Complete Question:
A volley ball is hit directly toward the ceiling in a gymnasium with a ceiling height of 10 m. If the initial vertical velocity is 13 m/s and the release height is 1.8 m will the ball hit the ceiling?
Answer:
The ball will hit the ceiling
Explanation:
Given;
Initial vertical Velocity U = 13 m/s
Height of the ceiling = 10 m
Released height of the volley ball = 1.8 m
Height traveled by the volley ball, is calculated as follows;
[tex]V^2 =U^2 -2gH[/tex]
where;
V is final vertical velocity
[tex]2gH =U^2\\\\H = \frac{U^2}{2g} = \frac{(13)^2}{2(9.8)} = 8.62 m[/tex]
Remember this ball was released from 1.8 m height and it traveled 8.62 m.
Total distance traveled = 1.8 + 8.62 = 10.42 m
Therefore, the ball will hit the ceiling
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
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 975 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 75 volts
Answer: 23.92 ns
Explanation:
The capacitance, C = (e(0) * A) / d
C = [8.85*10^-12 * 2*10^-2 * 10*10^-2] / 1*10^-3
C = (1.77*10^-14) / 1*10^-3
C = 1.77*10^-11
C = 17.7 pF
Vc = V * [1 - e^-(t/CR)]
75 = 100 * [1 - e^-(t/CR)]
75/100 = 1 - e^-(t/CR)
e^-(t/CR) = 1 - 0.75
e^-(t/CR) = 0.25
If we take the log of both sides, we then have
Log e^-(t/CR) = Log 0.25
-t/CR = In 0.25
t = -CR In 0.25
t = - 17.7*10^-12 * 975 * -(1.386)
t = 2.392*10^-8 s
t = 23.92 ns
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!
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.
Electroplating uses electrolysis to coat one metal with another. In a copper-plating bath, copper ions with a charge of 2ee move through the electrolyte from the copper anode to the cathode; the metal object to be plated.
If the current through the system is 1.2 A, how many copper ions reach the cathode each second?
Answer:
1.2 A of current will send (3.744 × 10¹⁸) ions to the cathode per second.
Explanation:
According to Faraday's second law of electrolysis, the amount of ions/mass of substance deposited at an electrode depends on its equivalent weight.
For a divalent ion, it will require 2F of electricity per mole.
1 F = 96500 C
Amount of electricity that passes through the electrolyte per second = (magnitude of current) × (time) = It = (1.2 × 1) = 1.2 C
2F (2×96500C) of electricity will deposit 1 mole of Copper ions
That is,
193000 C of electricity will deposit 1 mole of Copper.
1.2 C will deposit (1.2×1/193000); 0.0000062176 mole of Copper.
1 mole of Copper contains (6.022 × 10²³) ions according to the Avogadro's constant.
0.0000062176 mole of Copper will contain (0.0000062176 × 6.022 × 10²³) ions = (3.744 × 10¹⁸) ions.
Therefore, 1.2 A of current will send (3.744 × 10¹⁸) ions to the cathode per second.
Hope this Helps!!!
When a certain ideal gas thermometer is placed in water at the triple point, the mercury level in the right arm is 846 mm above the reference mark. Part A How far is the mercury level above the reference mark when this thermometer is placed in boiling water at a location where the atmospheric pressure is 1.00 atm
Answer:
P =1156mmHg
Explanation:
Given
Ptriple = 846mmHg
Ttriple = 273.16K
T = temperature of boiling water = 100°C = (273.16 +100)K = 373.16K
P/Ptriple = T/Ttriple
P = T/Ttriple × Ptriple
P = 373.16/×273.6 × 846
P = 1156mmHg.
The temperature used in this solution is the absolute temperature because the gas thermometer use the absolute temperature scale. On this scale (Kelvin temperature scale) the zero point, Absolute zero is 273.16K.
Answer:
The mercury level is [tex]1155.68 mm[/tex] above the reference levelExplanation:
Ideal gas obeys the law , [tex]PV = RT[/tex]
When volume remains constant
[tex]\frac{P2}{P1} = \frac{T2}{T1}[/tex]
Temperature of triple point of water is [tex]T1 = 273.16K[/tex]
Temperature of normal boiling point of water is [tex]T2 = 373.15 K[/tex]
therefore,
[tex]P2 = 846 * \frac{373.15}{273.16}\\\\P2 = 1155.68 mm[/tex]
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A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball?
