Answer:
Electric charge, Q = 55.69 C
Explanation:
It is given that,
Electric current drawn by the compressor, I = 23.7 A
Time taken, t = 2.35 s
We need to find the electric charge passes through the circuit during this period. The definition of electric current is given by total charge divided by total time taken.
[tex]I=\dfrac{q}{t}[/tex]
Where,
q is the electric charge
[tex]q=I\times t[/tex]
[tex]q=23.7\ A\times 2.35\ s[/tex]
q = 55.69 C
So, the electric charge passes through the circuit during this period is 55.69 C. Hence, this is the required solution.
A red train travelling at 72 km/h and a green train travelling at 144 km/h are headed toward each
other along a straight, level track. When they are 950 m apart, each engineer sees the other train
and applies the brakes, which slow each train at the rate of 1.0 m/s2. Is there a collision? If yes,
give the speed of the red train and the speed of the green train at impact, respectively. If no, give the
separation between the trains when they stop.
Answer:
Collision will occur.
Speed of red train when they collide = 0 m/s.
Speed of green train when they collide = 10 m/s.
Explanation:
Speed of red train = 72 km/h = 20 m/s
Speed of green train = 144 km/h = 40 m/s.
Deceleration of trains = 1 m/s²
For red train:-
Equation of motion v = u + at
u = 20 m/s
v = 0 m/s
a = -1 m/s²
Substituting
0 = 20 - 1 x t
t = 20 s.
Equation of motion s = ut + 0.5at²
u = 20 m/s
t = 20 s
a = -1 m/s²
Substituting
s = 20 x 20 - 0.5 x 1 x 20² = 200 m
So red train travel 200 m before coming to stop.
For green train:-
Equation of motion v = u + at
u = 40 m/s
v = 0 m/s
a = -1 m/s²
Substituting
0 = 40 - 1 x t
t = 40 s.
Equation of motion s = ut + 0.5at²
u = 40 m/s
t = 40 s
a = -1 m/s²
Substituting
s = 40 x 40 - 0.5 x 1 x 40² = 800 m
So green train travel 800 m before coming to stop.
Total distance traveled = 800 + 200 = 1000 m>950 m.
So both trains collide.
Distance traveled by green train when red train stops(t=20s)
Equation of motion s = ut + 0.5at²
u = 40 m/s
t = 20 s
a = -1 m/s²
Substituting
s = 40 x 20 - 0.5 x 1 x 20² = 600 m
Total distance after 20 s = 600 + 200 = 800 m< 950m . So they collide after red train stops.
Speed of red train when they collide = 0 m/s.
Distance traveled by green train when they collide = 950 - 200 = 750 m
Equation of motion v² = u² + 2as
u = 40 m/s
s= 750 m
a = -1 m/s²
Substituting
v² = 40² - 2 x 1 x 750 = 100
v = 10 m/s
Speed of green train when they collide = 10 m/s.
Final answer:
The red train traveling at 72 km/h and green train at 144 km/h will collide because their combined stopping distances exceed their initial separation. The speeds at impact are not provided, but the collision is inevitable due to their insufficient stopping distance.
Explanation:
To determine if a collision occurs between the red train traveling at 72 km/h and the green train traveling at 144 km/h, we need to convert their speeds into meters per second and calculate the stopping distance for both trains based on their deceleration.
The red train is traveling at 72 km/h, which is equivalent to 20 m/s (since 72 km/h / 3.6 = 20 m/s). The green train is traveling at 144 km/h, which is equivalent to 40 m/s (since 144 km/h / 3.6 = 40 m/s).
To calculate the stopping distance, use the equation d = v2 / (2a), where d is the stopping distance, v is the initial velocity, and a is the deceleration. So, for the red train, the stopping distance is (20 m/s)2 / (2 × 1.0 m/s2) = 200 m. For the green train, the stopping distance is (40 m/s)2 / (2 × 1.0 m/s2) = 800 m.
Adding both stopping distances, we get a total of 200 m + 800 m = 1000 m. Since the trains are only 950 m apart, their combined stopping distance exceeds the separation, meaning they will collide.
Just before the collision, both trains have been decelerating for the same amount of time. Given that the total stopping time can be found from v = at where v is the final velocity and t is time, we find that their time to stop (if unobstructed) would be t = v / a. Since the red train decelerates from 20 m/s, its stopping time is 20 m/s / 1 m/s2 = 20 s. For the green train, 40 m/s / 1 m/s2 = 40 s. Since they have not yet reached 20 s before collision, we know they will still be moving upon impact.
The collision occurs before either train can come to a complete stop, and thus we would use the physics of constant deceleration to determine their speeds at the moment of impact, but as the full calculation is not provided in this answer, it would require additional work to determine exact speeds.
At the instant a current of 0.40 A is flowing through a coil of wire, the energy stored in its magnetic field is 9.5 ✕ 10−3 J. What is the self-inductance of the coil (in H)?
Final answer:
The self-inductance of the coil is calculated using the energy stored in its magnetic field and the current flowing through it. By rearranging the formula W = (1/2) * L * I², the self-inductance L is found to be 0.11875 H.
Explanation:
To calculate the self-inductance of the coil, we can use the formula for the energy stored in the magnetic field of an inductor:
Energy (W) = (1/2) * L * I²
Where:
W is the energy stored in the magnetic field
L is the self-inductance of the coil
I is the current flowing through the coil
From the question, we're given that the energy (W) is 9.5 x 10⁻³ J, and the current (I) is 0.40 A. We can rearrange the formula to solve for L:
L = 2W / I²
Substitute the given values:
L = 2 * (9.5 x 10⁻³ J) / (0.40 A)²
L = 2 * (9.5 x 10⁻³ J) / (0.16 A²)
L = (19 x 10⁻³ J) / (0.16 A²)
L = 0.11875 H
Therefore, the self-inductance of the coil is 0.11875 H (henrys).
