Answers
The diagram of forces is shown below
We would express each vector in component form. We have
For F1 = 65 pounds,
F1 = <65Cos30, 65Sin30>
F1 = <56.2917, 32.5>
For F2 = 90 pounds,
F2 = <90Cos45, 90Sin45>
F2 = <63.6396, 63.6396>
For F3 = 125 pounds
F3 = <125Cos120, 125Sin120
F3 = <- 62.5, 108.2532>
Fnet = F1 + F2 + F3
Fnet = <56.2917, 32.5> + <63.6396, 63.6396> + <- 62.5, 108.2532>
Fnet = <56.2917 + 63.6396 - 62.5, 32.5 + 63.6396 + 108.2532>
Fnet = <57.4313, 204.3928>
Magnitude of resultant force = √(57.4313^2 + 204.3928^2)
Magnitude of resultant force = 212.3 pounds
Direction, θ = tan^-1(204.3928/57.4313)
θ = 74.3 degrees
Related Questions
What is the kinetic energy of a 763.91 N woman jogging at 1.10 m/s?
Answers
Givens.
• The weight force is 763.91 N.
,• The speed is 1.10 m/s.
First, find the mass using the formula for weight force.
[tex]w=mg[/tex]Where w = 763.91 N and g = 9.81 m/s^2.
[tex]\begin{gathered} 763.91N=m\cdot9.81\cdot\frac{m}{s^2} \\ m=\frac{763.91N}{9.81\cdot\frac{m}{s^2}} \\ m\approx77.87\operatorname{kg} \end{gathered}[/tex]Once we have the mass of the woman, we can find the kinetic energy.
Use the kinetic energy formula.
[tex]K=\frac{1}{2}mv^2[/tex]Where m = 77.87 kg and v = 1.10 m/s.
[tex]\begin{gathered} K=\frac{1}{2}\cdot77.87\operatorname{kg}\cdot(1.10\cdot\frac{m}{s})^2 \\ K=38.94\operatorname{kg}\cdot1.21\cdot\frac{m^2}{s^2} \\ K\approx47.12J \end{gathered}[/tex]Therefore, the kinetic energy is 47.12J.
as a 65kg skydiver falls at terminal velocity, the force of air resistance acting on her isa. 0 Nb. 65 Nc. 650 Nd. 6.5 N
Answers
c) 650 N
ExplanationTerminal velocity is defined as the highest velocity attained by an object falling through a fluid.in this case, the fluid is air,
the skydiver falls at a constant velocity as described by Newton's first law of motion. The constant vertical velocity is called the terminal velocity .
Step 1
To calculate terminal velocity: Multiply the mass of the object by the gravitational acceleration
[tex]\begin{gathered} terminal\text{ velocit is when} \\ Drag\text{ force= weight} \end{gathered}[/tex]so
a)let
[tex]\begin{gathered} mass=65\text{ kg} \\ weigth=mg=65\text{ kg*10}\frac{m}{s^2} \\ weigth=650\text{ Newtons} \end{gathered}[/tex]hence, the force of air resistance acting on her is the same ,so the answer is
c) 650 N
I hope this helps you
What is the flow of charge, measured in Amps called?A:voltage B:potential difference C:resistance D:current
Answers
Current:
The current is a scalar qunatity and can be defined as the flow of charges per unit time. The current can be measured or calculated in Amperes.
Hence, the correct answer is (d)
A 3.0-meter stick moves at a speed of 0.90c relative to an observer. The stick is aligned parallel to the direction ofmotion. What is the observed length of the stick?(a) 6.9 m(b) 4.2 m(C) 2.7 m(d) 1.3 m
Answers
In order to determine the observed length of the stick, use the following formula for the length contraction due to the relativistic speed of the stick (this formula can be used because the motion of the stick is in the same direction of the axis of the stick):
[tex]L=L_0\sqrt[]{1-\frac{v^2}{c^2}}[/tex]where:
L: observed length of the stick = ?
