Answer:
The soda is being sucket out at a rate of 3.14 cubic inches/second.
Explanation:
R= 2in
S= π*R²= 12.56 inch²
rate= 0.25 in/sec
rate of soda sucked out= rate* S
rate of soda sucked out= 3.14 inch³/sec
Final answer:
Soda is being pulled out of a cylindrical glass at the rate of -3.14 cubic inches per second.
Explanation:
The question asks at what rate soda is being absorbed from a cylindrical glass. Given the cylindrical glass's dimensions and the rate at which the depth of soda decreases, we need to find the volume rate of change.
He formula for the volume of a cylinder is V = πr2h, where r is the radius and h is the height of the cylinder. To find the rate at which volume changes, we differentiate with respect to time to obtain dV/dt = πr2(dh/dt). Plugging in the values, we have r = 2 inches and dh/dt = -0.25 inches/second (negative because the depth is decreasing).
Using the given values, the rate at which soda is being pulled out of the glass is
dV/dt = π(2 in)2(-0.25 in/sec) = π(4 in2)(-0.25 in/sec) = -3.14 in3/sec.
Therefore, the rate at which the soda is being pulled out of the glass is -3.14 in3/sec.
g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide box having mass m2 and they move together. Find the velocity of boxes.
Answer:Expression given below
Explanation:
Given mass of spring[tex]\left ( m_1\right )=0.5 kg[/tex]
Compression in the spring[tex]\left ( x\right )=20 cm[/tex]
Let the spring constant be K
Using Energy conservation
potential energy stored in spring =Kinetic energy of Block[tex]\left ( m_1\right )[/tex]
[tex]\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2[/tex]
[tex]v=x\sqrt{\frac{k}{m_1}}[/tex]
now conserving momentum
[tex]m_1v=\left ( m_1+m_2\right )v_0[/tex]
[tex]v_0=\frac{m_1}{m_1+m_2}v[/tex]
where [tex]v_0[/tex] is the final velocity
A rocket takes off from Earth and reaches a speed of 105 m/s in 18.0 s. If the exhaust speed is 1,200 m/sand the mass of fuel burned is 110 kg, what was the initial mass (in kg, including the initial fuel) of the rocket?
Answer:
The initial mass of the rocket is 526.2 kg.
Explanation:
Given that,
Final speed = 105 m/s
Time = 18.0 s
Exhaust initial speed = 1200 m/s
Mass of burned fuel = 110 kg
We need to calculate the initial mass
The velocity change of rocket under gravity is defined as,
[tex]v=u\ ln(\dfrac{m_{i}}{m})-gt[/tex]....(I)
We know that,
[tex]m=m_{i}-m_{bf}[/tex]
Put the value of m in equation (I)
[tex]v=u\ ln(\dfrac{m_{i}}{m_{i}-m_{bf}})-gt[/tex]
[tex]m_{i}=\dfrac{m_{bf}}{1-e^-{\dfrac{v+gt}{u}}}[/tex]
[tex]m_{i}=\dfrac{110}{1- e^-{\frac{105+9.8\times18}{1200}}}[/tex]
[tex]m_{i}=526.2 kg[/tex]
Hence, The initial mass of the rocket is 526.2 kg.
The initial mass of the rocket including the initial fuel is 1257.14 kg.
Conservation of linear momentum
The initial mass of the rocket is determined by applying the principle of conservation of linear momentum as follows;
m₁v₁ = m₂v₂
105m = 1200 x 110
105 m = 132,000
m = 132,000/105
m = 1257.14 kg
Thus, the initial mass of the rocket including the initial fuel is 1257.14 kg.
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Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter. The resistivity of copper is 1.68×10−8Ω⋅m and the resistivity of aluminum is 2.65×10−8Ω⋅m.
Answer: 0.258
Explanation:
The resistance [tex]R[/tex] of a wire is calculated by the following formula:
[tex]R=\rho\frac{l}{s}[/tex] (1)
Where:
[tex]\rho[/tex] is the resistivity of the material the wire is made of. For aluminium is [tex]\rho_{Al}=2.65(10)^{-8}m\Omega[/tex] and for copper is [tex]\rho_{Cu}=1.68(10)^{-8}m\Omega[/tex]
[tex]l[/tex] is the length of the wire, which in the case of aluminium is [tex]l_{Al}=12m[/tex], and in the case of copper is [tex]l_{Cu}=30m[/tex]
[tex]s[/tex] is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:
[tex]s=\pi{(\frac{d}{2})}^{2}[/tex] (2) Where [tex]d[/tex] is the diameter of the circumference.
