Answer:
a) 200m, 100m/s
b) 710.20m
c) -117.98 m/s
d) 26.24 s
Explanation:
To solve this we have to use the formulas corresponding to a uniformly accelerated motion problem:
[tex]V=Vo+a*t[/tex] (1)
[tex]X=Xo+Vo*t+\frac{1}{2}*a*t^2\\[/tex] (2)
[tex]V^2=Vo^2+2*a*X[/tex] (3)
where:
Vo is initial velocity
Xo=intial position
V=final velocity
X=displacement
a)
[tex]X=0+0*4+\frac{1}{2}*25*4^2[/tex]
the intial position is zero because is lauched from the ground and the intial velocitiy is zero because it started from rest.
[tex]X=200m[/tex]
[tex]V=0+25*4[/tex]
[tex]V=100m/s[/tex]
b)
The intial velocity is 100m/s we know that because question (a) the acceleration is -9.8[tex]\frac{m}{s^2}[/tex] because it is going downward.
[tex]0=100^2+2*(-9.8)*X\\X=510.20\\totalheight=200+510.20=710.20m[/tex]
c)
In order to find the velocity when it crashes, we can use the formula (3).
the initial velocity is 0 because in that moment is starting to fall.
[tex]V^2=0^2+2*(-9.8)*(710.20)\\V=-117.98 m/s[/tex]
the minus sign means that the object is going down.
d)
We can find the total amount of time adding the first 4 second and the time it takes to going down.
to calculate the time we can use the formula (2) setting the reference at 200m:
[tex]-200=0+100*t+\frac{1}{2}*(-9.8)*t^2[/tex]
solving this we have: time taken= 22.24 seconds
total time is:
total=22.24+4=26.24 seconds.
A room contains 75 kg of air at 100 kPa and 15°C. The room has a 250-W refrigerator (the refrigerator consumes 250 W of electricity when running), a 120-W TV, a 1.8-kW electric resistance heater, and a 50-W fan. During a cold winter day, it is observed that the refrigerator, the TV, the fan, and the electric resistance heater are running continuously but the air temperature in the room remains constant. The rate of heat loss from the room that day is _____ kJ/h.
Answer:
Rate of heat loss=7992 kJ/h
Explanation:
First, we can consider the room as a closed system, so we can use the first law of the thermodynamic:
[tex]Qnet,in-Wnet,out = Esystem[/tex]
The incoming net heat is the incoming heat less the leaving heat, the leaving net work is the leaving work less the incoming work and the system energy is only the change of the intern energy because we have a stationary system, so:
[tex](Qin-Qout)-(Wout-Win) =U2-U1[/tex]
From the problem we consider that there is no change of the temperature thus there is no change in the intern energy(U2=U1), moreover there is no incoming heat neither exit work, all the work is made to the system as a result the leaving heat is equal to the incoming work:
[tex](0-Qout)-(0-Win) =0[/tex]
[tex]-Qout+Win =0[/tex]
[tex]Qout=Win[/tex]
Substituting the known values we can get the rate of heat loss (exit heat):
[tex]Q_{out}=Q_{loss}=W_{ref}+W_{TV}+W_{res}+W_{fan}[/tex]
[tex]Q_{loss}=250 [W]+120 [W]+1800 [W]+50 [W][/tex]
[tex]Q_{loss}=2220 [W]=2220 [\frac{J}{s} ][/tex]
Converting to Joules/hours:
[tex]Q_{loss}=2220 [\frac{J}{s} ][\frac{3600 s}{1 h} ][/tex]
[tex]Q_{loss}=7992000 [\frac{J}{h} ][/tex]
Finally the rate of heat loss is:
[tex]Q_{loss}=7992 [\frac{kJ}{h} ][/tex]
Final answer:
The rate of heat loss from a room with several electrical appliances running and constant air temperature is 7992 kJ/h, calculated by summing the power of all appliances and converting to kJ/h.
Explanation:
To find the rate of heat loss from the room, we need to consider the energy input from all the electrical appliances and the fact that the air temperature in the room remains constant. The total power consumption of the appliances is the sum of the power of the refrigerator, TV, electric resistance heater, and fan, which is 250 W + 120 W + 1.8 kW + 50 W = 2.22 kW.
Since the temperature is constant, the rate of energy input equals the rate of heat loss. This input is the combined power consumption of the appliances and is entirely converted into heat (assuming 100% efficiency for simplicity).
To express this rate in kilojoules per hour (kJ/h), multiply the power in kilowatts by the number of hours and by 1000 to convert from kW to kJ since 1 kW = 1 kJ/s:
2.22 kW = 2.22 kJ/s
So, the rate of heat loss is:
2.22 kJ/s × 3600 s/h = 7992 kJ/h
A bird watcher meanders through the woods, walking 1.46 km due east, 0.123 km due south, and 4.24 km in a direction 52.4° north of west. The time required for this trip is 1.285 h. Determine the magnitudes of the bird watcher's (a)displacement and (b) average velocity.
Answer:
(a). The displacement is 3.11 m.
(b). The average velocity is 2.42 km/hr.
Explanation:
Given that,
A bird watcher meanders through the woods, walking 1.46 km due east, 0.123 km due south, and 4.24 km in a direction 52.4° north of west.
(a). We need to calculate the displacement
Using Pythagorean theorem
[tex]D=\sqrt{(OA)^2+(AB)^2}[/tex]
[tex]D=\sqrt{(1.46-4.24\sin52.4)^2+(0.123-4.24\cos52.4)^2}[/tex]
[tex]D=3.11\ km[/tex]
(b). We need to calculate the average velocity
Using formula of average velocity
[tex]v_{avg}=\dfrac{D}{T}[/tex]
Where, D = displacement
T = time
Put the value into the formula
[tex]v_{avg}=\dfrac{3.11}{1.285}[/tex]
[tex]v_{avg}=2.42\ km/hr[/tex]
Hence, (a). The displacement is 3.11 m.
(b). The average velocity is 2.42 km/hr.
An object moving on a horizontal, frictionless surface makes a glancing collision with another object initially at rest on the surface. In this case which of the following is true about momentum and kinetic energy? 4A) Momentum is always conserved, and kinetic energy may be conserved. (B) Kinetic energy is always conserved, and momentum may be conserved. (C) Momentum is always conserved, and kinetic energy is never conserved. (D) Both momentum and kinetic energy are always conserved. (E) Neither momentum nor kinetic energy is conserved.
Answer:
The correct answer is option 'a' 'The momentum is always conserved while as the kinetic energy may be conserved'
Explanation:
The conservation of momentum is a basic principle in nature which is always valid in an collision between 'n' number of objects if there are no external forces on the system. It is valid for both the cases weather the collision is head on or glancing or weather the object is elastic or inelastic.
The energy is only conserved in a collision that occurs on a friction less surface and the objects are purely elastic. Since in the given question it is mentioned that only the surface is friction less and no information is provided regarding the nature of the objects weather they are elastic or not hence we cannot conclusively come to any conclusion regarding the conservation of kinetic energy as the objects may be inelastic.
