We will have the following:
[tex]\frac{3.60m\ast1.50m}{0.50m}=10.8\approx11[/tex]So, the solution with the proper number of significant figures is approximately 11. [2 significant figures]
Amanda's Coffee Shop makes a blend that is a mixture of two types of coffee. A coffee costs Amanda $5.60 per pound, and type B coffee costs $4.55 per pound. This month, Amanda made 141 pounds of the blend, for a total cost of $728.70. How many pounds of type B coffee did she use?
We are given that Amanda made 141 pounds of coffee. If "x" is the pounds of type A coffee and "y" is the amount of type b coffee then we can write this mathematically as:
[tex]x+y=141,(1)[/tex]We are also given that the cost of type A is $5.6 per pound and that the cost of type B is $4.55 per pound and that the total cost is $728.70, this can be written mathematically as:
[tex]5.6x+4.55y=728.7,(2)[/tex]Now, we solve for "x", first by subtracting "y" from both sides:
[tex]y=141-x[/tex]Now, we substitute the value of "y" in equation (2):
[tex]5.6x+4.55(141-x)=728.7[/tex]Now, we apply the distributive property in the parenthesis:
[tex]5.6x+641.55-4.55x=728.7[/tex]Now, we add like terms:
[tex]1.05x+641.55=728.7[/tex]Now, we subtract 641.55 from both sides:
[tex]\begin{gathered} 1.05x=728.7-641.55 \\ 1.05x=87.15 \end{gathered}[/tex]Now, we divide both sides by 1.05:
[tex]x=\frac{87.15}{1.05}[/tex]solving the operations:
[tex]x=83[/tex]Now, we substitute the value of "x" in equation (1):
[tex]\begin{gathered} y=141-83 \\ y=58 \end{gathered}[/tex]Therefore, the amount of coffee type A is 83 pounds and type B is 58 pounds.
A dog barks with an intensity level of 80 decibels. Two barking dogs produce what intensity level?
We are given that a dog barks with an inetensity level of 80 dB and asked to find out the intensity level produced by two barking dogs.
The combined intensity level of both dogs is the sum of each dog's intensity level.
[tex]\begin{gathered} I_{net}=I_1+I_2 \\ I_{net}=2I_{} \\ I_{net}=2\cdot I_0\cdot10^{\frac{\beta}{10}}_{} \end{gathered}[/tex]Where β is 80 dB and I0 is the reference intensity (1x10^-12 W/m^2)
[tex]\begin{gathered} I_{net}=2\cdot10^{-12}\cdot10^{\frac{80}{10}}_{} \\ I_{net}=2\cdot10^{-12}\cdot10^8_{} \\ I_{net}=2\cdot10^{-12+8} \\ I_{net}=2\cdot10^{-4} \end{gathered}[/tex]The net β is given by
[tex]\begin{gathered} \beta_{\text{net}}=10\log (\frac{I_{net}}{I_0}) \\ \beta_{\text{net}}=10\log (\frac{2\cdot10^{-4}}{10^{-12}}) \\ \beta_{\text{net}}=10\log (2\cdot10^{-4+12}) \\ \beta_{\text{net}}=10\log (20^8) \\ \beta_{\text{net}}=10(8.301) \\ \beta_{\text{net}}=83\; dB \end{gathered}[/tex]Therefore, two barking dogs produce 83 dB intensity level.
. Car D is heading East at 25 m/s and Car T is heading West at 10 m/s. What is the relative velocity between car D and car T? How would the passenger in Car D describe Car T’s motion?
Cars D and T are travelling at a relative speed of 35 m/s.
Car D's velocity is u = + 25 m/s because it is travelling east at a speed of 25 m/s. (because it is travelling in the direction of positive x-)
Additionally, Car T is moving at v = -10 m/s and travelling west at a speed of 10 m/s. (because it is travelling opposite in the direction of positive x-)
Thus, relative velocity = velocity of D - velocity of T
relative velocity = 25 - (-10)
relative velocity = 35m/s
So, the relative velocity between Car D and Car T is 35 m/s.