Answer:
[tex]F=800N[/tex]
Explanation:
The average force is defined as the mass of the body multiplied by its average velocity over the contact time. According to the third Newton's law, the magnitude of the average force exerted on the wall by the ball is equal to the magnitude of the average force exerted on the ball by the wall. Thus:
[tex]F=m\frac{v_f-v_i}{t}\\F=0.80kg\frac{25\frac{m}{s}-(-25\frac{m}{s})}{0.05s}\\F=800N[/tex]
Answer:
800 N
Explanation:
Force: This can be defined as the product of mass and acceleration. The S.I unit of force is Newton(N).
From the question,
F = m(v-u)/t ....................... Equation 1
Where F = force exerted on the wall by the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time.
Note: Assuming the direction of the initial motion to be negative
Given: m = 0.80 kg, v = 25 m/s ( bounce back), u = -25 m/s, t = 0.05 s
Substitute into equation 1
F = 0.8[25-(25)]/0.05
F = 0.8(50)/0.05
F = 0.8(1000)
F = 800 N
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.
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.
The height h (in feet) of an object shot into the air from a tall building is given by the function h(t) = 650 + 80t − 16t2, where t is the time elapsed in seconds. (a) Write a formula for the velocity of the object as a function of time t.
Answer:
80 - 32t
Explanation:
The height, h, in terms of time, t, is given as:
h(t) = 650 + 80t − 16t²
Velocity is the derivative of distance with respect to time:
v(t) = dh(t)/dt = 80 - 32t
The velocity of the object as a function of time is given by the derivative of the height function, which is v(t) = 80 - 32t.
Explanation:The height h(t) of an object is given by the equation h(t) = 650 + 80t − 16t2. To find the velocity v(t), we need to take the derivative of h(t) with respect to time t. Using the power rule, we get:
v(t) = dh/dt = 0 + 80 - 32t.
So, the velocity of the object as a function of time t is v(t) = 80 - 32t.
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Four objects are situated along the y axis as follows: a 1.93-kg object is at +2.93 m, a 3.06-kg object is at +2.58 m, a 2.41-kg object is at the origin, and a 3.96-kg object is at -0.498 m. Where is the center of mass of these objects?
Answer: x, y (0, 1)
Explanation:
the X coordinate of the center mass is
X(c) = Σm(i)*x(i) / Σx(i)
X(c) = (0 + 0 + 0 + 0) / (1.93 + 3.06 + 2.41 + 3.96)
X(c) = 0 / 11.36
X(c) = 0
The y coordinate of the center mass is
Y(c) = Σm(i)y(i) / Σm(i)
Y(c) = [(1.93)(2.93) + (3.06)(2.58) + (2.41)(0) + (3.96)(-0.498)] / (1.93 + 3.06 + 2.41 + 3.96)
Y(c) = (5.6549 + 7.8948 + 0 - 1.97208) / 11.36
Y(c) = 11.57762 / 11.36
Y(c) = 1.02
Therefore, the center of masses is at x, y (0, 1)
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]
The system is in equilibrium and the pulleys are frictionless and massless. 5 kg 8 kg 9 kg T Find the force T. The acceleration of gravity is 9.8 m/s 2 . Answer in units of N.
Explanation:
It is given that mass at the right side of the pulley weighs 9 kg, mass at the center weighs 8 kg and mass at the left side weighs 5 kg.
Hence, starting from the right side
[tex]T_{1}[/tex] = 9 kg
Now, [tex]T_{2} = 2 \times 9 kg[/tex]
= 18 g
Hence, the force balance on 8 kg mass is as follows.
[tex]T_{2} + mg = T_{3}[/tex]
[tex]T_{3}[/tex] = [tex](18 + 8) \times g[/tex]
= 26 g
Now, force balance on 5 kg mass is as follows.
5 g + T = [tex]2 \times 26 g[/tex]
T = 47 g
= [tex]47 \times 9.8[/tex]
= 460.6 N
Therefore, we can conclude that value of force T is 460.6 N.
The value of force T is 460.6N
TensionTension is a force developed by a rope, string, or cable when stretched under an applied force.
Given:
Mass (m₁) = 9 kg, m₂ = 8 kg, m₃ = 5kg, g = acceleration due to gravity = 9.8 m/s².
