Find the net force and the direction of motion ? Mr Collin finally decided to take this seriously and put both hands on the rope and applied a 500 n force to the left While Jared and James still struggled with their 300n force to the right What is the net force and direction of motion?
Answers
Given:
The force applied by Mr Collin is F1 = 500 N towards the left.
The force applied by Jared and James is F2 = 300 N towards the right.
To find the net force and direction.
Explanation:
The net force can be calculated as
[tex]\begin{gathered} F_{net}=F1-F2 \\ =500-300 \\ =200\text{ N} \end{gathered}[/tex]The direction of force is towards F1 i.e. from left as F1> F2.
Related Questions
oxygen combines with hydrogen to form water, no atoms are left over from the reaction.
Answers
It is to be noted that when water is formed, two of each hydrogen molecule bond to one oxygen creating 2 hydrogens molecules and one oxygen molecule H₂O. This way, there is no atom left from the Chemical reaction.
What is a chemical reaction?A chemical reaction is a process that causes one group of chemicals to chemically change into another group of chemicals.
Any compound involves the connecting of distinct elements through the use of various bonds, which are as follows:
The covalent bondThe ionic bondThe sigma bondAccording to the question, oxygen atoms and hydrogen molecules link together in their empty space and connect together with the assistance of a covalent bond.
The link alters the physical state of the element, and they combine to create a liquid known as water.
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Full Question:
Explain how the model you developed shows that when oxygen combines with hydrogen to form water, no atoms are left over from the reaction.
See the attached image
Running at 2 m/s, Bruce the 45 kg quarterback, collides with Biff, the 90 kg tackle, who is traveling at 7 m/s in the other direction. Upon collision, Biff continues to travel forward at 1 m/s. How fast is Bruce knocked backwards? What is Bruce’s impulse?
Answers
Given data
*The mass of Bruce is m_1 = 45 kg
*The initial velocity of the Bruce is u_1 = 2 m/s
*The mass of the biff is m_2 = 90 kg
*The initial velocity of the Biff is u_2 = -7 m/s
*The final velocity of the first glider is v_2 = -1 m/s
According to the law of conservation of linear momentum, the total linear momentum of a system remains constant
Applying the law of conservation of momentum as
[tex]\begin{gathered} p_i=p_f \\ m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ v_1=\frac{m_1u_1+m_2u_2-m_2v_2_{}_{}_{}_{}}{m_1} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v_1=\frac{(45)(2)+(90)(-7)-(90)(-1)}{45} \\ =-10\text{ m/s} \end{gathered}[/tex]Hence, the speed of the bruce knock backwards is v_1 = -10 m/s
Question 9 of 25Which of the following is qualitative information?O A. 15 gramsB. 20 degreesO c. 25 secondsC.O D. Blue skySUBMIT
Answers
Answer: Blue sky
Explanation:
Qualitative information cannot be measured or quantified. Looking at the given options, we have temperature and weight that can be measured. Option D can't be measured. Thus, the correct option is
Blue sky
What is the angular displacement of a wheel that made 157.5revolutions? Round the answer to decimals.
Answers
To find the angular displacement, we have to transform 157.5 revolutions to radians. We know that 1 revolution equals 2pi radians
[tex]1\text{rev}=2\pi[/tex]Let's use the following proportion to find the angular displacement for 157.5 revolutions.
[tex]\frac{1\text{rev}}{157.5\text{rev}}=\frac{2\pi}{x}[/tex]Then, we solve for x
[tex]\begin{gathered} x=\frac{2\pi\cdot157.5}{1} \\ x=315\pi \end{gathered}[/tex]But, pi = 3.14, so
[tex]\begin{gathered} x=315\cdot3.14\text{rad} \\ x=989.1rad \end{gathered}[/tex]Therefore, the angular displacement is 989.1 radians.The angular displacement of a wheel that made 157.5 revolutions is equal to 989 radians.
What is angular displacement?Angular displacement can be described as the angle through which a point revolves around a center or a particular axis. When a body rotates about its axis, as in circular motion it undergoes a changing velocity and acceleration at any time (t).
When the separations between the particles remain constant throughout the motion then a body is considered rigid. Angular displacement can be measured in radians or degrees.
θ= s/r
θ = 2πr/r which easily simplifies to 2π. Therefore, 1 revolution is 2π radians. Using radians gives a relationship between the distance traveled around the circular path and the distance r from the center.
Given, the number of revolutions = 157.5 rev
The angular displacement = 157.5 × 2π = 315 × 3.14 rad = 989 rad
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Suppose the relative humidity of air was 96% on a cold morning when the temperature was 20 degree Celsius . In the afternoon, the air temperature has risen to 35 degree Celsius , but the actual amount of water vapor in air is the same as it was in the morning. The barometric pressure is also the same as it was in the morning. What is the approximate relative humidity in the afternoon?