Lo: length of the stick in its reference system = 3.0 m
v: speed of the stick = 0.90c
replace the previous values of the paramters into the formula for L:
[tex]\begin{gathered} L=(3.0m)\sqrt[]{1-\frac{(0.9c)^2}{c^2}} \\ L=(3.0m)\sqrt[]{1-0.81} \\ L=(3.0m)(0.43)=1.3m \end{gathered}[/tex]Hence, the observed length of the stick is 1.3m
A low sounding note has a frequency of 40 HZ and a speed in the air of 340 m/s How long in that wave?
Answers
Given,
The frequency of the sound, f=40 Hz
The speed of the sound in the air, c=340 m/s
The speed of a wave in a medium is given by the product of the frequency and the wavelength of the wave.
Thus the speed of the sound with the given frequency is calculated using the formula,
[tex]c=\lambda f[/tex]Where λ is the wavelength of the wave.
On rearranging the above equation,
[tex]\lambda=\frac{c}{f}[/tex]On substituting the known values,
[tex]\begin{gathered} \lambda=\frac{340}{40} \\ =8.5\text{ m} \end{gathered}[/tex]Thus the length of the wave or the wavelength of the given wave is 8.5 m
A baseball fan is 196.0 m from home plate. If the temperature at the ball field is 24.6°C, how much time
elapses between the instant the fan sees the batter hit the ball and the moment the fan hears the sound?
S
Answers
The time elapses between the instant the fan sees the batter hit the ball and the moment the fan hears the sound will be 6.5333333 x 10⁷ sec.
The given data:
Time is referred as the ratio of distance and speed of light. which can be expressed as:
t=d/c
Where, t is time and c is speed.
Distance =196m
c =speed of light .
Put the values of given data in given formula:
c= speed of light = 3 x 10⁸m/s
t= 196/[3 x 10⁸]
=6.5333333 x 10⁷ sec
Therefore, the time will be 6.5333333 x 10⁷ sec
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Calculate the volume of the following gas sample at STP. 2.3 mol He
Answers
Given,
The gas sample is at STP.
Thus, the temperature of the gas, T=273.15 K
The pressure of the gas, P=10⁵ Pa
The quantity of the gas, n=2.3 mol
The value of the gas constant is R=8.314 m³·Pa·K⁻¹·mol⁻¹
From the ideal gas equation,
[tex]\begin{gathered} PV=\text{nRT} \\ V=\frac{\text{nRT}}{P} \end{gathered}[/tex]Where V is the volume of the gas.
On substituting the known values,
[tex]\begin{gathered} V=\frac{2.3\times8.314\times273.15}{10^5} \\ =0.052m^3 \end{gathered}[/tex]Thus the volume of the given gas at STP is 0.052 m³
A car is headed east with a mass of 1,391 kg at a velocity of 17.4 m/s. It collides with acar headed west with a mass of 1,280 kg. After the collision, both cars have come to astop. What was the velocity of the second car before the collision?
Answers
According to the law of conservation of momentum, we have
[tex]p_{i1}+p_{i2}=p_{f1}+p_{f2}_{}[/tex]Where p = mv. So, using the definition of momentum, we have
[tex]m_1v_{i1}+m_2v_{i2}=m_1v_{f1}+m_2v_{f2}[/tex]Let's use the given magnitudes and replace them with their letters.
[tex]1,391\operatorname{kg}\cdot17.4(\frac{m}{s})+1,280\operatorname{kg}\cdot v_{i2}=1,391\operatorname{kg}\cdot0+1,280\operatorname{kg}\cdot0[/tex]Observe that both final velocities are null because they come to stop. Let's solve for v.
[tex]\begin{gathered} 24,203.4\operatorname{kg}\cdot\frac{m}{s}+1,280\operatorname{kg}\cdot v_{i2}=0 \\ 1,280\operatorname{kg}\cdot v_{i2}=-24,203.4\operatorname{kg}\cdot\frac{m}{s} \\ v_{i2}=\frac{-24,203.4\operatorname{kg}\cdot\frac{m}{s}}{1,280\operatorname{kg}} \\ v_{i2}\approx-18.91(\frac{m}{s}) \end{gathered}[/tex]Therefore, the velocity of the second car before the collision is -18.91(m/s).The negative sign shows that the second car is headed in the Western direction.
A light rod is being used as a lever as shown. The fulcrum is 1.20 m from the load and 2.40 m from the applied force. If the load has a mass of 18.0 kg, what force must be applied to lift the load?
