Answer:
[tex]Q = 1.45 \times 10^{-7} C[/tex]
Explanation:
Here flux passing through each surface is given
so total flux through whole cube is given by
[tex]\phi = 2060 + 1590 + 1690 + 3430 + 1870 + 5760[/tex]
[tex]\phi = 16400 Nm^2 C[/tex]
now we also know that total flux through a closed surface depends on the total charge enclosed in the surface
So we will have
[tex]\frac{Q}{\epsilon_0} = 16400[/tex]
[tex]Q = (8.85 \times 10^{-12})(16400)[/tex]
[tex]Q = 1.45 \times 10^{-7} C[/tex]
The total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].
The electric flux [tex](\( \Phi \))[/tex] through a closed surface is given by Gauss's Law:
[tex]\[ \Phi = \frac{Q}{\varepsilon_0} \][/tex]
where:
- [tex]\( \Phi \)[/tex] is the electric flux,
- [tex]\( Q \)[/tex] is the total charge enclosed by the closed surface,
- [tex]\( \varepsilon_0 \)[/tex] is the permittivity of free space [tex](\( \varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \))[/tex].
For a closed box, the total electric flux [tex](\( \Phi_{\text{total}} \))[/tex] is the sum of the electric flux through each of its six surfaces. Let's denote the electric flux through each surface as [tex]\( \Phi_i \) where \( i = 1, 2, \ldots, 6 \)[/tex].
[tex]\[ \Phi_{\text{total}} = \sum_{i=1}^{6} \Phi_i \][/tex]
Now, we can set up an equation using the given values:
[tex]\[ \Phi_{\text{total}} = 12060 \, \text{Nm}^2/\text{C} + 1590 \, \text{Nm}^2/\text{C} + 1690 \, \text{Nm}^2/\text{C} + 3430 \, \text{Nm}^2/\text{C} + 1870 \, \text{Nm}^2/\text{C} + 5760 \, \text{Nm}^2/\text{C} \][/tex]
[tex]\[ \Phi_{\text{total}} = 28800 \, \text{Nm}^2/\text{C} \][/tex]
Now, use Gauss's Law to find the total charge [tex](\( Q \))[/tex] enclosed by the closed surface:
[tex]\[ Q = \Phi_{\text{total}} \cdot \varepsilon_0 \][/tex]
[tex]\[ Q = 28800 \, \text{Nm}^2/\text{C} \cdot 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \][/tex]
[tex]\[ Q \approx 2.544 \times 10^{-4} \, \text{C} \][/tex]
So, the total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].
For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency (1.2×1010 Hz ) and wavelength (0.026 m ). Recall that the Avogadro constant is 6.022×1023 mol−1.
Answer:
4.79 J
Explanation:
The energy of a single photon is given by
[tex]E=hf[/tex]
where
[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck constant
f is the frequency of the photon
Here we have
[tex]f=1.2\cdot 10^{10} Hz[/tex]
so the energy of one photon is
[tex]E_1=(6.63\cdot 10^{-34})(1.2\cdot 10^{10})=7.96\cdot 10^{-24} J[/tex]
Here we have 1 mol of photons, which contains
[tex]N=6.022\cdot 10^{23}[/tex] photons (Avogadro number). So, the total energy of this mole of photons is:
[tex]E=NE_1 = (6.022\cdot 10^{23})(7.96\cdot 10^{-24})=4.79 J[/tex]
An energy efficient light bulb uses 15 W of power for an equivalent light output of a 60 W incandescent light bulb. How much energy is saved each month by using the energy efficient light bulb instead of the incandescent light bulb for 4 hours a day? Assume that there are 30 days in one month.
Answer:
5400 W-hr
Explanation:
Given:
Power used by energy efficient bulb = 15 W
Power used by incandescent bulb = 60 W
Total time of bulb used = 4 hours/day × 30 days = 120 hours
Now,
Energy used by the individual bulb by using them for 120 hours
we know,
Energy = Power × Time
thus,
Energy consumed by energy efficient bulb = 15 W × 120 hour = 1800 W-hr
Energy consumed by incandescent bulb = 60 W × 120 hour = 7200 W-hr
hence, the energy saved will be = 7200 - 1800 = 5400 W-hr
Final answer:
Calculating the monthly energy savings involves finding the difference in power consumption between a 60 W incandescent bulb and a 15 W energy efficient bulb, multiplying by the daily usage, and the number of days in a month. The savings are 45 W per hour, equating to 180 Wh/day and 5.4 kWh/month.
Explanation:
The question asks how much energy is saved each month by using an energy efficient light bulb instead of an incandescent light bulb for 4 hours a day. To calculate the energy saved, we need to determine the difference in power consumption between the two types of bulbs, multiply that difference by the number of hours used per day, and then by the number of days in a month.