Answers
Therelative humidity, temperature and dew point are approximately related by:
[tex]T_{dp}=T-\frac{100-RH}{5}[/tex]Plugging the values for the temperature and relative humidity given we have that:
[tex]\begin{gathered} T_{dp}=20-\frac{100-96}{5} \\ T_{dp}=20-\frac{4}{5} \\ T_{dp}=20-0.8 \\ T_{dp}=19.2 \end{gathered}[/tex]Once we know the dew point we can plug it in the equation with the afternoon temperature to find the relative humidity:
[tex]\begin{gathered} 19.2=35-\frac{100-RH}{5} \\ \frac{100-RH}{5}=35-19.2 \\ 100-RH=5(15.8) \\ 100-RH=79 \\ RH=100-79 \\ RH=21 \end{gathered}[/tex]Therefore, the relative humidity is 21%
Assume the sand bags are rectangular in shape and have a mass of 0.3kg and a length of 25 cm and a width of 11 cm. The terminal velocity is 14.77m/s if the drag coefficient is 0.8 and the density of the air is 1.225kg/m3. a) How long would it take for this object to reach its terminal velocity if we ignore drag until the sandbag reaches its terminal velocity? b) The sandbag reaches its terminal velocity just before it hits the ground. How tall is the tower?
Answers
ANSWER:
a) 1.51 seconds
b) 11.17 meters
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 0.3 kg
Length (l) = 25 cm = 0.25 m
Width (w) = 11 cm = 0.11 m
Terminal velocity (vt) = 14.77 m/s
Density of the air (d) = 1.225 kg/m³
Drag coefficient = 0.8
a)
We can determine the time using the following formula since we know the terminal velocity:
[tex]\begin{gathered} v_t=u+gt \\ \\ \text{ in this case the initial velocity is 0, therefore:} \\ \\ 14.77=0+9.8t \\ \\ t=\frac{14.77}{9.8} \\ \\ t=1.51\text{ sec} \end{gathered}[/tex]b)
Now, knowing the time, determine the height:
[tex]\begin{gathered} h=\frac{1}{2}gt^2 \\ \\ \text{ We replacing:} \\ \\ h=\frac{1}{2}(9.8)(1.51)^2 \\ \\ h=11.17\text{ m} \end{gathered}[/tex]Hello, I am having a difficult time understanding this question. Would it be possible to show me the steps to solve it?
Answers
In order to compare the weights, let's check the formula for the weight force between two objects:
[tex]F=\frac{G\cdot M_1\cdot M_2}{d^2}[/tex]Where G is the gravitational constant, M1 and M2 are the masses and d is the distance between them
We can see that the distance is in the denominator and it's squared.
That means if the distance doubles, the force will be reduced by a factor of 4.
The distance from the debris if it was on Earth is one radius, so the actual distance is twice the distance that it would be on Earth.
Therefore the weight of this debris is 4 times smaller than its weight on Earth.
Calculate the moment of inertia for a solid sphere with a mass of 10 kg and a radius of 0.2m.
Answers
Given:
the mass of the solid sphere is
[tex]m=10\text{ kg}[/tex]radius of the sphere is
[tex]r=0.2\text{ m}[/tex]Required: calculate the moment of inertia
Explanation:
to calculate the moment of inertia we will use the formula that is given by
[tex]I=\frac{2}{5}mr^2[/tex]Plugging all the values in the above formula, we get
[tex]\begin{gathered} I=\frac{2}{5}\times10\text{ kg}\times0.2\times0.2\text{ m}\times\text{ m } \\ I=0.16\text{ kg m}^2 \end{gathered}[/tex]Thus, the moment of inertia is 0.16 .
A species of extremely rare deep water fish has an extremely low reproduction rate . If there are 821 of this type of fish and their growth rate is 2% each month , how many will there be in 1/2 year ? In 10 years ? In 100 years ?
Answers
This can be model using the exponential growth formula:
[tex]\begin{gathered} y=a\cdot b^x \\ _{\text{ }}where\colon \\ b=1+r \end{gathered}[/tex]a = Initial amount = 821
r = Growth rate per time period = 2% = 0.02
x = time period
So:
[tex]y=821(1.02)^x[/tex]---
For x = 1/2 year = 6 months
[tex]\begin{gathered} y=821(1.02)^6 \\ y\approx925 \end{gathered}[/tex]-----
For x = 10 years = 120 months
[tex]\begin{gathered} y=821(1.02)^{120} \\ y\approx8838 \end{gathered}[/tex]-----------------------
For x = 100 years = 1200 months
[tex]\begin{gathered} y=821(1.02)^{1200} \\ y\approx1.7\times10^{13} \end{gathered}[/tex]a +80 microcoulombs point charge placed at the origin. calculate magnitude and direction of the Electric Field created by this point charge at the following locations a) at point p (5m, 0) and b) at point S (2m, 4m)
Answers
ANSWERS
(a) E = 28800 N/C, direction 0°
(b) E = 36000 N/C, direction 63.4°
EXPLANATION
Given:
• The charge, q = +80 μC
,• The location of the charge q, (0, 0)
,• The coordinates of points P(5m, 0) and S(2m, 4m)
Find:
• The magnitude and direction of the electric field, E, at points P and S
The magnitude of the electric field created by a charge q at a distance r from the charge is,
[tex]E=k\frac{q}{r^2}[/tex]Where k is Coulomb's constant and has an approximate value of 9x10⁹ Nm²/C².
