To find the electric field in the wire, we first find the cross-sectional area of the wire and then substitute the given values and the calculated area into the formula for electric field. The electric field in the gold wire is approximately 8.24 kN/C.
Explanation:The question asks for the electric field in a gold wire of given dimensions and current. We'll solve this using the formula for electric field (E) in terms of resistivity (ρ),current (I) and area (A), which is given by E = ρI/A.
First, we need to find the cross-sectional area (A) of the wire using the formula for the area of a circle, since the wire is cylindrical. The area of a circle is given by A = π*(d/2)^2, where d is the diameter. Substituting d = 1.8 mm or 1.8 * 10^-3 m, we find A ≈ 2.54 * 10^-6 m^2.
Next, we need to substitute the known values into the formula E = ρI/A. Using the given values ρ = 2.44 *10^-8 Ω•m, I = 860 mA or 860 * 10^-3 A, and the calculated A ≈ 2.54 * 10^-6 m^2, we find E ≈ 8.24 * 10^3 N/C or 8.24 kN/C.
So, the electric field in the wire is approximately 8.24 kN/C.
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The electric field within the gold wire can be calculated utilizing resistivity formula, determining the cross-sectional area of the wire, and then computing the current density. After these intermediate calculations, the ultimate electric field in the wire is found to be approximately 72 mV/m.
Explanation:
The electric field in the gold wire can be calculated by using the formula for
resistivity: ρ=E/J where
ρ (rho) is the resistivity of the material,
E is the electric field, and
J is the current density.
Before we apply this formula, let's calculate the cross-sectional area of the wire (A) using the formula A=πr², where r is the radius of the wire. The diameter is given as 1.8mm, therefore
r=0.9mm = 0.9x10^-3 m.
Substituting this into the formula gives us
A=π(0.9x10^-3)²≈2.54x10^-6 m².
Now we calculate current density: J=I/A. where I is current, given as
860mA=860x10^-3 A.
Substituting values of I and A into the formula gives
J=860x10^-3/2.54x10^-6≈3.39x10^8 A/m².
Finally, the electric field E=ρ/J=2.44x10^-8/3.39x10^8 ≈ 7.2x10^-2 V/m or 72mV/m.
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A particle is going into a nozzle at 10 m/s and then after 3m the particle is slowing down to 2 m/s. The particle travels through a straight path down the middle of the nozzle. How do you find average acceleration using eularian and then Lagrangian view point.
Answer:
[tex]a = -16 m/s^2[/tex]
Explanation:
As we know that the initial speed of the particle when it enters the nozzle is given as
[tex]v_i = 10 m/s[/tex]
after travelling the distance d = 3 m it will have its final speed as
[tex]v_f = 2 m/s[/tex]
now we know that when acceleration is constant the equation of kinematics is applicable
so we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]2^2 - 10^2 = 2(a)(3)[/tex]
[tex]a = -16 m/s^2[/tex]
Final answer:
To find the average acceleration from a fluid dynamics perspective, one can use the Eulerian viewpoint to observe velocity changes at a fixed point, or follow the particle through its motion using the Lagrangian viewpoint. However, the provided information is insufficient for precise calculation without the time variable or assumption of constant deceleration.
Explanation:
To find the average acceleration of a particle that slows down from 10 m/s to 2 m/s over a distance of 3 meters, we can use the concepts of Eulerian and Lagrangian viewpoints in fluid dynamics, though it is more common to apply these in the context of fluid flow rather than individual particles in motion.
In the Eulerian viewpoint, we observe the change of velocity at a fixed point in space as the particle passes through. Writing the kinematic equation δv = aδt, where δv is the change in velocity and a is the average acceleration, we can rearrange to find a = δv / δt.
The Lagrangian viewpoint, on the other hand, follows the particle along its path. Given the initial and final velocities (v₀=10 m/s and vf=2 m/s) and the distance traveled (3 m), we can use the kinematic equation vf^2 = v₀^2 + 2aδx to solve for a. In this case, δx is 3m, and vf and v₀ are given. From this equation, a = (vf^2 - v₀^2) / (2δx).