For aluminium wire the diameter is [tex]d_{Al}=2.5mm=0.0025m[/tex] and for copper is [tex]d_{Cu}=1.6mm=0.0016m[/tex]
So, in this problem we have two transversal areas:
For aluminium:
[tex]s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}[/tex]
[tex]s_{Al}=0.000004908m^{2}[/tex] (3)
For copper:
[tex]s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}[/tex]
[tex]s_{Cu}=0.00000201m^{2}[/tex] (4)
Now we have to calculate the resistance for each wire:
Aluminium wire:
[tex]R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}[/tex] (5)
[tex]R_{Al}=0.0647\Omega[/tex] (6) Resistance of aluminium wire
Copper wire:
[tex]R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}[/tex] (6)
[tex]R_{Cu}=0.250\Omega[/tex] (7) Resistance of copper wire
At this point we are able to calculate the ratio of the resistance of both wires:
[tex]Ratio=\frac{R_{Al}}{R_{Cu}}[/tex] (8)
[tex]\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}[/tex] (9)
Finally:
[tex]\frac{R_{Al}}{R_{Cu}}=0.258[/tex] This is the ratio
What is a gravational lens?
Final answer:
A gravitational lens occurs when a massive object like a galaxy causes the light from a more distant object to bend and magnify, much like how an optical lens works. Einstein predicted this based on general relativity, and its discovery has provided critical verification of this theory. The first gravitational lens was found in 1979, showing double images of a distant quasar.
Explanation:
A gravitational lens is a phenomenon where the light from a distant object, such as a quasar or galaxy, is bent and magnified by the gravity of a massive object, like a galaxy or galaxy cluster, that lies between the observer and the distant object. This effect is analogous to the way a conventional lens, like those found in glasses, bends light. Albert Einstein predicted this effect based on his theory of general relativity but thought it would be very unlikely for us to observe it. However, astronomers have since observed numerous instances of gravitational lensing, which provide significant evidence for general relativity. These observations include multiple images, arcs, or even rings of the same astronomical object, as gravity distorts the space around the massive intervening object.
The first gravitational lens was discovered in 1979 and presented dual images of the same distant object. This discovery and subsequent observations using powerful telescopes like the Hubble Space Telescope have allowed astronomers to study the universe's most distant objects and the distribution of dark matter. The bend of light due to massive objects can create spectacular effects such as Einstein rings, where a perfect alignment between the observer, massive object, and distant source results in a ring-like structure of light.
A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that when placed in the field the sphere is in a static situation (all the forces on the sphere cancel). If the thread is horizontal, find the magnitude and direction of the electric field. The sphere has a mass of 0.018 kg and contains a charge of + 6.80 x 103 C. The tension in the thread is 6.57 x 10-2 N. Show your work and/or explain your reasoning.
To find the magnitude and direction of the electric field, let us find the horizontal and vertical components of the field separately, then we will use those values to calculate the total magnitude and direction.
The tension in the thread is 6.57×10⁻²N and the thread is aligned horizontally, so the tension force is directed entirely horizontally. The sphere is in static equilibrium, therefore the horizontal component of the electrostatic force acting on the sphere, Fx, must act in the opposite direction of the tension and have a magnitude of 6.57×10⁻²N. We know this equation relating a charge, an electric field, and the force that the field exerts on the charge:
F = Eq
F is the electric force, E is the electric field, and q is the charge
Let us adjust the equation for only the horizontal components of the above quantities:
Fx = (Ex)(q)
Fx is the horizontal component of the electric force and Ex is the horizontal component of the electric field.
Given values:
F = 6.57×10⁻²N
q = 6.80×10³C
Plug in these values and solve for Ex:
6.57x10⁻² = Ex(6.80×10³)
Ex = 9.66×10⁻⁶N/C
Since the sphere is in static equilibrium, the vertical component of the electrostatic force acting on the sphere, Fy, must have the same magnitude and act in the opposite direction of the sphere's weight. If we assume the weight to act downwards, then Fy must act upward.
We know the weight of the sphere is given by:
W = mg
W is the weight, m is the mass, and g is the acceleration of objects due to earth's gravity field near its surface.
We also know this equation:
F = Eq
Let us adjust for the vertical components:
Fy = (Ey)(q)
Set Fy equal to W and we get:
(Ey)(q) = mg
Given values:
q = 6.80×10³C
m = 0.018kg
g = 9.81m/s²
Plug in the values and solve for Ey:
(Ey)(6.80×10³) = 0.018(9.81)
Ey = 2.60×10⁻⁵N/C
Let's now use the Pythagorean theorem to find the total magnitude of the electric field:
E = [tex]\sqrt{Ex^{2}+Ey^{2}}[/tex]
E = 2.77×10⁻⁵N/C
The direction of the electric field is given by:
θ = tan⁻¹(Ey/Ex)
θ = 20.4° off the horizontal
A catapult used by medieval armies hurls a stone of mass 32.0 kg with a velocity of 50.0 m/s at a 30.0 degree angle above the horizontal. What is the horizontal distance traveled when the stone returns to its original height? Ignore air resistance. Check your calculator deg/rad mode.