Calculate the heat flux (in W/m2) through a sheet of a metal 11-mm thick if the temperatures at the two faces are 350 and 110 ˚C. Assume steady-state heat flow and that the thermal conductivity of this metal is 53.0 W/m-K.
q = 1156363.6W/m².
To calculate the heat flux per unit area (W/m²) of a sheet made of metal:
q = -k(ΔT/Δx)
q = -k[(T₂ - T₁)/Δx]
Where, k is the thermal conductivity of the metal, ΔT is the temperature difference and Δx is the thick.
With Δx = 11 mm = 11x10⁻³m, T₂ = 350°C and T₁ = 110°C, and k = 53.0 W/m-K:
q = -53.0W/m-K[(110°C - 350°C)/11x10⁻³m
q = 1156363.6W/m²
An incandescent lightbulb emits 96 W of radiation. If the filanent is at a temperature of 3242 K, what is the area of the filament
Answer:
1.53 x 10^-5 m^2
Explanation:
use the Stefan's law
Energy per unit time = σ x A x T^4
σ = 5.67 x 10 -8 W/m^2 K^4
96 = 5.67 x 10^-8 x A x (3242)^4
A = 1.53 x 10^-5 m^2
Two cars approach an extremely icy four-way perpendicular intersection. Car A travels northward at 10 m/s and car B is traveling eastward. They collide and stick together, traveling at 41.5° north of east. What was the initial velocity of car B (in m/s)? (Enter the magnitude. Assume the masses of the cars are equal.)
Answer:
v = 11.3 m/s
Explanation:
Since there is no external force on the system of two cars
so here momentum of the system of two cars will remain constant
it is given here that two cars combined and move together at 41.5 degree North of East after collision
so here direction of final momentum is given as
[tex]tan\theta = \frac{P_y}{P_x}[/tex]
now we have initial momentum in the same direction
[tex]P_1 = m(10 m/s)\hat j[/tex]
[tex]P_2 = mv\hat i[/tex]
now we have
[tex]\frac{P_1}{P_2} = tan\theta[/tex]
[tex]\frac{m(10 m/s)}{mv} = tan41.5[/tex]
[tex]\frac{10}{v} = 0.88[/tex]
[tex]v = 11.3 m/s[/tex]
Using the principles of conservation of momentum and trigonometry, the initial velocity of car B can be calculated to be approximately 9.75 m/s.
Explanation:The problem can be solved using the principles of conservation of momentum. In a two-dimensional collision, the components of the momentum in the X and Y directions are separately conserved. Since we're given that the cars have equal mass and they stick together after colliding to make a single object, the magnitude of the velocity of the combined object after the collision must be equal to the vector sum of their initial velocities.
Car A is moving northward and thus its velocity is contributing only to the Y component of the total momentum. Car B, on the other hand, is moving eastward, so its velocity is contributing to the X component. When the cars collide and stick together, they move at an angle of 41.5° north of east, meaning there is both an X and Y component to their velocity.
Using trigonometry, the X component (eastward direction, i.e., car B's direction) can be found by VB = VA * tan(Θ). Given that VA is 10 m/s and Θ is 41.5 degrees, VB can be calculated to be approximately 9.75 m/s.
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How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to produce an electric field of 1500 N/C just outside the surface of the sphere?
Answer:
Required charge [tex]q=2.6\times 10^{9}C[/tex].
[tex]n=1.622\times 10^{10}\ electrons[/tex]
Explanation:
Given:
Diameter of the isolated plastic sphere = 25.0 cm
Magnitude of the Electric field = 1500 N/C
now
Electric field (E) is given as:
[tex]E =\frac{kq}{r^2}[/tex]
where,
k = coulomb's constant = 9 × 10⁹ N
q = required charge
r = distance of the point from the charge where electric field is being measured
The value of r at the just outside of the sphere = [tex]\frac{25.0}{2}=12.5cm=0.125m[/tex]
thus, according to the given data
[tex]1500N/C=\frac{9\times 10^{9}N\times q}{(0.125m)^2}[/tex]
or
[tex]q=\frac{0.125^2\times 1500}{9\times 10^{9}}[/tex]
or
Required charge [tex]q=2.6\times 10^{9}C[/tex].
Now,
the number of electrons (n) required will be
[tex]n=\frac{required\ charge}{charge\ of\ electron}[/tex]
or
[tex]n=\frac{2.6\times 10^{-9}}{1.602\times 10^{-19}}[/tex]
or
[tex]n=1.622\times 10^{10}\ electrons[/tex]
To produce an electric field of 1500 N/C just outside the surface of the plastic sphere, approximately 337.5 nC of excess charge must be distributed uniformly within its volume.
Explanation:To produce an electric field just outside the surface of the sphere, the excess charge on the surface of the sphere should be uniformly distributed within its volume. The electric field just outside a uniformly charged sphere is given by:
E = k * q / r2
Where E is the electric field, k is the Coulomb constant [tex](9 * 109 N m2/C^2)[/tex] q is the charge, and r is the distance from the center of the sphere to the point outside the surface. In this case, E is given as 1500 N/C and r is half the diameter of the sphere (12.5 cm).