Car D is travelling in the direction of the position axis' positive arrow, whereas Car T is moving in the direction of the axis' negative arrow, therefore the two automobiles are moving in the opposite direction.
As a result, car D is driving more quickly than car T, and both vehicles are travelling in the opposite direction.
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why short circuit are dangrous
Short circuits occur when a high-volume electrical current is passed through a low-resistance path that isn't meant to carry electricity. Short circuits occur when a live wire makes contact with a conductive object that it should not have.
The most dangerous aspect of a short circuit is that it will cause a sudden and dramatic change in electrical resistance. Electrical systems can cause a large amount of resistance and will be constantly sending electricity to different areas when appliances are plugged in or outlets are being used. This has the possibility to cause bodily harm and appliance damage to those who are using electrical appliances at the time a short circuit happens, and can also cause structural damage to your home through fires and burns
Wires with faulty or unstable connections will create a situation in which the electrical current can take a detour down an unintended path if it is the one of least resistance. This can be unpredictable, and the route that the current travels could be through flammable materials or a human being.
Short circuits are liable to cause fires, electrical burns, and electrocution. Stray electrical currents can also cause considerable damage to appliances and a home’s electrical system.
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Answer: Maybe B or D
Explanation:
When the length of a pendulum is doubled, its frequency will be cut in half. Is this true or false?
The period T of a pendulum with length L is given by the formula:
[tex]T=2\pi\cdot\sqrt[]{\frac{L}{g}}[/tex]Where g is the acceleration of gravity:
[tex]g=9.81\frac{m}{s^2}[/tex]If the length of the pendulum is doubled, the period will be:
[tex]\begin{gathered} T^{\prime}=2\pi\cdot\sqrt[]{\frac{2L}{g}} \\ =\sqrt[]{2}\times2\pi\cdot\sqrt[]{\frac{L}{g}} \\ =\sqrt[]{2}\times T \end{gathered}[/tex]On the other hand, the frequency is the reciprocal of the period. Then:
[tex]\begin{gathered} f=\frac{1}{T} \\ f^{\prime}=\frac{1}{T^{\prime}}=\frac{1}{\sqrt[]{2}\times T}=\frac{1}{\sqrt[]{2}}\times\frac{1}{T}=\frac{1}{\sqrt[]{2}}\times f \end{gathered}[/tex]Then, if the length of the pendulum is doubled, the frequency is cut by a factor of 1/√2:
[tex]f^{\prime}=\frac{1}{\sqrt[]{2}}\times f[/tex]This is not the same as if the frequency was cut by a factor of 1/2.
Therefore, the statement is:
[tex]\text{False}[/tex]Why is the sand on a dry beach not a fluid but when a wave comes in and out the sand does become a fluid and your feet sink when standing on it?
When the sand is mixed with the water, it will form a quicksand. It behaves like a solution, and when the water is removed from the sand, it will act as a solute.
When our feet sink into the sand, our movements will cause you to dig yourself deeper into it.
A ball is moving at 5.0 m/s and has a mass of 3.0 kg. It strikes a second ball with a mass of 4.0 kg that is sitting
motionless. What is the velocity of the second ball if the first ball stops immediately after the collision?
3.75 m/s is the velocity of the second ball if the first ball stops immediately after the collision
initial momentum=final momentum
m1v1=m2v2
3×5=4×v2
15=4v2
v2=15/4
v2=3.75 m/s
"The rate at which an item changes its position" is described by a vector number called velocity. Imagine someone walking swiftly, one step forward, one step back, and starting each step from the same location. Velocity is a vector quantity. Therefore, velocity is cognizant of direction. The direction must be taken into account when determining an object's velocity. A velocity of 55 miles per hour is not adequate information. It is necessary to account for the direction in order to accurately describe the object's velocity. Simply put, the direction of the velocity vector represents the motion of the object.
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A bullet of mass 18.8 grams is fired with a speed of 399 meters per second toward a wood block of mass 277 grams initially at rest on a very smooth surface. What is the change in momentum of the bullet if it ricochets in the opposite direction with a speed of 386 meters per second? The answer must be in 3 significant digits.