Answer:
The horizontal distance traveled when the stone returns to its original height = 220.81 m
Explanation:
Considering vertical motion of catapult:-
At maximum height,
Initial velocity, u = 50 sin30 = 25 m/s
Acceleration , a = -9.81 m/s²
Final velocity, v = 0 m/s
We have equation of motion v = u + at
Substituting
v = u + at
0 = 25 - 9.81 x t
t = 2.55 s
Time of flight = 2 x Time to reach maximum height = 2 x 2.55 = 3.1 s
Considering horizontal motion of catapult:-
Initial velocity, u = 50 cos30 = 43.30 m/s
Acceleration , a = 0 m/s²
Time, t = 5.10 s
We have equation of motion s= ut + 0.5 at²
Substituting
s= ut + 0.5 at²
s = 43.30 x 5.10 + 0.5 x 0 x 5.10²
s = 220.81 m
The horizontal distance traveled when the stone returns to its original height = 220.81 m
A 160-kg object and a 460-kg object are separated by 3.40 m.(a) Find the magnitude of the net gravitational force exerted by these objects on a 33.0-kg object placed midway between them.
Answer:
The net gravitational force exerted by these objects is [tex]5.712\times10^{-8}\ N[/tex]
Explanation:
Given that,
Mass of another object m = 33.0 kg
Mass M₁ = 160 kg
Mass M₂ = 460 kg
Distance = 3.40 m
We need to calculate the net gravitational force
Using formula,
[tex]F=\dfrac{GM_{2}m}{r^2}-\dfrac{GM_{1}m}{r^2}[/tex]
[tex]F=Gm(\dfrac{M_{2}-M_{1}}{r^2})[/tex]
Where, G = gravitational constant
m = mass of another object
[tex]M_{1}[/tex]= mass of first object
[tex]M_{2}[/tex]= mass of second object
r = distance
Put the value into the formula
[tex]F=6.67\times10^{-11}\times33.0\times\dfrac{460-160}{(3.40)^2}[/tex]
[tex]F=5.712\times10^{-8}\ N[/tex]
Hence, The net gravitational force exerted by these objects is [tex]5.712\times10^{-8}\ N[/tex]
The charge on a capacitor increases by 20 μC when the voltage across it increases from 84 V to 121 V. What is the capacitance of the capacitor?
The capacitance of a capacitor when the charge increases by 20 µC and the voltage increases from 84 V to 121 V is approximately 0.541 µF.
The question asks to determine the capacitance of a capacitor. To find the capacitance when the charge (ΔQ) increases by 20 µC as the voltage increases from 84 V to 121 V, we can use the capacitor charge formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. The change in voltage (ΔV) can be calculated as 121 V - 84 V = 37 V. The capacitance (C) can then be found using the change in charge and the change in voltage:
C = ΔQ / ΔV
Plugging in the values, we get:
C = 20 µC / 37 V ≈ 0.541 µF
Therefore, the capacitance of the capacitor is approximately 0.541 µF.
A boat has a mass of 6800 kg. Its engines generate a drive force of 4100 N due west, while the wind exerts a force of 800 N due east and the water exerts a resistive force of 1200 N due east. What are the magnitude and direction of the boat’s acceleration?
Answer:
The boat's acceleration are: a= 0.3 m/s² in west direction.
Explanation:
Fm= 4100 N
Fwi= 800 N
Fwa= 1200 N
Fm - Fwi - Fwa= m *a
clearing a:
a= 0.3 m/s²
A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between the sled and surface is 0.229. What is the speed of the sled after it has traveled a distance of 3.89 m?
The final speed of the sled is calculated by first calculating the frictional force using the coefficient of kinetic friction and weight of the sled. This force is then used to calculate the deceleration due to friction. Finally, an equation of motion is employed to calculate the sled's final speed after it has traveled the given distance.
Explanation:To answer this question, we first need to calculate the force of friction, which can be found using the formula F_friction = µN, where µ is the coefficient of kinetic friction and N is the normal force. In this case, as the sled is moving on a horizontal surface, the normal force is equal to the weight of the sled, N = mg. So the frictional force F_friction = µmg.
Next, due to Newton's second law (F = ma), the deceleration (negative acceleration) is calculated by a = F_friction/m. With this acceleration, we can now use one of the equations of motion, V_f = V_i + 2a*d, to calculate the final velocity, where V_f is the final velocity, V_i is the initial velocity, a is acceleration and d is the distance.
This method allows us to calculate the final speed of the sled after traveling a certain distance considering kinetic friction.
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The final speed of the sled is calculated by finding the work done by friction and applying the work-energy theorem to find the change in kinetic energy. The kinetic energy is equated to its initial value minus the work done by friction, and solved for the final speed.
Explanation:To tackle this question, we should first calculate the total work done by the friction force (kinetic friction) which is known to act in the opposite direction of the sled's movement. The work done by friction is given by the equation W = Fd in which 'F' stands for friction force and 'd' for distance. The friction force can be calculated by the formula F = μN, where 'μ' is the coefficient of friction and 'N' is the normal force. Since the sled moves across a horizontal surface, N is equal to the weight of the sled (mg, where 'm' is the mass and 'g' is the acceleration due to gravity).