Substituting the values into the equation, we can solve for q:
[tex]1500 N/C = (9 * 109 N m2/C2) * q / (0.125 m)^2[/tex]
Solving for q, we find that the excess charge on the surface of the sphere should be approximately [tex]3.375 x 10-7 C,[/tex] or about 337.5 nC.
An airplane is moving horizontally with a constant velocity of (+ 115 m/s) at altitude of 1050 m. The directions to the right and upward have been chosen as the positive directions. The plane releases a "care package' that falls to the ground along a curved trajectory. Ignoring air resistance, determine the time required for the package to hit the ground
Answer:
The time required for the package to hit the ground = 14.63 seconds.
Explanation:
Considering vertical motion of care package:-
Initial velocity, u = 0 m/s
Acceleration , a = 9.81 m/s²
Displacement, s = 1050 m
We have equation of motion s= ut + 0.5 at²
Substituting
s= ut + 0.5 at²
1050 = 0 x t + 0.5 x 9.81 x t²
t = 14.63 seconds
The time required for the package to hit the ground = 14.63 seconds.
To determine the time required for the care package to hit the ground, we can separate the vertical and horizontal components of the motion. By using the equation h = (1/2)gt^2 and plugging in the values h = 1050 m and g = 9.8 m/s^2, we find that it will take approximately 14.63 seconds for the care package to hit the ground.
Explanation:To determine the time required for the care package to hit the ground, we can separate the vertical and horizontal components of the motion. Since the horizontal velocity of the airplane does not change, it will not affect the time of flight. The vertical motion can be analyzed using the equation of motion for free fall, which is h = (1/2)gt^2, where h is the initial vertical displacement, g is the acceleration due to gravity, and t is the time of flight. In this case, the initial vertical displacement is 1050 meters and the acceleration due to gravity is approximately 9.8 m/s^2. Plugging these values into the equation, we can solve for t and find the time required for the care package to hit the ground.
By using the equation h = (1/2)gt^2 and plugging in the values h = 1050 m and g = 9.8 m/s^2, we have 1050 = (1/2)(9.8)t^2. Solving for t, we get t^2 = (1050) / (0.5)(9.8) = 214.29. Taking the square root of both sides, we find t ≈ 14.63 seconds. Therefore, it will take approximately 14.63 seconds for the care package to hit the ground.
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72) What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115 g of water? The molal freezing point depression constant for water is 1.86 °C/m. g
Answer: The freezing point of solution is -3.34°C
Explanation:
Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:
[tex]Ca(NO_3)_2(aq.)\rightarrow Ca^{2+}(aq.)+2NO_3^-(aq.)[/tex]
The total number of ions present in the solution are 3.
To calculate the molality of solution, we use the equation:[tex]Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}[/tex]
Where,
[tex]m_{solute}[/tex] = Given mass of solute [tex](Ca(NO_3)_2)[/tex] = 11.3 g
[tex]M_{solute}[/tex] = Molar mass of solute [tex](Ca(NO_3)_2)[/tex] = 164 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (water) = 115 g
Putting values in above equation, we get:
[tex]\text{Molality of }Ca(NO_3)_2=\frac{11.3\times 1000}{164\times 115}\\\\\text{Molality of }Ca(NO_3)_2=0.599m[/tex]
To calculate the depression in freezing point, we use the equation:[tex]\Delta T=iK_fm[/tex]
where,
i = Vant hoff factor = 3
[tex]K_f[/tex] = molal freezing point depression constant = 1.86°C/m.g
m = molality of solution = 0.599 m
Putting values in above equation, we get:
[tex]\Delta T=3\times 1.86^oC/m.g\times 0.599m\\\\\Delta T=3.34^oC[/tex]
Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.
[tex]\Delta T=\text{freezing point of water}-\text{freezing point of solution}[/tex]
[tex]\Delta T[/tex] = 3.34 °C
Freezing point of water = 0°C
Freezing point of solution = ?
Putting values in above equation, we get:
[tex]3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC[/tex]
Hence, the freezing point of solution is -3.34°C
Final answer:
The freezing point of a solution made by dissolving 11.3 g of Ca(NO3)2 in 115 g of water is -3.34 °C. To find this, we calculate molality, account for the dissociation of ions and use the freezing point depression constant for water.
Explanation:
To calculate the freezing point depression of a solution of Ca(NO3)2 in water, we first determine the molality of the solution. With the provided mass of Ca(NO3)2 (11.3 g) and its formula weight (164 g/mol), we find there are 0.0689 moles of Ca(NO3)2. Since we only have 115 g of water, to convert to kilograms, we have 0.115 kg. The molality (m) is then 0.0689 moles / 0.115 kg = 0.599 m. Since Ca(NO3)2 dissociates into three ions (Ca2+, 2NO3-), the van't Hoff factor (i) is 3.
The depression of the freezing point is determined using the formula ΔTf = i * Kf * m, where Kf is the molal freezing point depression constant for water (1.86 °C/m). So the depression is ΔTf = 3 * 1.86 °C/m * 0.599 m = 3.34 °C.
The freezing point of the solution is then 0 °C - 3.34 °C = -3.34 °C, which is the answer.
A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-go-round and Cameron is sitting right on the edge, 2.0 m from the center. What is the relationship between the rotational speeds of the two children?
Answer:
angular speed of both the children will be same
Explanation:
Rate of revolution of the merry go round is given as
f = 4.04 rev/min
so here we have
[tex]f = \frac{4.04}{60} =0.067 rev/s[/tex]
here we know that angular frequency is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi(0.067)[/tex]
[tex]\omega = 0.42 rad/s[/tex]
now this is the angular speed of the disc and this speed will remain same for all points lying on the disc
Angular speed do not depends on the distance from the center but it will be same for all positions of the disc
How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories.