Given data
*The given mass of the bullet is m = 18.8 g = 0.0188 kg
*The bullet is moving at a speed is u = 399 m/s
*The given final speed of the bullet is v = -386 m/s
*The mass of the wooden block is M = 277 g = 0.277 kg
The formula for the change in momentum is given as
[tex]\begin{gathered} \Delta p=mv_{}-mu \\ =m(v_{}-u_{}) \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \Delta p=0.0188(-386-399) \\ =-14.758\text{ kg.m/s} \end{gathered}[/tex]Hence, the change in momentum is -14.758 kg.m/s
Two equally charged, 4.982 g spheres are placed with 3.173 cm between their centers. When released, each begins to accelerate at 258.312 m/s2. What is the magnitude of the charge, in micro-Coulombs, on each sphere?
Given:
The mass of the spheres is m = 4.982 g
The distance between the center of spheres is d = 3.173 cm
The acceleration is a = 258.312 m/s^2.
To find the magnitude of charge in micro Coulomb on each sphere.
Explanation:
According to Newton's second law, the force will be
[tex]F\text{ =ma}[/tex]According to Coulomb's law, the force will be
[tex]F=\frac{kq^2}{r^2}[/tex]Here, k is the Coulomb's constant whose value is
[tex]k=9\times10^9\text{ N m}^2\text{ /C}^2[/tex]On equating the forces, the charge will be
[tex]\begin{gathered} ma=\frac{kq^2}{r^2} \\ q=\sqrt{\frac{mar^2}{k}} \end{gathered}[/tex]On substituting the values, the magnitude of charge will be
[tex]\begin{gathered} q=\sqrt{\frac{(4.982\times10^{-3})\times258.312\times(3.173\times10^{-2})^2}{9\times10^9}} \\ =3.79\text{ }\times10^{-7}\text{ C} \\ =0.379\text{ }\times10^{-6}\text{ C} \\ =0.379\text{ }\mu C \end{gathered}[/tex]The magnitude of the charge of each sphere is 0.379 microCoulomb
Two equally charged, 3.699 g spheres are placed with 3.592 cm between their centers. When released, each begins to accelerate at 297.727 m/s2. What is the magnitude of the charge on each sphere? Express your answer in microCoulombs.
Given:
The mass of the first sphere is: m1 = 3.699 g.
The mass of the second sphere is: m2 = 3.699 g
The distance between their centers is: d = 3.592 cm
The acceleration of each sphere is: a = 297.727 m/s^2
To find:
Since the spheres are identical in their masses, the force on each sphere is:
[tex]F=ma[/tex]Substitute the values in the above equation and simplify it, we get:
[tex]\begin{gathered} F=3.699\text{ g}\times297.727\text{ m/s}^2 \\ \\ F=3.699\text{ g }\times\frac{1\text{ kg}}{1000\text{ g}}\times297.727\text{ m/s}^2 \\ \\ F=3.699\times10^{-3}\text{ kg}\times297.727\text{ m/s}^2 \\ \\ F=1.1012\text{ N} \end{gathered}[/tex]This is the force experienced by each sphere and is has a magnitude equal to the magnitude of the electrostatic force.
The electrostatic force of attraction or repulsion between two charges is given by:
[tex]F=\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}[/tex]Substitute the values in the above equation and simplify it, we get:
[tex]\begin{gathered} 1.1012\text{ N}=\frac{9\times10^9\text{ N}\cdot m^2\text{/C}^2\times q^2}{(3.592\text{ cm})^2} \\ \\ 1.1012\text{ N}=\frac{9\times10^9\text{ N}\cdot m^2\text{ / C}^2\times q^2}{(3.592\text{ cm}\times\frac{1\text{ m}}{100\text{ cm}})^2} \\ \\ 1.1012\text{ N}=\frac{9\times10^9\text{ N}\cdot m^2\text{ /C}^2\times q^2}{(3.592\times10^{-2})^2\text{ m}^2} \\ \\ 1.1012\text{ N}=\frac{9\times10^9\text{ N.m}^2\text{/C}^2\times q^2}{1.2902\times10^{-3}\text{ m}^2\text{ }} \\ \\ 1.1012\text{ N}=6.9757\times10^{12}\text{ N/C}^2\times q^2 \\ \\ \end{gathered}[/tex]Rearranging the above equation and simplify it, we get:
[tex]\begin{gathered} q^2=\frac{1.1012\text{ N}}{6.9757\times10^{12}\text{ N/C}^2} \\ \\ q=\sqrt{1.5786\times10^{-13}\text{ C}^2} \\ \\ q=0.3973\times10^{-6}\text{ C} \\ \\ q=0.3973\text{ }\mu\text{C} \end{gathered}[/tex]Final answer:
The magnitude of the charge on each sphere is 0.3973 microcolumns.