If actual calculations are performed, please note that the given information in the problem statement is insufficient to solve for time, as neither the time taken to travel through the nozzle nor the deceleration time is provided. Assuming the particle moves at a constant deceleration, one can calculate the time through δv = aδt and then average acceleration using that time duration. However, without additional data or time measurement, we cannot find a precise numerical value for acceleration.
A capacitor is connected to a 9 V battery and acquires a charge Q. What is the charge on the capacitor if it is connected instead to an 18 V battery? A capacitor is connected to a 9 V battery and acquires a charge Q. What is the charge on the capacitor if it is connected instead to an 18 V battery? Q Q/2 2Q 4Q
Answer:
Option C is the correct answer.
Explanation:
We have charge stored in a capacitor
Q = CV
C is the capacitance and V is the voltage.
A capacitor is connected to a 9 V battery and acquires a charge Q.
That is
Q = C x 9 = 9C
What is the charge on the capacitor if it is connected instead to an 18 V battery
That is
Q' = C x 18 = 9C x 2 = 2Q
Option C is the correct answer.
When the voltage on a capacitor is doubled, the charge on the capacitor doubles as well. Hence, if a capacitor initially has a charge Q at 9V, it will have a charge of 2Q at 18V.
Explanation:The subject of the question is the behavior of a capacitor when it is connected to various voltage sources. The charge Q on a capacitor is given by the equation Q = CV, where C is the capacitance of the capacitor and V is the voltage across it. So if a capacitor is charged by a 9V battery (Q = C*9V), and then connected to an 18V battery, its new charge (lets call it Q1) would be C*18V. Therefore, Q1 = 2Q, since the voltage is doubled, hence, the charge on the capacitor also doubles.
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Determine the force between two long parallel wires, that are separated by a dis- tance of 0.1m the wire on the left has a current of 10A and the one on the right BT has a current of 15A, both going up. 31十131
Answer:
3 x 10^-4 N/m
Attractive
Explanation:
r = 0.1 m, i1 = 10 A, i2 = 15 A
The force per unit length between two parallel wires is given by
[tex]F = \frac{\mu _{0}}{4\pi }\times \frac{2i_{1}i_{2}}{r}[/tex]
[tex]F = \frac{10^{-7}\times 2\times 10\times 15}{0.1}[/tex]
F = 3 x 10^-4 N/m
Thus, the force per unit length between two wires is 3 x 10^-4 N/m, the force is attractive in nature because the direction of flow of current in both the wires is same.
A Nichrome wire is used as a heating element in a toaster. From the moment the toaster is first turned on to the moment the wire reaches it maximum temperature, the current in the wire drops by 20% from its initial value. What is the temperature change in the wire? The temperature coefficient for Nichrome is α = 0.0004 (°C)-1.
Answer:
Temperature change in the wire is 625°C
Explanation:
Let the initial current in the wire be I₀
Thus,
the final current in the wire will be, I = (1 - 0.20) × I₀ = 0.80I₀
Given α = 0.0004 1/°C
Now,
Resistance = (potential difference (E))/current (I)
also
R = R₀ [1 + αΔT]
Where ΔT is the increase in temperature , thus
E/I = E/I₀ [1 + α × ΔT)
on substituting the values in the above equation, we get
1/(0.80I₀) = 1/(I₀) × [1 + 0.0004 × ΔT]
or
1/(0.80) = [1 + 0.0004 × ΔT)]
or
ΔT = 625°C
The temperature change in the Nichrome wire, based on a 20% drop in current, is calculated to be approximately 625°C.
A Nichrome wire's resistance increases with temperature, primarily determined by its temperature coefficient.
Step-by-Step Solution :
Let I₀ be the initial current and I be the current after heating.E/I = E/I₀ [1 + α × ΔT)
on substituting the values in the above equation, we get
= 1/(0.80I₀) = 1/(I₀) × [1 + 0.0004 × ΔT]
= 1/(0.80) = [1 + 0.0004 × ΔT)]
= ΔT = 625°C
The temperature change in the Nichrome wire is approximately 625°C.