Answer:
The formula is Work = Force x distance moved
so that would be 0.600 x 5 = 3 Joules which is 0.003 kilojoules
The work done in by the attendant on a can of soup is 3 joules or 0.00072 Kilocalories
What is Work in physics?
Work in physics is defined as the dot product of force and the displacement produced by it.
Given is an attendant who pushed 0.600 m horizontally with a force of 5.00 N.
From the definition of work done, we can write -
W = F.d
W = Fd cosФ
Then angle between force and displacement is 0°. Therefore, cosФ is equal to 1.
W = Fd
W = 5 x 0.6
W = 3 joules = 3/4184 = 0.00072 Kilocalories
Therefore, the work done in by the attendant on a can of soup is 3 joules or 0.00072 Kilocalories
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Birds resting on high-voltage power lines are a common sight. The copper wire on which a bird stands is 2.3 cm in diameter and carries a current of 45 A. If the bird's feet are 3.9 cm apart, calculate the potential difference between its feet. (The resistivity of the wire is 1.7 10-8 Ω · m)
Explanation:
It is given that,
Diameter of the copper wire, d = 2.3 cm
Radius of copper wire, r = 1.15 cm = 0.0115 m
Distance between bird's feet, l = 3.9 cm = 0.039 m
The resistivity of the wire, [tex]\rho=1.7\times 10^{-8}\ \Omega-m[/tex]
We need to find the potential difference between bird's feet. The resistance of the wire is calculated by :
[tex]R=\rho\times \dfrac{l}{A}[/tex]
[tex]R=1.7\times 10^{-8}\ \Omega-m\times \dfrac{0.039\ m}{\pi(0.0115\ m)^2}[/tex]
R = 0.00000159 ohms
[tex]R=1.59\times 10^{-6}\ \Omega[/tex]
Let V is the potential difference between bird's feet. It can be calculated using Ohm's law as :
[tex]V=I\times R[/tex]
[tex]V=45\ A\times 1.59\times 10^{-6}\ \Omega[/tex]
V = 0.000071 volts
or
[tex]V=7.1\times 10^{-5}\ volts[/tex]
So, the potential difference between the bird's feet is [tex]7.1\times 10^{-5}\ volts[/tex]. Hence, this is the required solution.
Even with a large estimated resistance for the bird (say, R_bird = 1000 Ω), the potential difference (V_bird) would be very small (V_bird = 45 A * 1000 Ω ≈ 45000 V). However, in reality, most of the current will flow through the much lower resistance wire, making the actual potential difference between the bird's feet much closer to zero.
Calculation (for illustration purposes only):
Even though the potential difference is negligible, we can estimate its upper bound (worst-case scenario) by assuming all the current flows through the bird's body.
Wire Resistance:
Diameter (d) = 2.3 cm = 0.023 m
Radius (r) = d/2 = 0.0115 m
Wire length between feet (l) = 3.9 cm = 0.039 m
Resistivity (ρ) = 1.7 x 10^-8 Ω · m
Wire Resistance (R_wire) = ρ * (l / π * r^2) = (1.7 x 10^-8 Ω · m) * (0.039 m / (π * (0.0115 m)^2)) ≈ 4.2 x 10^-7 Ω (very small)
Assuming All Current Through Bird:
Current (I) = 45 A
Bird's Body Resistance (R_bird) = V_bird / I (Since we don't have V_bird, this is just a placeholder)
Total Circuit Resistance (ignoring wire resistance):
R_total = R_bird + R_wire ≈ R_bird (because R_wire is negligible)
Potential Difference Across Bird (upper bound):
V_bird = I * R_total = I * R_bird (since R_wire is negligible)
A 30.6 kg mass attached to a spring oscillates with a period of 3.45 s. Find the force constant of the spring.
Answer:
Force constant, K = 101.49 N/m
Explanation:
It is given that,
Mass, m = 30.6 kg
Time period of oscillation, T = 3.45 s
We need to find the force constant of the spring. The time period of the spring is given by :
[tex]T=2\pi\sqrt{\dfrac{m}{K}}[/tex]
[tex]K=\dfrac{4\pi^2m}{T^2}[/tex]
[tex]K=\dfrac{4\pi^2\times 30.6\ kg}{(3.45\ s)^2}[/tex]
K = 101.49 N/m
So, the force constant of the spring is 101.49 N/m. Hence, this is the required solution.