In a rectangular coordinate system, a positive point charge q = 6.50 nC is placed at the point x=0.190 m ,y=0, and an identical point charge is placed at x= -0.190 m , y=0. Find the x and y components and the magnitude and direction of the electric field at the following points. A) Find the x and y components of the electric field at the origin.
B) Find the magnitude of the electric field at the origin.
Answer:
A) x and y components of the electric field (Ep) at the origin.
Epx = -1620.5 N/C
Epy = -1620.5 N/C
Ep = (1620.5(-i)+1620.5(-j)) N/C
B) Magnitude of the electric field (Ep) at the origin.
Ep= 2291.7 N/C
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Equivalence
1nC= 10⁻⁹C
Data
K= 9x10⁹N*m²/C²
q₁ = q₂= +6.5nC=+6.5 *10⁻⁹C
d₁=d₂=0.190m
Graphic attached
The attached graph shows the field due to the charges:
Ep₁ : Electric Field at point P (x=0, y=0) due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge.
Ep₂: Electric Field at point P (x=0, y=0) due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge
Ep: Total field at point P due to charges q₁ and q₂.
Because q₁ = q₂ and d₁ = d₂, then, the magnitude of Ep₁ is equal to the magnitude of Ep₂
Ep₁ = Ep₂ = k*q/d² = 9*10⁹*6.5*10⁻⁹/0.190m² = 1620.5 N/C
Look at the attached graphic :
Epx = Ep₁= -1620.5 N/C
Epy = Ep₂= -1620.5 N/C
A) x and y components of the electric field (Ep) at the origin.
Ep = (1620.5(-i)+1620.5(-j)) N/C
B) Magnitude of the electric field (Ep) at the origin.
[tex]E_{p} =\sqrt{1620.5^{2}+1620.5^{2} } = 2291.7 \frac{N}{C}[/tex]
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 30.0 mph and half the distance at 70.0 mph . On her return trip, she drives half the time at 30.0 mph and half the time at 70.0 mph . What is Julie's average speed on the way to Grandmother's house?
What is her average speed on the return trip?
Answer:
Explanation:
Given
Distance to grandmother's house=100 mi
it is given that during return trip Julie spend equal time driving with speed 30 mph and 70 mph
Let Julie travel x mi with 30 mph and 100-x with 70 mph
[tex]\frac{x}{30}=\frac{100-x}{70}[/tex]
x=30 mi
Therefore
Julie's Average speed on the way to Grandmother's house[tex]=\frac{100}{\frac{50}{30}+\frac{50}{70}}[/tex]
=42 mph
On return trip
[tex]=\frac{100}{2\frac{30}{30}}=50 mph[/tex]
Final answer:
Julie's average speed to Grandmother's house is 42.02 mph. The average speed for the return trip cannot be calculated without the total time of travel or the division of time at each speed. Her average speed for the entire round trip is approximately 26.67 mph, and the average velocity is 0 mph since the starting and ending points are the same.
Explanation:
To calculate Julie's average speed on the way to Grandmother's house, we need to use the formula for average speed, which is total distance traveled divided by the total time taken. Since Julie drives half the distance at 30.0 mph and the other half at 70.0 mph for a total distance of 100 miles, we can calculate the time taken for each half. At 30 mph, for 50 miles, the time taken is ≈ 1.67 hours, and at 70 mph, for 50 miles, the time taken is ≈ 0.71 hours. The total time is ≈ 2.38 hours. Therefore, average speed is 100 miles ÷ 2.38 hours = 42.02 mph.
On the return trip, she drives half the time at 30.0 mph and half the time at 70.0 mph. However, without additional information, we cannot calculate the average speed for the return trip because we need the total time or the portion of time spent at each speed. The calculation is different from the first trip because this time it is dependent on time, not distance. For her entire trip, if Julie returned home 7 hours and 30 minutes after she left, and assuming the same 100-mile distance back, her average speed for the entire trip is the total distance (200 miles) divided by the total time (7.5 hours), which is ≈ 26.67 mph. However, since she returns to the starting point, her displacement is zero, and thus her average velocity for the entire trip is 0 mph, similar to the example in Figure 2.10.
Three equal charges are placed at the corners of an equilateral triangle 0.50 m on a side. What are the magnitude of the force on each charge if the charges are each -3.1 x 10^-9 C?
Answer:
the magnitude of the force is 192.29 N
Solution:
As per the question:
Charges present on the corner of the triangle are same, Q = [tex]- 3.1\times 10^{- 9} C[/tex]
Since, its an equilateral triangle, distance between the charges, l = 0.50 m
Now,
The Coulomb force on a charge due to the other is:
[tex]F_{C} = K\frac{Q^{2}}{l^{2}}[/tex]
where
K = Coulomb constant = [tex]9\times 10^{9} C^{2}/m^{2}[/tex]
[tex]F_{C} = (9\times 10^{9})\frac{(3.1\times 10^{- 9})^{2}}{0.5^{2}}[/tex]
[tex]F_{C} = 111.6 N[/tex]
The the net force on the charges in an equilateral triangle on all the charges due to each other:
[tex]F_{eq} = \sqrt{3}F_{C} = \sqrt{3}\times 111.6 = 193.29 N[/tex]
The magnitude of the force on each charge of -3.1 x 10^-9 C at the corners of an equilateral triangle 0.50 m on a side is 1.088 x 10^-7 N.
For one charge, the force due to the other two charges will be the vector sum of the forces due to each charge separately. Since the triangle is equilateral, the forces will have the same magnitude but different directions. Using Coulomb's law, we find the magnitude of the force between any two charges:
F = k × |q1 × q2| / r2
Where:
k is the Coulomb's constant (8.988 x 109 Nm2/C2)q1 and q2 are the magnitudes of the two chargesr is the distance between the chargesSubstituting the given values:
F = (8.988 x 109 Nm2/C2) × (3.1 x 10-9 C)2 / (0.50 m)2
This results in:
F = 1.088 x 10-7 N
As all sides are equal, and charges are equal, the net force will be the same in magnitude for each charge. The direction of the resulting force on each charge will be along the perpendicular bisector of the side opposite to each charge due to the symmetry of the configuration.
According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this principle to solve the following problems. (a) A wooden cylinder 30.0 cm high floats vertically in a tub of water (density 1:00 g/cm3). The top of the cylinder is 13.5 cm above the surface of the liquid. What is the density of the wood? (b) The same cylinder floats vertically in a liquid of unknown density. The top of the cylinder is 18.9 cm above the surface of the liquid. What is the liquid density? (c) Explain why knowing the length and width of the wooden objects is unnecessary in solving Parts (a) and (b). 2.15.
Answers:
a) [tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]
b) [tex]\rho_{liq}= 1.48 g/cm^{3}[/tex]
c) When we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.
Explanation:
a) According to Archimedes’ Principle:
A body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.
In this case we have a wooden cylinder floating (partially immersed) in water. This object does not completely fall to the bottom because the net force acting on it is zero, this means it is in equilibrium. This is due to Newton’s first law of motion, that estates if a body is in equilibrium the sum of all the forces acting on it is equal to zero.