Then, using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy, we can solve for the final speed. The kinetic energy is given by KE = 0.5mv², where 'm' is the mass and 'v' is the speed. So, the initial and the final kinetic energies of the sled are KE_initial = 0.5 * 2.12kg * (5.49 m/s)² and KE_final = KE_initial - Work_done_by_friction. We can solve the latter for the final speed of the sled, v_final = sqrt((2*KE_final)/m).
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In a performance test, each of two cars takes 9.0 s to accelerate from rest to 28 m/s. Car A has a mass of 1379 kg, and car B has a mass of 1902 kg. Find the net average force that acts on (a) car A and (b) car B during the test.
Explanation:
It is given that,
Mass of car A, [tex]m_A=1379\ kg[/tex]
Mass of car B, [tex]m_B=1902\ kg[/tex]
In a performance test, each of two cars takes 9.0 s to accelerate from rest to 28 m/s. Their acceleration is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{28\ m/s}{9}[/tex]
a = 3.12 m/s²
(a) The net force of car A, [tex]F_A=m_Aa[/tex]
[tex]F_A=1379\ kg\times 3.12\ m/s^2[/tex]
[tex]F_A=4302.48\ N[/tex]
(b) The net force of car B, [tex]F_B=m_Ba[/tex]
[tex]F_B=1902\ kg\times 3.12\ m/s^2[/tex]
[tex]F_A=5934.24\ N[/tex]
Hence, this is the required solution.
What determines how much induced current will flow through a conductor?
Answer:
Rate of change of magnetic flux
Explanation:
The induced current is equal to the ratio of induced emf to the resistance of the conductor.
According to the Faraday's law of electromagnetic induction, the induced emf is proportional to the rate of change of magnetic flux.
An EM wave has a wavelength A 500 nm and a peak electric field Eo 100 i. The wave propagates along the z axis. a) What is the magnitude and direction of the magnetic field? b) What is the frequency f? c) What is the angular frequency w? d) What is the wavenmber k?
Answer:
(a) 3.33 x 10^-7 T along Y axis
(b) 6 x 10^14 Hz
(c) 3.768 x 10^15 rad/s
(d) 2 x 10^6 m^-1
Explanation:
Wavelength = 500 nm = 500 x 10^-9 m
Electric Field, E = 100 i (along + X axis)
Wave is in +Z axis
So, the magnetic field is in + Y axis.
(a) c = E / B
B = E / c = 100 / (3 x 10^8) = 3.33 x 10^-7 T along Y axis
(b) Frequency = wave speed / wavelength = ( 3 x 10^8) / (500 x 10^-9)
frequency = 6 x 10^14 Hz
(c) w = 2 x 3.14 x f = 2 x 3.14 x 6 x 10^14 = 3.768 x 10^15 rad/s
(d) k = reciprocal of wavelength = 1 / (500 x 10^-9) = 2 x 10^6 m^-1
It is noted that a sample of 134La with a half-life of 6.5 minutes has an activity of 2.6 Ci. What was its activity 72 minutes ago?
Answer:
36//
Explanation:
(½)^t
(½)^72
72/2=36//
A 5.0-kg centrifuge takes 95 s to spin up from rest to its final angular speed with constant angular acceleration. A point located 6.00 cm from the axis of rotation of the centrifuge with a speed of 99 m/s when the centrifuge is at full speed moves (a) What is the angular acceleration (in rad/s2) of the centrifuge as it spins up? (b) How many revolutions does the centrifuge make as it goes from rest to its final angular speed?
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479
The heat of combustion of benzene, C6H6, is –41.74 kJ/g. Combustion of 4.50 g of benzene causes a temperature rise of 4.50°C in a certain bomb calorimeter. What is the heat capacity of this bomb calorimeter?
Answer:
Heat capacity this bomb calorimeter = [tex] 41.74 KJ/^oC [/tex]
Explanation:
Given data we have in the question:
The heat of combustion for the benzene, [tex]Q_{C_6H_6} = - 41.74 KJ/g[/tex] (Here the negative sign depicts the release of heat and only magnitude is considered fro the heat capacity)
The mass of benzene, [tex]M_{C_6H_6} = 4.50g[/tex]
The change in temperature due to the combustion, [tex]\Delta T = 4.50^oC[/tex]
Now, the formula for the heat capacity is given as:
Heat capacity (C) = [tex]\frac{Q_{C_6H_6}\times M_{C_6H_6}}{\Delta T }[/tex]
or
C = [tex]\frac{ 41.74 KJ/g\times 4.50g}{4.50^oC }[/tex]
or
C = [tex] 41.74 KJ/^oC [/tex]
The heat capacity of the bomb calorimeter, using the given values and the formula: (heat = heat capacity * change in temperature), is 41.74 kJ/°C.