T₁ = 9 kg * 9.8 m/s² = 88.2N
T₂ = 2 * m₁ * g = 2 * 9 kg * 9.8 = 176.4N
T₂ + m₂g = T₃
176.4 + 8*9.8 = T₃
T₃ = 254.8N
Force balance on 5 kg mass:
T + 5(9.8) = 2 * T₃
T + 5(9.8) = 2 * 254.8
T = 460.6N
The value of force T is 460.6N
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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
If blocks A and B of mass 10 kg and 6 kg respectively, are placed on the inclined plane and released, determine the force developed in the link. The coefficients of kinetic friction between the blocks and the inclined plane are mA = 0.1 and mB = 0.3. Neglect the mass of the link.
Answer:
The force developed in the link is 6.36 N.
Explanation:
Given that,
Mass of block A = 10 kg
Mass of block B = 6 kg
Coefficients of kinetic friction [tex]\mu_{A}= 0.1[/tex]
Coefficients of kinetic friction [tex]\mu_{B}= 0.3[/tex]
Suppose the angle is 30°
We need to calculate the acceleration
Using formula of acceleration
[tex]a=\dfrac{m_{A}g\sin\theta+m_{B}g\sin\theta-\mu_{A}m_{A}g\cos\theta-\mu_{A}m_{A}g\cos\theta}{m_{A}+m_{B}}[/tex]
Put the value into the formula
[tex]a=\dfrac{10\times9.8\sin30+6\times9.8\sin30-0.1\times10\times9.8\times\cos30-0.3\times6\times9.8\times\cos30}{16}[/tex]
[tex]a=3.415\ m/s^2[/tex]
We need to calculate the force developed in the link
For block A,
Using balance equation
[tex]ma=m_{A}g\sin\theta-\mu m_{A}g\cos\theta-T[/tex]
[tex]T=ma+\mu m_{A}g\cos\theta-m_{A}g\sin\theta[/tex]
Put the value into the formula
[tex]T=10\times3.415+0.1\times10\times9.8\times\cos30-10\times9.8\times\sin30[/tex]
[tex]T=-6.36\ N[/tex]
Negative sign shows the opposite direction of the force.
Hence, The force developed in the link is 6.36 N.
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]
A physics professor is pushed up a ramp inclined upward at an angle 33.0° above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 90.0 kg. He is pushed a distance 2.00 m along the incline by a group of students who together exert a constant horizontal force of 600 N. The professor's speed at the bottom of the ramp is 2.30 m/s.
A) Find velocity at the top of the ramp.
Answer:
2.51 m/s
Explanation:
Parameters given:
Angle, A = 33°
Mass, m = 90kg
Inclined distance, D = 2m
Force, F = 600N
Initial speed, u = 2.3m/s
From the relationship between work and kinetic energy, we know that:
Work done = change in kinetic energy
W = 0.5m(v² - u²)
We also know that work done is tẹ product of force and distance, hence, net work done will be the sum of the total work done by the force from the students and gravity.
Hence,
W = F*D*cosA - w*D*sinA
w = m*9.8 = weight
=> W = 600*2*cos33 - 90*9.8*2*sin33
W = 45.7J
=> 45.7 = 0.5*m*(v² - u²)
45.7 = 0.5*90*(v² - 2.3²)
45.7 = 45(v² - 5.29)
=> v² - 5.29 = 1.016
v² = 6.306
v = 2.51 m/s
The final velocity is 2.51 m/s
Answer: v = 3.26 m/s
Explanation:
Force applied (F) = 900 N,
mass of professor and chair (m) = 90kg,
g = acceleration due to gravity = 9.8 m/s²,
Fr = frictional force = 0 ( since from the question, the rollers the chair and professor are moving is frictionless),
θ = angle of inclination = 33°
a = acceleration of object =?
From newton's second law of motion, we have that
F - (mg sinθ +Fr) = ma
Where mg sinθ is the horizontal component of the weight of the mass of professor and chair due to the inclination of the ramp.
But Fr = 0
Hence, we have that
F - mg sinθ = ma
600 - (90×9.8×sin30) = 90 (a)
600 - 480.371= 90a
119.629 = 90a
a = 119.629/ 90
a = 1.33 m/s².