Hence:
[tex]W_(cylinder)=B[/tex] (1)
Where:
[tex]W_(cylinder)=m.g[/tex] is the weight of the wooden cylinder, where [tex]m[/tex] is its mass and [tex]g[/tex] gravity.
[tex]B[/tex] is the Buoyant force, which is the force the fluid (water in this situation) exert in the submerged cylinder, and is directed upwards.
We can rewrite (1) as follows:
[tex]m_{cylinder}g=m_{water}g[/tex] (2)
On the other hand, we know density [tex]\rho[/tex] establishes a relationship between the mass of a body andthe volume it occupies. Mathematically is expressed as:
[tex]\rho=\frac{m}{V}[/tex] (3)
isolating the mass:
[tex]m=\rho V[/tex] (4)
Now we can express (2) in terms of the density and the volume of cylinder and water:
[tex]\rho_{cylinder} V_{cylinder} g=\rho_{water} V_{water} g[/tex] (5)
In this case [tex]V_{water}[/tex] is the volume of water displaced by the wooden cylinder (remembering Archimedes's Principle).
At this point we have to establish the total volume of the cylinder and the volume of water displaced by the sumerged part:
[tex]V_{cylinder}=\pi r^{2} h[/tex] (6)
Where [tex]r[/tex] is the radius and [tex]h=30 cm[/tex] the total height of the cylinder.
[tex]V_{water}=\pi r^{2} (h-h_{top})[/tex] (7)
Where [tex]h_{top}=13.5 cm[/tex] is the height of the top of the cylinder above the surface of water and [tex](h-h_{top})[/tex] is the height of the sumerged part of the cylinder.
Substituting (6) and (7) in (5):
[tex]\rho_{cylinder} \pi r^{2} h g=\rho_{water} \pi r^{2} (h-h_{top}) g[/tex] (8)
Clearing [tex]\rho_{cylinder}[/tex]:
[tex]\rho_{cylinder}=\frac{\rho_{water}(h-h_{top})}{h}[/tex] (9)
Simplifying;
[tex]\rho_{cylinder}=\rho_{water}(1-\frac{h_{top}}{h}[/tex] (10)
Knowing [tex]\rho_{water}=1g/cm^{3}[/tex]:
[tex]\rho_{cylinder}=1g/cm^{3}(1-\frac{13.5 cm}{30cm})[/tex] (11)
[tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex] (12) This is the density of the wooden cylinder
b) Now we have a different situation, we have the same wooden cylinder, which density was already calculated ([tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]), but the density of the liquid [tex]\rho_{liq}[/tex] is unknown.
Applying again the Archimedes principle:
[tex]\rho_{cylinder} V_{cylinder} g = \rho_{liq} V_{liq} g[/tex] (13)
Isolating [tex]\rho_{liq}[/tex]:
[tex]\rho_{liq}= \frac{\rho_{cylinder} V_{cylinder}}{V_{liq}}[/tex] (14)
Where:
[tex]V_{cylinder}=\pi r^{2} h[/tex]
[tex]V_{liq}=\pi r^{2} (h-h_{top})[/tex]
Then:
[tex]\rho_{liq}= \frac{\rho_{cylinder} \pi r^{2} h}{\pi r^{2} (h-h_{top})}[/tex] (15)
[tex]\rho_{liq}= \frac{\rho_{cylinder} h}{h-h_{top}}[/tex] (16)
[tex]\rho_{liq}= \frac{0.55 g/cm^{3} (30 cm)} {30 cm - 18.9 cm}[/tex] (17)
[tex]\rho_{liq}= 1.48 g/cm^{3}[/tex] (18) This is the density of the liquid
c) As we can see, it was not necessary to know the radius of the cylinder (we did not need to knoe its length and width), we only needed to know the part that was sumerged and the part that was above the surface of the liquid.
This is because in this case, when we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.
A 1050 kg sports car is moving westbound at 13.0 m/s on a level road when it collides with a 6320 kg truck driving east on the same road at 12.0 m/s. The two vehicles remain locked together after the collision. What is the velocity (magnitude) of the two vehicles just after the collision?
At what speed should the truck have been moving so that it and car are both stopped in the collision?
Find the change in kinetic energy of the system of two vehicles for the situations of part A.
Find the change in kinetic energy of the system of two vehicles for the situations of part C.
Answer:
a) [tex]v_{3} =8.43 m/s[/tex]
b) [tex]v_{2}=2.15m/s[/tex]
c) ΔK=[tex]-28.18x10^4J[/tex]
d)ΔK=[tex]-10.33x10^4J[/tex]
Explanation:
From the exercise we know that there is a collision of a sports car and a truck.
So, the sport car is going to be our object number 1 and the truck object number 2.
[tex]m_{1}=1050kg\\v_{1}=-13m/s\\m_{2}=6320kg\\v_{2}=12m/s[/tex]
Since the two vehicles remain locked together after the collision the final mass is:
[tex]m_{3}=7370kg[/tex]
a) To find the velocity of the two vehicles just after the collision we must use linear's momentum principle
[tex]p_{1}=p_{2}[/tex]
[tex]m_{1}v_{1}+ m_{2}v_{2}=m_{3}v_{3}[/tex]
[tex]v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3}}=\frac{(1050kg)(-13m/s)+(6320kg)(12m/s)}{7370kg}[/tex]
[tex]v_{3}=8.43m/s[/tex]
b) To find the speed the truck should have had so both vehicles stopped in the collision we need to use the same principle used before
[tex]m_{1}v_{1}+ m_{2}v_{2}=0[/tex]
[tex]v_{2}=\frac{-m_{1}v_{1}}{m_{2} }=\frac{-(1050kg)(-13m/s)}{(6320kg)}=2.15m/s[/tex]
c) To find the change in kinetic energy we need to do the following steps:
ΔK=[tex]k_{2}-k_{1}=\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )[/tex]
ΔK=[tex]\frac{1}{2}(7370)(8.43)^{2}-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(12)^{2} )=-28.18x10^{4}J[/tex]
d) The change in kinetic energy where the two vehicles stopped in the collision is:
ΔK=[tex]k_{2}-k_{1}=0-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )[/tex]
ΔK=[tex]-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(2.15)^{2} )=-10.33x10^4J[/tex]
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a mass of m2 = 86 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.79 and the coefficient of kinetic friction between the two crates is μk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem). 1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?
2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?
3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?
4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s2
5)As the upper crate slides, what is the acceleration of the lower crate?
Answer:
The sum of all forces for the two objects with force of friction F and tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F
1) no sliding infers: a₁ = a₂= a
The two equations become:
m₂a = T - m₁a
Solving for a:
a = T / (m₁+m₂) = 2.1 m/s²
2) Using equation(i):
F = m₁a = 51.1 N
3) The maximum friction is given by:
F = μsm₁g
Using equation(i) to find a₁ = a₂ = a:
a₁ = μs*g
Using equation(ii)
T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N
4) The kinetic friction is given by: F = μkm₁g
Using equation (i) and the kinetic friction:
a₁ = μkg = 6.1 m/s²
5) Using equation(ii) and the kinetic friction:
m₂a₂ = T - μkm₁g
a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²
1) a = 2.13 m/s²
2) F = 51.12 N
3) T_max = 851.62 N
4)a₁ = 6.08 m/s²
5) a₂ = 12.11 m/s²
This is based on the concept of frictional motion with coefficients of friction.