Scenario 3: a clock attached to the wall.
1. State the object stationary.
2. State the outside, external, unbalanced force acting on the object.
answer: this object is stationary because it is in rest position until or unless the external force act on it we can say this by netwons 1st law
and the stationaty object has relatively zero velocity.
netwons 1st law: Newton's first law states that unless compelled to change its state by the action of an external force, every object will remain at rest or in uniform motion in a straight line.
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2. here the object has contact with the wall hence there is a normal force between clock and wall
the clock is at rest this means that the forcee is balanced and normal force is balanced by gravitational force .the normal force is always perpendicular to the surface of the object.
there is no external force is acting on the clock.
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If a ball is thrown with an initial horizontal velocity of 2.3m/s, from a tall building, how far away from thebuilding does the ball land if it takes 4s to land?Referring to the ball above, how tall is the building? (2 sig figs)
Given,
The initial velocity of the ball, u=2.3 m/s
The time interval in which the ball hits the ground, t=4 s
As the ball was thrown horizontally, the angle of projection is θ=0°
The distance traveled by the ball in the horizontal direction is given by,
[tex]\begin{gathered} x=u_xt \\ \Rightarrow x=u\cos \theta\times t \end{gathered}[/tex]Where uₓ is the x-component of the initial velocity.
On substituting the known values,
[tex]\begin{gathered} x=2.3\times\cos 0^{\circ}\times4 \\ =9.2\text{ m} \end{gathered}[/tex]Therefore the distance traveled by the ball in the x-direction is 9.2 m. That is the ball landed 9.2 meters away from the building.
Applying the equation of the motion in the y-direction,
[tex]\begin{gathered} y=y_0+u_yt+\frac{1}{2}gt^2 \\ =y_0+u\sin \theta\times_{}t+\frac{1}{2}gt^2 \end{gathered}[/tex]Where y is the final height of the ball which is zero meters, y₀ is the initial height of the ball, i.e., the height of the building, uy is the y-component of the initial velocity.
Let us consider that the upward direction is positive while the downward direction is negative. This makes the acceleration due to gravity, g, a negative value, and the height of the building a positive value.
On substituting the known values,
[tex]\begin{gathered} 0=y_0+2.3\times\sin (0^{\circ})\times4+\frac{1}{2}\times-9.8\times4^2 \\ \Rightarrow-y_0=0-4.9\times4^2 \\ y_0=78.4\text{ m} \\ \approx78\text{ m} \end{gathered}[/tex]Therefore the height of the building is 78 m
How much work is done to transfer 0.15 C of charge through a potential difference of 9.0 V?
Answer:
6.075J
Explanation:
The formula is 0.5X0.15X9X9. Hal of CxVxV. This results in 6.075J
what is the speed (m/s) of a caterpillar traveling 45m along a sidewalk in 50hr?
We will have the following:
[tex]v=\frac{45m}{50h}\cdot\frac{1h}{60s}\Rightarrow v=\frac{3}{200}m/s[/tex][tex]\Rightarrow v=0.015m/s[/tex]So, the speed is 0.015 m/s.
In the lab, a student is reading the amount of water in this graduated cylinder.
What is the amount of water in this cylinder?