Explanation:In bomb calorimetry, the heat capacity of the bomb calorimeter, or the calorimeter constant (C), can be calculated by using the formula: q = C * change in temperature. The heat (q) released by burning the benzene is the product of the mass burned and the heat of combustion, which is -41.74 kJ/g * 4.5 g, giving us -187.83 kJ. Since a negative q indicates that heat is released, we're actually dealing with 187.83 kJ of heat. The heat capacity (C) of the calorimeter can now be calculated by rearranging the formula to C = q / delta T = 187.83 kJ / 4.5°C = 41.74 kJ/°C. This means the heat capacity of the bomb calorimeter is 41.74 kJ/°C.
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A 1 kg ball is hung at the end of a rod 1 m long. If the system balances at a point on the rod one third of the distance from the end holding the mass, what is the mass of the rod?
Answer:
2 kg
Explanation:
Assuming the rod's mass is uniformly distributed, the center of mass is at half the length.
Sum of the moments at the balance point:
-(Mg)(L/3) + (mg)(L/2 − L/3) = 0
(Mg)(L/3) = (mg)(L/2 − L/3)
(Mg)(L/3) = (mg)(L/6)
2M = m
M = 1 kg, so m = 2 kg.
The mass of the rod is 2 kg.
By equating the torques due to the ball and the rod, we determine the mass of the rod is 6 kg.
To determine the mass of the rod, we need to use the concept of torque balance. Here, we have a 1 kg ball hanging at the end of a 1 m long rod.
The system balances at a point one-third of the distance from the end holding the mass. This means the pivot point is located 1/3 meter from the end with the ball.
The torque due to the ball:
Torque_ball = (1 kg) * (9.8 m/s²) * (1 m)The torque due to the rod:
The rod can be considered as having its mass concentrated at its center of mass, which is 0.5 m from either end. Since the pivot is 1/3 m from the ball end, the rod's center of mass is (0.5 m - 1/3 m) = 1/6 m from the pivot point. Thus, the torque due to the rod is:Torque_rod = (mass of the rod) * (9.8 m/s²) * (1/6 m)Since the system is in balance, the torques must be equal:
Torque_ball = Torque_rod
(1 kg) * (9.8 m/s²) * (1 m) = (mass of the rod) * (9.8 m/s²) * (1/6 m)
Solving for the mass of the rod:
(1 kg) * (9.8 m/s²) * (1 m) = (mass of the rod) * (9.8 m/s²) * (1/6)
mass of the rod = 6 kg
Thus, the mass of the rod is 6 kg.
One piece of copper jewelry at 111°C has exactly twice the mass of another piece, which is at 28°C. Both pieces are placed inside a calorimeter whose heat capacity is negligible. What is the final temperature inside the calorimeter (c of copper = 0.387 J/g·K)?
Answer:
83.33 C
Explanation:
T1 = 111 C, m1 = 2m
T2 = 28 C, m2 = m
c = 0.387 J/gK
Let the final temperature inside the calorimeter of T.
Use the principle of calorimetery
heat lost by hot body = heat gained by cold body
m1 x c x (T1 - T) = m2 x c x (T - T2)
2m x c X (111 - T) = m x c x (T - 28)
2 (111 - T) = (T - 28)
222 - 2T = T - 28
3T = 250
T = 83.33 C
Thus, the final temperature inside calorimeter is 83.33 C.
Araised flower bed requires 13 cubic feet of garden soil. You can get a good deal by importing the soil, but it is sold from the dealer in cubic meters. To the nearest hundredth, how many cubic meters of soil is needed? There are 3.28 feet in 1 meter.
Answer:
0.368 m^3
Explanation:
3.28 feet = 1 m
Cube on both the sides
35.29 feet^3 = 1 m^3
So, 13 cubic feet = 13 / 35.29 m^3
= 0.368 m^3
When performing the pupillary reflexes, describe the position of the subject and environment
the subject should be comfortable and relaxed in a well lit area
the subject should be seated, relaxed in a dimly lit area with one eye closed
the subject should be seated, relaxed in a well lit area with one eye closed
the subject should be comfortable and relaxed in a dimly lit area, the subject should shield their eyes by placing their hand vertically between the eye and nose
Answer:
The pupillary light reflex (PLR) or photopupillary reflex is a reflex that controls the diameter of the pupil, in response to the intensity (luminance) of light that falls on the retinal ganglion cells of the retina in the back of the eye, thereby assisting in adaptation of vision to various levels of lightness/darkness.
A greater intensity of light causes the pupil to constrict (miosis/myosis; thereby allowing less light in), whereas a lower intensity of light causes the pupil to dilate (mydriasis, expansion; thereby allowing more light in)
Explanation:
A counterclockwise current runs through a square wire loop in the xy plane centered at the origin. The length of each side is d. What is the magnetic field on the z axis?