We are given;
Mass of smaller top crate; m₁ = 24 kg
Mass of larger bottom crate; m₂ = 86 kg
Coefficient of static friction; μ_s = 0.79
Coefficient of kinetic friction; μ_k = 0.62
For the top crate, the sum of forces will be expressed as;
F = m₁a₁ - - - (eq 1)
For the bottom crate, the sum of forces will be expressed as;
T - F = m₂a₂ - - - (eq 2)
Where F is frictional force exerted by small crate on big crate and T is tension.
1) We are told that the rope is pulled with a tension T = 234 N. And that the top crate will not slide. Thus, the acceleration of the small crate will be gotten from a combination of equation 1 and 2 to get the formula;
T = m₂a + m₁a
Making a which is the acceleration the subject gives;
a = T/(m₁+m₂)
Plugging in the relevant values gives;
a = 234/(24 + 86)
a = 2.13 m/s²
2) From eq(1) above, we see that the formula for the frictional force the small crate exerts on the big one is;
F = m₁a₁
Thus, Plugging in the relevant values gives;
F = 24 × 2.12
F = 51.12 N
3) maximum tension that the lower crate can be pulled at before the upper crate begins to slide will occur at the maximum friction. The formula for mx friction is;
(F₂)_max = μ_s•m₂•g
Using the concept of eq(1) We can say that;
F₂ = m₂a
m₂a = μ_s•m₂•g
m₂ will cancel out to give;
a = μ_s • g
From first answer earlier, we saw that;
T = m₂a + m₁a
T = (m₂ + m₁)a
Plugging in the relevant values gives;
T_max = (m₁ + m₂)μ_s*g
T_max = (24 + 86) × 0.79 × 9.8
T_max = 851.62 N
4) To get the acceleration of the upper crate, we will make use of the formula for kinetic friction which is:
F = μ_k*m₁g
From earlier, we saw that; F = m₁a₁
Thus;
m₁a₁ = μ_k*m₁g
m₁ will cancel out to get;
a₁ = μ_k*g
a₁ = 0.62 × 9.8
a₁ = 6.08 m/s²
5) the acceleration of the lower crate as the upper crate slides will be gotten by putting μ_k*m₁*g for m₁a₁ in eq(2) to get;
m₂a₂ = T - (μ_k*m₁*g)
Making a₂ the subject gives us;
a₂ = (T - μ_k•m₁•g)/m₂
Plugging in the relevant values;
a₂ = (1187 - (0.62 × 24 × 9.8))/86
a₂ = 12.11 m/s²
Read more at; brainly.com/question/4853862
A boat sounds a fog horn on a day when both the sea water and the air temperature are 25.0° C. The speed of sound in sea water is 1,533 m/s. How much earlier (in s) does a dolphin 1000 m from the source hear the sound than a person in a boat that is also 1000 m distant? (Ignore the time it takes the sound to reach the water surface.) Please show work/ explanations
(A) 0.652
(B) 2.12
(C) 2.24
(D) 2.77
(E) 2.90
The sound would reach the dolphin in the water about 2.12 seconds earlier than the person in the boat, using given speeds of sound in air and water, and the distance provided.
Explanation:The time it takes for sound to travel is calculated using the formula Time = Distance/Speed. Given, distance is 1000m in both air and sea water. We need to find the speed of sound in the air to calculate how long it will take the sound to reach the person in the boat.
The speed of sound in the air at 25°C is approximately 346.13 m/s. Using the formula: Time = Distance/Speed, the time it takes for the sound to reach the person in the boat is about 2.889 seconds (1000m/346.13 m/s).
The speed of sound in seawater is given as 1,533 m/s. Using the same formula, the time it takes for the sound to reach the dolphin in the water is about 0.652 seconds (1000m/1,533 m/s).
Therefore, the difference in time it takes for the sound to travel to the person and the dolphin is approximately 2.237 seconds (2.889 seconds - 0.652 seconds). Hence, the correct answer would be (B) 2.12. Although, it's close to the calculated difference, none of the available options are an exact match.
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On average, an eye blink lasts about 100 ms. How far does a fighter jet travel during a pilot's blink if the plane's average velocity is 3650 km/h?
Answer:
Distance, d = 101.388 meters
Explanation:
It is given that,
The average velocity of plane, v = 3650 km/h = 1013.88 m/s
The time for which eye blinks, [tex]t=100\ ms=0.1\ s[/tex]
Let d is the distance covered by the jet. It can be calculated as :
[tex]d=v\times t[/tex]
[tex]d=1013.88\ m/s\times 0.1\ s[/tex]
d = 101.388 meters
So, the distance covered by a fighter jet during a pilot's blink is 101.388 meters. Hence, this is the required solution.
Final answer:
During a blink that lasts 0.1 seconds, a fighter jet traveling at an average velocity of 3650 km/h will cover approximately 101.39 meters.
Explanation:
To calculate the distance a fighter jet travels during a pilot's blink, we can use the formula distance = velocity × time. The pilot's blink lasts 100 milliseconds (ms) which is 0.1 seconds because 1000 ms equals 1 second. The jet's velocity is given as 3650 kilometers per hour (km/h).
First, we'll need to convert the jet's velocity to meters per second (m/s) since the time of the blink is given in seconds. 1 km equals 1000 meters, and there are 3600 seconds in an hour, so:
Velocity in m/s = (Velocity in km/h) × (1000 m/km) / (3600 s/h) = (3650) × (1000) / (3600) = 1013.89 m/s
Now, we will multiply the velocity in m/s by the time in seconds to get the distance:
Distance = Velocity × Time = (1013.89 m/s) × (0.1 s) = 101.39 meters
So, during the blink of an eye, which is 0.1 seconds, a fighter jet traveling at an average velocity of 3650 km/h will cover approximately 101.39 meters.
What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized light?
Answer:
a) [tex]R_s = 0.092[/tex]
b) [tex]R_p = 0.085[/tex]
Explanation:
given,
n =1.5 for glass surface
n = 1 for air
incidence angle = 45°
using Fresnel equation of reflectivity of S and P polarized light
[tex]R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2[/tex]
using snell's law to calculate θ t
[tex]sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}[/tex]
[tex]cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}[/tex]
a) [tex]R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2[/tex]
[tex]R_s = 0.092[/tex]
b) [tex]R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2[/tex]
[tex]R_p = 0.085[/tex]
A skateboarder starts up a 1.0-m-high, 30° ramp at a speed of 7.6 m/s . The skateboard wheels roll without friction. At the top, she leaves the ramp and sails through the air. How far from the end of the ramp does the skateboarder touch down?
The skateboarder's horizontal distance traveled after leaving the ramp can be calculated from her initial velocity and time in the air during projectile motion.
The skateboarder's motion can be analyzed using kinematic equations. When she leaves the ramp, she will follow a projectile motion trajectory. The horizontal distance she travels can be calculated using her initial vertical velocity and the time she is in the air.
