The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Therefore, the flux through surface A is three times larger than the flux through surface B.
The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Since the flux of electric field is proportional to the area of the surface, the flux through surface A is three times larger than the flux through surface B. Therefore, the correct answer is E) three times larger than the flux of electric field through Gaussian surface B.
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Final answer:
The correct answer is D) equal to the flux of the electric field through Gaussian surface B. The electric flux depends on the charge enclosed by the Gaussian surface, not on its size or shape, according to Gauss's law.
Explanation:
The question asks whether Gaussian surfaces A and B, enclosing the same positive charge +Q, with surface A having three times the area of surface B, have different electric fluxes. According to Gauss's law, the electric flux (Φ) through a closed surface is proportional to the charge enclosed (Φ = Q/ε0). Since both surfaces enclose the same charge, the flux through each surface must be the same, regardless of their respective areas.
Thus, the correct answer is D) equal to the flux of the electric field through Gaussian surface B because the flux depends only on the amount of enclosed charge, not on the size or shape of the Gaussian surface.
. The force exerted by the wind on the sails of a sailboat is 390 N north. The water exerts a force of 180 N east. If the boat (including its crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration?
Answer:
1.59 m/s^2, 65.2°
Explanation:
F1 = 390 N North
F2 = 180 N east
m = 270 kg
Net force is the vector sum of both the forces.
[tex]F = \sqrt{F_{1}^{2}+F_{2}^{2}}[/tex]
[tex]F = \sqrt{390^{2}+180^{2}}[/tex]
F = 429.53 N
Direction of force
tan∅ = F1 / F2 = 390 / 180 = 2.1667
∅ = 65.2°
The direction of acceleration is same as the direction of net force.
The magnitude of acceleration is
a = F / m = 429.53 / 270 = 1.59 m/s^2
Final answer:
The magnitude of the sailboat's acceleration is 1.59 m/s², and the direction is 25 degrees east of north. This is calculated using Newton's second law and vector addition of the orthogonal forces exerted by the wind and water.
Explanation:
To calculate the magnitude and direction of the sailboat's acceleration, we need to use Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Here, we combine the forces exerted by the wind and the water to obtain the net force. The net force can be calculated using vector addition, where the force due to the wind (390 N north) and the force due to the water (180 N east) are treated as orthogonal vectors.
The net force (Fnet) is the vector sum of the two individual forces. We calculate the net force using the Pythagorean theorem:
Fnet = √Fwind² + Fwater² = √390² + 180² = √152100 + 32400 = √184500 N
So, the magnitude of the net force is approximately 429.53 N. To find the acceleration (a), we use the formula:
a = Fnet / m
Substituting the mass of the sailboat (270 kg) and the net force, we get:
a = 429.53 N / 270 kg = 1.59 m/s²
To determine the direction of the acceleration, we take the arctangent of the ratio of the forces. Since arctan(180/390) equals approximately 25 degrees, this is the angle east of north.
The sailboat's acceleration has a magnitude of 1.59 m/s² and a direction of 25 degrees east of north.
A projectile of mass 100 kg is shot from the surface of Earth by means of a very powerful cannon. If the projectile reaches a height of 65,000 m above Earth's surface, what was the speed of the projectile when it left the cannon? (Mass of Earth 5.97x10^24 kg, Radius of Earth 6.37x10^6 m)
Answer:
[tex]v = 1.11 \times 10^3 m/s[/tex]
Explanation:
By energy conservation law we will have
[tex]U_i + KE_i = U_f + KE_f[/tex]
as we know that as the projectile is rising up then due to gravitational attraction of earth it will slow down
At the highest position the speed of the projectile will become zero
So here we will have
[tex]-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h} + 0[/tex]
[tex]-\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(100)}{6.37 \times 10^6} + \frac{1}{2}(100)v^2 = -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{-11})(100)}{(6.37 \times 10^{6} + 65000)}[/tex]
[tex] - 6.25 \times 10^7 + 0.5 v^2 = -6.19 \times 10^7[/tex]
[tex]v = 1.11 \times 10^3 m/s[/tex]
A wave has a frequency of 0.5 kHz and two particles with a phase difference of \pi /3 are 1.5 cm apart. Calculate: the time period of the wave.
Answer:
The time period of the wave is 0.002 sec.
Explanation:
Given that,
Frequency = 0.5 kHz
Phase difference [tex]\phi=\dfrac{\pi}{3}[/tex]
Path difference = 1.5 cm
We need to calculate the time period
Using formula of time period
The frequency is the reciprocal of time period.
[tex]f =\dfrac{1}{T}[/tex]
[tex]T=\dfrac{1}{f}[/tex]
Where, f = frequency
Put the value into the formula
[tex]T=\dfrac{1}{0.5\times10^{3}}[/tex]
[tex]T=0.002\ sec[/tex]
Hence, The time period of the wave is 0.002 sec.
In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km/h. Find the rotational kinetic energy of the pitcher’s arm given its moment of inertia is 0.720 kg m2 and the ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder.