Answers
ANSWER:
88.2
STEP-BY-STEP EXPLANATION:
In this case the torque about the fulcrum must be equal. Torque is force perpendicular distance. Therefore, the torque on both side must be equal, just like that:
[tex]\begin{gathered} F_g\cdot d_1=F_A\cdot d_2 \\ F_g=m\cdot g \\ d_1=1.2\text{ m} \\ d_2=2.4\text{ m} \\ m=18\text{ kg} \end{gathered}[/tex]Therefore, we replace and calculate the applied force, like this:
[tex]\begin{gathered} 18\cdot9.8\cdot1.2=F_A\cdot2.4 \\ F_A=\frac{211.68}{2.4} \\ F_A=88.2\text{ N} \end{gathered}[/tex]The applied force is 88.2 N
An isolated electron experiences an electric force of 6.4 x 10-14 N. What is the magnitude of the electric field at that location? (qe = 1.6 x 10-19 C)Question 3
Answers
Given,
The electric force experienced by the electron, F=6.4×10⁻¹⁴ N
The charge of an electron, q_e=1.6×10⁻¹⁹ C
The electric field is the physical field in which if a charged particle is placed, it experiences the electric force.
The electric field and electric force are related to each other using the formula,
[tex]F=Eq[/tex]On substituting the known values,
[tex]\begin{gathered} 6.4\times10^{-14}=E\times1.6\times10^{-19} \\ E=\frac{6.4\times10^{-14}}{1.6\times10^{-19}} \\ =400\times10^3\text{ N/C} \end{gathered}[/tex]Therefore the electric field at that location is 400×10³ N/C
A 4.0 -kg block is related from rest 50 m above ground when it has fallen 20m it's kinetic every is approximately
Answers
The mass of the block, m = 4 kg
The total height of the block from the ground = 50 m
When the block has only covered 20 m, the height above the ground,
h = 50 m - 20 m
h = 30 m
A weight of 17 N is located at a distance of 8 cm from the fulcrum of a simple balance beam. At what distance from the fulcrum should a weight of 22 N be placed on the opposite side in order to balance the system?
Answers
W1 = 17N
d1 = 8 cm
W2 = 22N
d2= x
Where:
w= weight
d= distance
W1*d1 = W2*d2
17N*8cm= 22N*x
(17M*8cm)/22N = x
x = 6.19 cm
A straight wire with a current placed in a uniform magnetic field experiences a force that is ______________.Group of answer choicesperpendicular to the fieldproportional to the size of the currentall of theseperpendicular to the current
Answers
To find:
The magnetic force.
Explanation;
The force experienced by the wire carrying current, placed in a uniform magnetic field is given by the formula,
[tex]F=\text{IBLsin}\theta[/tex]Where I is the current through the wire, B is the magnetic field, L is the length of the wire, and θ is the angle between the magnetic field and wire.
From the above equation, it is clear that the force experienced by the wire is directly proportional to the magnitude or size of the current passing through the wire.
The direction of the field is given by Fleming's left-hand rule. By this rule, the magnetic force acting on the wire will be perpendicular to both the field and the current.
Final answer:
Thus all of the given options are true.
Therefore the correct answer is option C, "All of these."
A 3.0 cm lighted candle is placed 20 cm from a concave spherical mirror with a radius of curvature of 30 cm.(a)Where should a screen be placed (in cm) in order to see the candle's image clearly? (Give your answer as a distance from the center of the mirror.) cm(b)What is the minimum height (in cm) of the screen in order to see the candle's complete image? cm
Answers
Given:
The object's height is h = 3 cm
The object's distance from the concave mirror is u = -20 cm
The radius of curvature is R = - 30 cm
Required:
(a) Image's distance from the mirror
(b) Image's height
Explanation:
The focal length of the mirror can be calculated as
[tex]\begin{gathered} f=\frac{R}{2} \\ =-\frac{30}{2} \\ =-15\text{ cm} \end{gathered}[/tex](a) The image's distance can be calculated using the mirror formula as
[tex]\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ =\frac{1}{-15}-\frac{1}{-20} \\ v=-60\text{ cm} \end{gathered}[/tex](b) The image's height can be calculated as
[tex]\begin{gathered} \frac{h^{\prime}}{h}=-\frac{v}{u} \\ h^{\prime}=-\frac{v}{u}\times h \\ =\frac{-(-60)}{-20}\times3 \\ =-9\text{ cm} \end{gathered}[/tex]Final Answer:
(a) Image's distance from the mirror is v = -60 cm
(b) Image's height is h' = -9 cm
units: centimeters cubed (cm³) or milliliters (mL) DensityMassvolume
Answers
The unit for density is kilogram per meter cubed (kg/m³) or gram per centimeter cubed (g/cm³).
