So the distance between the orbits of two satellites is 7.97 m.
The potential difference between the cloud and ground in a typical lightning discharge may be up to 100 MV (million volts). What is the gain in kinetic energy of an electron accelerated through this potential difference? Give your answer in both electron-volts and joules.
Answer:
Gain in kinetic energy is [tex]1.6 \times 10^{-11} J[/tex]
or [tex]10^{8} eV[/tex]
Explanation:
The potential difference between the cloud and ground is 100 MV (million volts)
Charge on the electron q = [tex]1.6 \times 10^{-19} C[/tex]
Kinetic energy of an electron accelerated through this potential difference is the work done to move the electron.
hence kinetic energy gained is
[tex]KE= \Delta V \times q\\KE= 100 \times 10^6 \times 1.6 \times 10^{-19}\\KE=1.6 \times 10^{-11} J[/tex]
In terms of electron volts, the conversion factor is 1 electron volt (eV) =
[tex]1.6 \times 10^{-19} J[/tex]
So,
[tex]\frac{1.6 \times 10^{-11} J}{1.6 \times 10^{-19} J} \\= 10^8 eV[/tex]
A charging bull elephant with a mass of 5240 kg comes directly toward youwith a speed of 4.55 m/s. You toss a 0.150- kg rubber ball at the elephant with a speed of 7.81 m/s. (a)When the ball bounces back toward you, what is its speed? (b) How do you account for the fact that theball’s kinetic energy has increased?
Answer:
Explanation:
mass of elephant, m1 = 5240 kg
mass of ball, m2 = 0.150 kg
initial velocity of elephant, u1 = - 4.55 m/s
initial velocity of ball, u2 = 7.81 m/s
Let the final velocity of ball is v2.
Use the formula of collision
[tex]v_{2}=\left ( \frac{2m_{1}}{m_{1}+m_{2}} \right )u_{1}+\left ( \frac{m_{2}-m_{1}}{m_{1}+m_{2}} \right )u_{2}[/tex]
[tex]v_{2}=\left ( \frac{2\times 5240}{5240+0.150} \right )(-4.55)+\left ( \frac{0.15-5240}{5240+0.150} \right )(7.81)[/tex]
v2 = - 16.9 m/s
The negative sign shows that the ball bounces back towards you.
(b) It is clear that the velocity of ball increases and hence the kinetic energy of the ball increases. This gain in energy is due to the energy from elephant.
A 70-kg circus performer is fired from a cannon that is elevated at an angle of 37° above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by 3.8 m from their unstrained length. He takes 4.1 s to travel between the launch point (where he is free from the bands) and the net into which he is shot. Assume the launch and landing points are at the same height and do not neglect the change in height during stretching.
The above is not complete. Couldn't find the it online but i have found a similar question which would help you solve the above
Answer:
Explanation:
From the question we are given that
Mass = 70kg
Angle of elevation = 40°
Length of stretch = 3.00 m
Height of performer above floor = Height of the net into the into which he is shot
Time traveled = 2.14 s
Distance traveled = 26.8 m
First obtain the horizontal velocity = [tex]\frac{Distance}{time} =\frac{26.8}{2.14} = 12.523 \ m/s[/tex]
To obtain the initial we would divide the final velocity by the cosine of the angle of elevation
Initial velocity = [tex]\frac{12.523}{cos\ 40} = 16.348m/s[/tex]
Next is to obtain the initial kinetic energy
This is equal to = [tex]\frac{1}{2} mv^2= \frac{1}{2} *70*16.348^2 = 9354 \ Joules[/tex]
Looking at the diagram in the second uploaded the cannon raised the performer before releasing him is [tex]30sin(40) = 19284 \ m[/tex]
So the potential energy given by the cannon is = mgh [tex]=70 *9.80*1.9284 = 1323 \ Joules[/tex]
Hence the total energy the band gives the performer i.e the total energy stored in the band is = 9354 +1323 = 10677 Joules
To obtain the Spring constant we would use the stored energy formula
i.e Stored Energy [tex]=\frac{1}{2}kx^2[/tex]
And we have calculated the stored energy as 10677
Substituting
[tex]10677 = \frac{1}{2} *k * 3.00^2[/tex]
[tex]k = 2373 \ N/m[/tex]
The circus performer's stunt includes aspects of elastic potential energy and projectile motion. The performer is launched due to the elastic potential energy in the cannon bands and then follows the laws of projectile motion. The performer's mass does not influence the time of flight.