The rotational kinetic energy of the pitcher's arm when throwing a softball at a speed of 139 km/h is calculated using the equation for rotational kinetic energy, and taking moment of inertia and angular velocity into account. The angular velocity is inferred from the linear speed of the ball and the distance it leaves the pitcher's hand from the pivot at the shoulder. The rotational kinetic energy is found to be approximately 1491 Joules.
Explanation:This question pertains to the rotational kinetic energy of the pitcher's arm during the act of throwing a softball. By definition, the kinetic energy associated with rotational motion (rotational kinetic energy) can be given by the equation: K_rot = 0.5 * I * ω² where 'I' is the moment of inertia and 'ω' is the angular velocity.
To calculate the angular velocity 'ω', we can infer it from the linear speed of the ball when it leaves the pitcher's hand, as 'ω = v/r', v = 139 km/h = 38.6 m/s, and r = 0.600 m (distance ball leaves hand from pivot at shoulder). Therefore, 'ω' is approximately 64.4 rad/s.
Substituting 'I' and 'ω' into the equation above, we get K_rot = 0.5 * 0.720 kg*m² * (64.4 rad/s)² = 1491 Joules, which is the rotational kinetic energy of the pitcher's arm.
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A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C is fixed to the x axis at x = +2.0 m. Find the location of the place(s) along the x axis where the electric field due to these two charges is zero.
Answer:
3.62 m and - 1.4 m
Explanation:
Consider a location towards the positive side of x-axis beyond the location of charge Q₂
x = distance of the location from charge Q₂
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
[tex]\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}[/tex]
[tex]\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}[/tex]
x = 1.62 m
So location is 2 + 1.62 = 3.62 m
Consider a location towards the negative side of x-axis beyond the location of charge Q₁
x = distance of the location from charge Q₁
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
[tex]\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}[/tex]
[tex]\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}[/tex]
x = - 1.4 m
Electric field is zero due to positive point charge Q1 and negative Q2 along the x axis is at the location of 3.62 meters.
What is electric field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
Given information-
The charge of the point 1 is [tex]2.5\times10^{-5} \rm C[/tex].
The charge of the point 2 is [tex]-5.0\times10^{-6} \rm C[/tex].
The distance between the point 1 and point 2 is 2 meters away from the x axis.
Let the position of the point 1 is at x.
Thus the electric force on point Q1 is,
[tex]F_1=\dfrac{kQ1}{x^2}[/tex]
As the distance between the point Q1 and point Q2 is 2 meters away from the x axis. Thus the position of it should be at (x+2). The electric force on point 2
[tex]F_2=\dfrac{kQ1}{(x+2)^2}[/tex]
As the force of two is equal and opposite thus,
[tex]\dfrac{kQ1}{x^2}=\dfrac{kQ2}{(x+2)^2}\\\dfrac{2.5\times10^{-5}}{x^2}=\dfrac{-5\times10^{-6}}{(x+2)^2}\\x=1.62[/tex]
Thus the position of point 2 is,
[tex]p_2=2+1.62\\p_2=3.62\rm m[/tex]
Thus, the location of the place along the x axis where the electric field due to these two charges is zero is 3.62 meters.
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A truck with 34-in.-diameter wheels is traveling at 55 mi/h. Find the angular speed of the wheels in rad/min, *hint convert miles to inches & hours to minutes:
The angular speed of the truck's wheels is approximately 3422.35 rad/min.
The angular speed [tex](\( \omega \))[/tex] is related to the linear speed [tex](\( v \))[/tex] and the radius [tex](\( r \))[/tex] of the wheels by the formula [tex]\( v = r \cdot \omega \)[/tex].
Firstly, we need to find the radius of the wheels, which is half of the diameter. So, [tex]\( r = \frac{34}{2} = 17 \) inches[/tex].
Now, convert the speed from miles per hour to inches per minute. There are 5280 feet in a mile, 12 inches in a foot, and 60 minutes in an hour.
[tex]\[ v = 55 \, \text{mi/h} \times \frac{5280 \, \text{ft}}{1 \, \text{mi}} \times \frac{12 \, \text{in}}{1 \, \text{ft}} \times \frac{1 \, \text{h}}{60 \, \text{min}} \][/tex]
[tex]\[ v \approx 58080 \, \text{in/min} \][/tex]
Now, use the formula [tex]\( v = r \cdot \omega \)[/tex] to find [tex]\( \omega \)[/tex]:
[tex]\[ \omega = \frac{v}{r} \][/tex]
[tex]\[ \omega = \frac{58080 \, \text{in/min}}{17 \, \text{in}} \][/tex]
[tex]\[ \omega \approx 3422.35 \, \text{rad/min} \][/tex]
Therefore, the angular speed of the truck's wheels is approximately 3422.35 rad/min.
An electron is released from rest in a uniform electric field. The electron accelerates, travelling 5.50 m in 4.00 µs after it is released. What is the magnitude of the electric field in N/C?