The temperature of a pot of water is 90.0 °C and the mass of the water is 112 g. A block with a mass of 21.0 g and a temperature of 20.0°C is dropped into the water. The final temperature of the water and the block is 88.3 °C. The specific heat of water is exactly 1 cal/g.°C. How much heat was lost by the water? O 1.90 x10^2 cal O 35.7 cal O 7650 cal O 1.70 cal How much heat was gained by the block? O 1430 cal O 190 x10^2 cal O 35.7 cal O 68.3 cal What is the specific heat of the block? O 0.133 cal/g.°C O 5.33 cal/g.°C O273000 cal/g.°C O0.0249 cal/g.°C
Answer:
1.90 x 10² cal
1.90 x 10² cal
0.133 cal/(g °C)
Explanation:
For water :
[tex]m_{w}[/tex] = mass of water = 112 g
[tex]c_{w}[/tex] = specific heat of water = 1 cal/(g °C)
[tex]T_{wi}[/tex] = initial temperature of water = 90.0 °C
[tex]T_{wf}[/tex] = final temperature of water = 88.3 °C
[tex]Q_{w}[/tex] = Heat lost by water
Heat lost by water is given as
[tex]Q_{w}= m_{w}c_{w}(T_{wi} - T_{wf})[/tex]
[tex]Q_{w}[/tex] = (112) (1) (90.0 - 88.3)
[tex]Q_{w}[/tex] = 1.90 x 10² cal
[tex]Q_{B}[/tex] = Heat gained by the block
As per conservation of energy
Heat gained by the block = Heat lost by water
[tex]Q_{B}[/tex] = [tex]Q_{w}[/tex]
[tex]Q_{B}[/tex] = 1.90 x 10² cal
For Block :
[tex]m_{B}[/tex] = mass of block = 21.0 g
[tex]c_{B}[/tex] = specific heat of block
[tex]T_{bi}[/tex] = initial temperature of block = 20.0 °C
[tex]T_{bf}[/tex] = final temperature of block = 88.3 °C
[tex]Q_{B}[/tex] = Heat gained by Block = 1.90 x 10² cal
Heat gained by water is given as
[tex]Q_{B}[/tex] = m_{B}c_{B}(T_{bf} - T_{bi})[/tex]
1.90 x 10² = (21.0) (88.3 - 20.0) c_{B}
c_{B} = 0.133 cal/(g °C)
If the earth were twice the distance from the sun that it is now, the gravitational force exerted on it by the sun would be: a) 1/4 what it is now
b) /2 what it is now
c) twice what it is now
d) 4 times what it is now
Answer:
a) 1/4 what it is now
Explanation:
As we know that force of gravitation between two planets at some distance "r" from each other is given as
[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]
now since we know that if the distance between Earth and Sun is changed
So the force of gravity will be given as
[tex]\frac{F_g'}{F_g} = \frac{r_1^2}{r_2^2}[/tex]
now we know that the distance between sun and earth is changed to twice the initial distance between them
so we have
[tex]r_2 = 2r_1[/tex]
so new gravitational force between sun and earth is given as
[tex]F_g' = \frac{r_1^2}{(2r_1)^2}F_g[/tex]
[tex]F_g' = \frac{1}{4}F_g[/tex]
An electric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an electric potential difference of 12 V is placed across its leads. What is the resistance of the heater wire?
Answer:
0.686 Ohm
Explanation:
Heat energy H = 50 cal = 50 x 4.2 J = 210 J, time t = 1 second, V = 12 V
Let R be the resistance.
Heat energy = V^2 x t / R
210 = 12 x 12 x 1 / R
210 = 144 / R
R = 144 / 210 = 0.686 Ohm
The resistance of the heater wire is approximately 0.688 ohms.
To calculate the resistance of the heater wire, we need to use the information that 50 cal of heat is generated per second when a potential difference of 12 V is applied across the leads. The first step is to convert the heat from calories to joules, which is the standard unit of energy in Physics. Since 1 calorie is equivalent to 4.184 joules, we can calculate the power (P) in joules per second (or watts) by multiplying the heat generated per second by this conversion factor:
P = 50 cal/s ×4.184 J/cal = 209.2 J/s = 209.2 W
Now, we can use the formula for electric power P = V^2/R, where P is the power, V is the potential difference, and R is the resistance. By rearranging the formula to solve for R, we get R = V^2/P.
Plugging our values into this equation gives us:
R = (12 V)^2 / (209.2 W) = 144 V^2 / 209.2 W = approx 0.688 ohms.
What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?
Answer:
Heat energy needed = 3036.17 kJ
Explanation:
We have
heat of fusion of water = 334 J/g
heat of vaporization of water = 2257 J/g
specific heat of ice = 2.09 J/g·°C
specific heat of water = 4.18 J/g·°C
specific heat of steam = 2.09 J/g·°C
Here wee need to convert 1 kg ice from -13°C to vapor at 100°C
First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.
Mass of water = 1000 g
Heat energy required to change ice temperature from -13°C to 0°C
H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ
Heat energy required to change ice from 0°C to water at 0°C
H₂ = mL = 1000 x 334 = 334 kJ
Heat energy required to change water temperature from 0°C to 100°C
H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ
Heat energy required to change water from 100°C to steam at 100°C
H₄ = mL = 1000 x 2257 = 2257 kJ
Total heat energy required
H = H₁ + H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ
Heat energy needed = 3036.17 kJ
A bluebird has mass of 34 grams. It flies from the ground to the top of an 8.5-m tree. What is the change in the bluebird's gravitational potential energy as it flies to the top of the tree? 2800J 290J 2.8J 280J 0.29J
Answer:
Gravitational potential energy, PE = 2.8 J
Explanation:
It is given that,
Mass of the bluebird, m = 34 g = 0.034 kg
It flies from the ground to the top of an 8.5-m tree, h = 8.5 m
We need to find the change in the bluebird's gravitational potential energy as it flies to the top of the tree. It can be calculated as :
[tex]PE=m\times g\times h[/tex]
[tex]PE=0.034\ kg\times 9.8\ m/s^2\times 8.5\ m[/tex]
PE = 2.83 J
or
PE = 2.8 J
So, the gravitational potential energy as it flies to the top of the tree is 2.8 J. Hence, this is the required solution.
A spring is hung from the ceiling. A 0.473 -kg block is then attached to the free end of the spring. When released from rest, the block drops 0.109 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.
