Answer:
2321 N
Explanation:
Let g = 9.807 m/s2
Assume this is a uniform beam and the center of mass it at the geometric center, which is half way between the 2 ends, or h = 13/2 = 6.5 m from the left end (support A).
For the system to remain at rest, then the total moments around a point (let pick support A) must be 0. Moments created by each force is the product of the force magnitude and the moment arm, aka distance from the force to support A
[tex]\sum M_A = 0[/tex]
[tex]M_w + M_b + M_B = 0[/tex]
[tex]F_wx + F_bh + F_BL = 0[/tex]
[tex]m_wgx + m_bgh = -F_BL[/tex]
If we take upward direction be the positive direction, that means all the gravity acting downward are negative
[tex]74*(-9.807)*7.5 + 388(-9.807)*6.5 = -F_B13[/tex]
[tex]-13F_B = -30176.139[/tex]
[tex]F_B = -30176.139/-13 = 2321 N[/tex]
So the force at support B has a magnitude of 2321 N acting upward
The force that support B must apply to keep the beam balanced when the worker sits at a distance of 7.5m from support A is approximately 1597.92 N.
Explanation:This problem involves understanding of statics, specifically the principle of moments or torques. Assuming the system is in equilibrium (i.e., the beam does not rotate or translate), the sum of the forces and the sum of the torques must both be zero.
We start by calculating the total weight of the beam, which is simply its mass multiplied by the acceleration due to gravity, g. This gives mb * g = 388 kg * 9.81 m/s^2 = 3806.68 N, this force acts in the middle of the beam, that is at a distance of L/2 = 13m/2 = 6.5m from each support.
Similarly, the weight of the worker is given by mw * g = 74 kg * 9.81 m/s^2 = 725.94 N, this force acts at a distance of x = 7.5m from the left support (A) and L - x = 13m - 7.5m = 5.5m from the right support (B).
Now we can calculate the net force (Fn) and the net torque (Tn). We choose the left support (A) as the pivot point. The total force acting on the beam is the sum of the weights of the beam and the worker, and this must be supported by the two supports. Since sum of the forces must be zero: Fn = F_A + F_B = mass_beam * g + mass_worker * g.
The net torque is calculated from the forces acting at a distance from the pivot point. So we get, Tn = beam_weight * distance_beam - worker_weight * distance_worker + F_B * L = 0. From this, we can solve for F_B (force at support B), we find: F_B = (beam_weight * distance_beam - worker_weight * distance_worker) / L.
Plugging into these formulas, we get the force that support B must apply to keep the system in balance: F_B = (3806.68 N * 6.5m - 725.94 N * 7.5m) / 13m = 1597.92 N.
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Oxygen (O 2 ) is confined to a cubic container, 15 cm on a side, at a temperature of 300 K. Compare the average kinetic energy of a molecule of the gas to the change in its gravitational potential energy if it falls from the top of the container to the bottom.
Answer: Average kinetic energy is greater than gravitational potential energy.
Explanation: The average kinetic energy formulae for a molecule of a gas relative to temperature (in Kelvin) is given below as
E = 3/2 (KT).
Where E = average kinetic energy =?
K = boltzman constant = 1.381×10^-23 m²kg/s²k
T = temperature = 300 k
By substituting the parameters in the formulae, we have that
E = 3/2 ×( 1.381×10^-23 × 300)
E = 1.5 × 4.143×10^-21
E = 6.2145×10^-21 J
To get the gravitational potential energy, we use the fact that
Gravitational potential energy = gravitational energy at the top - gravitational energy at bottom.
At the top, the height of cube is h= 15cm = 0.15m, g = acceleration dude to gravity = 9.8m/s², m = mass of molecule of oxygen = 1.661×10^-27 kg
Gravitational potential energy = mgh = 1.661×10^-27× 9.8 × 0.15 = 2.442×10^-27 J
At the bottom, height is zero, hence gravitational potential energy is also zero.
Hence the final gravitational potential energy = potential energy at top - potential energy at bottom =
2.442×10^-27 - 0 = 2.442×10^-27 J.
Gravitational potential energy = 2.442×10^-27 J
Average kinetic energy = 6.2145×10^-21 J
As we can see that the average kinetic energy is bigger than the gravitational potential energy.
Compare the average kinetic energy of oxygen molecules at 300 K to the change in their gravitational potential energy if they fall from the top to the bottom of a container.
Oxygen (O2) molecules at 300 K have an average kinetic energy of 3kT, where k is Boltzmann's constant. If an oxygen molecule falls from the top of a container to the bottom, its change in gravitational potential energy is mgΔh, where m is the mass of the molecule, g is the acceleration due to gravity, and Δh is the height change.
Two charged particles are separated by 10 cm. suppose the charge on each particle is doubled. By what factor does the electric force between the particles change?
Answer:
The electric force increases by a factor of 4.
Explanation:
The electric force between two charges [tex]q_1[/tex] and [tex]q_2[/tex] separated a distance d can be calculated using Coulomb's Law:
[tex]F=\frac{kq_1q_2}{d^2}[/tex]
where [tex]k=9\times10^9Nm^2/C^2[/tex] is the Coulomb constant.
If the value of each charge is doubled, then we will have a force between them which is:
[tex]F'=\frac{k(2q_1)(2q_2)}{d^2}=4\frac{kq_1q_2}{d^2}=4F[/tex]
So the new force is 4 times larger than the original force.
Doubling the charge on each particle increases the electric force between them by a factor of 4.
Explanation:The force between two charged particles is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. So, if we denote the electric force as F, the charges as q1 and q2, and the distance as r, we can write Coulomb's law as F = k* q1*q2/r^2, where k is a constant.