Answer:
[tex]B_{net} = \frac{2\sqrt2 \mu_0 i}{\pi d}[/tex]
Explanation:
Magnetic field due to straight current carrying wire is given by the formula
[tex]B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)[/tex]
now we will have for one side of the square at its center position given as
[tex]B = \frac{\mu_0 i}{4\pi (\frac{d}{2})}(sin45 + sin45)[/tex]
[tex]B = \frac{2\sqrt2 \mu_0 i}{4 \pi d}[/tex]
now for the we have for complete square loop it will become 4 times of the one side
[tex]B_{net} = 4 B[/tex]
[tex]B_{net} = 4 \frac{2\sqrt2 \mu_0 i}{4 \pi d}[/tex]
[tex]B_{net} = \frac{2\sqrt2 \mu_0 i}{\pi d}[/tex]
An ordinary glass is filled to the brim with water at 100oc. How much water could be added to the glass if the temperature is lowered to 20oC? Assume that the coefficient of volume expansion for glass is 2.7 x 10-5 K-1 and for water it is 2.1 x 10-4 K-1
Answer:
14.64% of the volume of the glass, more water can be added
Explanation:
given:
Initial temperature, = 100°C = 373K
Final temperature = 20°C = 293K
Coefficient of volume expansion for glass, = 2.7 x 10⁻⁵ K⁻¹
Coefficient of volume expansion for Water, = 2.1 x 10⁻⁴ K⁻¹
The apparent Coefficient of volume expansion for Water, γ = 2.1 x 10⁻⁴ K⁻¹ - 2.7 x 10⁻⁵ K⁻¹ = 1.83 × 10⁻⁴ K⁻¹
Change in temperature, ΔΘ = Final Temperature - Initial Temperature = 293K - 373K = -80K
Now, the Coefficient of volume expansion is given as:
[tex]\Delta V = -{\gamma}{V \Delta \theta}\\[/tex]
where,
V = initial Volume
ΔV = change in the volume
Thus,
[tex]\Delta V = -{1.83\times 10^{-4} K^{-1} }\times {V \times -80K}\\[/tex]
or
[tex]\Delta V = 146.4\times 10^{-4}\times V[/tex]
or
[tex]\frac{\Delta V}{V} = 146.4\times 10^{-4}[/tex]
Multiplying both sides by 100
we get
[tex]\frac{\Delta V}{V}\times 100 = 146.4\times 10^{-4}\times 100[/tex]
or
change in volume with respect to the initial volume = 1.464%
thus, 1.464 % of the original volume water can be added.
Upon cooling from 100 °C to 20 °C, the thermal contraction of the water in the glass will be greater than that of the glass itself due to higher coefficient of volume expansion for water, thereby creating some extra space in the glass. This extra space represents the amount of additional water that can be added.
Explanation:In order to find out how much more water can be added to the glass as it cools from 100 °C to 20 °C, we first need to know the volume of water and glass that will change with this temperature drop. The formula we use to calculate this is ΔV = βV₀ΔT, where β is the coefficient of volume expansion, V₀ is the initial volume, and ΔT is the change in temperature.
First, we calculate this for the glass and then for the water, using their respective coefficients of volume expansion. The difference in the volume changes for water and glass will give us the amount of water that can be added to the glass as it cools. Since the coefficient of volume expansion for water is greater than glass, its volume will contract more due to the decrease in temperature, leaving some extra space in the glass.
Note that we make an assumption here that the glass and water were initially at thermal equilibrium at 100 °C and hence have the same initial volume. This problem illustrates the principle of thermal expansion which is a property of matter to change its volume with a change in temperature, and is dependent on the material's coefficient of volume expansion.
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_________is defined as the how many cycles are completed in one second by an oscillating object. Frequency Period Linear Velocity Momentum
Answer:
Frequency
Explanation:
The number of cycles completed in one second by an oscillating object is called its frequency. The term frequency shows number of times an oscillation is occurring. It is denoted by [tex]\nu[/tex]. The reciprocal of frequency is called the time period of the wave i.e.
[tex]\nu=\dfrac{1}{T}[/tex]
T = time period
So, the correct option is (a) "frequency".
Final answer:
Frequency is the number of oscillations per second, measured in hertz (Hz), and is inversely related to the period (T), which is the time it takes to complete one oscillation.
Explanation:
Frequency is defined as the number of cycles completed in one second by an oscillating object. In physics, frequency (f) is the number of oscillations that take place per unit of time. The SI unit for frequency is the hertz (Hz), which is equivalent to one oscillation per second. The relationship between frequency and period (T) is given by f = 1/T, meaning that as the period increases, the frequency decreases, and vice versa.
Periodic motion, such as that of a mass suspended by a wire undergoing simple harmonic motion, has a constant time to complete one cycle, known as the period (T). Its units are usually seconds. Period refers to the time it takes to complete one oscillation, while frequency is the number of these oscillations within a given timeframe.
The car has a mass of 1.6 Mg and center of mass at G. If the coefficient of static friction between the shoulder of the road and the tires is μs = 0.41, determine the greatest slope θ the shoulder can have without causing the car to slip or tip over if the car travels along the shoulder at constant velocity.
The greatest slope the shoulder can have without causing the car to slip or tip over is 22.3⁰.