A driver parked his car on a steep hill and forgot to set the emergency brake. As a result, the car rolled down the hill and crashed into a parked truck. If the car was moving at 10 mph (4.5 m/s) when it hit the truck 7.0 seconds after it began to move, what was the car’s average acceleration while it rolled down the hill? Define the positive x direction to be down the hill.
Answer:0.642
Explanation:
Given
initially car is at rest i.e u=0
When car hits the truck it's velocity is 4.5 m/s
it hits the truck after 7 sec
Now average acceleration of car is =[tex]\frac{change\ in\ velocity}{time\ taken}[/tex]
average acceleration of car =[tex]\frac{\left ( final velocity\right )-\left ( initial velocity\right )}{time}[/tex]
average acceleration of car=[tex]\frac{\left ( 4.5\right )-\left ( 0\right )}{7}[/tex]
average acceleration of car=[tex]0.642 m/s^{2}[/tex]
Taking Down hill as positive x axis
average acceleration of car=[tex]0.642\hat{i}[/tex]
Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay model, which is initially motionless and has a mass of 0.350 kg. Both being soft clay, they naturally stick together. What is their final velocity?
Answer:
0.278 m/s
Explanation:
We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.
So we can write:
[tex]mu=(m+M)v[/tex]
where
m = 0.200 kg is the mass of the koala bear
u = 0.750 m/s is the initial velocity of the koala bear
M = 0.350 kg is the mass of the other clay model
v is their final combined velocity
Solving the equation for v, we get
[tex]v=\frac{mu}{m+M}=\frac{(0.200)(0.750)}{0.200+0.350}=0.278 m/s[/tex]
The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2 mm diameter copper wire where the internal electric field strength is 0.01 V/m?
Answer:
1.87 A
Explanation:
τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s
d = diameter of copper wire = 2 mm = 2 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
E = magnitude of electric field = 0.01 V/m
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of electron = 9.1 x 10⁻³¹ kg
n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³
[tex]i[/tex] = magnitude of current
magnitude of current is given as
[tex]i = \frac{Ane^{2}\tau E}{m}[/tex]
[tex]i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}[/tex]
[tex]i[/tex] = 1.87 A
A coil of self- induction 0.5 H, its ohmic resistance is 10 ⦠carrying a dc current of intensity 2 A, The voltage at its terminals isâ¦â¦ A. Zero
B. 20
C. 5
D. 0.2
Answer:
Option (B)
Explanation:
L = 0.5 H, I = 2 A, R = 10 Ohm
V = I R
In case of dc the inductor cannot works.
V = 2 x 10 = 20 V
While finding the spring constant, if X1 = 12 cm, X2 = 15 cm, and hanging mass = 22 grams, the value of spring constant K would be:________ (write your answer in newtons/meter)
Answer:
If x₁=12 cm then k=1.7985 N/m
If x₂=15 cm then k=1.4388 N/m
Explanation:
Hanging mass= 22 g=0.022 kg
Acceleration due to gravity g=9.81 m/s²
If x₁=displacement= 12 cm=0.12 m
k= spring constant
[tex]F=ma\\\Rightarrow F=0.022\times 9.81\\\Rightarrow F=0.21582\ N[/tex]
[tex]\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.012\\\Rightarrow k=1.7985\ N/m\\[/tex]
∴k = 1.7985 N/m
If x₂=15 cm=0.15 m
Force of the hanging mass is same however the spring constant will change
[tex]\text {For spring}\\F=kx\\\Rightarrow 0.21582=k\times 0.015\\\Rightarrow k=1.4388\ N/m\\[/tex]
∴k = 1.4388 N/m
As the mass is not changing the spring constant has to change. That means that here there are two spring one with k=1.7985 N/m and the other with k= 1.4388 N/m
A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 μC on each plate. The plates have a separation of 0.313 mm. What is the potential difference between the plates?
Answer:
The potential difference between the plates is 596.2 volts.