The skateboarder touches down approximately 4.87 meters from the end of the 1.0-meter-high, 30° ramp after being launched at a speed of 7.6 m/s.
First, we'll find the horizontal and vertical components of the skateboarder's initial velocity. The angle of the ramp is 30°, and her speed is 7.6 m/s:
Horizontal component ([tex]V_x[/tex]) = 7.6 * cos(30°)
or, [tex]V_x[/tex] = 7.6 * 0.866
or, [tex]V_x[/tex] = 6.58 m/s
Vertical component ([tex]V_y[/tex]) = 7.6 * sin(30°)
or, [tex]V_y[/tex] = 7.6 * 0.5
or, [tex]V_y[/tex] = 3.8 m/s
solving for [tex]{V_f}_y[/tex] in the horizontal and vertical components of velocity as the skater leaves the ramp :
[tex]V_f}_y^2 = V_y^2 - 2* g * \Delta x[/tex]
or, [tex]{V_f}_y[/tex] = [tex]\sqrt{V_y^2 - 2 * g * \Delta x }[/tex]
or, [tex]{V_f}_y[/tex] = [tex]\sqrt{|(3.8)^2 - 2 * 9.8 * 1.0|}[/tex]
or, [tex]{V_f}_y[/tex] = [tex]\sqrt{|14.44 - 19.6|}[/tex]
or, [tex]{V_f}_y[/tex] = [tex]\sqrt{|-5.16|}[/tex]
or, [tex]{V_f}_y[/tex] = √(5.16)
or, [tex]{V_f}_y[/tex] = 2.27 m/s
In the x-axis, there is no acceleration so, [tex]a_x[/tex] = 0 m/s²
[tex]{V_f}_x^{2}[/tex] = [tex]V_x^2[/tex]+ 2 * [tex]a_x[/tex] * Δx
or, [tex]{V_f}_x^{2}[/tex] = [tex]V_x^2[/tex] + 2 * 0 * Δx
or, [tex]{V_f}_x^{2}[/tex] = [tex]V_x^2[/tex]
or, [tex]{V_f}_x[/tex] = [tex]V_x[/tex]
or, [tex]{V_f}_x[/tex] = 6.58 m/s
The skateboarder reaches a height of 1.0 m. We will use the following kinematic equation to find the time (T) it takes for her to fall this distance:
y = [tex]{V_f}_y[/tex] * t + 0.5 * (-g) * t²
Here, y = -1.0 m (since she’s falling), [tex]{V_f}_y[/tex] = 2.27 m/s, and g = 9.8 m/s². Plugging in the values:
-1.0 = 2.27 * T - 0.5 * 9.8 * T²
or, 4.9 * T² - 2.27 * T - 1.0 = 0
Solving this quadratic equation (4.9 * T² - 2.27 * T - 1.0 = 0) for T gives:
T ≈ 0.74 seconds
Next, we calculate the horizontal distance traveled using this time and the horizontal velocity:
Distance = [tex]{V_f}_x[/tex] * T = 6.58 m/s * 0.74 s
Distance = 4.87 meters
Therefore, the skateboarder touches down approximately 4.87 meters from the end of the ramp.
A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your answer to three significant figures and include the appropriate units
Answer:
The kangaroo was 1.164s in the air before returning to Earth
Explanation:
For this we are going to use the equation of distance for an uniformly accelerated movement, that is:
[tex]x = x_{0} + V_{0}t + \frac{1}{2}at^2[/tex]
Where:
x = Final distance
xo = Initial point
Vo = Initial velocity
a = Acceleration
t = time
We have the following values:
x = 1.66m
xo = 0m (the kangaroo starts from the floor)
Vo = 0 m/s (each jump starts from the floor and from a resting position)
a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)
t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.
Now replace the values in the equation
[tex]x = x_{0} + V_{0}t + \frac{1}{2}at^2[/tex]
[tex]1.66 = 0 + 0t + \frac{1}{2}9.8t^2[/tex]
[tex]1.66 = 4.9t^2[/tex]
[tex]\frac{1.66}{4.9} = t^2[/tex]
[tex]\sqrt{0.339} = t\\ t = 0.582s[/tex]
It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is
t = 0.582s + 0.582s
t = 1.164s
The kangaroo was 1.164s in the air before returning to Earth
Coal is lifted out of a mine a vertical distance of 50 m by an engine that supplies 500 W to a conveyer belt. How much coal per minute can be brought to the surface? Ignore the effects of friction.
Answer:
mass = 61.16 kg
Explanation:
given data
distance = 50 m
energy = 500 W = 500 × 60sec = 30000 J/min
to find out
mass
solution
we will apply here energy equation
that is
energy = m×g×h .............1
put here all value m mass here and g = 9.81 and h is distance
energy = m×g×h
30000 = m×9.81×50
mass = 61.16 kg
Answer:
61 kg
Explanation:
Given data
distance (d): 50 m
power (P): 500 W = 500 J/s
mass (m): ?
time (t): 1 min = 60 s
The engine must do some work (w) to lift the coal. The work done in 1 minute is:
P = w / t
w = P . t = 500 J/s × 60 s = 3.0 × 10⁴ J
The work is equal to the force exerted (F) times the distance (d).
w = F × d
F = w / d = 3.0 × 10⁴ J / 50 m = 6.0 × 10² N
The force exerted is equal to the mass lifted (m) times the acceleration. Here, the acceleration to surpass is that of gravity (g = 9.8 m/s²).
F = m × g
m = F / g = 6.0 × 10² N / 9.8 m/s² = 61 kg
Jerry knocks a flowerpot off its third-story ledge, 9.5 m above the ground. If it falls freely, how fast is the flowerpot moving when it crashes to the sidewalk?
Answer:
13.65 m/s
Explanation:
t = Time taken
u = Initial velocity = 0
v = Final velocity
s = Displacement = 9.5 m
a = Acceleration = 9.81 m/s²
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 9.5-0^2}\\\Rightarrow v=13.65\ m/s[/tex]
The flower pot is moving at a speed of 13.65 m/s when it crashes the sidewalk.
A football is kicked straight up into the air; it hits the ground 4.0 s later. What was the greatest height reached by the ball? Assume it is kicked from ground level.Express your answer with the appropriate units.
Answer:
the greatest height at which the ball can reach is 19.62 m.
Explanation:
given,
time taken by the ball to reach the ground = 4 s
time at which ball reach at maximum height = 4/2 = 2 s
velocity of the ball at the top most point = 0 m/s
we know,
v = u + a t
0 = u + (-9.81 ) × 2
u = 19.62 m/s
maximum height achieved
[tex]s = u t + \dfrac{1}{2}at^2[/tex]
[tex]s = 19.62\times 2 + \dfrac{1}{2}(-9.81)\times 2^2[/tex]
s = 19.62 m
hence, the greatest height at which the ball can reach is 19.62 m.