But the body started the morning (at the bottom of the ramp) with a velocity of 2.30 m/s²
Hence u = initial velocity = 2.30 m/s², a = acceleration = 1.33m/s², v = final velocity =?, s = distance covered = 2m
By using equation of motion for a constant acceleration, we have that
v² = u² + 2as
v ² = (2.3)² +2(1.33)×(2)
v² = 5.29 + 5.32
v² = 10.61
v = √10.61
v = 3.26 m/s
A dielectric material is inserted between the charged plates of a parallel-plate capacitor. Do the following quantities increase, decrease, or remain the same as equilibrium is reestablished?
1. Charge on plates (plates remain connected to battery)
2. Electric potential energy (plates isolated from battery before inserting dielectric)
3.Capacitance (plates isolated from battery before inserting dielectric)
4. Voltage between plates (plates remain connected to battery)
5. Charge on plates (plates isolated from battery before inserting dielectric)
6. Capacitance (plates remain connected to battery)
7. Electric potential energy (plates remain connected to battery)
8. Voltage between plates (plates isolated from battery before inserting dielectric)
Answer:
1. Charge on plates (plates remain connected to battery) increases.
2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases.
3.Capacitance (plates isolated from battery before inserting dielectric) increases.
4. Voltage between plates (plates remain connected to battery) remains the same.
5. Charge on plates (plates isolated from battery before inserting dielectric) remains the same.
6. Capacitance (plates remain connected to battery) increases.
7. Electric potential energy (plates remain connected to battery) increases.
8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases.
Explanation:
When a dielectric material is inserted between the plates of a capacitor, the capacitance is increase by a factor K, the dielectric constant.
[tex]C = KC_0[/tex]
By the capacitance formula, the other factors change accordingly.
1. Charge on plates (plates remain connected to battery) increases, because charge and capacitance are directly proportional.
2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases, because potential is inversely proportional to capacitance, and potential energy is given by the following formula
[tex]U = \frac{1}{2}CV^2[/tex]
3.Capacitance (plates isolated from battery before inserting dielectric) increases.
4. Voltage between plates (plates remain connected to battery) stays the same, because the voltage is applied by the battery.
5. Charge on plates (plates isolated from battery before inserting dielectric) remains constant. If the plates isolated from the battery, then the total charge is conserved.
6. Capacitance (plates remain connected to battery) increases when a dielectric is inserted.
7. Electric potential energy (plates remain connected to battery) increases, because when plates remain connected to battery the voltage remains the same. But the capacitance increases. Therefore, electric potential energy increases.
8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases, because voltage is inversely proportional to capacitance.
Final answer:
When a dielectric is inserted into a capacitor, the capacitance increases regardless of whether the plates are connected to a battery or isolated. If connected to a battery, charge on the plates increases while voltage remains the same; if isolated, charge remains the same while voltage decreases.
Explanation:
When inserting a dielectric material into a parallel-plate capacitor, the effect on different parameters depends on whether the capacitor is connected to a battery or isolated:
Charge on plates (plates remain connected to battery): The charge on the plates will increase.
Electric potential energy (plates isolated from the battery before inserting the dielectric): The electric potential energy will decrease due to the polarization of the dielectric.
Capacitance (plates isolated from the battery before inserting the dielectric): The capacitance will increase.
Voltage between plates (plates remain connected to battery): The voltage will remain the same, as it is maintained by the battery.
Charge on plates (plates isolated from battery before inserting dielectric): The charge will remain the same as isolation prevents any charge exchange.
Capacitance (plates remain connected to battery): The capacitance will increase.
Electric potential energy (plates remain connected to the battery): The electric potential energy will increase, as more charge is accumulated due to the dielectric.
Voltage between plates (plates isolated from battery before inserting dielectric): The voltage will decrease because the same charge is now across a larger capacitance.
It is important to note that the introduction of a dielectric material affects the capacitance of a parallel-plate capacitor by reducing the effective electric field and permitting greater charge storage capability for a given voltage.
My Notes (a) A 41 Ω resistor is connected in series with a 6 µF capacitor and a battery. What is the maximum charge to which this capacitor can be charged when the battery voltage is 6 V? (When entering units, use micro for the metric system prefix µ.)
Explanation:
The give data is as follows.
C = 6 [tex]\mu F[/tex] = [tex]6 \times 10^{-6} F[/tex]
V = 6 V
Now, we know that the relation between charge, voltage and capacitor for series combination is as follows.
Q = CV
= [tex]6 \times 10^{-6} F \times 6 V[/tex]
= [tex]36 \times 10^{-6} C[/tex]
or, = 36 [tex]\mu C[/tex]
Thus, we can conclude that maximum charge of the given capacitor is 36 [tex]\mu C[/tex].