Now if you double the charges (q1 and q2 become 2q1 and 2q2), and use these values in the formula, we get Fnew = k*(2q1) *(2q2)/r^2 = 4 * k*q1*q2/r^2 = 4F.
So, by doubling the charge on each particle, the electric force between them is multiplied by the factor of 4. So, the force increases fourfold.
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A flat circular loop of radius 0.10 m is rotating in a uniform magnetic field of 0.20 T. Find the magnetic flux through the loop when the plane of the loop and the magnetic field vector are perpendicular.'
Answer:
[tex]\phi=6.28\times10^{-3}\;\;weber[/tex]
Explanation:
Given,
Magnetic field [tex]B=0.2\;\;T\\[/tex]
Radius [tex]r=0.1\;\;m[/tex]
The angle between the area vector and magnetic field is 0 degree, because the direction of area vector is always perpendicular to the plane.
[tex]\phi=BAcos\theta\\\phi=o.2\times\pi (0.1)^2\times cos0^o\\\phi=0.00628\;\;weber\\\phi=6.28\times10^{-3}\;\;weber[/tex]
The magnetic flux through a flat circular loop rotating in a uniform magnetic field is zero when the plane of the loop and the magnetic field vector are perpendicular.
Explanation:To find the magnetic flux through the loop when the plane of the loop and the magnetic field are perpendicular, we will need to use the formula for magnetic flux:
Φm = BA cos θ
where Φm is the magnetic flux through the surface, B is the magnitude of the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field direction and the normal (perpendicular) to the surface. In this case, if the loop and the magnetic field are perpendicular, θ = 90°, and cos 90° = 0. Hence, the magnetic flux will be zero.
However, if the provided problem included a different angle, we would adjust the equation to accommodate that. For example, if the field were parallel to the loop (θ = 0), then cos θ would equal 1 and the magnetic flux would just be the product of B and A (A = πr²)
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Three identical resistors, when connected in series, transform electrical energy into thermal energy at a rate of 12 W (4.0 W per resistor). Part A Determine the power consumed by the resistors when connected in parallel to the same potential difference. Express your answer with the appropriate units.
Answer:
108 Watts
Explanation:
The total circuit resistance when the resistors are connected in series is
R + R + R = 3R
When he resistors are connected in parallel, the resistance reduces from 3R in the series circuit to become;
[tex]\frac{1}{R} + \frac{1}{R} + \frac{1}{R}[/tex]
= [tex]\frac{R}{3}[/tex] Ω
[tex]Power = \frac{V^{2}}{R}[/tex]
The voltage supply was given to be constant for both the series and parallel circuits. This implies that V² is constant and power is inversely proportional to resistance.
Therefore;
Power for the parallel connected circuit = [tex]\frac{3R}{\frac{R}{3} } * 12 W[/tex]
= 9 × 12 W = 108 Watts
Answer:
Explanation:
Potential difference, V and let each resistance, R
Resistors are in series, total resistance, Rₓ = R1 + R2 + R3
= R + R + R
= 3R
Power, P = V²/Rₓ
12 = V²/3R
V²/R = 36
Resistors are in parallel, total resistance, 1/Rₓ = 1/R1 + 1/R2 + 1/R3
Rₓ = R/3
P = V²/Rₓ
P = V²/(R/3)
P = 3(V²/R)
= 3(36)
= 108 W.
A horizontal power line carries a current of 7250 A from south to north. Earth's magnetic field (65.7 µT) is directed toward the north and is inclined downward at 70.0° to the horizontal. Find the (a) magnitude and (b) direction of the magnetic force on 140 m of the line due to Earth's field.
Answer:
a) [tex]||\vec F_{B}||=62.664\,N[/tex], b) From east to west.
Explanation:
Vectorially, the magnetic force can be calculed by the following formula:
[tex]\vec F_{B} = i\cdot \vec l\, \times \, \vec B[/tex]
The cross product is:
[tex]\vec F_{B} = \left|\begin{array}{ccc}i&j&k\\1015000\,A\cdot m&0\,A\cdot m&0\,A\cdot m\\22.471\times 10^{-6}\,T&-61.738\times 10^{-6}\,T&0\,T\end{array}\right|[/tex]
[tex]\vec F_{B} = - 62.664\,N\cdot k[/tex]
a) The magnitude of the magnetic force is:
[tex]||\vec F_{B}||=62.664\,N[/tex]
b) The direction of the magnetic force is:
From east to west.
A conducting loop is placed in a magnetic field. What must be true for there to be a current induced in the loop? (1 point) There must be a source of charge The magnetic field must be changing Potential energy must change into kinetic energy The loop must be surrounded by insulating material
Answer:
EMF will induce in the conducting loop when
The magnetic field must be changing
Explanation:
As per Farady's law of EMI we know that the magnetic flux linked with the closed conducting loop must be changed with time
so we have
[tex]EMF = \frac{d\phi}{dt}[/tex]
so we know that
[tex]\phi = BAcos\omega t[/tex]
now if magnetic field is changing with time then we have
[tex]EMF = Acos\theta \frac{dB}{dt}[/tex]
so EMF will induce when magnetic field will change with time
Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the same time by pushing on box A with a horizontal pushing force FP = 8.9 N. Here, A has a mass mA = 10.2 kg and B has a mass mB = 7.0 kg. The contact force between the two boxes is FC. The coefficient of kinetic friction between the boxes and the floor is 0.04. (Assume FP acts in the +x direction.)