(a) We have to find the distance from the charge to point P,
Both the charge and the point are on the x-axis, so the distance is the horizontal distance between them: 5m.
The magnitude of the electric field is,
[tex]E=k\cdot\frac{q}{r^2}=9\cdot10^9\frac{Nm^2}{C^2}\cdot\frac{80\cdot10^{-6}C}{5^2m^2}=28800\frac{N}{C}[/tex]Positive charges create fields that point radially away from them. Since q is a positive charge, the direction of its electric field at point P is in the positive x-direction, which is 0 degrees.
(b) For point S we will have to use the Pythagorean theorem to find the distance to the charge,
With the diagram above we will find the distance from point S to the charge and the direction of the electric field, θ.
The distance squared is,
[tex]r^2=2^2+4^2=4+16=20[/tex]So the magnitude of the electric field is,
[tex]E=k\cdot\frac{q}{r^2}=9\cdot10^9\frac{Nm^2}{C^2}\cdot\frac{80\cdot10^{-6}C}{20m^2}=36000\frac{N}{C}[/tex]As explained in part a, positive charges create electric fields pointing radially away from them, so the direction of this electric field is given by the angle θ,
[tex]\tan\theta=\frac{4m}{2m}[/tex]Solving for θ,
[tex]\theta=\tan^{-1}2\approx63.4\degree[/tex]I need help with this question, could you help me? :)))
Answers
Answer: Structure A produces heat, which turns water into steam within structure B, which turns a turbine in structure C.
Explanation:
Structure A has heating columns/rods that help transfer heat more efficiently to the water, heating the water in the tank of structure B. This pressurized steam enters structure C through a pipe, which serves as a turbine.
A short piece of metal that melts when a large current passes through it is a:1) circuit breaker2) surge suppressor3) relay4) fuse
Answers
A short peice of metal that act as an electrical safety device by melting when the large current passes through it. The melting of the metal stops the flow of current and provide safety.
This metal peice is known as fuse.
Hence, option 4 is the correct answer.
Roger Maris swung a bat which had a mass of 2 Kg at a velocity of 45 m/s. How many joules of kinetic energy could he give to a ball?
Answers
ANSWER:
2025 joules
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 2 kg
Velocity (v) = 45 m/s
The kinetic energy is given by the following equation:
[tex]K_E=\frac{1}{2}mv^2[/tex]We replacing:
[tex]\begin{gathered} K_E=\frac{1}{2}\cdot2\cdot45^2 \\ K_E=2025\text{ J} \end{gathered}[/tex]The kinetic energy is equal to 2025 joules.
An electric motor transforms 125 J of electrical energy into mechanical energy in 3 minutes. What is the power of the motor?
Answers
Given:
The electric motor transforms electrical energy,
[tex]E=125\text{ J}[/tex]time taken in this transformation is,
[tex]\begin{gathered} t=3\text{ m} \\ =3\times60\text{ s} \\ =180\text{ s} \end{gathered}[/tex]To find:
The power of the motor
Explanation:
Power is the rate of change in energy. So,
[tex]\begin{gathered} P=\frac{E}{t} \\ =\frac{125}{180} \\ =0.69\text{ W} \end{gathered}[/tex]Hence, the power is 0.69 W.
Resistors of values 6.0Ω, 4.0Ω, 10.0Ω and 7.0Ω are combined as shown. What is the equivalent resistance for this combination? Answer: _________ Ω
Answers
ANSWER
[tex]10.67Ω[/tex]EXPLANATION
First, we have to find the total resistance for the 4Ω and 10Ω resistors connected in series.
To do this, find the sum of the resistances:
[tex]4+10=14Ω[/tex]This total resistance is connected in parallel to the 7Ω resistor. Find the total resistance of the two of them:
[tex]\begin{gathered} \frac{1}{\frac{1}{14}+\frac{1}{7}} \\ \\ \frac{1}{\frac{3}{14}}=\frac{14}{3} \\ \\ 4.67Ω \end{gathered}[/tex]The equivalent resistance is the total resistance of the 6Ω resistor and the 4.67Ω resistance. Since they are connected in series, find the sum:
[tex]\begin{gathered} 6+4.67 \\ \\ 10.67Ω \end{gathered}[/tex]That is the equivalent resistance.