Answer:
3.91 N/C
Explanation:
u = 0, s = 5.50 m, t = 4 us = 4 x 10^-6 s
Let a be the acceleration.
Use second equation of motion
s = u t + 1/2 a t^2
5.5 = 0 + 1/2 a (4 x 10^-6)^2
a = 6.875 x 10^11 m/s^2
F = m a
The electrostatic force, Fe = q E
Where E be the strength of electric field.
So, q E = m a
E = m a / q
E = (9.1 x 10^-31 x 6.875 x 10^11) / ( 1.6 x 10^-19)
E = 3.91 N/C
A 1200 W microwave oven transforms 1.8 x10^5 J of energy while reheating some food. Calculate how long the food was in the microwave. Answer in minutes.
Answer:
2.5 min
Explanation:
Hello
by definition the power is the amount of work done per unit of time. in this case, we know the total power and the work developed
[tex]1 Watt= \frac{Joule }{sec}[/tex]
Let
[tex]P=1200 W\\E=1.8 *10^{5}\\X=\frac{1.8 *10^{5} j }{1200 W}\\X= unknown\ time\ (sec) \\\\we\ need\ the\ answer\ in\ minutes,\\ 1\ min=60\ sec\\\\x=150 sec \\150\ sec*(\frac{1 min}{ 60 sec})=2.5 min\\[/tex]
hence , the data is not altered, we did it like this to eliminate the sec units.
Answer 2.5 min
I hope it helps
The food was in the 1200 W microwave for 150 seconds, which is equivalent to 2.5 minutes.
To calculate how long the food was in the microwave, we use the formula for power, which is the rate at which work is done or energy is transferred:
Power (P) = Energy (E) / Time (t)
First, rearrange this formula to solve for time (t):
t = E / P
Plugging in the given values:
So:
t = (1.8 times 10⁵J) / (1200 W)
Calculate this to find the time in seconds, and then convert it to minutes as follows:
t = 150 seconds
Divide by 60 to get minutes:
t = 2.5 minutes
A 40-W lightbulb is 1.9 m from a screen. What is the intensity of light incident on the screen? Assume that a lightbulb emits radiation uniformly in all directions (i.e., over 4π steradians). Express your answer with the appropriate units.
Answer:
[tex]INTNESITY = 0.88W/m^{2}[/tex]
Explanation:
given data:
power P = 40W
Distance of light bulb from screen is R = 1.9 m
The light intensity considered as the light energy falling on the surface per unit area per unit time
Intensity of light can be determined by using following formula
[tex]intesnity = \frac{P} {4\pi R^{2}}[/tex]
[tex]intesnity = \frac{40} {4\pi *1.9^{2}}[/tex]
[tex]INTNESITY = 0.88W/m^{2}[/tex]
A juggler tosses balls vertically to height H. To what height must they be tossed if they are to spend twice as much time in the air?
Answer:
The ball must be tossed to a height of 4 times the initial height H
Explanation:
We have equation of motion S = ut + 0.5at²
A juggler tosses balls vertically to height H.
That is
H = 0 x t + 0.5 x a x t²
H = 0.5at²
To what height must they be tossed if they are to spend twice as much time in the air.
H' = 0 x 2t + 0.5 x a x (2t)²
H' = 2at² = 4 H
So the ball must be tossed to a height of 4 times the initial height H
In a certain cyclotron a proton moves in a circle of radius 0.740 m. The magnitude of the magnetic field is 0.960 T. (a) What is the oscillator frequency? (b) What is the kinetic energy of the proton?
Answer:
Part a)
[tex]f = 1.46 \times 10^7 Hz[/tex]
Part b)
[tex]KE = 3.87 \times 10^{-12} J[/tex]
Explanation:
Part a)
As we know that radius of circular path of a charge moving in constant magnetic field is given as
[tex]R = \frac{mv}{qB}[/tex]
now we have
[tex]v = \frac{qBR}{m}[/tex]
now the frequency of oscillator is given as
[tex]f = \frac{v}{2\pi R}[/tex]
[tex]f = \frac{qB}{2\pi m}[/tex]
[tex]f = \frac{(1.6 \times 10^{-19})(0.960)}{2\pi(1.67\times 10^{-27})}[/tex]
[tex]f = 1.46 \times 10^7 Hz[/tex]
PART b)
now for kinetic energy of proton we will have
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}m(\frac{qBR}{m})^2[/tex]
[tex]KE = \frac{q^2B^2R^2}{2m}[/tex]
[tex]KE = \frac{(1.6 \times 10^{-19})^2(0.960)^2(0.740)^2}{2(1.67\times 10^{-27})}[/tex]
[tex]KE = 3.87 \times 10^{-12} J[/tex]
A body is projected downward at an angle of 30° with the horizontal from the top of a building 170 m high. Its initial speed is 40 m/s. 2.49 (c) At what angle with the horizontal will it strike? (c) 60
Answer:
The body will strike at angle 60.46°
Explanation:
Vertical motion of body:
Initial speed, u = 40sin30 = 20m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 170 m
We have equation of motion, v² = u² + 2as
Substituting
v² = 20² + 2 x 9.81 x 170
v = 61.12 m/s
Final vertical speed = 61.12 m/s
Final horizontal speed = initial horizontal speed = 40cos30= 34.64m/s
Final velocity = 34.64 i - 61.12 j m/s
Magnitude
[tex]v=\sqrt{34.64^2+(-61.12)^2}=70.25m/s[/tex]
Direction
[tex]\theta =tan^{-1}\left ( \frac{-61.12}{34.64}\right )=-60.46^0[/tex]
The body will strike at angle 60.46°
A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13 N, what is the strength of the magnetic field? (Express your answer using two significant figures.)