Explanation:This problem demonstrates the principles of projectile motion and elastic potential energy applied to a real-world situation. The 70-kg circus performer is propelled out of the cannon because of the elastic potential energy stored in the stretched bands. Once launched, the performer's motion is an example of projectile motion.
In projectile motion, the horizontal and vertical motions are independent of each other. Therefore, it's possible to find the horizontal and vertical components of the performer's velocity separately and use them to construct the overall motion. The angle of 37° is used to calculate these components.
While the performer is in flight, the vertical motion is under constant acceleration due to gravity, while the horizontal motion has a constant velocity because air resistance is negligible. This allows us to calculate various aspects of the motion, such as the time it takes for the performer to land in the net.
Lastly, the performer's mass does not influence the time of flight in a vacuum. This illustrates an important principle of physics known as the independence of motion.
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suppose that throughout the united states, 350.0 x 10^6 suck braking processes occur in the course of a given day. calculate the average rate (megawatts) at which energy us being
Full Question:
Consider an automobile with a mass of 5510 lbm braking to a stop from a speed of 60.0 mph.
How much energy (Btu) is dissipated as heat by the friction of the braking process?
______Entry field with incorrect answer Btu
Suppose that throughout the United States, 350.0 x 10^6 such braking processes occur in the course of a given day. Calculate the average rate (megawatts) at which energy is being dissipated by the resulting friction.
______Entry field with incorrect answer MW
Answer:
Energy dissipated as heat = 852.10 Btu
Average rate of energy dissipation = 3.64 MW
Explanation:
1 lb = 0.453592 Kg
Mass = 5510 lb = 0.453592 * 5510 Kg = 2499.29 Kg
1 mph = 0.44704 m/s
Velocity = 60 mph = 60 * 0.44704 m/s = 26.82 m/s
Let E be equal to energy acquired
[tex]E = 1/2 MV^{2}[/tex]
[tex]E = 0.5 * 2499.29 * 26.88^{2} \\E = 899.01 * 10^{3} Joules[/tex]
Energy dissipated as heat:
[tex]E= 8.99*10^{3} = 0.0009478 * 8.99*10^{3}\\E = 852.10 tu[/tex]
E = 852.10 Btu
Energy dissipated throughout the United States
[tex]E = 350 * 10^{6} * 899.01 *10^{3} \\E=3.1466 * 10^1^4 Joules\\[/tex]
Average power rate in MW, P
[tex]P = (3.1466*10^{14} )/(24*60*60)\\P=3.64*10^{9} \\P=3.64MW[/tex]
ou slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the charges on the plates remain constant. What effect does adding the dielectric have on the energy stored in the capacitor?
Answer:
Explanation:
Energy stored in a capacitor is given by the expression
E = Q² / 2C , where Q is charge on it and C is its capacitance . As we put dielectric slab between the plates , capacitance increases , Due to it energy decreases as capacitance form the denominator in the formula above and Q is constant .
Hence energy decreases.
A satellite is in a circular orbit very close to the surface of a spherical planet. The period of the orbit is 2.17 hours. What is density of the planet
Explanation:
Lets consider
Circumference of orbit = T
as it is mentioned in the question that a satellite is in orbit that is very close to the surface of planet. so
circumference of orbit = circumference of planet
Time period = T
radius of planet = R
orbital velocity = V
gravitational constant = G
mass of planet = m
Solution:
Time period for a uniform circular motion of orbit is,
T = [tex]\frac{2\pi R}{V}[/tex]
[tex]T = \frac{2\pi R }{\sqrt{\frac{GM}{R} } }[/tex]
[tex]T= 2\pi \sqrt{(\frac{R^3}{GM} )}[/tex]
[tex]M = \frac{4}{3} \pi R^{3}p[/tex]
where p = density
[tex]T = 2\pi \sqrt{\frac{R^{3} }{G\frac{4}{3}\pi R^{3} p } }[/tex]
[tex]T = \sqrt{\frac{3\pi }{Gp} }[/tex]
T = 2.17 hours = 7812 sec
(7812)² = [( 3×3.14)/6.67×[tex]10^{-11}[/tex]×ρ)]
ρ = 6.28/6.67×[tex]10^{-11}[/tex]×6.10×[tex]10^{-7}[/tex]
ρ = 6.28/40.687×[tex]10^{-18}[/tex]
ρ = 0.1543×[tex]10^{18}[/tex]kg/m³
ρ = 15.43×[tex]10^{16}[/tex]kg/m³
Final answer:
Calculating a planet's density based on a satellite's orbital period involves physics concepts like gravitational laws and Kepler's third law, applying them to given data about the satellite's orbit. Direct calculation requires additional information that is not provided in the question.