Explanation:
Given that,
Capacitance [tex]C=260\ pF[/tex]
Charge [tex]q=0.155\ \mu\ C[/tex]
Separation of plates = 0.313 mm
We need to calculate the potential difference between the plates
Using formula of potential difference
[tex]V= \dfrac{Q}{C}[/tex]
Where, Q = charge
C = capacitance
Put the value into the formula
[tex]V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}[/tex]
[tex]V=596.2\ volts[/tex]
Hence,The potential difference between the plates is 596.2 volts.
A force of 20 N holds an ideal spring with a 10-N/m spring constant in compression. The potential energy stored in the spring is: a. 0.5 J b. 2.5 J c. 5J d. 20 J e. 200 J
The potential energy stored in the spring is calculated using the formula U = 1/2kx^2. With a force of 20 N and a spring constant of 10 N/m, the displacement is found to be 2 m, and thus the potential energy is 20 J.
Explanation:The question is related to the calculation of the potential energy stored in an ideal spring. To find this, we use the formula for elastic potential energy, which is U = 1/2kx2, where U is the potential energy, k is the spring constant, and x is the displacement (or compression) of the spring from its rest position.
In this case, since the force holding the spring in compression is 20 N and the spring constant is 10 N/m, we first need to find the displacement x. The force F applied to a spring is also given by F=kx, so rearranging for x gives us x = F/k = 20 N / 10 N/m = 2 m. Plugging this into the potential energy formula, we get U = 1/2 * 10 N/m * (2 m)2 = 1/2 * 10 * 4 = 20 J.
Therefore, the potential energy stored in the spring is 20 J.
Dinosaur fossils are too old to be reliably dated using carbon-14, which has a half-life of about 5730 years. Suppose we had a 68 million year old dinosaur fossil. How much of the living dinosaur's 14C would be remaining today? (Round your answer to five decimal places.)
Answer:
0.00000
Explanation:
The half-life of a radioisotope, in this case carbon-14, is the time that a sample requires to reduce its amount to half, and it is a constant for every radioisotope (it does not change with the amount of sample).
Then, the formula for the remaining amount of a radioisotope is:
A / A₀ = (1/2)ⁿWhere:
A is the final amount of the element,A₀ is the initial amount of the element, A/A₀ is ratio of remaining amount to the original amount, andn is the number of half-lives elapsedThe number of half-lives for carbon-14 elapsed for the dinosaur fossil is:
n = 68 million years / 5730 years ≈ 11,867Then, A / A₀ = (1/2)ⁿ = (1/2)¹¹⁸⁶⁷ ≈ 0.00000 .
The number is too small, and when you round to five decimal places the result is zero. That is why carbon-14 cannot be used to date dinosaur fossils, given that they are too old.
C-14 dating is not applicable to a 68 million years old dinosaur fossil, because the half-life of C-14 (5730 years) only allows for accurate dating up to about 50,000-57,000 years. After such time period, the remaining C-14 would be too small to measure accurately. Therefore, for a 68 million year old fossil, the remaining C-14 would be effectively zero.
Explanation:The question is asking about the amount of Carbon-14 (C-14) remaining in a dinosaur fossil that is approximately 68 million years old. However, C-14 dating is not applicable to such old samples. This is because C-14 decays back to Nitrogen-14 (N-14) with a half-life of approximately 5,730 years. Because of this half-life, the best-known method for determining the absolute age of fossils with C-14 allows for reliable dating of objects only up to about 50,000 years old. After 10 half-lives, which would be about 57,000 years, any original C-14 existing in a sample would become too small to measure accurately
For very old samples like a 68 million year old dinosaur fossil, isotopes with much longer half-lives are used for dating, such as uranium-235 and potassium-40. Therefore, for a 68 million year old dinosaur fossil, the remaining C-14 would effectively be zero.
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An elevator car, with a mass of 450 kg is suspended by a single cable. At time = 0s, the elevator car is raised upward. The tension on the cable is constant at 5000N during the first 3 seconds of operation. Determine the magnitude of the velocity of the elevator at time = 3 seconds.