Answer:
Explanation:
capacitance, C = 6 μF
Voltage, V = 6 V
Let the maximum charge is Q.
Q = C x V
Q = 6 x 6 = 36 μC
A loop of wire in the shape of a rectangle rotates with a frequency of 284 rotation per minute in an applied magnetic field of magnitude 6 T. Assume the magnetic field is uniform. The area of the loop is A = 4 cm2 and the total resistance in the circuit is 9 Ω.(a) Find the maximum induced emf.(b) Find the maximum current through the bulb.
Answer:
a) Magnitude of maximum emf induced = 0.0714 V = 71.4 mv
b) Maximum current through the bulb = 0.00793 A = 7.93 mA
Explanation:
a) The induced emf from Faraday's law of electromagnetic induction is related to angular velocity through
E = NABw sin wt
The maximum emf occurs when (sin wt) = 1
Maximum Emf = NABw
N = 1
A = 4 cm² = 0.0004 m²
B = 6 T
w = (284/60) × 2π = 29.75 rad/s
E(max) = 1×0.0004×6×29.75 = 0.0714 V = 71.4 mV
Note that: since we're after only the magnitude of the induced emf, the minus sign that indicates that the induced emf is 8n the direction opposite to the change in magnetic flux, is ignored for this question.
b) Maximum current through the bulb
E(max) = I(max) × R
R = 9 ohms
E(max) = 0.0714 V
I(max) = ?
0.0714 = I(max) × 9
I(max) = (0.0714/9) = 0.00793 A = 7.93 mA
Hope this Helps!!
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.
(c)How much work does the gravitational force do on the ball while it is compressing the mattress?
(d)How much work does the mattress do on the ball?
(e)Now model the mattress as a single spring with an unknown spring constant k, and consider the whole system formed by the ball, the earth and the mattress. By how much does the potential energy of the mattress increase as it compresses?
(f)What is the value of the spring constant k?
Answer:
(c) 10.29 J
(d) 113.19 J
(e) 113.19 J
(f) 10061 N/m
Explanation:
15 cm = 0.15 m
Let g = 9.8 m/s2
(c) The work done by gravitational force is the product of gravity force and the distance compressed
[tex]E_p = mgx = 7*9.8*0.15 = 10.29 J[/tex]
(d) By using law of energy conservation with potential energy reference being 0 at the maximum compression point. As the ball falls and come to a stop at the compression point, its potential energy is transferred to elastic energy, which is the work that the mattress does on the ball:
[tex]E_p = E_e[/tex]
[tex]E_e = mgh[/tex]
where h = 1.5 + 0.15 = 1.65 m is the vertical distance that it falls.
[tex]E_e = 7*9.8*1.65 = 113.19 J[/tex]
(e) Before the compression, the potential energy of the mattress is 0. After the compression, the potential energy is 113.19J. So it has increased by 113.19J due to the potential energy transferred from the falling ball.
(f) [tex]E_e = 113.19 = kx^2/2[/tex]
[tex]k0.15^2/2 = 113.19[/tex]
[tex]k = 10061 N/m[/tex]
Answer:
(C) Wg = 113.2J
(D) Wm = 10.3J
(E) E = 1/2kx² + mgh where h is the height above the mattress and x is the compressed distance in the mattress.
(F) k = 457N/m.
Explanation:
See attachment below.
To understand the terms in Faraday's law and to be able to identify the magnitude and direction of induced emf. Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux. Mathematically, it can be written as
Answer: [tex]V_{\epsilon}\propto \frac{d\phi_{B}}{dt}[/tex]
Explanation:
A direct proportionality means a linear relationship between two variables and rate of change means an application of derivatives. Hence, the mathematical model is:
[tex]V_{\epsilon}\propto \frac{d\phi_{B}}{dt}[/tex]
Faraday's Law of Induction, primarily focused on Physics, reveals that the induced electromotive force in a coil is proportional to the rate of change of magnetic flux through the coil and is also dependent on the number of coil turns.