The unit for mass is kilogram (kg) or gram (g).
The unit for volume is meter cubed (m³) or liters (L) or centimeters cubed (cm³) or milliliters (mL).
Therefore, volume is the correct choice.
draw a free body diagram of a 1.5kg book on a desk that is being pushed across the desk to the right with a force of 0.5N
Answers
Here ,
m= 1.5Kg
N= normal reaction = mg
mg= 1.5×9.8 N
F= 0.5N
which of the following is most accurate ? 1)the speed of sound is constant as altitude increases 2)the speed of sound decreases as altitude increases 3)the speed of sound increases as altitude increases
Answers
As the number of molecules in the air decreases with the increase in height or altitude.
The sound wave needs molecules in the air to travel from one point to other.
Thus, with the increase in altitude, the speed of the sound wave decreases.
Hence, the second option is the correct answer.
Objects float because
Answers
The objects float on a fluid, like air or water, due to the floatation principle. It says that an object floats when the fluid displaced by the object has the same weight as the object
help me on these am struggling hard on them
Answers
3.1 ) Equation that links power, work done and time, P = W / t
3.2 ) Power output of the car to extend the spring = 8 W
3.3 ) Spring constant of the spring = - 4 N / m
3.1 ) Equation that links power, work done and time
P = W / t
3.2 ) Power output of the car to extend the spring,
W = 4 J
t = 0.5 s
P = 4 / 0.5
P = 8 W
3.3 ) Spring constant of the spring,
k = - F / x
W = F d
x = 40 cm = 0.04 m
4 = F * 0.04
F = 0.16 N
k = - 0.16 / 0.04
k = - 4 N / m
3.4 ) Total input power of the car,
Efficiency = Output power / Input power * 100
Efficiency = 62.5 %
62.5 = 8 / Input power * 100
Input power = 800 / 62.5
Input power = 12.8 W
3.5 ) Energy is transferred to a thermal energy store as the tyres are in contact with the floor. So the heat is lost as frictional energy.
Therefore,
3.1 ) Equation that links power, work done and time, P = W / t
3.2 ) Power output of the car to extend the spring = 8 W
3.3 ) Spring constant of the spring = - 4 N / m
3.4 ) Total input power of the car = 12.8 W
3.5 ) Energy is transferred to a thermal energy store as friction.
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5. A light ray travels through the air (n =1.00) at 0 = 35° until it enters a lake(n2=1.33).a. What is the angle of the refracted light?b. Given the index of refraction for the air and the speed of light constant, what is thevelocity of light in air?C. Given the index of refraction for the lake and the speed of light constant, what is thevelocity of the light in the lake?
Answers
a.
In order to calculate the angle of the refracted light, we can use the law of refraction below:
[tex]n_1\cdot\sin\theta_1=n_2\cdot\sin\theta_2[/tex]Where n1 and n2 are the index of refraction and theta1 and theta2 are the incident angle and angle of refraction.
So, using n1 = 1, n2 = 1.33 and theta1 = 35°, we have:
[tex]\begin{gathered} 1\cdot\sin\mleft(35°\mright)=1.33\cdot\sin\theta_2\\ \\ 0.5735764=1.33\cdot\sin\theta_2\\ \\ \sin\theta_2=\frac{0.5735764}{1.33}\\ \\ \sin\theta_2=0.43126\\ \\ \theta_2=25.55° \end{gathered}[/tex]b.