Answer:
B = 0.75 T
Explanation:
As we know that the force on a moving charge in magnetic field is given by the formula
[tex]F = qvB[/tex]
here we have
[tex]B = \frac{F}{qv}[/tex]
here we know that
[tex]F = 4.8 \times 10^{-13} N[/tex]
[tex]q = 1.6 \times 10^{-19} C[/tex]
[tex]v = 4 \times 10^6 m/s[/tex]
now from above equation we have
[tex]B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}[/tex]
[tex]B = 0.75 T[/tex]
Two point masses a 6 Kg mass and an 18 kg mass are connected by a mass less rod 6 meters long. Calculate the distance of the center of mass from the 18 kg mass. Calculate the moment of inertial about an axis located at the center of mass that is perpendicular to the rod.
The center of mass is given by:
∑mx/∑m
m is the mass of each object
x is the position of each object
We will assign x = 0m to the 18kg mass, therefore x = 6m for the 6kg mass.
∑mx/∑m = (18×0+6×6)/(18+6) = 1.5m
The center of mass is located 1.5m away from the 18kg mass.
The total moment of inertia of the system about the center of mass is given by:
I = ∑mr²
I is the moment of inertia
m is the mass of each object
r is the distance of each object from the center of mass
We know r = 1.5m for the 18kg mass and the rod is 6m long, therefore the 6kg mass must be r = 4.5m from the center of mass.
I = 18(1.5)² + 6(4.5)²
I = 162kg×m²
A projectile is launched vertically from the surface of the Moon with an initial speed of 1360 m/s. At what altitude is the projectile's speed two-fifths its initial value?
Using the kinematic equations of motion, we can calculate that a projectile launched at an initial speed of 1360 m/s from the moon's surface will reach two-fifths its initial speed at an altitude of roughly 680 kilometers.
Explanation:A projectile launched vertically from the moon's surface decelerates due to the moon's gravity at a rate of 1.6 m/s². Given the initial speed of the projectile is 1360 m/s, we want to find out the altitude at which the speed is two-fifths of this initial velocity. To find out, we use the kinematic equation v² = u² + 2as.
Substituting the values into the equation, we get (2/5 * 1360)² = (1360)² - 2 * 1.6 * s. Solving this equation gives us the distance s, where s comes out to be approximately 680000 meters or 680 km.
This means that the projectile's speed is two-fifths its initial value at an altitude of approximately 680 km from the moon's surface.
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To find the altitude at which the projectile's speed is two-fifths its initial value, the loss of kinetic energy must equal the gain in potential energy. Solving for the altitude (h), we find the answer.
Explanation:This problem involves physics concepts of projectile motion and gravitational acceleration. Given that the projectile is launched with an initial velocity of 1360 m/s, and we want to determine the altitude at which it has slowed to two-fifths of that speed, we will be considering the effects of the Moon's gravity on the deceleration of the projectile.
Using the formula for kinetic energy K.E = 1/2 m v^2, when the projectile's velocity slows to two-fifths of its original speed, the kinetic energy will be four-fifths less since kinetic energy is proportional to the square of the speed. This loss of kinetic energy must equal the gain in potential energy (since energy is conserved), which is given by the formula P.E = mgh (where m is mass, g is gravitational acceleration, and h is height or altitude).
Solving P.E = K.E for h, we get the altitude at which the projectile's speed is two-fifths its initial value.
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Calculate the internal energy (in J) of 86 mg of helium at a temperature of 0°C.
Answer:
The internal energy is 73.20 J.
Explanation:
Given that,
Weight of helium = 86 mg
Temperature = 0°C
We need to calculate the internal energy
Using formula of internal energy
[tex]U =nc_{v}T[/tex]
Where, [tex]c_{v}[/tex] = specific heat at constant volume
He is mono atomic.
So, The value of [tex]c_{v}= \dfrac{3}{2}R[/tex]
now, 1 mole of Helium = 4 g helium
n =number of mole of the 86 mg of helium
[tex]n = \dfrac{86\times10^{-3}}{4}[/tex]
[tex]n =2.15\times10^{-2}\ mole[/tex]
T = 0°C=273 K
Put the value into the formula
[tex]U = 2.15\times10^{-2}\times\dfrac{3}{2}\times8.314\times273[/tex]
[tex]U = 73.20\ J[/tex]
Hence, The internal energy is 73.20 J.