Firstly, we find the power saved per hour:
60 W (incandescent) - 15 W (efficient) = 45 W saved per hour
Next, we calculate the daily savings:
45 W x 4 hours/day = 180 Wh/day
Finally, we calculate the monthly savings:
180 Wh/day x 30 days/month = 5400 Wh/month
Since 1 kWh = 1000 Wh, the saving is 5.4 kWh/month
This is the amount of energy saved by using the energy efficient bulb instead of the incandescent bulb for 4 hours each day over a month.
A curve ball is a type of pitch in which the baseball spins on its axis as it heads for home plate. If a curve ball is thrown at 34.5 m/s (77 mph ) with a spin rate of 26 rev/s , how many revolutions does it complete before reaching home plate? Assume that home plate is 18.3 m (60 ft) from the pitching mound and that the baseball travels at a constant velocity.
Answer:
13.79 revolutions
Explanation:
Velocity of ball = v = 34.5 m/s
Spin rate of ball = 26 revolutions/s
Distance between home plate and pitching mound = s = 18.3 m
Time taken by the ball to reach the home plate = t
[tex]t=\frac{18.3}{34.5}\ s[/tex]
Number of revolutions the ball completes is
[tex]26\times \frac{18.3}{34.5}=13.79[/tex]
∴Number of revolutions the ball completes before reaching home plate is 13.79
Answer:
The number of revolutions is 13.78
Explanation:
Given data:
v = speed = 34.5 m/s
d = distance = 18.3 m
spin rate = 26 rev/s
The time taken is equal to:
[tex]t=\frac{d}{v} =\frac{18.3}{34.5} =0.53s[/tex]
The number of revolutions is equal:
N = 0.53 * 26 = 13.78 rev
Upper A 16-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 feet/second, how fast is the top of the ladder moving down when the foot of the ladder is 5 feet from the wall?
Answer:
The ladder is moving at the rate of 0.65 ft/s
Explanation:
A 16-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 feet/second. We need to find the rate at which the top of the ladder moving down when the foot of the ladder is 5 feet from the wall.
The attached figure shows whole description such that,
[tex]x^2+y^2=256[/tex].........(1)
[tex]\dfrac{dx}{dt}=2\ ft/s[/tex]
We need to find, [tex]\dfrac{dy}{dt}[/tex] at x = 5 ft
Differentiating equation (1) wrt t as :
[tex]2x.\dfrac{dx}{dt}+2y.\dfrac{dy}{dt}=0[/tex]
[tex]2x+y\dfrac{dy}{dt}=0[/tex]
[tex]\dfrac{dy}{dt}=-\dfrac{2x}{y}[/tex]
Since, [tex]y=\sqrt{256-x^2}[/tex]
[tex]\dfrac{dy}{dt}=-\dfrac{2x}{\sqrt{256-x^2}}[/tex]
At x = 5 ft,
[tex]\dfrac{dy}{dt}=-\dfrac{2\times 5}{\sqrt{256-5^2}}[/tex]
[tex]\dfrac{dy}{dt}=0.65[/tex]
So, the ladder is moving down at the rate of 0.65 ft/s. Hence, this is the required solution.
This is a calculation of a related rates problem in calculus, using the Pythagorean Theorem. The calculation finds the rate of descent (db/dt) for the top of a ladder sliding down a wall. The final answer found is -1.5 feet/second, meaning the top of the ladder is moving downward at a rate of 1.5 feet per second when the ladder is 5 feet away from the building.
Explanation:The question involves a concept in calculus known as related rates problem. It's a practical extension of the Pythagorean theorem where the sides of the right triangle formed by the ladder, building, and the ground are changing over time. In this case, a 16-feet ladder is sliding down a wall and we are to find the rate at which the top of the ladder is moving down when the foot of the ladder is 5 feet from the wall. First, we label the bottom (horizontal) side of our triangle 'a', the side along the wall (vertical) 'b', and the hypotenuse (ladder) 'c'. The Pythagorean theorem gives us a² + b² = c². Differentiating both sides with respect to time (t) gives us 2a(da/dt) + 2b(db/dt) = 2c(dc/dt). Here, we know that da/dt = 2 feet/sec (rate at which the ladder is moving away from the wall), c = 16 feet (length of the ladder), and a (distance from the wall) = 5 feet. Also, the ladder's length is not changing, so dc/dt = 0. Substituting these values, we can solve for db/dt (rate of descent of the ladder). The resultant rate of descent of the top of the ladder is -1.5 feet/second.
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A 13.7 N force is applied to a cord wrapped around a disk of radius 0.43 m. The disk accelerates uniformly from rest to an angular speed of 30.3 rad/s in 3.43 s. Determine the angular acceleration of the disk. Hint: the average acceleration is the change in angular speed over time.