Answer:
Explanation:
The force of friction acting on the system
= .04 x 9.8 ( 10.2 + 7 )
= 6.74 N
Net force = 8.9 - 6.74
= 2.16 N
Acceleration in the system
= 2.16 / ( 10.2 + 7 )
= .12558 m / s ²
Contact force between boxes = FP
Considering force on box A
Net force = 8.9 - FP
Applying Newton's law on box A
8.9 - FP = 10.2 x .12558
= 1.28
FP = 8.9 - 1.28
= 7.62 N
In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors A⃗ =(1,0,−3), B⃗ =(−2,5,1), and C⃗ =(3,1,1). Calculate and express your answers as ordered triplets of values separated by commas
(1) A - B
(2) B - C
(3) - A + B - C
(4) 3A - 2C
(5) - 2A + 3B - C
(6) 2A - 3 (B - C)
Answer:
(1) (3,-5,-4)
(2) (-5, 4, 0)
(3) (-6, 4, 3)
(4) (-3, -2, -11)
(5) (-11, 14, 8)
(6) (17, -12, -6)
Explanation:
A⃗ =(1,0,−3)
B⃗ =(−2,5,1)
C⃗ =(3,1,1)
Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.
1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)
2) (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)
3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)
4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)
5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)
6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)
Tarik winds a small paper tube uniformly with 183 turns 183 turns of thin wire to form a solenoid. The tube's diameter is 9.49 mm 9.49 mm and its length is 2.09 cm 2.09 cm . What is the inductance, in microhenrys, of Tarik's solenoid?
143μH
Explanation:The inductance (L) of a coil wire (e.g solenoid) is given by;
L = μ₀N²A / l --------------(i)
Where;
l = the length of the solenoid
A = cross-sectional area of the solenoid
N= number of turns of the solenoid
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
From the question;
N = 183 turns
l = 2.09cm = 0.0209m
diameter, d = 9.49mm = 0.00949m
But;
A = π d² / 4 [Take π = 3.142 and substitute d = 0.00949m]
A = 3.142 x 0.00949² / 4
A = 7.1 x 10⁻⁵m²
Substitute these values into equation (i) as follows;
L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]
L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209
L = 143 x 10⁻⁶ H
L = 143 μH
Therefore the inductance in microhenrys of the Tarik's solenoid is 143
The inductance, in microhenrys, of Tarik's solenoid should be considered as the 143μH.
Calculation of the inductance:Since
The inductance (L) of a coil wire (e.g solenoid) should be provided by
L = μ₀N²A / l --------------(i)
here,
l = the length of the solenoid
A = cross-sectional area of the solenoid
N= number of turns of the solenoid
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
So,
N = 183 turns
l = 2.09cm = 0.0209m
diameter, d = 9.49mm = 0.00949m
So,
A = π d² / 4
= 3.142 x 0.00949² / 4
= 7.1 x 10⁻⁵m²
Now
L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]
L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209
L = 143 x 10⁻⁶ H
L = 143 μH
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A Global Positioning System (GPS) functions by determining the travel times for EM waves from various satellites to a land-based GPS receiver. If the receiver is to detect a change in travel distance of the order of 3 m, what is the associated travel time in (in ns) that must be measured
Answer:
10ns
Explanation:
Suppose the EM wave travels at light speed [tex]c = 3\times10^8 m/s[/tex]. A change in travel distance of the order of 3 m would result of a change in travel time of
[tex] \Delta t = \frac{\Delta s}{c} = \frac{3}{3\times10^8} = 10^{-8} s = 10 ns[/tex]
To detect a change in travel distance of about 3 m, a GPS receiver must measure a change in travel time of approximately 10 nanoseconds. This is calculated using the speed of light and the time = distance/speed formula.
Explanation:To determine the associated travel time that must be measured for a change in travel distance of 3 meters, we need to know the speed of electromagnetic waves, which is the speed of light, typically denoted 'c'. The speed of light is approximately 3.00 x 108 meters per second (m/s).
Now we can calculate the time it takes for light to travel a given distance using the formula time (t) = distance (d) / speed (s). Substituting for our given distance (3 m) and the speed of light, we get t = 3 m / 3.00 x 108 m/s. This results in a time of approximately 1 x 10-8 seconds or 10 nanoseconds (ns).
Therefore, to detect a change in travel distance of about 3 m, the GPS receiver must be able to measure a change in travel time of roughly 10 nanoseconds.
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A certain monatomic gas inside a cylinder is at a temperature of 22°C. It takes 353 J of work done on the gas to compress it and increase the temperature to 145°C. If there are originally 8.2 moles of gas inside the cylinder, calculate the quantity of heat flowing into or out of the gas. (Indicate the direction with the sign of your answer. Let "into the gas" be positive, and "out of the gas" be negative.)
The calculated heat Q is [tex]\( 1.22 \times 10^4 \mathrm{~J} \),[/tex] and the positive sign indicates that heat is being added to the system.
Let's break down the given answer step by step:
1. Given Values:
Temperature of gas, [tex]\( T = 22^{\circ} \mathrm{C} = 295 \mathrm{~K} \)[/tex]
Work done on the gas to compress it, [tex]\( W = -353 \mathrm{~J} \)[/tex](negative because work is done on the gas)
Final temperature, [tex]\( T_f = 145^{\circ} \mathrm{C} = 418 \mathrm{~K} \)[/tex]
Number of moles,[tex]\( n = 8.2 \) moles[/tex].
2. First Law of Thermodynamics:
The first law of thermodynamics is given by [tex]\( Q = \Delta U + W \),[/tex] where Q is the heat added to the system, [tex]\( \Delta U \)[/tex] is the change in internal energy, and W is the work done on the system.