A rock is suspended by a light string. When the rock is in air, the tension in the string is 38.7 N . When the rock is totally immersed in water, the tension is 32.0 N . When the rock is totally immersed in an unknown liquid, the tension is 19.0 N .
What is the density of the unknown liquid?
Answer:
[tex]\rho = 2940 kg/m^3[/tex]
Explanation:
When pendulum is in air then the tension in the string is given as
[tex]T = mg[/tex]
[tex]T = 38.7 N[/tex]
now when it is submerged in water then tension is decreased due to buoyancy force of water
so we will have
[tex]T_{water} = mg - F_b[/tex]
[tex]32 = 38.7 - F_b[/tex]
[tex]F_b = 6.7 N[/tex]
When rock is immersed in unknown liquid then tension is given by
[tex]T = mg - F_b'[/tex]
[tex]19 = 38.7 - F_b'[/tex]
[tex]F_b' = 19.7 N[/tex]
from buoyancy force of water we can say
[tex]F_b = 1000(V)(9.8) = 6.7[/tex]
[tex]V = 6.84 \times 10^{-4} m^3[/tex]
now for other liquid we will have
[tex]19.7 = \rho(6.84 \times 10^{-4})(9.8)[/tex]
[tex]\rho = 2940 kg/m^3[/tex]
To find the density of the unknown liquid, we can use Archimedes' principle and the difference in tension when the rock is immersed in water and in the unknown liquid. The formula for buoyant force can be used to calculate the density of the unknown liquid.
Explanation:To find the density of the unknown liquid, we can use Archimedes' principle. Archimedes' principle states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
In this case, the difference in tension when the rock is in air and when it is totally immersed in water represents the buoyant force acting on the rock. We can use the formula:
Buoyant force = weight of the fluid displaced
We know that the tension in the string when the rock is in water is 32.0 N and when it is in the unknown liquid is 19.0 N. Subtracting the tension in water from the tension in the unknown liquid, we get:
Tension in unknown liquid - Tension in water = Buoyant force
19.0 N - 32.0 N = -13.0 N
The negative sign indicates that the buoyant force is acting upward on the rock. Since the buoyant force is equal to the weight of the fluid displaced, we can equate the buoyant force to the weight of the unknown liquid displaced by the rock:
-13.0 N = -(density of unknown liquid) x (gravitational acceleration) x (volume of rock)
To find the density of the unknown liquid, we can rearrange the equation as follows:
Density of unknown liquid = (-13.0 N) / (9.8 m/s^2) x (volume of rock)
Plugging in the values, the density of the unknown liquid can be calculated.
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A0.350 kg iron horseshoe that is initially at 600°C is dropped into a bucket containing 21.9 kg of water at 21.8°C. What is the final equilibrium temperature (in °C)? Neglect any heat transfer to or from the surroundings. Do not enter units.
Answer: [tex]22.8^0C[/tex]
Explanation:-
[tex]Q_{absorbed}=Q_{released}[/tex]
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]
where,
[tex]m_1[/tex] = mass of iron horseshoe = 0.35 kg = 350 g (1kg=1000g[/tex]
[tex]m_2[/tex] = mass of water = 21.9 kg = 21900 g
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_1[/tex] = temperature of iron horseshoe = [tex]600^oC[/tex]
[tex]T_2[/tex] = temperature of water = [tex]21.8^oC[/tex]
[tex]c_1[/tex] = specific heat of iron horseshoe = [tex]0.450J/g^0C[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]
[tex]350\times 0.450\times (T_{final}-600)^0C=-[21900g\times 4.184\times (T_{final}-21.8)][/tex]
[tex]T_{final}=22.8^0C[/tex]
Therefore, the final equilibrium temperature is [tex]22.8^0C[/tex].
How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius 120 m at a speed of 119 km/h 7 Express your answer using two significant figures
Answer:
Coefficient of static friction between the tires and the road is 0.92.
Explanation:
It is given that,
Radius of the curve, r = 120 m
Speed, v = 119 km/h = 33.05 m/s
We need to find the coefficient of static friction between the tires and the road. In a curved road the safe velocity is given by :
[tex]v=\sqrt{\mu rg}[/tex]
[tex]\mu[/tex] is the coefficient of static friction
g is acceleration due to gravity
[tex]\mu=\dfrac{v^2}{rg}[/tex]
[tex]\mu=\dfrac{(33.05\ m/s)^2}{120\ m\times 9.8\ m/s^2}[/tex]
[tex]\mu=0.92[/tex]
So, the coefficient of static friction between the tires and the road is 0.92. Hence, this is the required solution.
The angular position of a point on the rim of a rotating wheel is given by θ(t) = 4.0t - 3.0t2 + t3, where θ is in radians and t is in seconds. (a) What is θ(0)? What are the angular velocities at (b) t = 2.0 s and (c) t = 4.0 s? (d) What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s? What are the instantaneous angular accelerations at (e) the beginning and (f) the end of this time interval?