Answer:
8.83 rad/s²
Explanation:
w₀ = initial angular speed of the disk = 0 rad/s
w = final angular speed of the disk = 30.3 rad/s
t = time period of rotation of the disk = 3.43 sec
[tex]\alpha[/tex] = Angular acceleration of the disk
Angular acceleration of the disk is given as
[tex]\alpha = \frac{w-w_{o}}{t}[/tex]
inserting the values
[tex]\alpha[/tex] = (30.3 - 0)/3.43
[tex]\alpha[/tex] = 8.83 rad/s²
A molecule of roughly spherical shape has a mass of 6.10 × 10-25 kg and a diameter of 0.70 nm. The uncertainty in the measured position of the molecule is equal to the molecular diameter. What is the minimum uncertainty in the speed of this molecule? (h = 6.626 × 10-34 J · s)
To calculate the minimum uncertainty in the speed of the molecule, we can use the Heisenberg uncertainty principle. The uncertainty in position is equal to the molecular diameter, and the uncertainty in momentum can be calculated using the uncertainty principle equation. Finally, the uncertainty in velocity can be found by dividing the uncertainty in momentum by the mass of the molecule.
Explanation:To calculate the minimum uncertainty in the speed of the molecule, we can use the Heisenberg uncertainty principle. The uncertainty in position (Ax) is equal to the molecular diameter, which is given as 0.70 nm. The uncertainty in momentum (Ap) can be calculated using the equation AxAp ≥ h/4. Once the uncertainty in momentum is found, the uncertainty in velocity can be found from Ap = mΔv, where m is the mass of the molecule.
Using the given values, the uncertainty in position (Ax) is 0.70 nm. Plugging this into the uncertainty principle equation and solving for Ap, we get Ap ≥ h/4Ax. Substituting the values for h and Ax, we have Ap ≥ (6.626 × 10-34 J·s)/(4 × 0.70 × 10-9 m).
Next, we can find the uncertainty in velocity (Δv) by using Ap = mΔv. Rearranging the equation, we have Δv = Ap/m. Plugging in the values for Ap (obtained from the previous calculation) and the mass of the molecule, we can calculate the minimum uncertainty in the speed of the molecule.
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Which of the following is the kinetic rate equation for the addition-elimination mechanism of nucleophilic aromatic substitution? rate = k[aryl halide] rate = k[nucleophile] rate = k[aryl halide][nucleophile] rate = k[aryl halide][nucleophile]2
Answer:
Rate = k[aryl halide][nucleophile]
Explanation:
The simple aryl halides are almost inert to usual nucleophilic reagents but considerable activation on the ring can be produced by the addition of strongly electron-attracting substituents on either the ortho or para positions, or both. These groups deactivate the ring to allow the attack of the nucleophille on the ring.
Thus, these reactions can occur by following addition-elimination mechanism in which the nucleophille first attacks the aryl halide and then the elimination of the leaving group takes place.
Kinetic studies of this type of mechanism demonstrate that the reactions are of second-order kinetics– first order w.r.t. nucleophile and also, first-order w.r.t. aromatic substrate. The rate determining step (r.d.s.) is the formation of the addition intermediate.
Thus,
Rate = k[aryl halide][nucleophile]
A long wire is known to have a radius greater than 10.0 mm and less 20 mm, carry a current uniformly distributed over its cross section. If the magnitude of the magnetic field is 3 mT at a point 6.0 mm from the axis of the wire and 1.50 mT at a point 20 mm from the axis, what is the radius of the wire?
Answer:
Radius = 7.75 mm
Explanation:
magnetic field inside a point in the wire is given as
[tex]B = \frac{\mu_0 i r}{2\pi R^2}[/tex]
now we have
[tex]3mT = \frac{\mu_0 i (6mm)}{2\pi R^2}[/tex]
now outside wire magnetic field is given as
[tex]B = \frac{\mu_0 i}{2\pi r}[/tex]
now we have
[tex]1.50 mT = \frac{\mu_0 i}{2\pi(20 mm)}[/tex]
now divide above two equations
[tex]2 = \frac{6 mm\times 20 mm}{R^2}[/tex]
[tex]R = 7.75 mm[/tex]
What is the effective resistance of a car’s starter motor when 150 A flows through it as the car battery applies 12.0 V to the motor?
Answer:
From ohms law,
V=IR
R=V/I =12.0/150 =0.08 ohm.
A plank 2.65 m long is supported by a cable, and a small ledge. The cable attaches 79.7 cm from the ledge, and makes an angle of 40.3 with the plank. It can hold 2648 N. How close (in meters) to the end of the plank can a 76.4 kg person walk before the cable breaks? Ignore the mass of the plank itself.
Answer:
0.51 m
Explanation:
Given :
Plank length = 2.65 m
Distance from the ledge from which the cable is attached = 79.7 cm = 0.797 m
Angle = 40.3 degree
Mass of the man, m = 76.4 kg
Now the net torque acting on the plank is zero.
Therefore, maximum torque applied by cable is
= 2648 x 0.797 x cos (40.3)
= 1609.5 N-m
We know that the man should apply the same torque,
Therefore, x [tex]\times[/tex]76.4 [tex]\times[/tex] g = 1609.5
x =2.14 m
Therefore, length to the end of the plank is = 2.65-2.14
=0.51 m
The electric field everywhere on the surface of a thin, spherical shell of radius 0.750 m is of magnitude 890 N/C and points radially toward the center of the sphere. What is the net charge in nano coulomb within the sphere’s surface?