Explanation:Understanding Faraday's Law of Induction
Faraday's Law of Induction is a critical concept in electromagnetism, underlying the working principles of various electrical devices. The induced electromotive force (emf) is directly proportional to the time rate of change of magnetic flux. In simpler terms, when the magnetic environment of a circuit changes, an emf is induced. Faraday's Law is given by the equation EMF = -N(ΔΦ/Δt), where 'N' represents the number of turns in a coil, and (ΔΦ/Δt) is the rate of change of magnetic flux over time. This equation highlights three main factors influencing the induced EMF:
The induced EMF is proportional to the number of loops in the coil.It is proportional to the change in magnetic flux, ΔΦ.It is inversely proportional to the time interval, Δt, over which the change occurs.The negative sign in Faraday's Law is a reflection of Lenz's Law, which indicates that the induced emf always works to oppose the change in magnetic flux that produces it.
The current supplied by a battery in a portable device is typically about 0.151 A. Find the number of electrons passing through the device in five hours.
Answer:
n = 1.7*10²² electrons.
Explanation:
As the current, by definition, is the rate of change of charge, assuming that the current was flowing at a steady rate of .151 A during the 5 hours, we can find the total charge that passed perpendicular to the cross-section of the circuit, as follows:[tex]I =\frac{\Delta q}{\Delta t} \\ \\ \Delta q = I* \Delta t \\ \\ \Delta t = 5hs*\frac{3600s}{1h} = 18000 s[/tex]
⇒ Δq = I * Δt = 0.151 A * 18000 s = 2718 C
As this charge is carried by electrons, we can express this value as the product of the elementary charge e (charge of a single electron) times the number of electrons flowing during that time, as follows:Δq = n*e
Solving for e:[tex]n = \frac{\Delta q}{e} =\frac{2718C}{1.6-19C} = 1.7e22 electrons.[/tex]
Explanation:
Below is an attachment containing the solution.
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
A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in our atmosphere is ozone, O3. In particular, ozone absorbs radiation with frequencies around 9.38×1014 Hz . What is the wavelength λ lambda of the radiation absorbed by ozone?
Answer:
[tex]3.2\times 10^{-7}\ m[/tex] or 0.32 μm.
Explanation:
Given:
The radiations are UV radiation.
The frequency of the radiations absorbed (f) = [tex]9.38\times 10^{14}\ Hz[/tex]
The wavelength of the radiations absorbed (λ) = ?
We know that, the speed of ultraviolet radiations is same as speed of light.
So, speed of UV radiation (v) = [tex]3\times 10^8\ m/s[/tex]
Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:
[tex]v=f\lambda[/tex]
Now, expressing the above equation in terms of wavelength 'λ', we have:
[tex]\lambda=\frac{v}{f}[/tex]
Now, plug in the given values and solve for 'λ'. This gives,
[tex]\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m[/tex]
Therefore, the wavelength of the radiations absorbed by the ozone is nearly [tex]3.2\times 10^{-7}\ m[/tex] or 0.32 μm.
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]
(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
Learn more about parallel circuit here:
https://brainly.com/question/32539613
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A wire is stretched just to its breaking point by a force F . A longer wire made of the same material has the same diameter. The force that will stretch the longer wire to its breaking point is much smaller than F?
Answer: No
Explanation:
The force F required is equal to the Force exerted in stretching the first material since conditions are the same
Answer:
No.
Explanation:
Both wires are made up of the same material which means they will have the same young modulus.
Y = F/A × L/ΔL
From the equation above, the force applied per cross sectional (same diameter) are is the same for both materials and is constant.
So
L1/ΔL1 = L2/ΔL2
Or L2/L1 = ΔL2/ΔL1
So the force needed to bring about a stretching the the breaking point is the same. The only difference is in by how much the longer wire would have to stretch before reaching its breaking point.
Human eyes detect only a very small band of the electromagnetic spectrum. However, some animals and insects see in higher and lower frequency bands than humans do. For this discussion, first do some research to find three examples of animals or insects that see in higher or lower frequency bands.Then, in your initial discussion post, address the following:What were the examples that you found? Did they see in lower or higher frequency bands? List the specific wavelengths used by the animals/insects you chose.Were you surprised by the information that you found? Why or why not?
Answer:
Squids = 450 - 490 nm (Moderate Frequency) (Blue)
Bees = 300 - 650 nm (Lower Frequency Bands)
Frogs = 280 - 580 nm (Very Low Frequency)
Explanation:
All of the above mentioned ranges are compared to that of humans.
I'm just surprised a little bit in the imagination that how these organisms see the world through their unique eyes. On the other hands, they are evolved like this just like we do so that may not be surprising enough. SIKE
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