The given parameters;
mass of the car, m = 1.6 Mgcoefficient of static friction, [tex]\mu_ s[/tex] = 0.41The greatest slope of the shoulder is determined by taking net force on the car as shown below;
[tex]\Sigma F= ma \\\\Wsin\theta - \mu_s Wcos \theta = ma[/tex]
at constant velocity, acceleration, a, = 0
[tex]Wsin\theta - \mu_s W cos \theta = 0\\\\W sin\theta = \mu_s Wcos \theta \\\\\mu_s = \frac{Wsin\theta }{W cos \theta } \\\\\mu_s = tan \theta \\\\\theta = tan^{-1}(\mu_s)\\\\\theta = tan^{-1}(0.41)\\\\\theta =22.3\ ^0[/tex]
Thus, the greatest slope the shoulder can have without causing the car to slip or tip over is 22.3⁰.
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The greatest slope the car can be without slipping or tipping over is approximately 22.4 degrees where weight due to gravity is balanced by static friction.
Explanation:This problem involves the concept of static friction, which prevents an object from sliding down a slope. Here, the car is held in place on a slope due to static friction between the tires and the road. First, we need to calculate the force of gravity acting on the car, which is the weight of the car mg. The mass of the car is given as 1.6 Mg = 1.6 x 10^6 grams. Therefore, the weight of the car is 1.6 x 10^6 g * 9.8 m/s² = 1.568 x 10^7 N.
Next, the force of static friction is given by the equation μsN, where N is the normal force and μs is the coefficient of static friction which is 0.41. Since the car is not moving, the net force on it must be zero. This implies that the force of static friction must balance out the force of gravity. So, we equate the forces and solve for the slope angle θ, getting a relationship: tan θ = μs.
So, θ = tan^-1 (0.41) ≈ 22.4°. Therefore, the greatest slope the shoulder can have without causing the car to slip or tip over is approximately 22.4 degrees.
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A potter's wheel of radius 11 cm starts from rest and rotates with constant angular accel- eration until at the end of 35 s it is moving with angular velocity of 19 rad/s. What is the angular acceleration? Answer in units of rad/s2.
Final answer:
The angular acceleration of the potter's wheel is approximately 0.543 rad/s^2, calculated by dividing the final angular velocity by the time taken to reach it.
Explanation:
To calculate the angular acceleration of a potter's wheel that starts from rest and reaches an angular velocity of 19 rad/s in 35 seconds, we can use the following kinematic equation for rotational motion:
\(\omega = \omega_0 + \alpha \times t\)
Where:
\(\omega\) is the final angular velocity.
\(\omega_0\) is the initial angular velocity (0 rad/s since the wheel starts from rest).
\(\alpha\) is the angular acceleration.
\(t\) is the time taken to reach the final angular velocity.
Substituting the given values:
19 rad/s = 0 rad/s + \(\alpha\) \times 35 s
To solve for \(\alpha\), we rearrange the equation:
\(\alpha = \frac{19 rad/s}{35 s}\)
\(\alpha = \frac{19}{35} rad/s^2\)
\(\alpha \approx 0.543 rad/s^2\)
Therefore, the angular acceleration of the potter's wheel is approximately 0.543 rad/s2.
A solenoid has 100 turns per cm and a total resistance of 60 Ω center of the solenoid when it is connected to a 12 volt battery Determine the magnetic field in the 21 A flat c
Answer:
2.5 x 10^-3 T
Explanation:
n = 100 turns / cm = 10000 turns / m, R = 60 ohm, V = 12 V
i = V / R = 12 / 60 = 0.2 A
Magnetic field due to solenoid is
B = μ0 x n x i
B = 4 x 3.14 x 10^-7 x 10000 x 0.2
B = 2.5 x 10^-3 T
(I) What is the capacitance of a pair of circular plates with a radius of 5.0 cm separated by 3.2 mm of mica?
As per the details given, the capacitance of the pair of circular plates is approximately [tex]\(1.11 \times 10^{-11}\) Farads.[/tex]
A capacitor is a device made to store electric charge, and one of its basic electrical characteristics is capacitance.
When a voltage difference (potential difference) exists between a capacitor's two conducting plates, the capacity of the capacitor to store and hold electrical energy is measured.
Capacitance is a measurement of how much charge can be held on a capacitor's plates for a specific voltage.
The capacitance (C) of a parallel plate capacitor is given by the formula:
[tex]\[ C = \dfrac{\varepsilon_0 \cdot A}{d} \][/tex]
Where:
[tex]\( \varepsilon_0 \)[/tex] is the vacuum permittivity ([tex]\(8.85 \times 10^{-12} \, \text{F/m}\)[/tex])
A is the area of the plates (in square meters)
d is the distance between the plates (in meters)
Given:
Radius of the circular plates (r) = 5.0 cm
= 0.05 m
Distance between the plates (d) = 3.2 mm
= 0.0032 m
Calculate the area of one of the plates:
[tex]\[ A = \pi \cdot r^2 \\\\= \pi \cdot (0.05 \, \text{m})^2 \][/tex]
Calculate the capacitance (C):
[tex]\[ C = \dfrac{\varepsilon_0 \cdot A}{d} \\\\= \dfrac{8.85 \times 10^{-12} \, \text{F/m} \cdot \pi \cdot (0.05 \, \text{m})^2}{0.0032 \, \text{m}} \][/tex]
[tex]\[ C \approx 1.11 \times 10^{-11} \, \text{F} \][/tex]
Thus, the capacitance of the pair of circular plates is approximately [tex]\(1.11 \times 10^{-11}\)[/tex] Farads.