Answer:
55.6 nC
Explanation:
The electric field at the surface of a charged sphere has the same expression of the electric field produced by a single point charge located at the centre of the sphere and having the same charge of the sphere, so it is given by
[tex]E=k\frac{Q}{r^2}[/tex]
where
[tex]k=9\cdot 10^9 N m^2 C^{-2}[/tex] is the Coulomb's constant
Q is the charge on the sphere
r is the radius of the sphere
In this problem we know
E = 890 N/C is the magnitude of the electric field on the sphere
r = 0.750 m is the radius of the sphere
So by re-arranging the equation we can find the net charge on the sphere:
[tex]Q=\frac{Er^2}{k}=\frac{(890)(0.750)^2}{9\cdot 10^9}=5.56\cdot 10^{-8} C=55.6 nC[/tex]
To find the net charge within the surface of the sphere, we can use Gauss's Law. The net electric flux through a closed surface enclosing a charge is equal to the charge divided by the electric constant, ε₀. Given the electric field and the radius of the sphere, we can calculate the net charge using the equation derived from Gauss's Law.
Explanation:The electric field everywhere on the surface of a thin, spherical shell is radially inward and has a magnitude of 890 N/C. To find the net charge within the surface of the sphere, we can use Gauss's Law. The net electric flux through a closed surface enclosing a charge is equal to the charge divided by the electric constant, ε₀. In this case, the closed surface is the spherical shell and the electric field is constant on the surface, so the net electric flux is equal to the electric field multiplied by the surface area of the sphere. The surface area of the sphere is given by 4πr², where r is the radius of the sphere. Setting the net electric flux equal to the charge divided by ε₀, we can solve for the charge.
Given that the electric field is 890 N/C, the radius of the sphere is 0.750 m, and the electric constant, ε₀, is 8.99 x 10⁹ Nm²/C², we can plug in these values to solve for the charge.
Let Q be the net charge within the sphere's surface:
Net electric flux = Electric field * Surface area
Q / ε₀ = Electric field * Surface area
Q / (8.99 x 10⁹ Nm²/C²) = (890 N/C) * (4π(0.750 m)²)
Q = (890 N/C) * (4π(0.750 m)²) * (8.99 x 10⁹ Nm²/C²)
Solving this equation will give us the net charge within the sphere's surface.
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A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal ( s = 0.900 J/g·°C), find the final temperature of the block and the water
Answer:
34.17°C
Explanation:
Given:
mass of metal block = 125 g
initial temperature [tex]T_i[/tex] = 93.2°C
We know
[tex]Q = m c \Delta T[/tex] ..................(1)
Q= Quantity of heat
m = mass of the substance
c = specific heat capacity
c = 4.19 for H₂O in [tex]J/g^{\circ}C[/tex]
[tex] \Delta T[/tex] = change in temperature
Now
The heat lost by metal = The heat gained by the metal
Heat lost by metal = [tex]125\times 0.9\times (93.2-T_f)[/tex]
Heat gained by the water = [tex]100\times 4.184\times(T_f -18.3)[/tex]
thus, we have
[tex]125\times 0.9\times (93.2-T_f)[/tex] = [tex]100\times 4.184\times(T_f -18.3)[/tex]
[tex]10485-112.5T_f = 418.4T_f - 7656.72[/tex]
⇒ [tex]T_f = 34.17^oC[/tex]
Therefore, the final temperature will be = 34.17°C
Which of the following statements are true about Electric Force, Electric Field, Electric Potential and Electric Potential Energy?
(i) Electric Force, Electric Field, and Electric Potential are vectors; Electric Potential Energy is a scalar value.
(ii) The Electric Field reports the Electric Force on a 1C test charge placed at a particular location.
(iii) The Electric Potential reports the Electric Potential energy of a 1C test charge place at a particular location.
(iv) The spatial derivative of the electric potential at a particular location reports the electric field at that location.
Answer:
(ii) The Electric Field reports the Electric Force on a 1C test charge placed at a particular location.
(iii) The Electric Potential reports the Electric Potential energy of a 1C test charge place at a particular location.
(iv) The spatial derivative of the electric potential at a particular location reports the electric field at that location.
ii, iii and iv are true.
Explanation:
Electric force per unit charge is the electric field. It is a vector. E = F/q.
Electric potential is the electric potential energy per unit charge and is a scalar quantity, like the electric potential. V = U/q where U is the electric potential energy and q the charge.
Electric field exists due to a difference of potential, between two different points.E =dV/dx .
Option (ii),(iii) & (iv) is correct.
The question posed deals with the concepts of Electric Force, Electric Field, Electric Potential, and Electric Potential Energy in the context of physics. Let's analyze each statement:
Electric Force, Electric Field, and Electric Potential are vectors; Electric Potential Energy is a scalar value. This statement is partially incorrect. Electric Force and Electric Field are indeed vectors because they have both magnitude and direction. However, Electric Potential is a scalar quantity, not a vector, because it only has magnitude without direction. Electric Potential Energy is correctly identified as a scalar value.