A) 20.5 mL
B) 25mL
C) 250 mL
D) 24mL
D. O499.917 JE. O 1358.602 J4. An object of mass 5 kg is sliding downa friction less inclined plane of length3 m that makes an angle of10 deg. Calculate its speed just before it hits the ground. (1 point)A. 03.195 m/sB. O 1.958 m/sC. O5.58 m/sD. O2.525 m/sE. 04.225 m/s5. A roller coaster extends to the groundfrom a height of 39 m (point A) and then risesAn obiect of mass 12 kg
Given
Mass of the object, m=5 kg
Length of the inclined plane, s=3m
Angle of inclination,
[tex]\theta=10^o[/tex]To find
Calculate its speed just before it hits the ground.
Explanation
Here the acceleration is
[tex]\begin{gathered} a=gsin\theta \\ \Rightarrow a=9.8sin10^o \end{gathered}[/tex]The initial velocity is zero.
So the final velocity is
[tex]\begin{gathered} v=\sqrt{2as} \\ \Rightarrow v=\sqrt{2\times9.8sin10^o\times3} \\ \Rightarrow v=3.19\text{ m/s} \end{gathered}[/tex]Conclusion
The velocity before it hit the ground, A.3.19 m/s
In the table below place the different forms of electromagnetic radiation and visible light in order of lowest energy to highest energy
Answer:
Step-by-step explanation:
The electromagnetic spectrum is the range of all types of EM radiation. I will write the electromagnetic spectrum from lowest energy/longest wavelength to highest energy/shortest wavelength:
*Visible light refers to the visible region of the electromagnetic spectrum, that is the range of wavelengths that trigger brightness and color perception in humans. It lies between Ultraviolet and Infrared radiation
What is the momentum of a 0.144kg baseball thrown at a speed of 46m/s? Give your answer in kgm/s.
ANSWER:
6.624 kgm/s
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 0.144 kg
Speed (v) = 46 m/s
We can calculate momentum using the following formula:
[tex]\begin{gathered} P=m\cdot v \\ \\ \text{ We replacing} \\ \\ P=0.144\cdot46 \\ \\ P=6.624\text{ kgm/s} \end{gathered}[/tex]The momentum is 6.624 kgm/s
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Answer:
Thermal energy flows from the flask on the left to the flask on the right
Explanation:
As the flask on the left is hot water the temperature of it will be higher meaning that the metal bar will conduct the heat at a faster rate than the flask on the left. Due to this, the metal bar will gradually begin to increase in temperature slowly throughout the whole bar. This means that the heat will travel through the bar reaching tho whole flask.
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A rocket weighing 220,000 N is taking off from Earth with a total thrust of 500,000 N at an angle of 20 degrees, as shown in the image below. What is the approximate vertical component of the net force that is moving the rocket away from Earth?A. 400,000 NB. 350,000 NC. 250,000 ND. 300,000 N
ANSWER
[tex]C.250,000N[/tex]EXPLANATION
Parameters given:
Weight of rocket, W = 220,000 N
Thrust, F = 500,000N
Angle of thrust, θ = 20°
The vertical forces acting on the rocket are the thrust (acting at an angle) and its weight. This means that the sum of vertical forces is:
[tex]F_y=F\cos \theta-W[/tex]Note: W is negative since it acts opposite the motion of the rocket.
Therefore, we have that the net vertical force (vertical component of forces) is:
[tex]\begin{gathered} F_y=500000\cos 20-220000 \\ F_y=469,846.31-220000 \\ F_y=249,846.31N \\ F_y\approx250,000N \end{gathered}[/tex]The answer is option C.
What is a limitation to using nuclear fusion for energy?
Answer:
Explanation:
The radiation of components in a fusion reactor is not much enough for the materials to be reused or recycled within centuries
A force gives energy to an object and that energy can:set the object in motion.stop its motion.change the speed and direction of its motion.All of the choices are correct.
To find:
Which of the given statements is the correct answer?
Explanation:
The work done is the measure of the transfer of energy through the application of force.
The force thus applied can either change the state of motion of the object or the speed and direction of the motion.
That is, when a force is applied to an object, the force can bring the object in motion to rest or set the object in rest, into motion. Or it can also change the speed or direction of motion of the object.
Final answer:
The correct answer is option D, All of the choices are correct.
In which of the following processes does the Sun play little to no role inproviding energy?A. Plate tectonicsB. Ocean currentsC. SeasonsD. Trophic levels
Plate tectonics are influenced by the earth's mantle and its variations, therefore, the role of the sun is less in comparison with the other processes described.