Answer:
a)
85.05 N/m
b)
179.81 rad/s
Explanation:
a)
k = spring constant of the spring
m = mass of the block = 0.473 kg
x = stretch caused in the spring = 0.109 m
h = height dropped by the block = 0.109 m
Using conservation of energy
Spring potential energy gained by the spring = Potential energy lost by the block
(0.5) k x² = mgh
(0.5) k x² = mgx
(0.5) k x = mg
(0.5) k (0.109) = (0.473) (9.8)
k = 85.05 N/m
b)
angular frequency is given as
[tex]w = \sqrt{\frac{k}{m}}[/tex]
[tex]w = \sqrt{\frac{85.05}{0.473}}[/tex]
[tex]w [/tex] = 179.81 rad/s
The spring constant of the spring is 42.54 N/m, and the angular frequency of the block's vibrations is 4.88 rad/s.
Explanation:To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. In this case, the weight of the block is equal to the force provided by the spring at the equilibrium position.
Using the equation F = kx, where F is the force, k is the spring constant, and x is the displacement, we can solve for k. Since the block momentarily comes to rest after dropping 0.109 m, we can set the force provided by the spring equal to the weight of the block and solve for k.
Given:
Mass of the block (m) = 0.473 kg
Displacement of the block (x) = 0.109 m
Using the equation F = kx, we can rewrite it as k = F/x. The weight of the block is equal to its mass multiplied by the acceleration due to gravity (9.8 m/s^2), so the force provided by the spring is 0.473 kg * 9.8 m/s^2 = 4.6354 N. Substituting these values into the equation, we find the spring constant (k) to be:
k = 4.6354 N / 0.109 m = 42.54 N/m
To find the angular frequency of the block's vibrations, we can use the equation:
ω = sqrt(k/m)
Substituting the values of k and the mass of the block (m) = 0.473 kg into the equation, we can calculate the angular frequency (ω):
ω = sqrt(42.54 N/m / 0.473 kg) = 4.88 rad/s
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Earth moves in an elliptical orbit with the sun at one of the foci. The length of half of the major axis is 149,598,000 kilometers, and the eccentricity is 0.0167. Find the minimum distance (perihelion) and the maximum distance (aphelion) of Earth from the sun.
Answer:
147,099,713.4 km
152,096,286.6 km
Explanation:
a = 149598000 km
e = 0.0167
The formula to find the perihelion
Rp = a ( 1 - e) = 149598000 ( 1 - 0.0167) = 147,099,713.4 km
The formula for aphelion
Ra = a ( 1 + e) = 149598000 ( 1 + 0.0167) = 152,096,286.6 km
To find the minimum and maximum distances from Earth to the Sun (perihelion and aphelion), we calculate them using Earth's semi-major axis of 149,598,000 kilometers and the eccentricity of 0.0167. The perihelion is 147,099,014 kilometers, and the aphelion is 152,096,986 kilometers.
Explanation:The student's question revolves around finding the minimum (perihelion) and maximum (aphelion) distances from the Earth to the Sun, given the length of half of the major axis — also known as the semi-major axis — and the eccentricity of Earth's orbit. The semi-major axis (a) is 149,598,000 kilometers and the eccentricity (e) is 0.0167. The distance from the center of the ellipse, where Earth's orbit is, to the focus (c) is equal to the product of the semi-major axis and the eccentricity (c = ae).
The perihelion distance is the semi-major axis minus the distance c, resulting from the Earth being at the closest point to the Sun in its orbit. Conversely, the aphelion distance is the semi-major axis plus the distance c, when Earth is farthest from the Sun. Therefore, the perihelion (rp) can be calculated as rp = a - c, and the aphelion (ra) as ra = a + c.
Using the formula c = ae, we find that c is approximately 2,498,986 kilometers (149,598,000 km * 0.0167). Thus:
What is a primitive solid?
Answer:
A primitive solid is a 'building block' that you can use to work with in 3D. Rather than extruding or revolving an object, AutoCAD has some basic 3D shape commands at your disposal.
Explanation:
A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.870 m/s at an angle of 35.0° above the table, and it lands on the magazine 0.0770s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.
Answer:
Thickness of magazine = 9.42 mm.
Explanation:
Considering vertical motion of spider:-
Initial velocity, u = 0.870 sin 35 = 0.5 m/s
Acceleration , a = -9.81 m/s²
Time, t = 0.077 s
We have equation of motion s= ut + 0.5 at²
Substituting
s= 0.5 x 0.077 - 0.5 x 9.81 x 0.077²
s = 9.42 x 10⁻³ m = 9.42 mm
Thickness of magazine = 9.42 mm.
The thickness of the magazine is determined using the vertical component of the spider's initial velocity and kinematic equations for constant acceleration. By calculating and converting the vertical displacement to millimeters, we can find the thickness of the magazine the spider lands on.
Given that the initial velocity in the vertical direction (Vy) can be found using the formula Vy = V * sin(θ), where V is the initial velocity of the spider and θ is the angle of projection, we can calculate Vy = (0.870 m/s) * sin(35.0°).
The vertical displacement (y) can then be calculated using the kinematic equation for constant acceleration, y = Vy * t + (1/2) * g * t^2, where t is the time of flight, and g is the acceleration due to gravity (approximately -9.81 m/s^2). Since we want to find the displacement in the upward direction and the spider lands on the magazine after 0.0770 s, we are looking for the magnitude of y when the spider lands.
Substituting the values we have: y = (0.870 m/s) * sin(35.0°) * 0.0770 s + (0.5) * (-9.81 m/s^2) * (0.0770 s)^2. The negative sign in the acceleration term accounts for the direction of gravity, which is downwards. After calculating this expression, we convert the result from meters to millimeters (by multiplying by 1000) to obtain the thickness of the magazine.