To find the velocity, we can use the formula below:
[tex]\begin{gathered} n=\frac{c}{v}\\ \\ 1=\frac{3\cdot10^8}{v}\\ \\ v=3\cdot10^8\text{ m/s} \end{gathered}[/tex]c.
Using the same formula from item b, but now using n2, we have:
[tex]\begin{gathered} n=\frac{c}{v}\\ \\ 1.33=\frac{3\cdot10^8}{v}\\ \\ v=\frac{3\cdot10^8}{1.33}\\ \\ v=2.256\cdot10^8\text{ m/s} \end{gathered}[/tex]I am very confused about how to do this problem. I know I multiply the first three numbers but I am unsure what to do nextA rectangular waterbed is 8 ft long, 4 ft wide, and 2 ft tall.How many gallons of water are needed to fill the waterbed?Assume 1 gallon is 0.13 cu ft. Round to the nearest whole gallon.
Answers
ANSWER:
492 gallons
STEP-BY-STEP EXPLANATION:
Given:
length: 8 ft
width: 4 ft
height: 2 ft
The first thing is to calculate the volume of the waterbed, since it is a rectangular prism, the volume is the product of its length, width and height.
[tex]\begin{gathered} V=8\cdot4\cdot2 \\ V=64ft^3 \end{gathered}[/tex]Knowing the volume in cubic feet, we can convert this value to gallons thanks to the conversion factor, where 1 gallon is equal to 0.13 cubic feet.
We perform the operation, just like this:
[tex]\begin{gathered} 64ft^3\cdot\frac{1\text{ gal}}{0.13ft^3}=492.30\text{ gal} \\ 492.30\text{ gal }\cong492\text{ gal} \end{gathered}[/tex]Therefore a total of 492 gallons are needed.
A box slides down an incline plane with initial speed 5.0 m/s. The box has mass 1.50 kg. Consider thebox when it reaches the bottom of the incline with final velocity 6.5 m/s.a) How much work does the gravitational force do on the box? (2 pts)b) How much work does the normal force do on the box? (2 pts)c) What is the size of the friction force acting on the box (3 pts)?
Answers
Given:
Initial speed = 5.0 m/s
Mass of box = 1.50 kg
Final velocity = 6.5 m/s
Vertical distance = 1.5 m
Let's solve for the following:
• A) How much work does the gravitational force do on the box?
To find the amount of work done by the gravitational force, we have:
[tex]W=mgh[/tex]Where:
m is the mass = 1.50 kg
g is acceleration due to gravity = 9.8 m/s²
h is the distance = 1.5 m
[tex]\begin{gathered} W=1.50*9.8*1.5 \\ \\ W=22.05\text{ J} \end{gathered}[/tex]Therefore, the work done by the gravitational force on the box is 22.05 J.
• (b). How much work does the normal force do on the box?
The normal force is perpendicular to the surface.
As the block slides down the surface, there will be no normal force parallel to the motion.
Therefore, no work is done by the normal force.
• (c). What is the size of the friction force acting on the box?
To determine the frictional force acting on the box, let's first find the angle:
[tex]\begin{gathered} \theta=sin^{-1}(\frac{1.5}{2.5}) \\ \\ \theta=36.87^o \end{gathered}[/tex]Now, find the downward force using the formula below:
[tex]\begin{gathered} F=mgsin\theta \\ \\ F=1.5*9.8sin36.87 \\ \\ F=8.82\text{ N} \end{gathered}[/tex]The box moves a distance of 2.5 m.
Let's find the acceleration using the motion formula:
[tex]\begin{gathered} v^2=u^2+2as \\ \\ a=\frac{v^2-u^2}{2s} \\ \\ a=\frac{6.5-5.0}{2(2.5)} \\ \\ a=0.3\text{ m/s}^2 \end{gathered}[/tex]The acceleration is 0.3 m/s².
The net force is the force pushing the box downwards - the frictional force.
Hence, we have:
[tex]\begin{gathered} F_f=F-ma \\ \\ F_f=8.82-1.5*0.3 \\ \\ F_f=8.82-0.45 \\ \\ F_f=8.37\text{ N} \end{gathered}[/tex]Therefore, the frictional force is 8.37 N
ANSWER:
• (a). 22.05 J
,• (b) 0
,• (c). 8.37 N
What is the frequency of yellow light which has a wavelength of 569 nm?