For a monoatomic gas,[tex]\( C_v = \frac{3}{2} R \),[/tex]
where R is the universal gas constant [tex](\( R = 8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \)). So, \( Q = n C_v \Delta T + W \).[/tex]
3. Calculation:
Substituting the given values into the equation:
[tex]\[ \begin{aligned} Q & = \left(8.2 \times \frac{3}{2} \times 8.314 \times (418 - 295)\right) + (-353) \\ & = 12225 \mathrm{~J} = 1.22 \times 10^4 \mathrm{~J} \end{aligned} \][/tex]
4. Interpretation:
Since Q is positive, it means heat is flowing inside the cylinder. A positive value of Q indicates that the system is gaining heat, and in this case, it's due to both the compression work done on the gas and the increase in internal energy.
In summary, the calculated heat Q is [tex]\( 1.22 \times 10^4 \mathrm{~J} \),[/tex] and the positive sign indicates that heat is being added to the system.
Two particles, one with charge − 7.97 μC and one with charge 3.55 μC, are 6.59 cm apart. What is the magnitude of the force that one particle exerts on the other?
Answer:
58.6 N
Explanation:
We are given that
[tex]q_1=-7.97\mu C=-7.97\times 10^{-6} C[/tex]
[tex]q_2=3.55\mu C=3.55\times 10^{-6} C[/tex]
Using [tex]1\mu C=10^{-6} C[/tex]
[tex]r=6.59 cm=6.59\times 10^{-2} m[/tex]
[tex]1 cm=10^{-2} m[/tex]
The magnitude of force that one particle exerts on the other
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Where [tex]k=9\times 10^9[/tex]
Substitute the values
[tex]F=\frac{9\times 10^9\times 7.97\times 10^{-6}\times 3.55\times 10^{-6}}{(6.59\times 10^{-2})^2}[/tex]
F=58.6 N
If the potential in a region is given by the function V = 2 x − y 2 − cos(z), what is the y-component of the electric field at the point P = (x ′ , y ′ , z ′ )?
Answer:
2y
Explanation:
Electric field in terms of Electric potential is given as:
E = dV/dr(x, y, z)
Where r(x, y, z) = position in x, y, z plane
The y component of the Electric field will be:
Ey = -dV/dy
Given that
V = 2x - y² - cos(z)
dV/dy = -2y
=> E = - (-2y)
E = 2y
The maximum allowed power dissipation for a 23.9 Ω 23.9 Ω resistor is stated to be 10.0 W . 10.0 W. Calculate the largest current that this resistor can take safely without burning out.
Answer:
0.65 A.
Explanation:
Given:
Pmax = 10 W
R = 23.9 Ω
Formula for calculating power,
P = I × V
= I^2 × R
I^2 = 10/23.9
I = 0.65 A.
A steel bucket contains 4 liters of water at 128C. An electric immersion heater rated at 1400 Watts is placed in the * Indicates an advanced level in solving. Problems 61 62 CHAPTER 1 Introduction bucket. Determine how long it will take for water to heat to 708C. Assume that the empty bucket weighs 1.1 kg. The specific heat of steel is 0.46 kJ/(kg 8C). Use an average specific heat of water of 4.18 kJ/(kg 8C). Disregard any heat loss to the surroundings.
Answer:
The time is 0.713 sec.
Explanation:
Given that,
Weight of water = 4 liters
Initial temperature = 128°C
Power = 1400 Watts
Final temperature = 708°C
Weight = 1.1 kg
Specific heat of steel = 0.46 kJ/kg°C
Specific heat of water = 4.18 kJ/kg°C
We need to calculate the heat gained by bucket
Using formula of heat
[tex]Q_{b}=mc\Delta T[/tex]
Put the value into the formula
[tex]Q_{b}=1.1\times0.46\times(70-12)[/tex]
[tex]Q_{b}=29.348\ kJ[/tex]
We need to calculate the heat gained by water
Using formula of heat
[tex]Q_{w}=mc\Delta T[/tex]
Put the value into the formula
[tex]Q_{w}=4\times4.18\times(70-12)[/tex]
[tex]Q_{w}=969.76\ kJ[/tex]
We need to calculate the total heat
Using formula of heat
[tex]Q=Q_{b}+Q_{w}[/tex]
Put the value into the formula
[tex]Q=29.348+969.76[/tex]
[tex]Q=999.108\ kJ[/tex]
We need to calculate the time
Using formula of time
[tex]t=\dfrac{Q}{P}[/tex]
Put the value into the formula
[tex]t=\dfrac{999.108}{1400}[/tex]
[tex]t=0.713\ sec[/tex]
Hence, The time is 0.713 sec.
You sit at the middle of a large turntable at an amusement park as it begins to spin on nearly frictionless bearings, and then spins freely. When you crawl toward the edge of the turntable, does the rate of the rotation increase, decrease, or remain unchanged, and why
Answer:
Explanation:
As there is no external torque is applied so the angular momentum remains constant.
L = I ω = constant
Where, I is the moment of inertia of the system and ω is the angular velocity
As we move towards the edge, the moment of inertia increases, hence the angular velocity decreases.
A 6861 kg car traveling at 48 m/s is at the foot of a hill that rises 142 m in 2.3 km. At the top of the hill, the speed of the car is 10 m/s. The acceleration of gravity is 9.81 m/s 2 . Assuming constant acceleration, find the average power delivered by the car’s engine, neglecting any internal frictional losses.