The Electric Field reports the Electric Force on a 1C test charge placed at a particular location. This statement is true. The Electric Field is defined as the Electric Force per unit charge, and it indeed represents the force that a 1 coulomb test charge would experience at that point.
The Electric Potential reports the Electric Potential energy of a 1C test charge placed at a particular location. This statement is true. Electric Potential is the amount of Electric Potential Energy per unit charge that a test charge would have in a particular location in the electric field.
The spatial derivative of the electric potential at a particular location reports the electric field at that location. This statement is true. According to electrostatics, the electric field at a point is the negative gradient (spatial derivative) of the electric potential at that point.
It is essential to understand that electric potential is different from an electric field: the former is a scalar that represents potential energy per unit charge, while the latter is a vector representing the force per unit charge. Calculating the scalar electric potential is generally simpler than calculating the vector electric field.
A 310-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,190 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable
Answer:
t = 141.55 years
Explanation:
As we know that the radius of the wire is
r = 2.00 cm
so crossectional area of the wire is given as
[tex]A = \pi r^2[/tex]
[tex]A = \pi(0.02)^2[/tex]
[tex]A = 1.26 \times 10^{-3} m^2[/tex]
now we know the free charge density of wire as
[tex]n = 8.50 \times 10^{28}[/tex]
so drift speed of the charge in wire is given as
[tex]v_d = \frac{i}{neA}[/tex]
[tex]v_d = \frac{1190}{(8.50 \times 10^{28})(1.6 \times 10^{-19})(1.26\times 10^{-3})}[/tex]
[tex]v_d = 6.96 \times 10^{-5} m/s[/tex]
now the time taken to cover whole length of wire is given as
[tex]t = \frac{L}{v_d}[/tex]
[tex]t = \frac{310 \times 10^3}{6.96 \times 10^{-5}}[/tex]
[tex]t = 4.46 \times 10^9 s[/tex]
[tex]t = 141.55 years[/tex]
What is the length of an aluminum rod at 65°C if its length at 15°C is 1.2 meters? A. 0.00180 meter B. 1.201386 meters C. 1.214855 meters D. 0.001386 meter
Answer:
Option B is the correct answer.
Explanation:
Thermal expansion
[tex]\Delta L=L\alpha \Delta T[/tex]
L = 1.2 meter
ΔT = 65 - 15 = 50°C
Thermal Expansion Coefficient for aluminum, α = 24 x 10⁻⁶/°C
We have change in length
[tex]\Delta L=L\alpha \Delta T=1.2\times 24\times 10^{-6}\times 50=1.44\times 10^{-3}m[/tex]
New length = 1.2 + 1.44 x 10⁻³ = 1.2014 m
Option B is the correct answer.
Answer:
B. 1.201386 meters
Explanation:
The length of an aluminum rod at 65°C if its length at 15°C is 1.2 meters is 1.201386 meters.
A wire placed on the plane of this screen carries a current toward the top of the screen. The wire feels a magnetic force toward the right. The direction of the magnetic field causing this force is which of the following? (a) outward out of the screen (b) inward into the screen (c) in the plane of the screen and toward the left edge (d) in the plane of the screen and toward the bottom edge.
Answer:
option (a)
Explanation:
The direction of magnetic field is given by Fleming's left hand rule.
If we spread the fore finger, middle finger and the thumb of our right hand such that they are mutually perpendicular to each other, then the direction of force is in the direction of thumb, direction of magnetic field is given by the direction of fore finger and the middle finger indicates the direction of current .
Here, current is upwards, magnetic force is rightwards, so the direction of magnetic field is given in outwards from the screen.
A ball is released at a velocity of 400 m/s at an angle of 85 degrees from the horizontal. How far does it travel horizontally before the ball lands on the ground? (use a-10m/s2 if needed, round answer to a whole number-no decimals). Do not put the units in the answer. QUESTION 5 angle of 85 degrees from the horizontal. How long will it take for the ball to get if needed, round answer to a whole number-no decimals). Do not put the units in the answer. A ball is released at a velocity of 400 m/s at an to its apex? (use a-10m/s QUESTION 6 A ball is released at a velocity of 400 m/s at an angle of 45 degrees from the horizontal. What is the vertical component of the velocity? (use a--10m/s if needed, round answer to a whole number-no decimals), Do not put the units in the answer.
Answer:
8000
Explanation:
V²=U²+2as
V=400m/s
U=0
a=10
s=?
(400)²=(0)²+2*10*a
160,000=20s
s=160,000/20
s=8000
A 40 g block of ice is cooled to -69°C and is then added to 590 g of water in an 80 g copper calorimeter at a temperature of 22°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g·°C = 2090 J/kg°C.
The final temperature of the system consisting of the ice, water, and calorimeter is approximately 6.8°C.