Answers
The frequency is related to the speed and wavelength by the formula:
[tex]f=\frac{v}{\lambda}[/tex]In this case, since we are talking of light, the speed is c which has a value of 3e8 m/s. Plugging this and the wavelength given we have:
[tex]\begin{gathered} f=\frac{3\times10^8}{569\times10^{-9}} \\ f=5.27\times10^{14} \end{gathered}[/tex]Therefore, the frequency is:
[tex]f=5.27\times10^{14}\text{ Hz}[/tex]A circuit with a battery, a 5 Ω resistor, a 14 Ω resistor, and a 21 Ω resistor in parallel. The total voltage is the system is 4.7 V. What is the current through the 5 Ω resistor?
Answers
ANSWER
0.94A
EXPLANATION
We have the following circuit,
All the resistors of the circuit are connected in parallel, which means that they all have the same voltage drop. Since they are connected in parallel with the voltage source, the voltage drop across all of them is 4.7V.
Using Ohm's law, we can find the current through the 5Ω resistor,
[tex]I_{5\Omega}=\frac{V_{5\Omega}}{R_{5\Omega}}=\frac{4.7V}{5\Omega}=0.94A[/tex]Hence, the current through the 5Ω resistor is 0.94A.
Which simple machine is best used to split apart an object?a screwa levera wedgea pulley
Answers
A wedge is best used to split apart an object; it can also be used to hold objects in place, or lift objects from the ground. It consists of two inclined planes joined together
Explain why an airplane flying fromNew York to Los Angeles has both kinetic energy and potential energy.(finished) Inter what other forms of energy are present in the airplane. Words in the book: radiant energySound energyThermal energyMechanical energyElectric energy Gravitational potential energyChemical potential energy Nuclear energy
Answers
gravitational potential energy- and
Mechanical energy, because mechanical energy is = Potential energy + kinetic energy.
During another immunity challenge, a box (m = 26 kg) is positioned on an inclined surface (angle = 33°). A rope is attached to the box and the rope is placed through a pulley and another box(m = 32 kg) filled with puzzle pieces and is suspended from the other end of the rope (see figure below). Several tribemates are positioned on the inclined surface holding onto the first box andwaiting for the rest of their tribe to position themselves below the box with the puzzle pieces in order to catch it and complete the rest of the challenge. When the tribemates on the inclinedsurface release the first box from rest, how long does it take the other tribemates to catch the box of puzzle pieces if the box falls from a height of 4.2 m one the system is released and thecoefficient of kinetic friction between the first box and the inclined surface is 0.25? What is the tension in the rope connecting the two boxes?
Answers
Here's a diagram of the situation with all the forces acting on the boxes.
To calculate the tension in the rope, we need to calculate the sum of forces pulling on the rope.
Let's call Fg1 the force of gravity of the first box (on the incline) and Fg2 the force of gravity on the second.
Remember that F = mg, where m is mass and g is gravitational acceleration (9.81 m/s^2)
Fg2 = 32*9.81 = 313.92 N
Fg1 = 26*9.81 = 255.06 N
Since the first box is on an incline, the gravitational force can be broken up into two components: the component parallel (tangent) to the incline, and the other perpendicular (normal) to the incline. Let's call those forces FT and FN respectively.
Let's rearrange the force arrows to form a right triangle:
Since we know Fg1 and an angle measure in this triangle, we can calculate FN and FT using trigonometry.
FN = Fg1*cos(33) = 255.06*cos(33) = 213.911 N
FT = Fg1*sin(33) = 255.06*sin(33) = 138.916 N
Now that we have FN, we can calculate the force of friction, let's call that Ff.
Ff = μ*FN, where μ is the coefficient of friction.
Ff = 0.25*213.911 = 53.478 N
In relation to the whole pulley system, the boxes pull each other toward their respective directions. For example, the first box pulls the second box to the left of the pulley, and the second box pulls the first to the right. Imagine the boxes on opposite sides of a tug-of-war. The box that pulls with the most force will win the tug-of-war and pull the entire pulley system in its direction.
The second box is tugging with a force of Fg2, its gravitational force.