Answer: P = 25050.8w
Explanation:
total energy at top = K.E + P.E
= (1/2)(6861)(100) + 6861(9.81)(142)
total energy at bottom
= (1/2)(6861)(48)^2
work done = energy at top - energy at bottom
average velocity = (48+10)/2
time = 2300/average velocity
power = work done/time
plus potential) at the base and the top; is the energy input from the engine
the ascent time is the average speed, (top + bottom) / 2; divided by the 2.3 km distance
energy / time equals power
The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion. Assume that the North American continent can be represented by a slab of rock 4700 km on a side and 25 km deep and that the rock has an average mass density of 2880 kg/m3 . The continent is moving at the rate of about 3.2 cm/year. What is the mass of the continent
The mass of the North American continent is calculated by multiplying the volume of the continent by the density of the rock it is made of. The volume is determined based on the given dimensions, converted to meters. Multiplying the resulting volume by the rock's density provides the estimated mass of 1.59712e21 kilograms.
Explanation:In order to find the mass of a continent, we need to multiply the volume of the continent by the density of the rock that it's made of. First, we need to convert the dimensions of the North American continent into meters from kilometers since the density is in kg/m³. That's 4700km x 1000m/km =4.7e6m for length & width, and 25km x 1000m/km = 2.5e4m for depth. Thus, the volume of the continent becomes 4.7e6m x 4.7e6m x 2.5e4m = 5.54e17m³.
Now, we multiply by the rock's density, which is 2880 kg/m³. So, mass = volume * density = 5.54e17 m³ x 2880 kg/m³ = 1.59712e21 kg. So, the theoretical mass of the North American continent is approximately 1.59712e21 kilograms considering its simplified shape as a slab of rock.
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A proton moves through an electric potential created by a number of source charges. Its speed is 2.5 x 105 m/s at a point where the potential is 1500 V. What will be the proton’s speed a short time later when it reaches a point where the potential is −500 V? The mass of a proton is 27 m 1.67 10 kg.
Answer:
Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s
Explanation:
Given;
initial speed of proton, u = 2.5 x 10⁵ m/s
initial potential, V = 1500 V
mass of proton = 1.67 x 10⁻²⁷ kg
Work done, W = eV= ΔK.E = ¹/₂mu²
eV = ¹/₂mu² (J)
where;
e is the charge of the proton in coulombs
V is the electric potential in volts
m is the mass of the proton in kg
u is the speed of the proton in m/s
[tex]m =\frac{2eV_1}{u_1{^2}} = \frac{2eV_2}{u_2{^2}} = \frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}}[/tex]
[tex]\frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}} = \frac{1500}{(2.5*10^5)^2} = \frac{500}{u_2{^2}} \\\\u_2{^2} =\frac{500*(2.5*10^5)^2}{1500} = 0.333*6.25*10^{10}\\\\u_2 = \sqrt{0.333*6.25*10^{10}} =1.4 *10^5 \ m/s[/tex]
Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s
Air enters a horizontal, well-insulated nozzle operating at steady state at 12 bar, 500 K, with a velocity of 50 m/s and exits at 7 bar, 440 K. The mass flow rate is 1 kg/s. Determine the net force, in N, exerted by the air on the duct in the direction offlow.Answer:
Answer:
The net force is 300.8 N
Explanation:
∆H = Cp(T1 - T2) = 1/2(V2^2 - V1^2)
Cp is the heat capacity of air at constant pressure = 1005 J/kg
T1 is initial temperature of air = 500 K
T2 is the exit temperature of air = 440 K
V1 is the initial velocity of air = 50 m/s
V2 is the exit velocity of air
1005(500 - 440) = 1/2(V2^2 - 50^2)
60,300×2 = V2^2 - 2,500
V2^2 = 120,600 + 2,500 = 123,100
V2 = sqrt(123,100) = 350.8 m/s
Net force (F) = mass flow rate × change in velocity = 1 kg/s × (350.8 - 50)m/s = 1 kg/s × 300.8 m/s = 300.8 kgm/s^2 = 300.8 N
The net force exerted by the air on the duct in the direction of flow is approximately [tex]\( 0.00293 \) N[/tex].
Given:
- Inlet conditions (state 1):
- [tex]\( P_1 = 12 \) bar = \( 12 \times 10^5 \) Pa[/tex]
- [tex]\( T_1 = 500 \) K[/tex]
- [tex]\( v_1 = 50 \) m/s[/tex]
- Exit conditions (state 2):
- [tex]\( P_2 = 7 \) bar = \( 7 \times 10^5 \) Pa[/tex]
- [tex]\( T_2 = 440 \) K[/tex]
- Mass flow rate, [tex]\( \dot{m} = 1 \) kg/s[/tex]
1. Calculate specific volumes:
[tex]\( v_1 = \frac{RT_1}{P_1} = \frac{287 \times 500}{12 \times 10^5} \approx 0.02379 \) m^3/kg[/tex]
[tex]\( v_2 = \frac{RT_2}{P_2} = \frac{287 \times 440}{7 \times 10^5} \approx 0.02086 \) m^3/kg[/tex]
2. Calculate the change in specific volume:
[tex]\( \Delta v = v_2 - v_1 = 0.02086 - 0.02379 = -0.00293 \) m^3/kg[/tex]
3. Calculate the net force in the direction of flow:
[tex]\( F = \dot{m} \cdot \Delta v = 1 \cdot (-0.00293) = -0.00293 \) N[/tex]
Since the force is negative, it indicates that the force is acting opposite to the direction of flow.
Therefore, the net force exerted by the air on the duct in the direction of flow is approximately [tex]\( 0.00293 \) N[/tex].