Here's how you can calculate it step by step:
1. Calculate the heat absorbed by the ice to reach 0°C:
[tex]\[ Q_1 = mc\Delta T \]\[ Q_1 = 40 g \times 2090 J/kg^\circ C \times (0^\circ C - (-69^\circ C)) \]\[ Q_1 = 40 g \times 2090 J/kg^\circ C \times 69^\circC \]\[ Q_1 = 571560 J \][/tex]
2. Calculate the heat absorbed by the ice to melt completely:
[tex]\[ Q_2 = mL_f \]\[ Q_2 = 40 g \times 334 J/g \]\[ Q_2 = 13360 J \][/tex]
3. Calculate the heat absorbed by the melted ice to reach the final temperature (assuming the final temperature is T):
[tex]\[ Q_3 = mc\Delta T \]\[ Q_3 = 40 g \times 4186 J/kg°C \times (T - 0°C) \]\[ Q_3 = 167440T J \][/tex]
4. Calculate the heat absorbed by the water and calorimeter to reach the final temperature (T):
[tex]\[ Q_4 = mc\Delta T \]\[ Q_4 = (590 g + 80 g) \times 4186 J/kg°C \times (T - 22°C) \]\[ Q_4 = 270104T - 114208 \][/tex]
Now, since heat gained = heat lost (assuming no heat loss to the surroundings):
[tex]\[ Q_1 + Q_2 + Q_3 = Q_4 \]\[ 571560 + 13360 + 167440T = 270104T - 114208 \]\[ 571560 + 13360 + 114208 = 270104T - 167440T \]\[ 699128 = 102664T \]\[ T \aprrox 6.8°C \][/tex]
The final temperature of the system is approximately 6.8°C.
Complete Question:
A 40 g block of ice is cooled to -69°C and is then added to 590 g of water in an 80 g copper calorimeter at a temperature of 22°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g·°C = 2090 J/kg°C.
The magnetic force which a particle feels is: A) inversely proportional to the velocity B) inversely proportional to the magnetic field C) directly proportional to the electric field D) inversely proportional to the charge E) directly proportional to the charge
Answer:
Magnetic force is directly proportional to the charge.
Explanation:
The magnetic force of a charged particle is given by :
[tex]F=qvB[/tex]..... (1)
Where
q = charged particle
v = velocity of particle
B = magnetic field
From equation (1) it is clear that the magnetic force is directly proportional to the charge. Hence, the correct option is (E).
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.503 X 104 T) at a distance of 15 cm from the wire, what is the maximum current the wire can carry?
Answer:
37.725 A
Explanation:
B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T
r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m
i = magnitude of current flowing through the wire = ?
Magnetic field by a long wire is given as
[tex]B = \frac{\mu _{o}}{4\pi }\frac{2i}{r}[/tex]
Inserting the values
[tex]0.503\times 10^{-4} = (10^{-7})\frac{2i}{0.15}[/tex]
i = 37.725 A
A pipe has a diameter of 20cm. What is the cross-sectional area of the pipe with units m^2.
Answer:
[tex]Area = 0.126 m^2[/tex]
Explanation:
Since the crossection of cylinder is of circular shape
so here the crossectional area will be given as
[tex]A = \pi r^2[/tex]
here we know that radius of the cylinder will be
[tex]R = 20 cm[/tex]
so in SI units it is given as
[tex]R = 0.20 m[/tex]
now we have
[tex]Area = \pi (0.20)^2[/tex]
[tex]Area = 0.126 m^2[/tex]
What is the velocity at discharge if the nozzle of a hose measures 68 psi? 100.25 ft./sec 10.25 ft./sec 125.2 ft./sec 11.93 ft./sec
Answer:
The velocity at discharge is 100.46 ft/s
Explanation:
Given that,
Pressure = 68 psi
We need to calculate the pressure in pascal
[tex]P=68\times6894.74\ Pa[/tex]
[tex]P=468842.32\ Pa[/tex]
We need to calculate the velocity
Let the velocity is v.
Using Bernoulli equation
[tex]P=\dfrac{1}{2}\rho v^2[/tex]
[tex]468842.32=0.5\times1000\times v^2[/tex]
[tex]v=\sqrt{\dfrac{468842.32}{0.5\times1000}}[/tex]
[tex]v=30.62\ m/s[/tex]
Now, We will convert m/s to ft/s
[tex]v =30.62\times3.281[/tex]
[tex]v=100.46\ ft/s[/tex]
Hence, The velocity at discharge is 100.46 ft/s
The speed of water discharged from a hose depends on the nozzle pressure and the constriction of the flow, but the specific speed cannot be determined from pressure alone without additional parameters.
Explanation:The question is asking about the velocity or speed achieved by water when it is forced out of a hose with a nozzle pressure of 68 psi. To understand this, we need to know that the pressure within the hose is directly correlated with the speed of the water's exit. This is due to the constriction of the water flow by the nozzle, causing speed to increase.
However, the specific velocity at discharge can't be straightforwardly calculated from pressure alone without knowing more details, such as the dimensions of the hose and nozzle, and the properties of the fluid. Therefore, based on the provided information, a specific answer in ft/sec can't be given.
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Find the speed of light in carbon tetrachlorideethyl alcohol. The refraction index is 1.461 using 3 x 10^8 m/s as the speed of light in vacuum. Answer in units of m/s.