Answer:
33.33 m/s
Explanation:
m = 450 kg. T = 5000 N, t = 3 seconds,
let the net acceleration is a.
T = m a
a = 5000 / 450 = 11.11 m/s^2
u = 0 , v = ?
Let v be the velocity after 3 seconds.
Use first equation of motion
v = u + a t
v = 0 + 11.11 x 3 = 33.33 m/s
Can the dielectric permittivity of an insulator be negative?
A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.45 rad at t=0. What is the angle of the speck at t = 8.0 s ? Your answer should be between 0 and 2Ï rad.
Answer:
64.08 rad
Explanation:
f = 78 rpm = 76 / 60 rps
θ = 0.45 rad, t = 8 s
(θ2 - θ1) / t = ω
θ2 - θ1 = 2 x 3.14 x 76 x 8 / 60 = 63.63
Angle turns from initial position = 0.45 + 63.63 = 64.08 rad
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.80 s for the boat to travel from its highest point to its lowest, a total distance of 0.600 m. The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart.
Part A How fast are the waves traveling? Express the speed v in meters per second using three significant figures.
Answer:
Very fast!
Explanation:
The waves are traveling very fast. I know this because they're not traveling slow for sure.
If this is incorrect, they're probably traveling moderately fast.
A 36-in-diameter pipeline carries oil (SG = 0.89) at 1 million barrels per day (bbl/day). The friction head loss is 13 ft/1000 ft of pipe. It is planned to place pumping stations every 10 mi along the pipe. Estimate the horsepower which must be delivered to the oil by each pump.
Answer:
144540.16735 hp
Explanation:
Diameter of pipe = 36 in
Specific gravity of oil = 0.89
ρ = Density of oil = 0.89×62.2407
Q = Flow rate of oil = 1×10⁶ bbl/day = 1×10⁶×5.61458/86400 = 64.9836033565 ft³/s
hf = 13 ft/1000ft = 0.013 ft/ft
for 10 miles
hf = 0.013×10×5280 = 686.4 ft
g = Acceleration due to gravity = 32.17405 ft/s²
Power
P = ρQghf
⇒P = 0.89×62.2407×64.9836033565×32.17405×686.4
⇒P = 79497097.342 ft lbf/s
1 ft lbf/s = 0.0018181817 hp
79497097.342 ft lbf/s = 144540.16735 hp
∴ Horsepower which must be delivered to the oil by each pump is 144540.16735 hp
A 640-N hunter gets a rope around a 3200-N polar bear. They are stationary, 20m apart, on frictionless level ice. When the hunter pulls the polar bear to him, the polar bear will move: A. 1:0m B. 3:3m C. 10m D. 12m E. 17m
Answer:
When the hunter pulls the polar bear to him, the polar bear will move 3.3 meters.
Explanation:
It is given that, a 640-N hunter gets a rope around a 3200-N polar bear. They are stationary, 20 m apart, on friction less level ice.
The given system is an isolated system which is initially at rest and they both will meet at the center of mass. Then finding the center of mass of the system as :
[tex]X_{com}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}[/tex]
Multiplying "g" to the numerator and denominator both as :
[tex]X_{com}=\dfrac{gm_1x_1+m_2x_2g}{m_1g+m_2g}[/tex]
[tex]X_{com}=\dfrac{640(0)+3200\times 20}{640+3200}=16.67\ m[/tex]
i.e. 640 N hunter moves at a distance of 16.67 meters and when the hunter pulls the polar bear to him, the polar bear will move, 20 m - 16.67 m = 3.3 meters. Hence, this is the required solution.
A 1000 kg train car traveling at 12 m/s on a level track has 400 kg of sand dumped, vertically, into it. What is the final speed of the train car after the sand is dumped into it? Ignore friction
Answer:
The final speed of the train car is 8.57 m/s.