Two coils of wire are placed close together. Initially, a current of 1.80 A exists in one of the coils, but there is no current in the other. The current is then switched off in a time of 4.02 x 10-2 s. During this time, the average emf induced in the other coil is 4.38 V. What is the mutual inductance of the two-coil system?
Answer: the mutual inductance is - 0.0978H
Explanation:
Detailed explanation and calculation is shown in the image below
Answer:
M = 0.1H
Explanation:
Please see attachment below.
Suppose that at a price of $2.60, the quantity of output demanded is 17, and at a price of $6.30, the quantity of output demanded is 8. What is the elasticity of demand? (Ignore the negative sign.)
Answer: elastic (e = 2.43)
Explanation:
The price elasticity formulae is given below as
Elasticy of price = change in quantity demanded / change in price.
P1 =$2.60, P2 = $6.30, q1 =17 and q2 = 8
e = q2 - q1/ P2 - P1
e = 8 - 17/ 6.30 - 2.60
e = - 9 /3.7
e = - 2.43
We take the modulus of e to have a positive value. Hence e = 2.43
Since e is greater than 1, then the elasticity of demand is elastic
Final answer:
The elasticity of demand between the given prices is calculated using the midpoint formula. It results in a value of -0.865 (ignoring the minus sign), indicating that the demand is elastic.
Explanation:
To calculate the elasticity of demand, we can use the midpoint formula, which is defined as the percentage change in quantity demanded divided by the percentage change in price. The formula for computing elasticity is:
Price Elasticity of Demand (Ed) = ([(Q2 - Q1) / ((Q2 + Q1)/2)] / [(P2 - P1) / ((P2 + P1)/2)]) * 100
Using the information provided: At a price of $2.60, quantity demanded is 17, and at a price of $6.30, quantity demanded is 8. We can fill in the values:
(Q2 - Q1) is (8 - 17) = -9
(Q2 + Q1)/2 is (8 + 17)/2 = 12.5
(P2 - P1) is ($6.30 - $2.60) = $3.70
(P2 + P1)/2 is ($6.30 + $2.60)/2 = $4.45
Now putting all these into the formula:
Ed = [(-9 / 12.5) / (3.70 / 4.45)] * 100
The percentage change in quantity is -72%, and the percentage change in price is 83.15%. Therefore:
Ed = (-0.72 / 0.8315) * 100
Ed = -0.865 (ignoring the minus sign)
This value is greater than 1, which indicates that the demand is elastic between these two prices; meaning, the quantity demanded is quite responsive to price changes.
Remember, we ignore the negative sign and use the absolute value when discussing elasticity, even though it is understood that price and quantity demanded move in opposite directions.
after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF capacitor from a camera flash unit retains a voltage of 150 V when an unwary student removes it from the camera. If the student accidentally touches the two terminals with his hands, and if the resistance of his body between his hands is 1.8 kΩ, for how long will the current across his chest exceed the danger level of 50 mA?
Answer:
The time is [tex]110.16\times10^{-3}\ sec[/tex]
Explanation:
Given that,
Capacitor = 120 μF
Voltage = 150 V
Resistance = 1.8 kΩ
Current = 50 mA
We need to calculate the discharge current
Using formula of discharge current
[tex]i_{0}=\dfrac{V_{0}}{R}[/tex]
Put the value into the formula
[tex]i_{0}=\dfrac{150}{1.8\times10^{3}}[/tex]
[tex]i_{0}=83.3\times10^{-3}\ A[/tex]
We need to calculate the time
Using formula of current
[tex]i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}[/tex]
Put the value into the formula
[tex]50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}[/tex]
[tex]\dfrac{50}{83.3}=e^{\frac{-t}{RC}}[/tex]
[tex]\dfrac{-t}{RC}=ln(0.600)[/tex]
[tex]t=0.51\times1.8\times10^{3}\times120\times10^{-6}[/tex]
[tex]t=110.16\times10^{-3}\ sec[/tex]
Hence, The time is [tex]110.16\times10^{-3}\ sec[/tex]
The current across the student's chest will exceed the danger level for approximately 216 seconds.
Explanation:The current that flows through the student's chest when he accidentally touches the terminals of the 120 mF capacitor can be calculated using Ohm's law: I = V/R, where I is the current, V is the voltage, and R is the resistance. In this case, the voltage is 150 V and the resistance of the student's body is 1.8 kΩ. Plugging in these values, we can find the current.
I = (150 V)/(1.8 kΩ) = 0.0833 A
To find how long the current will exceed the danger level of 50 mA, we need to calculate the time it takes for the charge on the capacitor to dissipate through the student's body. Since the current is constant, we can use the formula Q = I*t, where Q is the charge and t is the time. Rearranging the formula, we can solve for t.
t = Q/I = (120 mF * 150 V) / (0.0833 A) = 216 s
Therefore, the current will exceed the danger level of 50 mA for approximately 216 seconds.
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Two long, parallel wires are separated by a distance of 3.30 cm. The force per unit length that each wire exerts on the other is 5.00×10−5 N/m, and the wires repel each other. The current in one wire is 0.620 A.
a. What is the current in the second wire?
b. Are the two currents in the same direction or in opposite directions?
Answer with Explanation:
We are given that
Distance between two parallel long wires=r=3.3 cm=[tex]\frac{3.3}{100}=0.033m[/tex]
1 m=100 cm
[tex]\frac{F}{l}=5\times 10^{-5} N/m[/tex]
[tex]I_1=0.62 A[/tex]
a.We have to find the current in the second wire.