What is the acceleration of a car that starts at rest and achieves a final velocity of 33.5 m/s in a time of 12 seconds?
Variables we are given and that we know:
vi: initial velocity; vi = 0 m/s
vf: final velocity; vf = 33.5 m/s
t: time; t = 12 s
We need to find a: acceleration
We can use the formula:
vf = vi + a*t
Let's substitute the values we know:
33.5 = 0 + a*12
33.5 = 12a
Let's solve for a:
a = 33.5/12
a = 2.7917 m/s^2
2 similar metal spheres, A and B, have charges of +2.0×10^-6 coulomb and +1.0×10^-6 coupons, respectively, as shown in the diagram below. The magnitude of the electrostatic force on fear a do to sphere B is 2.4 in period what is the magnitude of the electrostatic force on sphere B due to spear A?
Given data:
The magnitude of the electrostatic force on sphere A due to sphere B is Fab = 2.4 N
The magnitude of the electrostatic force on sphere A due to sphere B is equal to the magnitude of the electrostatic force on sphere B due to sphere A.
Therefore, the magnitude of the electrostatic force on sphere B due to sphere A is given as:
[tex]F_{ba}=2.4\text{ N}[/tex]Thus, the magnitude of the electrostatic force on sphere B due to sphere A is 2.4 N.
I need help filling out the acceleration for gravity due to the numbers in this chart
In order to calculate the acceleration of gravity, we can use the free fall formula:
[tex]d=\frac{gt^2}{2}\to g=\frac{2d}{t^2}[/tex]Where d is the distance travelled and t is the time.
So for each distance in the chart, we have:
[tex]\begin{gathered} d=0.2\colon \\ g=\frac{0.4}{0.184^2}=11.81 \\ \\ d=0.4\colon \\ g=\frac{0.8}{0.258^2}=12.02 \\ \\ d=0.6\colon \\ g=\frac{1.2}{0.4^2}=7.5 \\ \\ d=0.8\colon \\ g=\frac{1.6}{0.236^2}=28.73 \\ \\ d=1\colon \\ g=\frac{2}{0.33^2}=18.37 \end{gathered}[/tex](All gravity units in m/s²)
Now, calculating the average, we have:
[tex]g=\frac{11.81+12.02+7.5+28.73+18.37}{5}=15.69[/tex]Finally, calculating the percent error, we have:
[tex]\text{Percent error}=\frac{|9.81-15.69|}{9.81}=0.5994=59.94\text{\%}[/tex]How much force must I lift with to lift a 30kg object off the ground?
The force required to lift the object is 294 N.
Given data:
The mass of object is m=30 kg.
The force applied to lift an object will be equal to the weight of the object.
The weight of the object can be calculated as,
[tex]\begin{gathered} W=mg \\ W=(30)(9.8) \\ W=294\text{ N} \end{gathered}[/tex]The weight of the object is 294 N; therefore, the force required to lift the object is 294 N.
Write 8.0090 x 10^2 in standard form . (Example of standard form: 4550)Blank 1:
We are given the following number in scientific notation:
[tex]8.0090\times10^2[/tex]To convert this number into standard form we will move the decimal point a number of times equal to the exponent of the ten. Since the exponent is positive we will move the point to the right, like this:
[tex]8.0090\times10^2=800.90[/tex]Therefore, the standard form of the number is 800.90
Create a table comparing and contrasting constructive and destructive interference .
Constructive interference:-
1. The constructive interference takes place, when the amplitude of the two or more waves combines to form the resultant wave with more amplitude.
2. The phase difference between the waves during the constructive interference is,
[tex]\varnothing=\frac{n\pi}{2}[/tex]where n is the even number.
3. In the constructive interefernce, the bright fringes are observed.
Destructive interference:-
1. The destructive interference takes place when the ampltude of two or more wave combines to form the resultant wave with less amplitude.
2. The phase difference between the waves during the destructive interference is,
[tex]\varnothing=\frac{n\pi}{2}[/tex]where n is the odd number.
3. In the destructive interference, the dark fringes are formed.