Answer:
(a) 0 rad
(b) 4 rad/s
(c) 28 rad/s
(d) 12 rad/s^2
(e) 6 rad/s^2
(f) 18 rad/s^2
Explanation:
[tex]\theta (t)=4t-3t^{2}+t^{3}[/tex] .... (1)
(a) here, we need to find angular displacement when t = 0 s
Put t = 0 in equation (1), we get
[tex]\theta (t=0)=0[/tex]
(b) Angular velocity is defined as the rate of change of angular displacement.
ω = dθ / dt
So, differentiate equation (1) with respect to t.
[tex]\omega =\frac{d\theta }{dt}=4-6t+3t^{2}[/tex] .... (2)
Angular velocity at t = 2 s
Put t = 2 s in equation (2), we get
ω = 4 - 6 x 2 + 3 x 4 = 4 rad/s
(c) Angular velocity at t = 4 s
Put t = 4 s in equation (2), we get
ω = 4 - 6 x 4 + 3 x 16 = 4 - 24 + 48 = 28 rad/s
(d) Average angular acceleration,
[tex]\alpha =\frac{\omega (t=4s)-\omega (t=2s)}{4-2}[/tex]
α = (28 - 4) / 2 = 12 rad/s^2
(e) The rate of change of angular velocity is called angular acceleration.
α = dω / dt
α = - 6 + 6 t
At t = 2 s
α = - 6 + 12 = 6 rad/s^2
(f) At t = 4 s
α = - 6 + 24 = 18 rad/s^2
A thin coil has 16 rectangular turns of wire. When a current of 3 A runs through the coil, there is a total flux of 4 × 10-3 T·m2 enclosed by one turn of the coil (note that , and you can calculate the proportionality constant ). Determine the inductance in henries.
Answer:
The inductance is 0.021 H.
Explanation:
Given that,
Number of turns = 16
Current = 3 A
Total flux [tex]\phi=4\times10^{-3}\ Tm^2[/tex]
We need to calculate the inductance
Using formula of total flux
[tex]N\phi = Li[/tex]
[tex]L=\dfrac{N\phi}{i}[/tex]
Where, i = current
N = number of turns
[tex]\phi[/tex] = flux
Put the value into the formula
[tex]L=\dfrac{16\times4\times10^{-3}}{3}[/tex]
[tex]L=0.021\ H[/tex]
Hence, The inductance is 0.021 H.
Find the x-value of all points where the function below has any relative extrema. Find the value(s) of any relative extrema. G(x)equalsx cubed minus 3 x squared minus 24 x plus 2
Answer:
[tex]x_{1}=-2,x_{2}=4[/tex]
value of g(x) at these points are as follows
g(-2)=30
g(4)=-78
Explanation:
Given
g(x)=[tex]x^{3}-3x^{2}-24x+2[/tex]
Differentiating with respect to x we get
[tex]g'(x)=3x^{2}-6x-24[/tex]
to obtain point of extrema we equate g'(x) to zero
[tex]g'(x)=3x^{2}-6x-24\\\\\therefore 3x^{2}-6x-24=0\\\\\Rightarrow x^{2}-2x-8=0\\\\x^{2}-4x+2x-8=0\\x(x-4)+2(x-4)=0\\(x+2)(x-4)=0[/tex]
Thus the critical points are obtained as [tex]x_{1}=-2,x_{2}=4[/tex]
The values at these points are as
[tex]g(-2)=(-2)^{3}-3(-2)^{2}-24(-2)+2=30\\\\g(4)=(4)^{3}-3(4)^{2}-24(4)+2=-78[/tex]
When an electron enters a magnetic field, it will accelerate up the field. True OR False
Answer:
True
Explanation:
The force on the electron when it enters in a magnetic field is given by
F = q ( v x B)
F = -e x V x B x Sin∅
here, F is the force vector, B be the magnetic field vector and v be the velocity vector.
If the angle between the velocity vector and the magnetic field vector is 0 degree, then force is zero.
When the electrons enters in the magnetic field at any arbitrary angle, it experiences a force and hence it accelerate up.
Two blocks of masses mA and mB are connected by a massless spring. The blocks are moved apart, stretching the spring, and subsequently released from rest. Find (a) the ratio of velocities 7of the blocks at any point of their ensuing motion (when their velocities are non-zero) and (b) the ratio of the kinetic energies of the blocks.