Answer:
2.05 x 10^8 m /s
Explanation:
c = 3 x 10^8 m/s
μ = c / v
where, μ is the refractive index, c be the velocity of light in air and v be the velocity of light in the medium.
μ = 1.461
1.461 = 3 x 10^8 / v
v = 3 x 10^8 / 1.461
v = 2.05 x 10^8 m /s
A 2.50-kg textbook is forced against a horizontal spring of negligible mass and force constant 250 N/m, compressing the spring a distance of 0.250 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \mu_kμ k = 0.30. Use the work–energy theorem to find how far the text-book moves from its initial position before coming to rest.
Answer:
L = 1.06 m
Explanation:
As per work energy theorem we know that work done by all forces must be equal to change in kinetic energy
so here we will have
[tex]W_{spring} + W_{friction} = KE_f - KE_i[/tex]
now we know that
[tex]W_{spring} = \frac{1}{2}kx^2[/tex]
[tex]W_{friction} = -\mu mg L[/tex]
initial and final speed of the book is zero so initial and final kinetic energy will be zero
[tex]\frac{1}{2}kx^2 - \mu mg L= 0 - 0[/tex]
here we know that
k = 250 N/m
x = 0.250 m
m = 2.50 kg
now plug in all data in it
[tex]\frac{1}{2}(250)(0.250)^2 = 0.30(2.50)(9.81)L[/tex]
now we have
[tex]7.8125 = 7.3575L[/tex]
[tex]L = 1.06 m[/tex]
The distance that the textbook moves from its initial position before coming to rest is 1.0629m
According to the work-energy theorem, the work done on the spring is equal to the work done by friction
Work done by the spring = [tex]\frac{1}{2}kx^2[/tex]
Work done by friction = [tex]\mu mgL[/tex]
k is the spring constant = 250 N/m
x is the distance moved by the spring = 0.250m
[tex]\mu[/tex] is the coefficient of friction = 0.30
m is the mass of the textbook = 2.50kg
g is the acceleration due to gravity = 9.8m/s²
L is the distance from its initial position before coming to rest
Using the formula
[tex]\frac{1}{2} kx^2=\mu mgL[/tex]
Substitute the given parameters into the formula as shown;
[tex]\frac{1}{2} (250)(0.25)^2=0.3 (2.5)(9.8)L\\7.8125=7.35L\\L=\frac{7.8125}{7.35}\\L= 1.0629m[/tex]
Hence the distance that the textbook moves from its initial position before coming to rest is 1.0629m
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A 19 kg rock slides on a rough horizontal surface at 9.25 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.63. What average thermal power (in Watts) is produced as the rock stops?
Answer:
541.89 Watt
Explanation:
Consider the motion of rock on rough surface
μ = Coefficient of kinetic friction = 0.63
acceleration caused to kinetic friction is given as
a = - μg
a = - (0.63) (9.8)
a = - 6.2 m/s²
v₀ = initial velocity of the rock = 9.25 m/s
t = time taken to stop
v = final velocity = 0 m/s
Using the equation
v = v₀ + a t
0 = 9.25 + (- 6.2) t
t = 1.5 sec
m = mass of the rock = 19 kg
Energy lost due to friction is given as
E = (0.5) m (v₀² - v²)
E = (0.5) (19) (9.25² - 0²)
E = 812.84 J
Average thermal power is given as
[tex]P = \frac{E}{t}[/tex]
[tex]P = \frac{812.84}{1.5}[/tex]
P = 541.89 Watt
An object moves on a trajectory given by Bold r left parenthesis t right parenthesis equals left angle 10 cosine 6 t comma 10 sine 6 t right angle for 0 less than or equals t less than or equals pi. How far does it travel?
The object is moving along a circular path. Given the time restriction, it completes a half revolution, which equates to half the circumference of the circle. Therefore, the object travels approximately 31.4 units.
Explanation:The object is moving on a circular trajectory defined by r(t) = <10 cos 6t, 10 sin 6t>. This equation describes a circular path with a radius of 10 units. To find out how far the object travels, we can use the formula for the circumference of a circle, which is 2πr. However, since the time t varies from 0 to π, the object only makes a half revolution along the circle. So, the total distance the object travels would be equal to half the circumference of the circle, which is πr.
Substituting r = 10, we get, distance traveled = π*10 = 31.4 units approx.