Explanation:
Given that,
Mass of train car = 1000 kg
Mass of sand = 400 kg
Speed of train car = 12 m/s
We need to calculate the final speed of the train car after the sand is dumped
Using conservation of momentum
[tex]m_{1}u_{1}=(m_{1}+m_{2})v[/tex]...(I)
Put the value in the equation (I)
[tex]1000\times12=(1000+400)v[/tex]
[tex]v=\dfrac{1000\times12}{1400}[/tex]
[tex]v=8.57\ m/s[/tex]
Hence, The final speed of the train car is 8.57 m/s.
Suppose that you time this pendulum, and it completes exactly 68.0 complete cycles in 100 s. [Hint: think carefully about all given distances.] 5. (5 pts) Which choice best represents the value of g in the vicinity of this pendulum?
Answer:
Explanation:
Number of oscillations in 100 s is 68.
time taken to complete one oscillation = 100 / 68 s = 1.47 s
Use the formula for the time period of an oscillating body
[tex]T = 2\pi \sqrt{\frac{L}{g}}[/tex]
[tex]T^{2} = 4\pi^{2} \frac{L}{g}}[/tex]
In this equation substitute the value of T amd L and then find the value of g.
A singly charged ion of 7Li (an isotope of lithium which lost only one electron) has a mass of 1.16 ×10^-26 kg. It is accelerated through a potential difference of 224 V and then enters a magnetic field with magnitude 0.724 T perpendicular to the path of the ion. What is the radius of the ion’s path in the magnetic field? (Give your answer in decimal using "mm"(millimeter) as unit)
Explanation:
It is given that,
Mass of lithium, [tex]m=1.16\times 10^{-26}\ kg[/tex]
It is accelerated through a potential difference, V = 224 V
Uniform magnetic field, B = 0.724 T
Applying the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=qV[/tex]
[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]
q is the charge on an electron
[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}[/tex]
v = 78608.58 m/s
[tex]v=7.86\times 10^4\ m/s[/tex]
To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :
[tex]qvB=\dfrac{mv^2}{r}[/tex]
[tex]r=\dfrac{mv}{qB}[/tex]
[tex]r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}[/tex]
r = 0.0078 meters
So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.
A box rests on the back of a truck. The coefficient of friction between the box and the surface is 0.32. (3 marks) a. When the truck accelerates forward, what force accelerates the box? b. Find the maximum acceleration the truck can have before the box slides.
Answer:
Part a)
here friction force will accelerate the box in forward direction
Part b)
a = 3.14 m/s/s
Explanation:
Part a)
When truck accelerates forward direction then the box placed on the truck will also move with the truck in same direction
Here if they both moves together with same acceleration then the force on the box is due to friction force between the box and the surface of the truck
So here friction force will accelerate the box in forward direction
Part b)
The maximum value of friction force on the box is known as limiting friction
it is given by the formula
[tex]F = \mu mg[/tex]
so we have
[tex]F = ma = \mu mg[/tex]
now the acceleration is given as
[tex]a = \mu g[/tex]
[tex]a = (0.32)(9.8) = 3.14 m/s^2[/tex]
Final answer:
The force that accelerates the box when the truck moves forward is the force of static friction. The maximum acceleration the truck can have before the box slides can be found using the equation: max acceleration = coefficient of friction x acceleration due to gravity.
Explanation:
When the truck accelerates forward, the force that accelerates the box is the force of static friction between the box and the surface of the truck bed. This force opposes the motion of the box and allows it to stay in place on the truck.
The maximum acceleration the truck can have before the box slides can be found using the equation:
max acceleration = coefficient of friction x acceleration due to gravity
Using the given coefficient of friction between the box and the surface, you can substitute the value and solve for the maximum acceleration.
PHYSICS: Can someone help me identify a concept/hypothesis and possible experiment to run? THis is supposed to be a mock experiment and not an actual lab completed.
Look around and find a certain pattern. Describe the pattern / phenomena that you have selected. Pretend that you have just performed your observation or observational experiment. You may need to repeat the observation at a later time, if there is a possibility. Illustrate it with a photo or a sketch.
Devise a hypothesis of what created this pattern or what caused this particular pattern/phenomenon. Record it with as much details as possible.