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The capacitance of a pair of circular plates with a radius of 5.0 cm separated by 3.2 mm of mica is 10.3 pF. This is calculated using the formula for the capacitance of a parallel plate capacitor and the area of a circle.
Explanation:To determine the capacitance of a pair of circular plates of a parallel plate capacitor with a given radius, separated by a specific dielectric material, we use the formula:
C = ε0 εr (A/d)
Where:
C is the capacitance,
ε0 is the vacuum permittivity (8.85 × 10-12 F/m),
εr is the relative permittivity of the dielectric (for mica, it's approximately 6),
A is the area of one of the plates, and
d is the separation between the plates.
To find the area A of a circular plate with radius r, we use the formula A = πr2. For a radius of 5.0 cm (or 0.05 m), A = π(0.05 m)2 = 7.85 × 10-3 m2.
Substituting into the capacitance formula, with d = 3.2 mm (or 0.0032 m), we get:
C = (8.85 × 10-12 F/m) × 6 × (7.85 × 10-3 m2/0.0032 m) = 1.03 × 10-11 F
This result is usually expressed in picofarads (pF), so the capacitance is 10.3 pF.
A charge alters the space around it. What is this alteration of space called? Electric ether Electric Force Electric field Charged space
Explanation:
A charge alters the space around it. This alteration of space is called the electric field. It is also defined as the electric force acting on a charged particle per unit test charge. It is given by :
[tex]E=\dfrac{F}{q}[/tex]
Where
F is the electric force, [tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
The direction of electric field is in the direction of electric force. For a positive charge, the direction of electric field lines are outwards and for a negative charge, the direction of field lines are inwards.
Hence, the correct option is (c) "electric field".
A certain helium-neon laser pointer, emitting light with a wavelength of 632 nm, has a beam with an intensity of 825 W/m2 and a diameter of 2.80 mm. How many photons are emitted by the laser pointer every second?
Answer:
[tex]N = 1.62 \times 10^{15}[/tex]
Explanation:
Energy of each photon is given as
[tex]E = \frac{hc}{\lambda}[/tex]
here we will have
[tex]\lambda = 632 nm[/tex]
now we will have
[tex]E = \frac{(6.626\times 10^{-34})(3 \times 10^8)}{632 \times 10^{-9}}[/tex]
[tex]E = 3.14 \times 10^{-19} J[/tex]
now let say there is N number of photons per second
so power due to photons is
[tex]P = N(3.14 \times 10^{-19})[/tex]
now intensity is given as power received per unit area
so we have
[tex]I = \frac{P}{A}[/tex]
[tex]825 = \frac{N(3.14 \times 10^{-19})}{\pi (1.40 \times 10^{-3})^2}[/tex]
[tex]825 = N(5.11 \times 10^{-14})[/tex]
[tex]N = 1.62 \times 10^{15}[/tex]
Aluminum has a magnetic susceptibility χ = 1.7 × 10-5 at T = 300 K. What will be the magnetization of a small sample of aluminum placed in a 1.5 T magnetic field at T = 150 K?
Answer:
The magnetization of a small sample of aluminum is 40.6 A/m
Explanation:
Given that,
Magnetic susceptibility [tex]\chi_{1}=1.7\times10^{-5}[/tex] at 300 K
Magnetic field = 1.5 T
We need to calculate the magnetic susceptibility at 150 K
[tex]\dfrac{\chi_{2}}{\chi_{1}}=\dfrac{T_{1}}{T_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{\chi_{2}}{1.7\times10^{-5}}=\dfrac{300}{150}[/tex]
[tex]\chi_{2}=\dfrac{300\times1.7\times10^{-5}}{150}[/tex]
[tex]\chi_{2}=3.4\times10^{-5}[/tex]
We need to calculate the magnetization
Using formula of magnetization
[tex]I=\chi_{2}\times H[/tex]
Where, H = magnetic intensity
Formula of magnetic intensity
[tex]H=\dfrac{B}{\mu_{0}}[/tex]
Where, B = magnetic field
Put the value of H into the formula of magnetization
[tex]I=\chi_{2}\times\dfrac{B}{\mu_{0}}[/tex]
[tex]I=3.4\times10^{-5}\times\dfrac{1.5}{4\pi\times10^{-7}}[/tex]
[tex]I=40.6\ A/m[/tex]
Hence, The magnetization of a small sample of aluminum is 40.6 A/m