A 0.8 g object is placed in a 159 N/C uniform electric field. Upon being released from rest, it moves 72 m in 2.9 s. Determine the object's acceleration & charge magnitude. Assume the acceleration is due to the E-field (i.e., ignore all other forces). a =
Answer:
The acceleration and charge are 17.122 m/s² and [tex]8.6\times10^{-5}\ C[/tex]
Explanation:
Given that,
Mass of object = 0.8 g
Electric field = 159 N/C
Distance = 72 m
Time = 2.9 s
We know that,
The electric force is
[tex]F = Eq[/tex]....(I)
The newton's second law
[tex]F=ma[/tex]
Put the value of F in the equation (I)
[tex]ma=Eq[/tex]...(II)
We calculate the acceleration
Using equation of motion
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
[tex]a =\dfrac{2s}{t^2}[/tex]
[tex]a=\dfrac{2\times72}{(2.9)^2}[/tex]
[tex]a=17.122\ m/s^2[/tex]
From equation (II)
[tex]q=\dfrac{ma}{E}[/tex]
[tex]q=\dfrac{0.8\times10^{-3}\times17.122}{159}[/tex]
[tex]q=0.000086148427673\ C[/tex]
[tex]q=8.6\times10^{-5}\ C[/tex]
Hence, The acceleration and charge are 17.122 m/s² and [tex]8.6\times10^{-5}\ C[/tex]
A 100-W light bulb generates 95W of heat, which is dissipated through a glass bulb that has a radius of 3.0 cm and is 0.50 mm thick. What is the temperature difference between the inner and outer surfaces of the glass? Use 'deg C' as your units.
To calculate the temperature difference between the inner and outer surfaces of the glass bulb in a 100 W bulb that generates 95W of heat, we use a formula from the principles of heat conduction where we input parameters including the heat generated by the bulb, the thermal conductivity of the glass, the surface area of the glass and the thickness of the glass.
Explanation:In order to calculate the temperature difference between the inner and outer surfaces of the glass bulb, we need to use the formula for heat conduction, which is given by the formula Q = (k*A*ΔT)/d, where Q is the heat generated by the bulb, k is the thermal conductivity of the glass, A is the surface area of the glass, ΔT is the temperature difference, and d is the thickness of the glass. In this case, we know that the light bulb generates 95W of heat, the radius of the bulb is 3.0 cm, the thickness of the glass is 0.50 mm. Assuming the thermal conductivity of the glass to be 0.8 W/m.K, we can substitute these values into the formula to calculate the temperature difference ΔT = Qd / (k*A). Note that here, the surface area of the glass, A = 4πr².
As you can see, the calculation requires a clear grasp of the concepts of heat conduction, thermal conductivity, and physical constants of materials. Understanding how these factors interact is key to solving problems about heat transfer in Physics.
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A power P is required to do work W in a time interval T. What power is required to do work 3W in a time interval 5T? (a) 3P (b) 5P (c) 3P/5 (a) P (e) 5P/3
Answer:
(c) 3P/5
Explanation:
The formula to calculate the power is:
[tex]P=\frac{W}{T}[/tex]
where
W is the work done
T is the time required for the work to be done
In the second part of the problem, we have
Work done: 3W
Time interval: 5T
So the power required is
[tex]P=\frac{3W}{5T}=\frac{3}{5}\frac{W}{T}=\frac{3}{5}P[/tex]
A wheel that was initially spinning is accelerated at a constant angular acceleration of 5.0 rad/s^2. After 8.0 s, the wheel is found to have made an angular displacement of 400 radians. (a) How fast was the wheel spinning initially? (b) What is the final angular velocity of the wheel?
Answer:
a) Initial angular speed = 30 rad/s
b) Final angular speed = 70 rad/s
Explanation:
a) We have equation of motion s = ut + 0.5at²
Here s = 400 radians
t = 8 s
a = 5 rad/s²
Substituting
400 = u x 8 + 0.5 x 5 x 8²
u = 30 rad/s
Initial angular speed = 30 rad/s
b) We have equation of motion v = u + at
Here u = 30 rad/s
t = 8 s
a = 5 rad/s²
Substituting
v = 30 + 5 x 8 = 70 rad/s
Final angular speed = 70 rad/s
Kinematics
A car starts from rest and reaches a speed of 80 ft/s after traveling 500 ft along a straight road.
Determine its constant acceleration and the time of travel.
Answer:
The acceleration and time are 1.95 m/s and 12.5 s.
Explanation:
Given that,
Speed = 80 ft/s =24.384 m/s
Distance = 500 ft =152.4 m
We need to calculate the acceleration
Using third equation of motion
[tex]v^2-u^2= 2as[/tex]
[tex]a = \dfrac{v^2-u^2}{2s}[/tex]
Where, u = initial velocity
v = final velocity
a = acceleration
s = distance
Put the value in the equation
[tex]a=\dfrac{(24.384)^2-0}{2\times152.4}[/tex]
[tex]a=1.95\ m/s^2[/tex]
We need to calculate the time
Using first equation of motion
[tex]v=u+at[/tex]
[tex]t =\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{24.384-0}{1.95}[/tex]
[tex]t =12.5\ s[/tex]
Hence, The acceleration and time are 1.95 m/s and 12.5 s.
Two objects which have a total mass of 12.0 kg are placed adjacent to one another. The object on the left is 7.4 kg, and it pushed to the right with a force of 37 Newtons. With what force does the left object push against the right object?
Answer:
22.89 N
Explanation:
F = 37 N
Let the acceleration in the system is a and f be the force between the tewo blocks.