The first box is tugging with FT+Ff, its tangent gravitational force and the force of friction.
The total tension in the rope is the sum of all of these tugging forces.
Total tension = Fg2 + FT + Ff = 313.92+138.916+53.478
Total tension = 506.314 N (answer to second part)
Now, the second box tugs with a force of 313.92 N, and the first box tugs with a force of 138.916+53.478 = 192.394 N, and since the tugging force of the second box is greater, the whole pulley system will move in that direction. The resultant force is the tugging force of the second box minus that of the first box. Let's call this FR.
FR = 313.92 - 192.394 = 121.526 N
The pulley now pulls the mass of both of these boxes combined with this force. Let's call this m.
m = 26kg + 32kg = 58 kg
Remember again that F = m*a. Let's substitute FR for F, then calculate a.
121.526 = 58a; a = 2.0953 m/s^2
So the pulley system accelerates at this rate, so therefore the second box falls at this rate as well.
We also have that it falls a distance d = 4.2m before it's caught, and the pulley system starts at rest (up until the second box is released), so the initial velocity vi = 0.
We need to calculate the time it takes for the box to fall this far, t.
Given the variables we know and need, we can use the following equation:
d = vi*t + (a/2)*t^2
We can substitute for d, vi, and a, then solve for t.
4.2 = 0 + (2.0953/2)*t^2
Solving for t,
t = 2.0022 s (answer to first part)
2. A capacitor has a capacitance of 1.50 x 10-10 F. How much charge is stored when the potential difference is 1.75 V?
Answers
The charge in a capacitor is given by:
[tex]Q=CV[/tex]plugging the values given we have that:
[tex]\begin{gathered} Q=(1.5\times10^{-10})(1.75) \\ Q=2.625\times10^{-10} \end{gathered}[/tex]Therefore the charge in the capacitor is:
[tex]2.625\times10^{-10}\text{ C}[/tex]A source of sound is directed along a straight line perpendicular to a large brick wall. A girl walks along the line towards the wall and notices that the intensity of sound decreases to a minimum every 50.0 cm. What is the frequency of the sound? (air temp is 20oC)
Answers
Answer:
686 Hz
Explanation:
Here we must remember the following relationship.
[tex]v=\lambda f[/tex]where v = speed of sound, λ = wavelength, and f = frequency.
Now in our case, we know that the speed of sound at 20°C is 343 m/s.
Furthermore, we also know that sound intensity decreases to a minimum every 50.0 cm; this means that the wavelength of the sound wave must be 50 cm. Why? because the wavelength is defined as the distance between two minima or two maxima.
Now, since we know that
v = 343 m/s and λ = 50.0 cm = 0.50 m, our formula gives
[tex]\begin{gathered} v=\lambda f \\ \Rightarrow343m/s=(0.50m)f \end{gathered}[/tex]Dividing both sides by 0.50 m gives
[tex]f=\frac{343\; m/s}{0.50\; m}[/tex][tex]f=686/s[/tex][tex]\boxed{f=686\; Hz\text{.}}[/tex]Hence, the frequency of sound is 686 Hz.
2.3 kg wooden block is at rest on a frictionless surface. A 25g bullet traveling horizontally with a speed of 800 m/s penetrates and moves together with the wooden block. What is their velocity in m/s?
Answers
ANSWER
[tex]\begin{equation*} 8.60\text{ }m\/s \end{equation*}[/tex]EXPLANATION
Parameters given:
Mass of the block, M = 2.3 kg
Mass of the bullet, m = 25 g = 0.025 kg
Initial velocity of the block, U = 0 m/s
Initial velocity of the bullet u = 800 m/s
Final velocity of the bullet and the block = v
The bullet and the block will have the same final velocity since they move together. To find the final velocity of the bullet and the block, apply the conservation of momentum:
[tex]mu+MU=(m+M)v[/tex]Solving for v, we have that the final velocity of the block and the bullet is:
[tex]\begin{gathered} (0.025*800)+(2.3*0)=(0.025+2.3)v \\ 20=2.325v \\ \Rightarrow v=\frac{20}{2.325} \\ v=8.60\text{ }m\/s \end{gathered}[/tex]That is the answer.