Explanation:
One can use Kepler's third law in combination with the formula for gravitational force to find the planet's density based on the period of a satellite in a circular orbit close to the planet's surface.
The key relationship to use is T^2 = (4π^2/GM)r^3, where T is the period of the satellite, G is the gravitational constant, M is the mass of the planet, and r is the radius of the orbit, which is approximately the radius of the planet if the satellite is very close to the surface. However, to find the density (ρ) of the planet, we use the formula ρ = M/(4/3πr^3).
Given that we do not have the mass (M) or radius (r) of the planet explicitly, we cannot directly calculate the density without additional information such as the gravitational constant (G) and the exact radius of the planet's orbit.
The solution involves using the period (T) given to manipulate these equations to express mass in terms of the period and then apply it to the density formula. This complex physics problem requires an understanding of orbital mechanics and gravitational laws.
A generator has a square coil consisting of 269 turns. The coil rotates at 113 rad/s in a 0.222-T magnetic field. The peak output of the generator is 79.2 V. What is the length of one side of the coil?
Answer:
length of one side of the coil is 0.1083 m
Explanation:
given data
no of turn = 269
coil rotates = 113 rad/s
magnetic field = 0.222-T
peak output = 79.2 V
solution
we will apply here formula for maximum emf induced in a coil that is
εo = N × A × B × ω ........................1
and here Area A = l²
so
l = [tex]\sqrt{A}[/tex]
put here value of A from equation 1
I = [tex]\sqrt{\frac{\epsilon _o}{N\times B\times \omega} }[/tex]
put here value and we will get
l = [tex]\sqrt{\frac{79.2}{269\times 0.222\times 113}}[/tex]
l = 0.1083 m
Answer:
Explanation:
number of turns, n = 269
angular velocity, ω = 113 rad/s
magnetic field, B = 0.222 T
Voltage, V = 79.2 V
Let the area of the coil is A and the side of the square is s.
The maximum emf generated by the coil is
V = n x B x A x ω
79.2 = 269 x 0.222 x A x 113
A = 0.01174 m²
so, s x s = 0.01174
s = 0.108 m
Thus, the side of the square coil is 0.108 m.
A wire with current I = 15 A flows at an angle of 30 degrees to the direction of a uniform magnetic field of 0.9 T. What is the magnitude of the force exerted on a 10 m length of wire? (in units of N)
Answer:
67.5
Explanation:
[tex]F= BILsin\theta[/tex] where F is the magnitude of the force, B is the magnetic field, I is current, [tex]\theta[/tex] is the angle of inclination, L is the length of the wire.
Substituting 0.9 T for magnetic field B, 10 m for length, 15 A for current I and 30 degrees for [tex]\theta[/tex] then
[tex]F=0.9\times 15\times 10\times sin 30^{\circ}=67.5[/tex]
The electric potential difference between two locations can exist when there is only a source charge. Even when there is no TEST charge q, the potential difference can exist between two points in the space.a. trueb. false
Answer:
False
Explanation:
The potential difference between two points exists solely because of a source charge and depends on the source charge distribution. For a potential energy to exist, there must be a system of two or more charges. The potential energy belongs to the system and changes only if a charge is moved relatively to the rest of the system.
(a) All our household circuits are wired in parallel. Below is an example of a simple household circuit with 3 identical bulbs with resistance of 100 Ω each. Assume that this is the standard household 120 V circuit. What is the current running through each bulb? bulb 1 bulb 2 bulb 3
Explanation:
Below is an attachment containing the solution.
(a) In a parallel circuit, each bulb receives the full voltage of the household circuit. Therefore, the current running through each bulb is 1.2A.
Explanation:(a) In a parallel circuit, the voltage across each component remains the same. Therefore, each bulb in the circuit will receive the full 120V supplied by the household circuit.
Using Ohm's Law (V = IR), we can calculate the current running through each bulb. The resistance of each bulb is given as 100 Ω. Therefore, the current through each bulb is:
I = V/R = 120V / 100 Ω = 1.2A
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In preparing to shoot an arrow, an archer pulls a bow string back 0.424 m by exerting a force that increases uniformly from 0 to 223 N. What is the equivalent spring constant of the bow?