Apply the Newton's second law
By the free body diagrams
F - f = 7.4 x a .... (1)
f = 12 x a ...... (2)
Adding both of them
37 = 19.4 x a
a = 1.9 m/s^2
Put in equation (2)
f = 12 x 1.9 = 22.89 N
When the particle with charge q reaches the center of the original square, it is, as stated in the problem, momentarily at rest. Is the particle at equilibrium at that moment?
Answer:
NO
Explanation:
The answer is NO because when the particle reaches the center of square there is a net force acting on the particle dew to various other charges and this net force gives acceleration to the particle. Moreover, For particle or object to be in equilibrium the net force acting on it should be zero and hence no acceleration. Although velocity can be zero or non zero at equilibrium state.
Answer:
The context is missing here, but ill try to explain a general case.
Something is in equilibrium if it is in a valley of the potential energy, this is because things in life try to be in the minimal energy state possible. Think for example in a thing that is away from the ground, the object will try to reach the ground, in this way minimizing the potential energy.
Now, if once the particle reaches the center of the square it remains at rest, it means that the total forces acting on the particle are zero and this is why the particle stays at rest, this would mean that the particle is in equilibrium, and if someone moves it a little bit of the center, some of the forces will increase and others will decrease, and then the equilibrium will be broken and the particle will move again.
In another case, if the particle is momentarily at rest (just for a few seconds) it may be because the forces acting on it are affecting the particle in such way that is moving is fully stopped in one direction, and the new forces are accelerating the particle in the opposite direction (in the same way that if you throw something upside when it reaches the maximum height it has for a brief moment a velocity equal to zero)
A solenoid with n1 = 1200 turns/m and a current I1 = 2.5 A is filled with a paramagnetic material at a temperature T1 = 320 K. A second solenoid with n2 = 1000 turns/m and a current I2= 0.85 A is filled with the same paramagnetic material at a different temperature T2. The magnetizations are the same in both cases. What is the value of T2?
Answer:
[tex]T_2[/tex] = [tex]90.667K[/tex]
Explanation:
Given:
For the first solenoid
Number of turns, n₁ = 1200 turns/m
Current, I₁ = 2.5 A
Paramagnetic material temperature, T₁ = 320 K
Now for the second solenoid
Number of turns, n₂ = 1000 turns/m
Current, I₂ = 0.85 A
Paramagnetic material temperature = T₂
The magnetic flux (B) is given as
[tex]B=\frac{c\mu_onI}{T}[/tex]
where,
c = curie's constant
μ₀ = arithmetic constant
also it is given that the magnetization in both the cases are same
therefore the magnetic flux will also be equal
thus,
[tex]\frac{c\mu_on_1I_1}{T_1}[/tex] = [tex]\frac{c\mu_on_2I_2}{T_2}[/tex]
or
[tex]\frac{n_1I_1}{T_1}[/tex] = [tex]\frac{n_2I_2}{T_2}[/tex]
or
[tex]\frac{1200\times 2.5}{320}[/tex] = [tex]\frac{1000\times 0.85}{T_2}[/tex]
or
[tex]9.375[/tex] = [tex]\frac{850}{T_2}[/tex]
or
[tex]T_2[/tex] = [tex]\frac{850}{9.375}[/tex]
or
[tex]T_2[/tex] = [tex]90.667K[/tex]
As you stand by the side of the road, a car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear a frequency of 65 Hz. What is the speed of the car? The speed of sound in the air is 343 m/s.
Answer:
26.8 m/s
Explanation:
[tex]v[/tex] = constant speed of the car
[tex]V[/tex] = speed of sound = 343 m/s
[tex]f[/tex] = actual frequency of the horn
[tex]f_{app}[/tex] = frequency heard as the car approach = 76 Hz
frequency heard as the car approach is given as
[tex]f_{app}=\frac{vf}{V - v}[/tex]
[tex]76 =\frac{vf}{343 - v}[/tex] eq-1
[tex]f_{rec}[/tex] = frequency heard as the car recedes = 65 Hz
frequency heard as the car goes away is given as
[tex]f_{rec}=\frac{vf}{V + v}[/tex]
[tex]65 =\frac{vf}{343 + v}[/tex] eq-2
dividing eq-1 by eq-2
[tex]\frac{76}{65}=\frac{343+v}{343-v}[/tex]
[tex]v[/tex] = 26.8 m/s
Final answer:
To determine the car's speed using the Doppler Effect, we calculate the difference in observed sound frequencies as the car approaches and moves away. Applying formulas for Doppler Effect calculations, the speed of the car comes out to be approximately 14.6 m/s.
Explanation:
The question revolves around the phenomenon known as the Doppler Effect, which is observed when a sound source moves relative to an observer. To calculate the speed v of the car, we use the Doppler Effect equations for sound frequencies heard when the source is moving towards and then away from the observer:
For the source approaching:
f' = f * ((v + vo) / (v - vs))
, where:
f' is the observed frequency when the source is approaching (76 Hz)
f is the original frequency emitted by the source
v is the speed of sound (343 m/s)
vo is the speed of the observer (0 m/s, since the observer is stationary)
vs is the speed of the source (the car's speed, what we are solving for)
For the source receding:
f'' = f * ((v - vo) / (v + vs))
, where:
f'' is the observed frequency when the source is receding (65 Hz)
To find the car's speed, we need to solve for vs in both equations. By eliminating f (since it's the same for both equations), we can solve for vs. Using these equations, we determine that the speed of the car is approximately 14.6 m/s.