Answer:
Part a)
[tex]\frac{v_A}{v_B} = -\frac{m_B}{m_A}[/tex]
Part b)
[tex]\frac{K_A}{K_B} = \frac{m_B}{m_A}[/tex]
Explanation:
Part a)
As we know that initially the two blocks are connected by a spring and initially stretched by some amount
Since the two blocks are at rest initially so its initial momentum is zero
since there is no external force on this system so final momentum is also zero
[tex]m_Av_{1i} + m_Bv_{2i} = m_Av_A + m_Bv_B[/tex]
now for initial position the speed is zero
[tex]0 = m_Av_A + m_Bv_B[/tex]
now we have
[tex]\frac{v_A}{v_B} = -\frac{m_B}{m_A}[/tex]
Part b)
now for ratio of kinetic energy we know that the relation between kinetic energy and momentum is given as
[tex]K = \frac{P^2}{2m}[/tex]
now for the ratio of energy we have
[tex]\frac{K_A}{K_B} = \frac{P^2/2m_A}{P^2/2m_B}[/tex]
since we know that momentum of two blocks are equal in magnitude so we have
now we have
[tex]\frac{K_A}{K_B} = \frac{m_B}{m_A}[/tex]
Final answer:
To calculate the ratios of velocities and kinetic energies for two blocks connected by a massless spring released from rest, use conservation of momentum and the kinetic energy formula. The velocity ratio is mB/mA and the kinetic energy ratio is (mB/mA)².
Explanation:
The question is about two blocks connected by a massless spring on a frictionless surface. When they are released from rest after stretching the spring, we want to find the ratio of their velocities and kinetic energies during their motion.
Ratio of Velocities
The ratio of velocities is obtained through conservation of momentum. Since no external forces are involved and the spring does not have mass, the momentum of the system is conserved. Assuming block A and block B move in opposite directions after being released, their velocities will be in different directions but they will have the same magnitude of momentum.
Momentum conservation dictates that mAvA = mBvB, where vA and vB are their speeds respectively. Thus, the ratio of their velocities vA/vB = mB/mA.
Ratio of Kinetic Energies
The kinetic energy of each block is given by KE = (1/2)[tex]mv^2[/tex]. The ratio of their kinetic energies KEA/KEB is therefore [tex](m_Av_A^2)/(m_Bv_B2^)[/tex]. Inserting the ratios found in the velocities section, we find that the ratio of the kinetic energies is [tex](mB/mA)^2[/tex].
A cylindrical resistor has a length of 2 m and a diameter of 0.1 m. If I hook up a 12 V battery to the resistor and notice that the current flowing through the resistor is 3.2 A, what is the resistivity of the resistor? A. 4.12 x 103 ? m B. 1.15 x 103 ? m C. 6.10 x 102 ? m D. 1.47 x 102 m
Answer:
D . 1.47 x 10⁻² Ω-m
Explanation:
L = length of the cylindrical resistor = 2 m
d = diameter = 0.1 m
A = Area of cross-section of the resistor = (0.25) [tex]\pi[/tex] d² = (0.25) (3.14) (0.1)² = 0.785 x 10⁻² m²
V = battery Voltage = 12 volts
[tex]i [/tex] = current flowing through the resistor = 3.2 A
R = resistance of the resistor
Resistance of the resistor is given as
[tex]R = \frac{V}{i}[/tex]
[tex]R = \frac{12}{3.2}[/tex]
R = 3.75 Ω
[tex]\rho[/tex] = resistivity
Resistance is also given as
[tex]R = \frac{ \rho L}{A}[/tex]
[tex]3.75 = \frac{ \rho (2)}{(0.785\times 10^{-2})}[/tex]
[tex]\rho[/tex] = 1.47 x 10⁻² Ω-m
Calculate your weight on the International Space Station (ISS), which orbits at roughly 400 km about the surface of Earth. What is "g at the ISS?
Answer:
517.6 N
8.63 m/s²
Explanation:
M = mass of earth = 5.98 x 10²⁴ kg
m = mass of the person = 60 kg
W = weight of the person
R = radius of earth = 6.4 x 10⁶ m
d = distance of ISS above the surface of earth = 400 km = 4 x 10⁵ m
Weight of person on the ISS is given as
[tex]W = \frac{GMm}{(R+d)^{2}}[/tex]
[tex]W = \frac{(6.67\times 10^{-11})(5.98\times 10^{24})(60)}{((6.4\times 10^{6})+(4\times 10^{5}))^{2}}[/tex]
W = 517.6 N
Acceleration due to gravity is given as
[tex]g = \frac{GM}{(R+d)^{2}}[/tex]
[tex]g = \frac{(6.67\times 10^{-11})(5.98\times 10^{24})}{((6.4\times 10^{6})+(4\times 10^{5}))^{2}}[/tex]
g = 8.63 m/s²
A Styrofoam box has a surface area of 0.73 m and a wall thickness of 2.0 cm. The temperature of the inner surface is 5.0°C, and the outside temperature is 25°C. If it takes 8.4 h for 5.0 kg of ice to melt in the container, determine the thermal conductivity of the Styrofoam W/m-K
Answer:
[tex]K = 0.076 W/m K[/tex]
Explanation:
Heat required to melt the complete ice is given as
[tex]Q = mL[/tex]
here we have
m = 5.0 kg
[tex]L = 3.35 \times 10^5 J/kg[/tex]
now we have
[tex]Q = (5 kg)(3.35 \times 10^5)[/tex]
[tex]Q = 1.67 \times 10^6 J[/tex]
now the power required to melt ice in 8.4 hours is
[tex]P = \frac{Q}{t} = \frac{1.67\times 10^6}{8.4 \times 3600 s}[/tex]
[tex]P = 55.4 Watt[/tex]
now by formula of conduction we know
[tex]P = \frac{KA(\Delta T)}{x}[/tex]
now we have
[tex]55.4 = \frac{K(0.73 m^2)(25 - 5)}{0.02}[/tex]
[tex]K = 0.076 W/m K[/tex]
A curve of radius 28 m is banked so that a 990 kg car traveling at 41.1 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. You are commissioned to tell the local police the range of speeds at which a car can travel around this curve without skidding. Neglect the effects of air drag and rolling friction. If the coefficient of static friction between the road and the tires is 0.300, what is the range of speeds you tell them?