Answer:
525.94N/m
Explanation:
According to Hooke's law, the extension or compression of an elastic material is proportional to an applied force provided that the elastic limit of the material is not exceeded.
[tex]F=ke..................(1)[/tex]
where F is the applied force or load, k is the elastic constant or stiffness of the material and e is the extension.
In this problem, the bow string is assumed to behave like an elastic material that is stretched not beyond the elastic limit.
Given;
F = 223N;
e = 0.424m
k = ?
We make substitutions into equation (1) and then solve for k.
[tex]223=k*0.424\\k=\frac{223N}{0.424m}\\k = 525.94N/m[/tex]
The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it is elevated from rest at A by the action of the constant 58-N force in the cord. The cord is guided by the small fixed pulleys.
Answer:
6.21 m/s
Explanation:
Using work energy equation then
[tex]U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})[/tex]
where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity
Substituting 1 Kg for m, 0.4m for h, [tex]v_a[/tex] as 0, 9.81 for g then
[tex]58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s[/tex]
A certain elastic conducting material is stretched into a circular loop of 15.0 cm radius. It is placed with its plane perpendicular to a uniform 0.600 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 68.0 cm/s. What emf is induced in the loop at that instant?
The emf induced in the loop is 0.38 V.
Explanation:
The emf induced in any coil is directly proportional to the rate of change in flux linked with the coil as stated by Faraday's law.
Flux can be obtained by the product of magnetic lines of force with cross sectional area. As flux is the measure of magnetic lines of force crossing a certain cross sectional area.
As here the region is considered as circular loop, so the area will be area of the circle.
Then, [tex]emf=\frac{d(phi)}{dt}[/tex]
phi = Magnetic field * Area
[tex]emf = \frac{d(BA)}{dt} =B*2\pi*r\frac{dr }{dt}[/tex]
[tex]emf = 0.6*2*3.14*0.15*0.68=0.38 V[/tex]
So, the emf induced in the loop is 0.38 V.
The turd is launched at a speed of 72 m/s at an angle of 12 derees above the horizontal. at exactly what time after launch should the ground be covered in a portal to intercept the turd before it hits the ground?
The time the projectile (or 'turd') takes to hit the ground can be calculated using the formula (T = 2u*sin(θ)/g). This formula uses the initial velocity, the projection angle, and gravitational acceleration.
Explanation:This question relates to the physics concept of projectile motion. Given the launch speed, in this case, 72 m/s, and the angle of projection, 12 degrees, you can calculate the time of flight, i.e., the time taken by the projectile (in this humorous context, a 'turd') to hit the ground.
The time of flight (T) for a projectile can be calculated using the formula: T = 2u*sin(θ)/g, where u is the initial velocity, θ is the projection angle, and g is the gravitational acceleration (approximately 9.81 m/s² on Earth). Substituting the given values, we get T = 2*72*sin(12)/9.81, which gives the time after which the ground should be covered by a hypothetical 'portal' to intercept the turd just before it hits the ground.
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You hear 2 beats per second when two sound sources, both at rest, play simultaneously. The beats disappear if source 2 moves toward you while source 1 remains at rest. The frequency of source 1 is 500 Hz. The frequency of source 2 is...
A) 498 Hz.
B) 496 Hz.
C) 502 Hz.
D) 500 Hz.
E) 504 Hz.
Answer:
Option A
498 Hz
Explanation:
Beat frequency is given by F1-F2 where F is the frequency of source while F2 is frequency that disappear as one moves towards the source. when a source moves towards observer the frequency increases
Substituting 500 for F1 and 2 for F2 then
Beat frequency is 500-2=498 Hz
Option A
They are originally neutral. A positively charged rod is brought near (but not touching) the far end of A. While the charged rod is still close, A and B are separated. The charged rod is then withdrawn. Is the sphere then positively charged, negatively charged, or neutral?
Answer:
The sphere returns back to the neutral.
Explanation:
When a positively charged rod is brought near the far end of the rod A a redistribution of the charges occurs in the rod A. The free electrons in the rod A are attracted towards the positively charged rod. This causes positive charges to be shifted towards the other end of the rod A. The sphere B is in contact with the positively charged end of rod A. So it also acquires a positive charge.
When the positively charged rod is removed from the faf end of rod A, the electrons return to their original state and the charges redistributes themselves and a neutral state is re-established .