Think of how you can check if your hypothesis is valid (design a testing experiment) and record the prediction of this experiment. It is preferable to provide accurate details of your reasoning , assumptions, and the reasoning behind your prediction. Why you think the suggested cause-effect link would work? Are you making any assumptions? If so, what are they?
Identify what quantities you will be keeping track of and how they will connect the predicted cause to the observed effect. What are these quantities? How would you measure these quantities?
Perform the experiment. If it is possible at all, video-record your experiment to analyze your actions, including the measurements. You have to trust me for now on this, but the more information you get on your thought, work, and measurements, the easier it would be for you to pinpoint the reason for the prediction to be wrong, if that should happen. There is no need in attaching the movie to your submission, but I would recommend you to keep it among the files on Blackboard, so it is within reach if you will need to defend your reasoning.
In one paragraph, explain the results of your experiment. Did they match the prediction?
In the summary, please reflect on the whole process. If the prediction holds true, what would be the better way to further confirm your idea of the cause-effect connection? If not, revise your hypothesis or suggest a different observational/testing experiment to get a better insight into the pattern/phenomenon
A cool experiment with a testable hypothesis is testing an apple and if it will be able to oxidize in different temperatures with the inside exposed to Oxygen.
Hypothesis: If the cut piece of an apple is exposed to warm air then the apple will turn brown and oxidize faster than a cut piece of apple in a cooler temperature.
Supplies: Three thermometers, An apple, an apple cutter, a plate, etc.
Directions place the three thermometers in a cool cold area,
A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15 A. If it is lying flat on the ground, and the Earth’s magnetic field points due north, has a magnitude of 5.8 x 10-5 T, and makes a downward angle of 25° with the vertical, what is the torque on the loop?
Answer:
0.01663 Nm
Explanation:
N = 1000, r = 12 cm = 0.12 m, i = 15 A, B = 5.8 x 10^-5 T, θ = 25
Torque = N i A B Sinθ
Torque = 1000 x 15 x 3.14 x 0.12 x 0.12 x 5.8 x 10^-5 x Sin 25
Torque = 0.01663 Nm
A piece of wood with density of 824 kg/m^3 is tied to the bottom of a pool and the wood does not move. The volume of the wood is 1.2 m^3. What is the tension is in the rope?
Answer:
2069.76 N
Explanation:
density of wood = 824 kg/m^3
density of water = 1000 kg/m^3
Volume of wood = 1.2 m^3
True weight of wood = volume of wood x density of wood x gravity
True weight of wood = 1.2 x 824 x 9.8 = 9690.24 N
Buoyant force acting on wood = volume of wood x density of water x gravity
Buoyant force acting on wood = 1.2 x 1000 x 9.8 = 11760 N
Tension in the rope = Buoyant force - True weight
tension in rope = 11760 - 9690.24 = 2069.76 N
Starting from rest, the boy runs outward in the radial direction from the center of the platform with a constant acceleration of 0.5 m/s2 . The platform rotates at a constant rate of 0.2 rad/s. Determine his velocity (magnitude) when t = 3 sec
Answer:
His velocity when t= 3 sec is V= 1.56 m/s
Explanation:
a= 0.5 m/s²
ω= 0.2 rad/s
t= 3 sec
Vr= a*t
Vr= 1.5 m/s
r= a*t²/2
r= 2.25m
Vt= w*r
Vt= 0.45 m/s
V= √(Vr²+Vt²)
V= 1.56 m/s
The boy's tangential velocity, after accelerating radially for 3 seconds at 0.5 m/s^2 while the platform rotates at 0.2 rad/s, is 1.5 m/s.
Explanation:To determine the boy's velocity (magnitude) at t = 3 sec while he runs outward radially from the center of a rotating platform, we must consider both his radial (tangential) velocity due to his acceleration and the rotational movement of the platform.
The boy starts from rest with a constant radial acceleration of 0.5 m/s2. Using the kinematic equation v = u + at (where u is the initial velocity, a is the acceleration, and t is the time), we can calculate his radial velocity after 3 seconds as:
vradial = 0 + (0.5 m/s2)(3 s) = 1.5 m/s
The platform's constant rotation rate is given as 0.2 rad/s. This rotational movement does not change the boy's radial (tangential) velocity directly, as the question only asks for his velocity in the radial direction. Therefore, the total magnitude of the boy's velocity after 3 seconds is 1.5 m/s tangentially.
