Answer:
a. Increases
Explanation:
Conceptually, when a block of certain mass slides on the rough ramp kept on the earth which happens under the influence of gravity.The block is initially at rest but as the acceleration due to gravity acts on the block at the ramp. The ramp is rough so it applies kinetic friction on the moving block as the block slides down the slope.By the definition we know that the acceleration is the rate of change in velocity and here we have acceleration component in the direction of motion of the block.
Mathematically:
when the body is moving down:
[tex]v=u+gt[/tex]
where:
[tex]v=[/tex] final velocity of the block
[tex]u=[/tex] initial velocity of the block
[tex]g=[/tex] acceleration due to gravity
[tex]t=[/tex] time of observation during the instance of motion
From above it is clear that the velocity of the block will increase as the time passes during the motion.As we know that kinetic energy is given as:
[tex]KE=\frac{1}{2} \times m.v^2[/tex]
where:
[tex]m=[/tex] mass of the block which remains constant (macroscopically)
[tex]v=[/tex] velocity of the block (which increases here as the body descends)
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
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 specific surface energy for aluminum oxide is 0.90 N/m, and the elastic modulus is 393 GPa. Compute the critical stress, in MPa, required for propagation of a surface crack of length 0.25 mm. Round answer to 3 significant figures and report in the format: 12.3 MPa
Answer:
42.4 Npa
Explanation:
Explanation is attached in the picture below
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)
a heavy box is pulled across the floor with a rope. The rope makes an angle of 60 degrees with the floor. A force of 75 N is exerted on the rope. What is the component of the force parallel to the floor
Answer:
37.5 N
Explanation:
Horizontal component is represented by Fx and is given as
Fx = F cos θ
Here,
θ = 60 degrees
F= 75 N
So,
Fx= 75 cos (60°)
==> Fx = 75 × 0.5 =37.5 N
Final answer:
The horizontal component of the force acting parallel to the floor when a 150 N force is applied at a 60-degree angle to the horizontal is 75 N, calculated using the cosine of the angle.
Explanation:
The question involves resolving a force into its components, which is a common problem in physics. When a force is applied at an angle to the horizontal, it has both horizontal and vertical components. The horizontal (parallel) component ([tex]F_{parallel}[/tex]) is found using the cosine function of the angle Θ , which in this case is 60 degrees. The formula is [tex]F_{parallel}[/tex] = F * cos(Θ), where F is the magnitude of the force.
Given that a force of 150 N is applied at an angle of 60 degrees to the horizontal, the horizontal component of the force can be calculated as follows:
[tex]F_{parallel}[/tex] = 150 N * cos(60°)
[tex]F_{parallel}[/tex] = 150 N * 0.5
[tex]F_{parallel}[/tex] = 75 N
The horizontal component of the force acting parallel to the floor is 75 N. To find the net force acting on the box and thus the acceleration, you would subtract the frictional force from the parallel component of the applied force and then use Newton's second law, F = m * a, where F is the net force, m is the mass of the object and a is the acceleration. However, the frictional force is not needed for this particular question about the horizontal component.
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.
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.
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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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.
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].
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
Ask Your Teacher Given the displacement vectors A with arrow = (5.00 î − 7.00 ĵ + 5.00 k) m and B with arrow = (3.00 î + 7.00 ĵ − 3.00 k) m, find the magnitudes of the following vectors and express each in terms of its rectangular components.
Answer: [tex]||\vec A|| \approx 9.95 m,||\vec B|| \approx 8.19 m[/tex]
Explanation:
Magnitudes can be calculated by using Pythagorean theorem:
[tex]||\vec A|| =\sqrt{(5m)^{2}+(-7m)^{2}+(5m)^{2}}\\||\vec A|| \approx 9.95 m\\||\vec B|| =\sqrt{(3m)^{2}+(7m)^{2}+(-3m)^{2}}\\||\vec B|| \approx 8.19 m\\[/tex]
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.
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 human being can be electrocuted if a current as small as 55 mA passes near the heart. An electrician working with sweaty hands makes good contact with the two conductors he is holding. If his resistance is 2300 Ω, what might the fatal voltage in volts be?
Answer:
The lethal voltage for the electrician under those conditions is 126.5 V.
Explanation:
To discover what is the lethal voltage to the electrician we need to find out what is the voltage that produces 55 mA = 0.055 A when across a resistance of 2300 Ohms (Electrician's body resistancy). For that we'll use Ohm's Law wich is expressed by the following equation:
V = i*R
Where V is the voltage we want to find out, i is the current wich is lethal to the electrician and R is his body resistance. By applying the given values we have:
V = 0.055*2300 = 126.5 V.
The lethal voltage for the electrician under those conditions is 126.5 V.
Answer:
126.5 V
Explanation:
Using Ohm's Law,
V = IR............................. Equation 1
Where V = Voltage, I = current, R = resistance.
Note: The current needed to bring about the fatal voltage is equal to the current that will cause human being to be electrocuted.
Given: I = 55 mA = 55/1000 = 0.055 A, R = 2300 Ω
Substitute into equation 1
V = 0.055×2300
V = 126.5 V.
Hence the fatal voltage = 126.5 V
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
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.
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.
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.
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.
(e) Imagine the Moon, with 27.3% of the radius of the Earth, had a charge 27.3% as large, with the same sign. Find the electric force the Earth would then exert on the Moon.
Explanation:
Below is an attachment containing the solution.
To find the electric force the Earth would exert on the Moon, we can use Coulomb's law. The ratio of the electric force between the Earth and the Moon compared to the electric force between two charges with the same sign is 0.273^3.
Explanation:To find the electric force the Earth would exert on the Moon, we can use Coulomb's law. Coulomb's law states that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for electric force is:
F = k * (q1 * q2) / r^2
Where F is the electric force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges. In this case, the charge of the Moon is 27.3% of the charge of the Earth, and the radius of the Moon is also 27.3% of the radius of the Earth. Using these values, we can calculate the electric force.
Let's assume the charge of the Earth is q1 and the charge of the Moon is q2. Since the charge of the Moon is 27.3% as large as the charge of the Earth, we can write q2 = 0.273 * q1. Similarly, the radius of the Moon is 27.3% of the radius of the Earth, so we can write r = 0.273 * R, where R is the radius of the Earth. Plugging these values into Coulomb's law formula:
F = k * (q1 * (0.273 * q1)) / (0.273 * R)^2
Simplifying the equation, we get:
F = k * (q1^2 * 0.273) / (0.273^2 * R^2)
The ratio of the electric force between the Earth and the Moon compared to the electric force between two charges with the same sign is 0.273^3.
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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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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
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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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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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
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.
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.
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.