The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.7 m/s in 2.50 s and can continue to accelerate to reach a top speed of 28.7 m/s. Assume the acceleration is constant until the top speed is reached and is zero thereafter. 1) Express the cheetah's top speed in mi/h. (Express your answer to three significant figures.) 2) Starting from a crouched position, how long does it take a cheetah to reach its top speed? (Express your answer to three significant figures.) 3) How far does it travel in that time? (Express your answer to three significant figures.) 4) If a cheetah sees a rabbit 120 m away, how long will it take to reach the rabbit, assuming the rabbit does not move?(Express your answer to three significant figures.)
Answer:
1) 64.2 mi/h
2) 3.31 seconds
3) 47.5 m
4) 5.26 seconds
Explanation:
t = Time taken = 2.5 s
u = Initial velocity = 0 m/s
v = Final velocity = 21.7 m/s
s = Displacement
a = Acceleration
1) Top speed = 28.7 m/s
1 mile = 1609.344 m
[tex]1\ m=\frac{1}{1609.344}\ miles[/tex]
1 hour = 60×60 seconds
[tex]1\ s=\frac{1}{3600}\ hours[/tex]
[tex]28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h[/tex]
Top speed of the cheetah is 64.2 mi/h
Equation of motion
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2[/tex]
Acceleration of the cheetah is 8.68 m/s²
2)
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s[/tex]
It takes a cheetah 3.31 seconds to reach its top speed.
3)
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m[/tex]
It travels 47.5 m in that time
4) When s = 120 m
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s[/tex]
The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds
The cheetah's top speed of 28.7 m/s is approximately 64.2 mi/h. It takes a cheetah 3.31 seconds to reach its top speed, traveling a distance of 47.4 meters during this acceleration. To reach a stationary rabbit 120 meters away, it would take the cheetah a total of 5.84 seconds.
Explanation:The question involves converting speeds from meters per second to miles per hour, finding the time taken to achieve a certain speed, calculating the distance traveled in that time, and determining the time required to reach a target.
To convert the cheetah's top speed from meters per second (m/s) to miles per hour (mi/h), we use the conversion factor of 1 m/s = 2.23694 mi/h. The top speed of 28.7 m/s is equivalent to 64.2 mi/h to three significant figures.The time taken to reach the top speed can be found using the acceleration formula: a = (v - u) / t, where 'v' is final velocity, 'u' is initial velocity, and 't' is time. Since the cheetah starts from rest (u = 0), we know it reaches a speed of 21.7 m/s in 2.5 s, implying an acceleration of 21.7 m/s / 2.5 s = 8.68 m/s². To then reach the top speed of 28.7 m/s, the time t = (28.7 m/s) / (8.68 m/s²), which is approximately 3.31 s.The distance traveled while accelerating can be found using the kinematic equation: d = ut + (1/2)at². As the cheetah starts from rest (u = 0), the distance is d = (1/2)(8.68 m/s²)(3.31 s)², which equals 47.4 m.If the cheetah sees a rabbit 120 m away and accelerates towards it, we need to find the time taken to cover this distance. The cheetah covers 47.4 m while reaching top speed, which takes 3.31 s. The remaining distance at top speed is 120 m - 47.4 m = 72.6 m. The time taken to cover this at 28.7 m/s is 72.6 m / 28.7 m/s = 2.53 s. The total time to reach the rabbit is 3.31 s + 2.53 s = 5.84 s.If a point charge of -30 µC experiences an electrostatic upward force of 27 mN at a certain location in the laboratory, what are the magnitude and direction of the electric field at that location? Express your answer with appropriate units. Enter positive value if the field is upward and negative value if the field is downward.
Explanation:
Charge, [tex]q=-30\ \mu C=-30\times 10^{-6}\ C[/tex]
Electric force, [tex]F=27\ mN=27\times 10^{-3}\ N[/tex]
We need to find the magnitude and direction of electric field at that location. The relation between the electric field and electric force is given by :
[tex]E=\dfrac{F}{q}[/tex]
[tex]E=\dfrac{27\times 10^{-3}\ N}{-30\times 10^{-6}\ C}[/tex]
[tex]E=-900\ N/C[/tex]
For a negative charge, the direction of electric field is inward. The direction of electric force and electric field is same. So, the direction of electric field in this case is in upward direction. Hence, this is the required solution.
The magnitude of the electric field is 900 N/C, and the direction is downward. Therefore, the electric field at that location is -900 N/C (negative value indicating downward).
To determine the magnitude and direction of the electric field at the location where the point charge experiences an electrostatic force, we can use the relationship between the force [tex]\mathbf{F}[/tex] experienced by a charge q in an electric field [tex]\mathbf{E}[/tex] :
[tex]\mathbf{F} = q \mathbf{E}[/tex]
Given:
- The charge q is -30 µC (microcoulombs).
- The electrostatic force [tex]\mathbf{F}[/tex] is 27 mN (millinewtons) upward.
First, we convert the given quantities to standard SI units:
- Charge [tex]q = -30 \, \mu \text{C} = -30 \times 10^{-6} \, \text{C}[/tex]
- Force [tex]\mathbf{F} = 27 \, \text{mN} = 27 \times 10^{-3} \, \text{N}[/tex]
Next, we use the equation [tex]\mathbf{F} = q \mathbf{E}[/tex] to solve for the electric field [tex]\mathbf{E}[/tex] :
[tex]\mathbf{E} = \frac{\mathbf{F}}{q}[/tex]
Substituting the given values:
[tex]\mathbf{E} = \frac{27 \times 10^{-3} \, \text{N}}{-30 \times 10^{-6} \, \text{C}} \\\\\mathbf{E} = \frac{27 \times 10^{-3}}{-30 \times 10^{-6}} \, \text{N/C} \\\\\mathbf{E} = \frac{27}{-30} \times 10^{3} \, \text{N/C} \\\\\mathbf{E} = -0.9 \times 10^{3} \, \text{N/C} \\\\\mathbf{E} = -900 \, \text{N/C}[/tex]
The negative sign indicates the direction of the electric field. Since the force on a negative charge is upward, the electric field must be directed downward (opposite to the direction of the force on a negative charge).
You mount a ball launcher to a car. It points toward the back of the car 39 degrees above the horizontal and launches balloons at 9.7 m/s. When you drive the car straight ahead at 2.2 m/s. Assume the ball is launched from ground level. For the ball, what is the separation between the ball and the car in m when the ball hits the ground?
Answer:
9.4 m
Explanation:
We can use a moving frame of reference with the same speed as the car. From this frame of reference the car doesn't move. The origin is at the back of the car, the positive X axis points back and the positive Y axis points up.
If the ballon is launched at 9.7 m/s at 39 degrees of elevation.
Vx0 = 9.7 * cos(39) = 7.5 m/s
Vy0 = 9.7 * sin(39) = 6.1 m/s
If we ignore air drag, the baloon will be subject only to the acceleration of gravity. We can use the equation of position under constant acceleration.