Answer:
For no friction condition there is no range of speed only on possible speed is 41.1 km/h
while for the case of 0.300 friction coefficient the range of speed is from 6.5 m/s to 15.75 m/s
Explanation:
When there is no friction on the turn of road then the centripetal force is due to the component of Normal force only
So we will have
[tex]Ncos\theta = mg[/tex]
[tex]Nsin\theta = \frac{mv^2}{R}[/tex]
so we have
[tex]tan\theta = \frac{v^2}{Rg}[/tex]
[tex]\theta = tan^{-1}\frac{v^2}{Rg}[/tex]
v = 41.1 km/h = 11.42 m/s
[tex]\theta = tan^{-1}\frac{11.42^2}{28(9.8)}[/tex]
[tex]\theta = 25.42^o[/tex]
so here in this case there is no possibility of range of speed and only one safe speed is possible to take turn
Now in next case if the coefficient of static friction is 0.300
then in this case we have
[tex]v_{max} = \sqrt{(\frac{\mu + tan\theta}{1 - \mu tan\theta})Rg}[/tex]
[tex]v_{max} = \sqrt{(\frac{0.3 + 0.475}{1 - (0.3)(0.475)})(28 \times 9.8)}[/tex]
[tex]v_{max} = 15.75 m/s[/tex]
Similarly for minimum speed we have
[tex]v_{min} = \sqrt{(\frac{\mu - tan\theta}{1 + \mu tan\theta})Rg}[/tex]
[tex]v_{min} = \sqrt{(\frac{-0.3 + 0.475}{1 + (0.3)(0.475)})(28 \times 9.8)}[/tex]
[tex]v_{min} = 6.5 m/s[/tex]
So the range of the speed is from 6.5 m/s to 15.75 m/s
A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upward. What is the change in momentum of the ball?
The change in momentum of the ball is 24 kg·m/s.
Explanation:The change in momentum of the ball can be found by subtracting the initial momentum from the final momentum. The initial momentum is given by the formula p1 = m * v1, where m is the mass of the ball and v1 is its initial velocity. In this case, the initial velocity is 12 m/s downward, so the initial momentum is -12 kg·m/s. The final momentum is given by the formula p2 = m * v2, where v2 is the velocity of the ball after the bounce. Since the motion after the bounce is the mirror image of the motion before the bounce, the final velocity is 12 m/s upward. Thus, the final momentum is 12 kg·m/s. Subtracting the initial momentum from the final momentum, we get the change in momentum to be 24 kg·m/s.
Learn more about momentum here:https://brainly.com/question/30677308
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The change in momentum of a 1.0-kg ball that bounces with a velocity of 12 m/s upward after striking the ground with a velocity of 12 m/s downward is 24 kg·m/s. This is calculated using the formula for momentum, p = mv, and finding the difference between initial and final momentum.
Momentum (old{p}old{) is calculated by the formula:
p = mv
Step-by-Step Explanation:
The initial momentum just before striking the ground (p₁) is:
p₁ = mass × velocity
p₁ = 1.0 kg × (-12 m/s)
p₁ = -12 kg·m/s
The final momentum just after bouncing up (p₂) is:
p₂ = mass × velocity
p₂ = 1.0 kg × 12 m/s
p₂ = 12 kg·m/s
The change in momentum (Δp) is:
Δp = p₂ - p₁
Δp = 12 kg·m/s - (-12 kg·m/s)
Δp = 12 kg·m/s + 12 kg·m/s
Δp = 24 kg·m/s
So, the change in momentum of the ball is 24 kg·m/s.
Two very small spheres are initially neutral and separated by a distance of 0.66 m. Suppose that 5.7 × 1013 electrons are removed from one sphere and placed on the other. (a) What is the magnitude of the electrostatic force that acts on each sphere? (b) Is the force attractive or repulsive?
Answer:
1.718 N , attractive
Explanation:
r = 0.66 m, n = 5.7 x 10^13
q1 = 5.7 x 10^13 x 1.6 x 10^-19 = 9.12 x 10^-6 C
q2 = - 5.7 x 10^13 x 1.6 x 10^-19 = - 9.12 x 10^-6 C
F = K q1 q2 / r^2
F = 9 x 10^9 x 9.12 x 10^-6 x 9.12 x 10^-6 / (0.66)^2
F = 1.718 N
As both the charges are opposite in nature, so the force between them is attractive.