In case of a damped oscillation. (a) Amplitude increases (b) Remains constant (c) None (d) Amplitude reduces
Answer:
option (d)
Explanation:
There are two types of oscillations.
1. Damped oscillations: The oscillations in which the amplitude of a wave goes on decreasing continuously are called damped oscillations.
For example, the oscillations of a seismic wave.
2. Undamped oscillations: The oscillations in which the amplitude o a wave remains constant are called undamped oscillations.
for example, oscillations of light waves.
A 0.45 m radius, 500 turn coil is rotated one-fourth of a revolution in 4.01 ms, originally having its plane perpendicular to a uniform magnetic field. Find the magnetic field strength in T needed to induce an average emf of 10,000 V.
Answer:
B = 0.126 T
Explanation:
As per Faraday's law we know that rate of change in magnetic flux will induce EMF in the coil
So here we can say that EMF induced in the coil is given as
[tex]EMF = \frac{\phi_2 - \phi_1}{\Delta t}[/tex]
initially the coil area is perpendicular to the magnetic field
and after one fourth rotation of coil the area vector of coil will be turned by 90 degree
so we can say
[tex]\phi_1 = NBAcos 0 = BA[/tex]
[tex]\phi_2 = NBAcos 90 = 0[/tex]
now we will have
[tex]EMF = \frac{NBA}{t}[/tex]
[tex]10,000 V = \frac{(500)(B)(\pi \times 0.45^2)}{4.01\times 10^{-3}}[/tex]
[tex]B = 0.126 T[/tex]
An insulated cup contains 250g of water at 20° C. What will the water temperature be after 100g of ice at 0° C is added to the cup and allowed to melt?
answer
50°C
Explanation:
Mw*©=Mi*©
250*20=100*©
©=250*20/100
©=50°C//
To find the final temperature after mixing ice and water, one must calculate the heat required to melt the ice and compare it to the heat the water loses in the process, using the specific heat of water and the enthalpy of fusion of ice.
Explanation:To determine the final temperature of water after adding 100g of ice at 0°C to 250g of water at 20°C, we must consider the heat transfer involved in the process of melting ice and the subsequent equilibrium between the melted ice and the initial water. The specific heat of water (4.184 J/(g°C)) and the enthalpy of fusion of ice (6.01 kJ/mol) are crucial to this calculation.
The process involves calculating the heat required to melt the ice (Q = mfusion * ΔHfusion) and equating it to the heat lost by the water as it cools (Q = mcwater*ΔT), and solving for the final temperature. However, in this scenario, it's important to note that without the specific values or further calculations included in the question details, we can't provide a precise numerical answer but can outline the method to find it.
A projectile is fired vertically with an initial velocity of 192 m/s. Calculate the maximum altitude h reached by the projectile and the time t after firing for it to return to the ground. Neglect air resistance and take the gravitational acceleration to be constant at 9.81 m/s2.
Answers:
h = m
t = s
Answer:
a) Maximum height reached = 1878.90 m
b) Time of flight = 39.14 seconds.
Explanation:
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g [tex]m/s^2[/tex] and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile , [tex]t=\frac{2usin\theta }{g}[/tex]
Vertical motion (Maximum height reached, H) :
We have equation of motion, [tex]v^2=u^2+2as[/tex], where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H
[tex]0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}[/tex]
In the give problem we have u = 192 m/s, θ = 90° we need to find H and t.
a) [tex]H=\frac{u^2sin^2\theta}{2g}=\frac{192^2\times sin^290}{2\times 9.81}=1878.90m[/tex]
Maximum height reached = 1878.90 m
b) [tex]t=\frac{2usin\theta }{g}=\frac{2\times 192\times sin90}{9.81}=39.14s[/tex]
Time of flight = 39.14 seconds.