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A talented PHY210 student is tuning her car. She increase the engine speed from 660 rpm very (69.1 rad/s) to 4500 rpm (471 rad/s). If the flywheel has a moment of inertia of 0.525 kg-m2 calculate the work done by the engine on the flywheel during this change in angular velocity. Select one: o a. 105 b, 1.43 x 104 J c. 1.44 x 103 O d. 5.70 x 104 J e. 1.14 x 105J
Answer:
d. 5.7 x 10⁴ J
Explanation:
I = moment of inertia of the flywheel = 0.525 kg-m²
w₀ = initial angular speed of the flywheel = 69.1 rad/s
w = final angular speed of the flywheel = 471 rad/s
W = work done by the engine on the flywheel
Work done by the engine is given as
W = (0.5) I (w² - w₀²)
Inserting the values
W = (0.5) (0.525) (471² - 69.1²)
W = 5.7 x 10⁴ J
he density of copper is 8.96g/cm^3 and the density of water is 1 g/cm^3. When a copper is submerged in a cylindrical beaker whose bottom has surface area 10 cm^2 the water level rises by 2 cm. The volume of the cylinder is the area of its base times its height. a) What is the specific gravity of copper?
b) What is the buoyant force on the copper object?
c) What is the buoyant force on the copper object?
d) What is the mass of the copper object?
Answer:
(a) 8.96
(b) 19600 dyne
(c) 19600 dyne
(d) 20 g
Explanation:
dcu = 8.96 g/cm^3, dw = 1 g/cm^3, A = 10 cm^2
Water level rises by 2 cm.
(a) The specific gravity of copper = density of copper / density of water
8.96 / 1 = 8.96
(b) According to the Archimedes's principle, the buoyant force acting on the body is equal to the weight of liquid displaced by the body.
Weight of water displaced by the copper = Area of beaker x rise in water level
x density of water x gravity
= 10 x 2 x 1 x 980 = 19600 dyne
(c) Same as part b
(d) Let mass of copper be m.
For the equilibrium condition,
the true weight of copper = Buoyant force acting on copper
m x g = 19600
m = 19600 / 980 = 20 g
Heated lithium atoms emit photons of light with an energy of 2.961 × 10−19 J. Calculate the frequency and wavelength of one of these photons. What is the total energy in 1 mole of these photons? What is the color of the emitted light?
Answer:
4.5 x 10¹⁴ Hz
666.7 nm
1.8 x 10⁵ J
The color of the emitted light is red
Explanation:
E = energy of photons of light = 2.961 x 10⁻¹⁹ J
f = frequency of the photon
Energy of photons is given as
E = h f
2.961 x 10⁻¹⁹ = (6.63 x 10⁻³⁴) f
f = 4.5 x 10¹⁴ Hz
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of photon
Using the equation
c = f λ
3 x 10⁸ = (4.5 x 10¹⁴) λ
λ = 0.6667 x 10⁻⁶ m
λ = 666.7 x 10⁻⁹ m
λ = 666.7 nm
n = number of photons in 1 mole = 6.023 x 10²³
U = energy of 1 mole of photons
Energy of 1 mole of photons is given as
U = n E
U = (6.023 x 10²³) (2.961 x 10⁻¹⁹)
U = 1.8 x 10⁵ J
The color of the emitted light is red
The frequency and wavelength of a photon with energy of 2.961 × 10-19 J are approximately 4.468 x 10¹4 s¯¹ and 656.3 nm, respectively. The energy of one mole of these photons is approximately 1.823 x 10⁵ J mol-¹. The color of the light emitted by these photons is red.
Explanation:The energy of a photon of light can be calculated into both a frequency and wavelength using the Planck-Einstein relation E = hf, and the speed of light relation c = λf, where h is Planck's constant, c is the speed of light, and λ is wavelength.
Let's first find the frequency. Given that the energy E is 2.961 × 10-19 J and the value of Planck's constant h is 6.63 × 10-³4 J·s (approximately), you can solve for the frequency f = E/h, which results in approximately 4.468 × 10¹4 s¯¹.
To find the wavelength, you can use the light speed equation c = fλ. Given the speed of light c is approximately 3.00 × 10⁸ m/s and using the frequency calculated above, solve for λ = c/f, which results in approximately 656.3 nm.
To find the energy of one mole of these photons, use Avogadro's number (6.022 x 10²³ photons/mole) and the provided energy of a single photon. The total energy is then calculated as E = (2.961 × 10-19 J/photon) x (6.022 x 10²³ /mole) = 1.823 × 10⁵ J mol-¹.
Lastly, the color of the photon can be inferred from its wavelength. As the wavelength is 656.3 nm, this falls within the range of the visible light spectrum for the color red.
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Suppose that you determine the density of a mineral by measuring its mass (m) (4.635±0.002) g and its volume (1.13±0.05) mL. d = m/V What is the uncertainty in the computed density?
Answer:
[tex]\Delta \rho =0.18/mL326gm[/tex]
Explanation:
we have error in division of 2 quantities is related as
[tex]z=\frac{a}{b}\\\\\frac{\Delta z}{z}=\frac{\Delta a}{a}+\frac{\Delta b}{b}[/tex]
where [tex]\Delta a,\Delta b[/tex] are errors in quantities a,b
Thus error in density becomes
[tex]\Delta \rho =\rho _{0}\times(\frac{\Delta m}{m}+\frac{\Delta V}{V})[/tex]
Applying values we get
[tex]\Delta \rho =\frac{4.635}{1.13} \times(\frac{0.002}{4.635}+\frac{0.05}{1.13})[/tex]
thus [tex]\Delta \rho =0.18326gm/mL[/tex]