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 = 0
a = -9.81 m/s^2
It will fall when Y(t) = 0
0 = 6.1 * t - 4.9 * t^2
0 = t * (6.1 - 4.9 * t)
t1 = 0 (this is when the balloon was launched)
0 = 6.1 - 4.9 * t2
4.9 * t2 = 6.1
t2 = 6.1 / 4.9 = 1.25 s
The distance from the car will be the horizonta distance it travelled in that time
X(t) = X0 + Vx0 * t
X(1.25) = 7.5 * 1.25 = 9.4 m
The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 23.0 m/s . The drops pass through a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length D0 = 1.80 cm , where there is a uniform vertical electric field with magnitude E = 8.20×10^4 N/C If a drop is to be deflected a distance d = 0.290 mm by the time it reaches the end of the deflection plate, what magnitude of charge q must be given to the drop? Assume that the density of the ink drop is 1000 kg/m^3
The magnitude of the charge q that must be given to the ink drop is [tex]7.5 \times 10^{-14}[/tex] coulombs.
The vertical displacement of the ink drop as it passes between the deflecting plates is due to the electric force acting on the charged drop. The force of gravity on the drop is negligible compared to the electric force.
Given:
Mass of the ink drop [tex](m) = 1.00 \times 10^{-11}\ kg[/tex]
Vertical displacement[tex](d) = 0.290\ mm[/tex]
Horizontal velocity [tex](v) = 23.0\ m/s[/tex]
Electric field magnitude [tex](E) = 8.20 \times 10^4\ N/C[/tex]
Plate length [tex](D_o) = 1.80\ cm[/tex]
The electric force on a charged particle in an electric field is given by Coulomb's law:
[tex]F_{electric} = q \times E[/tex]
Where:
[tex]F_{electric}[/tex] is the electric force,
q is the charge of the particle,
E is the magnitude of the electric field.
The vertical displacement (d) of the drop is caused by this electric force and is given by the equation:
[tex]d = (1/2) \times a \times t^2[/tex]
Since the drop is initially moving horizontally and is not influenced by gravity, the vertical acceleration (a) is due solely to the electric force
[tex]F_{electric} = m \times a[/tex]
Equate the two expressions for the electric force:
[tex]q \times E = m \times a\\a = (q \times E) / m\\d = (1/2) \times ((q \times E) / m) \times t^2[/tex]
The time it takes to travel through the plates using the horizontal distance [tex](D_o)[/tex] and the horizontal velocity:
[tex]t = D_o/ v[/tex]
[tex]d = (1/2) \times ((q \times E) / m) \times (D_o / v)^2\\q = (2 \times m \times d \times v^2) / (E \times D_o^2)\\q = (2 \times (1.00 \times 10^{-11}) \times (0.290 \times 10^{-3} ) \times (23.0 )^2) / ((8.20 \times 10^4) \times (1.80 \times 10^{-2})^2)\\q = 7.5 \times 10^{-14}[/tex]
Therefore, the magnitude of the charge q that must be given to the ink drop is [tex]7.5 \times 10^{-14}[/tex] coulombs.
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To find the charge necessary for an ink drop to be deflected a certain distance in an electric field, apply the equation q = Fe/E, where Fe is derived from the kinematic equations for the ink drop's motion in the field.
Explanation:To calculate the magnitude of charge q that must be given to an ink drop such that it is deflected by a distance d = 0.290 mm by the time it reaches the end of the deflection plate of length D0 = 1.80 cm in a uniform electric field with magnitude E = 8.20×104 N/C, we can use the concepts of electric force and motion.
Firstly, the electric force Fe acting on the ink drop is equal to qE. This force will cause the ink drop to accelerate. Since the drops are moving horizontally at velocity v = 23.0 m/s and they need to be deflected by a certain distance d while travelling through the electric field, we can use the kinematic equations of motion to find the vertical displacement due to acceleration.
The time t it takes for the ink drop to pass through the plates is t = D0/v. From this, the vertical displacement d can be given as d = 0.5 * a * t^2 where a is the acceleration due to the electric force. Knowing d, t, and E, we can find a and subsequently use Fe = mdrop × a to solve for the charge q as q = Fe/E.
Using the given mass m = 1.00×10−11 kg for the ink drop, we can rearrange and solve the appropriate equations to find the value of the charge q necessary to cause the required deflection d.
During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing the lighting and hearing the sound was 5 second. Calculate the distance of the colliding cloud from the observer.
Answer:
d = 1700 meters
Explanation:
During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds
Speed of sound, v = 340 m/s (say)
Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :
[tex]d=v\times t[/tex]
[tex]d=340\ m/s\times 5\ s[/tex]
d = 1700 meters
So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.
A cannon tilted up at a 33.0° angle fires a cannon ball at 74.0 m/s from atop a 18.0 m -high fortress wall. What is the ball's impact speed on the ground below?
Answer:
The cannon ball hits the ground with a speed of 76.35 m/s.
Explanation:
We shall use the conservation of energy principle to solve the problem
The initial energy of the cannon ball is the sum of the potential and the kinetic energies.
We know that potential energy = [tex]mgh[/tex]
Applying the given values we obtain the initial potential energy as
[tex]P.E=m\times g\times 18.0[/tex]
Similarly the initial kinetic energy of the ball equals
[tex]K.E_{initial}=\frac{1}{2}mv_{o}^{2}[/tex]
Applying values we get
[tex]K.E_{initial}=\frac{1}{2}m(74.0)^{2}[/tex]
Thus the initial energy is thus [tex]E_{initial}=m\times g\times 18.0+\frac{1}{2}\times m\times (74.0)^{2}[/tex]
Now upon hitting the ground the only energy that the cannon ball will posses is kinetic energy since the potential energy of any object upon touching the surface of earth equals zero
Thus we have
[tex]Energy_{final}=\frac{1}{2}mv_{f}^{2}[/tex]
Equating initial and final energies we get
[tex]m\times g\times 18+\frac{1}{2}\times m\times (74.0)^{2}=\frac{1}{2}\times m\times v_{f}^{2}\\\\v_{f}^{2}=36g+(74.0)^{2}\\\\\therefore v_f=\sqrt{36\times 9.81+(74.0)^{2}}\\\\v_f=76.35m/s[/tex]
A string (L = 0.76 m) was used to generate standing waves. If the number of antinodes generated in the string was 3, the determine the wavelength.
Answer: 0.506 m
Explanation: To solve this we use the relationship for the harmonic in the string which are given by the following expression:
λ=2*L/3
λ=2*0.76 m/3= 0.506 m
A bug lands on a windshielf wiper. Explain why the bug is more likely to be dislodged when the wipers are turned on at the high rather than low setting.
Answer and Explanation:
The motion of the viper is circular and it completes a semi- circle forming an arc and then retraces its path back.
Thus the force experienced by the bug is centripetal force thus centripetal acceleration and in the absence of this force the bug will get dislodged.
This centripetal force is mainly provided by the static friction between the blades and the bug.
When the wipers are moving with high velocity when turned on, larger centripetal force is required to keep the bug moving with the wiper on the arc than at low setting.
Thus there are more chances for the bug to be dislodged at higher setting than at low setting.
An airplane is heading due south at a speed of 690 km/h . A) If a wind begins blowing from the southwest at a speed of 90 km/h (average), calculate the velocity (magnitude) of the plane relative to the ground.B) Calculate the velocity (direction) of the plane relative to the ground.C) Calculate how far from its intended position will it be after 11 min if the pilot takes no corrective action.
Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h east. c) 16,5 km NE of the desired position
Explanation:
Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where
see fig 1
Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then
90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,
this because we have an isosceles triangle, then the cathetus length is
hypotenuse/[tex]\sqrt{2}[/tex]
using Pythagoras, here the hypotenuse is 90, then the cathetus are of length
90/[tex]\sqrt{2}[/tex] km/h= 63,6396 km/h.
Now the total speed of the plane is
690km/h south + 63,6396 km/h north +63,6396 km/h east,
this is 626,3604 km/h south + 63,6396 km/h east, here north is as if we had -south.
then using again Pythagoras we get the magnitude of the total speed it is
[tex]\sqrt{626,3604 ^2+63,6396^2} km/h=629,5851km/h [/tex],
the direction is calculated with respect to the south using trigonometry, we know the
sin x= cathetus opposed / hypotenuse,
then
x= [tex]sin^{-1}'frac{63,6396}{629,5851}[/tex]=5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form
626,6396 km/h south + 63,6396km/h east.
Finally since the detour is caused by the west speed component plus the slow down caused by the north component of the wind speed, we get
Xdetour{east}= 63,6396 km/h* (11 min* h)/(60 min)=11,6672 km=Xdetour{north} ,
since 11 min=11/60 hours=0.1833 hours.
Then the total detour from the expected position, the one it should have without the influence of the wind, we get
Xdetour=[/tex]\sqrt{2* 11,6672x^{2} }[/tex] = 16,5km at 45 degrees from east pointing north
The situation is sketched as follows see fig 2
A charge of +3.00 μC is located at the origin, and a second charge of -2.00 μC is located on the x-y plane at the point (50.0 cm, 60.0 cm). 1) Determine the x-component of the electric force exerted by the -2.00 μC charge on the +3.00 μC charge. (Express your answer to three significant figures.)
Answer:
0.0567 N
Explanation:
q1 = 3 micro coulomb
q2 = - 2 micro coulomb
OB = 50 cm
AB = 60 cm
By using Pythagoras theorem in triangle OAB
[tex]OA^{2}=OB^{2}+AB^{2}[/tex]
[tex]OA^{2}=50^{2}+60^{2}=6100[/tex]
OA = 78.1 cm
By using the Coulomb's law, the force on 3 micro coulomb due to -2 micro coulomb is
[tex]F=\frac {Kq_{1}q_{2}}{OA^{2}}= \frac{9 \times 10^{9}}\times 3\times 10^{-6} \times 2 \times 10^{-6}{0.781^{2}}[/tex]
F = 0.0885 N
The horizontal component of force is
= F CosФ = [tex]F\times \frac{OB}{OA}[/tex]
= 0.0885 x 50 / 78.1 = 0.0567 N
A tortoise and a hare engage in a race. A tortoise can run with a speed of 0.15 m/s. A hare can run 25 times as fast as the tortoise. In the race, they start at the same time; however, the hare being very proud of his much higher speed stops for a nap for 5.0 minutes. The tortoise wins the race by 35 cm. (a) What is the length of the race? (b) How long does the race take?
Answer:
Explanation:
The speed of hare = .15 x 25 = 3.75 m /s . Let tortoise took t second to complete the race .
Distance traveled by it = .15 t
Distance traveled by hare = .15 t - .35 m
Time taken by hare to complete this distance
= t - 5 x 60 s
Speed of hare
= Distance / time
( .15t-.35 ) / t - 300 , so
[tex]\frac{.15t-.35}{t-300} = 3.75[/tex]
t = 312.40
= 5 minutes 12.4 seconds
Distance of race
312.4 x speed of tortoise
= 312.4 x .15
= 46.85 m
(a) The length of the race is 91.1 meters.
(b) The race takes 605.0 seconds.
The tortoise moves at a speed of 0.15 m/s while the hare moves at a speed of 25 * 0.15 = 3.75 m/s.
The hare stops for 5 minutes, which is equal to 5 * 60 = 300 seconds.
In the 300 seconds that the hare was taking a nap, the tortoise was able to move a distance of 0.15 * 300 = 45 meters.
When the hare woke up, he started running at his top speed. He was able to catch up to the tortoise and surpass him by 35 cm. This means that the hare was able to run a distance of 45 + 0.35 = 45.35 meters.
Therefore, the length of the race is 45.35 meters.
The total time it took for the race to finish is 300 + 605 = 905 seconds.
So the answer is (a) 91.1 meters (b) 605.0 seconds
To learn more about speed, here
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Beach sand grains are approximately spherical with radius of 50 microns (μm), and are mostly silicon dioxide, meaning their density is 2600 kg/m^3. Find the mass of sand that would have the same surface area (total surface area of all spheres) as a cube of side length 1.0 m.
Answer:
260 g
Explanation:
Given:
r = radius of the spherical sand = 50 micron = [tex]5\times 10^{-5}\ m[/tex]a = edge length of the cube = 1.0 m\rho = density of sand = [tex]2600\ kg/m^3[/tex]Let n be the number of such silicon dioxide that would give the desired mass of sand.
According to the question, the surface area of all the sand particles will be equal to the surface area of the cube.
[tex]\therefore n\times \textrm{Area of a sand particle}=\textrm{Surface area of a cube }\\\Rightarrow n\times 4\pi r^2= 6a^2\\\Rightarrow n = \dfrac{6a^2}{4\pi r^2}[/tex]
Let the total mass of sand required be M.
[tex]\textrm{Total mass of sand} = \textrm{Mass of all the required sand particle}\\\Rightarrow M = n\times \textrm{mass of one sand particle}\\\Rightarrow M = n\times \textrm{Density of sand particle}\times \textrm{Volume of a sand particle}\\\Rightarrow M = n\times\rho \times \dfrac{4}{3}\pi r^3\\[/tex]
[tex]\Rightarrow M = \dfrac{6a^2}{4\pi r^2}\times\rho \times \dfrac{4}{3}\pi r^3\\\Rightarrow M = 2\times\rho \times a^2r\\\Rightarrow M = 2\times2600 \times (1)^2\times 5\times 10^{-5}\\\Rightarrow M =0.260\ kg\\\Rightarrow M =260\ g\\[/tex]
Hence, the total mass of the sand required will be equal to the 260 g.
Represent 7468 N with SI units having an appropriate prefix. Express your answer to four significant figures and include the appropriate units. CHAROE ? Value Units 7468 N =
Answer:
7.468 kN
Explanation:
Here the force of 7468 Newton is given.
Some of the prefixes of the SI units are
kilo = 10³
Mega = 10⁶
Giga = 10⁹
The number is 7468.0
Here, the only solution where the number of significant figures is kilo
1 kilonewton = 1000 Newton
[tex]1\ Newton=\frac{1}{1000}\ kilonewton[/tex]
[tex]\\\Rightarrow 7468\ Newton=\frac{7468}{1000}\ kilonewton\\ =7.468\ kilonewton[/tex]
So 7468 N = 7.468 kN