We know that
[tex]\frac{F}{l}=\frac{2\mu_0I_1I_2}{4\pi r}[/tex]
Using the formula
[tex]5\times 10^{-5}=\frac{2\times 10^{-7}\times 0.62\times I_2}{0.033}[/tex]
Where [tex]\frac{\mu_0}{4\pi}=10^{-7}[/tex]
[tex]I_2=\frac{5\times 10^{-5}\times 0.033}{2\times 10^{-7}\times 0.62}[/tex]
[tex]I_2=13.3 A[/tex]
Hence, the current in the second wire=13.3 A
b.We are given that the wires repel each other.When the current carrying in the wires in opposite direction then, the wires repel to each other.
Hence,the two currents in opposite directions.
During a very quick stop, a car decelerates at 6.2 m/s2. Assume the forward motion of the car corresponds to a positive direction for the rotation of the tires (and that they do not slip on the pavement). Randomized Variables at = 6.2 m/s2 r = 0.275 m ω0 = 93 rad/s
a. What is the angular acceleration of its tires in rad/s2, assuming they have a radius of 0.275 m and do not slip on the pavement?
b. How many revolutions do the tires make before coming to rest, given their initial angular velocity is 93 rad/s ?
c. How long does the car take to stop completely in seconds?
d. What distance does the car travel in this time in meters?
e. What was the car’s initial speed in m/s?
Answer:
a) [tex]-22.5 rad/s^2[/tex]
b) 30.6 revolutions
c) 4.13 s
d) 52.9 m
e) 25.6 m/s
Explanation:
a)
The relationship between linear acceleration and angular acceleration for an object in circular motion is given by
[tex]a=\alpha r[/tex]
where
[tex]a[/tex] is the linear acceleration
[tex]\alpha[/tex] is the angular acceleration
r is the radius of the motion of the object
For the tires of the car in this problem, we have:
[tex]a=-6.2 m/s^2[/tex] is the linear acceleration (the car is slowing down, so it is a deceleration, therefore the negative sign)
r = 0.275 m is the radius of the tires
Solving for [tex]\alpha[/tex], we find the angular acceleration:
[tex]\alpha = \frac{a}{r}=\frac{-6.2}{0.275}=-22.5 rad/s^2[/tex]
b)
To solve this part of the problem, we can use the suvat equation for the rotational motion, in particular:
[tex]\omega^2 - \omega_0^2 = 2\alpha \theta[/tex]
where:
[tex]\omega[/tex] is the final angular velocity
[tex]\omega_0[/tex] is the initial angular velocity
[tex]\alpha[/tex] is the angular acceleration
[tex]\theta[/tex] is the angular displacement
Here we have:
[tex]\omega=0[/tex] (the tires come to a stop)
[tex]\omega_0 = 93 rad/s[/tex]
[tex]\alpha = -22.5 rad/s^2[/tex]
Solving for [tex]\theta[/tex], we find the angular displacement:
[tex]\theta=\frac{\omega^2-\omega_0^2}{2\alpha}=\frac{0^2-(93)^2}{2(-22.5)}=192.2 rad[/tex]
And since 1 revolution = [tex]2\pi rad[/tex],
[tex]\theta=\frac{192.2}{2\pi}=30.6 rev[/tex]
c)
To solve this part, we can use another suvat equation:
[tex]\omega=\omega_0 + \alpha t[/tex]
where in this case, we have:
[tex]\omega=0[/tex] is the final angular velocity, since the tires come to a stop
[tex]\omega_0 = 93 rad/s[/tex] is the initial angular velocity
[tex]\alpha=-22.5 rad/s^2[/tex] is the angular acceleration
t is the time
Solving for t, we can find the time required for the tires (and the car) to sopt:
[tex]t=\frac{\omega-\omega_0}{\alpha}=\frac{0-93}{-22.5}=4.13 s[/tex]
d)
The car travels with a uniformly accelerated motion, so we can find the distance it covers by using the suvat equations for linear motion:
[tex]s=vt-\frac{1}{2}at^2[/tex]
where:
v = 0 is the final velocity of the car (zero since it comes to a stop)
t = 4.13 s is the time taken for the car to stop
[tex]a=-6.2 m/s^2[/tex] is the deceleration for the car
s is the distance covered during this motion
Therefore, substituting all values and calculating s, we find the distance covered:
[tex]s=0-\frac{1}{2}(-6.2)(4.13)^2=52.9 m[/tex]
e)
The relationship between angular velocity and linear velocity for a rotational motion is given by
[tex]v=\omega r[/tex]
where
v is the linear velocity
[tex]\omega[/tex] is the angular speed
r is the radius of the circular motion
In this problem:
[tex]\omega_0 = 93 rad/s[/tex] is the initial angular speed of the tires
r = 0.275 m is the radius of the tires
Therefore, the initial velocity of the car is:
[tex]u=\omega_0 r = (93)(0.275)=25.6 m/s[/tex] is the initial velocity of the car
There is a spot of paint on the front wheel of the bicycle. Take the position of the spot at time t=0 to be at angle θ=0 radians with respect to an axis parallel to the ground (and perpendicular to the axis of rotation of the tire) and measure positive angles in the direction of the wheel's rotation. What angular displacement θ has the spot of paint undergone between time 0 and 2 seconds? Express your answer in radians using three significant figures.
Answer:
[tex]\int\limits^2_0 {w} \, dt[/tex]
Explanation:
If Angular velocity w (omega ) is given, which is defined as, w = (change in angle)/(change in unit time).
then simply taking integral from 0 to 2 gives us the answer.
Answer:
Change in theta = 0.793rad
Explanation:
Please see attachment below.
A 1.83 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.442 and the coefficient of kinetic friction is 0.240 . How much force is needed to begin moving the book?
Answer:7.92 N
Explanation:
Given
mass of book [tex]m=1.83\ kg[/tex]
coefficient of static friction [tex]\mu _s=0.442[/tex]
coefficient of kinetic friction [tex]\mu _k=0.240[/tex]
To move the book, one need to overcome the static friction
Static friction [tex]F_s=\mu _sN[/tex]
[tex]F_s=\mu _s\times 1.83\times 9.8[/tex]
[tex]F_s=0.442\times 1.83\times 9.8[/tex]
[tex]F_s=7.92\ N[/tex]
After overcoming the Static friction , Force needed to move the block is
[tex]F_k=\mu _kN[/tex]
[tex]F_k=0.240\times 1.83\times 9.8[/tex]
[tex]F_k=4.30\ N[/tex]
Final answer:
The force needed to begin moving a 1.83 kg book on a flat desk, given a coefficient of static friction of 0.442, is approximately 7.90 N. This is calculated using the book's weight and the static friction formula fs(max) = μs * N.
Explanation:
To calculate the force needed to begin moving the book, we need to use the coefficient of static friction and the book's weight. The formula to find the maximum static frictional force (fs max) that must be overcome to start moving the object is fs(max) = μs * N, where μs is the coefficient of static friction and N is the normal force.
The book's weight is given by W = m * g, where m is the mass of the book (1.83 kg) and g is the acceleration due to gravity (9.81 m/s²). The weight of the book is equivalent to the normal force (N) exerted by the desk on the book since the book is resting on a flat surface and there are no other vertical forces acting on it. This simplifies the normal force to N = W = m * g.
Using the given coefficient of static friction (0.442) and the calculated normal force, the force needed to begin moving the book is fs(max) = 0.442 * (1.83 kg * 9.81 m/s²) ≈ 7.90 N. Therefore, a force slightly greater than 7.90 N is required to overcome static friction and start moving the book.
Describe how an uncharged pith ball suspended from a string can be used to test whether an object is charged. Predict what will happen when an uncharged pith ball is brought near one of the poles of the magnet. Explain.
Answer:
Pithball electroscope is used to determine if any object has static charge.
It consists of one or two small balls of a lightweight non-conductive substance. When this material is moved near an object having static charge, polarization will be induced in the atoms of pithballs which will either attract or repel the object depending on the nature of the charge in it.
It will not move in case it is brought near to a neutral object.
Similarly, the pith ball will move when it will brought into the magnetic field of the magnet as it will also induce polarization within the atoms of the pithballs.
If the momentum of a 1000 kg car travelling at 10 m/s was transferred completely to a 20.0 kg traffic barrier, what would the final speed of the barrier be
Answer:
500 m/s
Explanation:
Momentum, p is a product of mass and velocity. From law of conservation of momentum, the initial momentum equals final momentum
[tex]m_1v_1=m_2v_2\\1000*10=20v_2\\v_2=10000/20=500 m/s[/tex]
Here m and v represent mass and velocity respectively and subscripts 1 and 2 represent car and barrier respectively
Therefore, the velocity of barrier will be 500 m/s
An object of mass m is lowered at constant velocity at the end of a string of negligible mass. As it is lowered a vertical distance h, its gravitational potential energy changes by ∆Ug = −m g h. However, its kinetic energy remains constant, so that if we define E = K + Ug, we find ∆E = −m g h. Why isn’t the total energy E conserved? 1. Because the universe is accelerating in its expansion, the object is actually at rest and not descending ... the earth moves away as fast as it moves "down." 2. An external force is doing work on the system. 3. In reality, all objects are massless, so that m = 0 and ∆E = 0. 4. The acceleration of the system is zero. 5. The net force on the system is not zero. 6. Ug is defined incorrectly as if gravity were a constant force. 7. The total energy is indeed conserved, since ∆E = ∆Ug. 8. E is useless in real-world examples like this.
Answer:
Mechanical would have been conserved if only the force of gravity (the weight of the object does work on the system). The tension force does work also on the system but negative work instead. The net force acting of the system is zero since the upward tension in the string suspending the object is equal to the weight of the object but acting in the opposite direction. As a result they cancel out. In the equation above the effect of the tension force on the object has been neglected or not taken into consideration. For the mechanical energy E to be conserved, the work done by this tension force must be included into the equation. Otherwise it would seem as though energy has been generated in some manner that is equal in magnitude to the work done by the tension force.
The conserved form of the equation is given by
E = K + Ug + Wother.
In this case Wother = work done by the tension force.
In that form the total mechanical energy is conserved.
Final answer:
The correct answer is: An external force is doing work on the system.
Explanation:
When the object is lowered at constant velocity, an external force (in this case, the force exerted by the person lowering the object) is doing work against gravity to keep the velocity constant. This work done by the external force results in a change in the object's gravitational potential energy (∆Ug = -mgh).
Since work is being done on the system externally, the total mechanical energy of the system (the sum of kinetic and potential energies) is not conserved.
However, in the scenario described, an external force is doing work on the system. Work is being done to counteract gravity and maintain a constant velocity. This work done by the external force results in a change in the object's gravitational potential energy (∆Ug = -mgh), where "m" is the mass of the object, "g" is the acceleration due to gravity, and "h" is the vertical distance through which the object is lowered.
As a result, the total mechanical energy of the system (the sum of kinetic and potential energies) is not conserved. The work done by the external force manifests as a change in the object's gravitational potential energy, causing a decrease in potential energy as the object is lowered. This decrease in potential energy is exactly balanced by the work done by the external force, so the total mechanical energy of the system remains constant, despite the fact that kinetic energy remains constant throughout the process.