Two uniform solid spheres have the same mass, but one has twice the radius of the other. The ratio of the larger sphere'smoment of inertia to that of the smaller sphere isSelect one:O a. 2O b. 4O C. 1/2O d. 8/5O e.' 4/5
Answers
For the smaller sphere with mass M and radius, R the moment of inertia is
[tex]I_{\text{small}}=\frac{2MR^2}{5}[/tex]For the larger sphere with mass M and radius 2R the moment of Intertia is
[tex]I_{\text{larger}}=\frac{2M(2R)^2}{5}[/tex]The ratio between the larger and the small spheres can be calculated as
[tex]\frac{I_{\text{larger}}}{I_{s\text{mall}}}=\frac{\frac{2M(2R)^2}{5}}{\frac{2M(R)^2}{5}}[/tex]we simplify
[tex]\frac{I_{\text{larger}}}{I_{s\text{mall}}}=\frac{\frac{2M(4)(R)^2}{5}}{\frac{2M(R)^2}{5}}[/tex][tex]\frac{I_{\text{larger}}}{I_{s\text{mall}}}=\frac{4}{1}=4[/tex]ANSWER
The ratio is 4
b. 4
A circuit with blank-diameter(small/large) connecting wires at a blank temperature(low/high) will have the least electrical resistance
Answers
Resistance is directly proportional to temperature, so with high temperature, resistance is also high.
Resistance is inversely proportional to area of cross-section, and so the diameter. Thus, with large diameter resistance will be less.
Hence the answer is large diameter and low temperature will have least electrical resistance.
In a series circuit, what does the sum of all the voltage drops across each resistor equal?Select one:a.the voltage of the power supplyb.a voltage greater than the power supply c.a voltage less than the power supply d.a voltage that’s a multiple of the power supply and total resistance
Answers
Answer:
a. the voltage of the power supply
Explanation:
In a series circuit, the voltage of the power supply is equal to the sum of the voltage on each resistor, so the sum of all the voltage drops across each resistor equals the voltage of the power supply.
Therefore, the answer is
a. the voltage of the power supply
Current is:A.measured in ohms.B.an excess accumulation of charge.C.the flow of electric charge.D.measured in volts.
Answers
Choice A is incorrect because current is measured in amps, not ohms.
Choice B is incorrect because current measures the flow of charge, not the accumulation of it.
Choice C is correct because current measures the flow of electric charge.
Choice D is incorrect because current is measured in amps, not volts.
Answer: C. the flow of electric charge
A body moves along one dimension with a constant acceleration of 4.35 m/s2 over a time interval. At the end of this interval it has reached a velocity of 13.8 m/s.(a)If its original velocity is 6.90 m/s, what is its displacement (in m) during the time interval? m (b)What is the distance it travels (in m) during this interval? m(c) A second body moves in one dimension, also with a constant acceleration of 4.35 m/s2 , but over some different time interval. Like the first body, its velocity at the end of the interval is 13.8 m/s, but its initial velocity is −6.90 m/s. What is the displacement (in m) of the second body over this interval? m (d) What is the total distance traveled (in m) by the second body in part (c), during the interval in part (c)?
Answers
Given that the acceleration of the object is
[tex]a=4.35m/s^2[/tex]The final velocity is
[tex]v_f=13.8\text{ m/s}[/tex]The initial velocity is
[tex]v_o=6.9\text{ m/s}[/tex]We have to find displacement and distance.
Let the displacement be denoted by S and the distance by d.
First, we need to calculate time in order to find distance and displacement.
The time taken will be
[tex]\begin{gathered} t=\frac{(v_f-v_o)}{a} \\ =\frac{13.8-6.9}{4.35} \\ =1.59\text{ s} \end{gathered}[/tex](a) The displacement will be
[tex]\begin{gathered} S=v_ot+\frac{1}{2}at^2 \\ =6.9\times1.59+\frac{1}{2}\times4.35\times(1.59)^2 \\ =10.971+5.498 \\ =16.469\text{ m} \end{gathered}[/tex](b) The initial speed will be equal to the initial velocity.
Thus, the distance will be 16.469 m.
(c) The initial velocity of the second object is
[tex]v^{\prime}_o=\text{ -6.9 m/s}[/tex]The final velocity of the second object is
[tex]v^{\prime}_f=13.8\text{ m/s}[/tex]The acceleration of the second object is
[tex]a^{\prime}=\text{ 4.35 m/s}[/tex]The time taken will be
[tex]\begin{gathered} t^{\prime}=\frac{(v_f-v_o)}{a} \\ =\frac{13.8-(-6.9)_{}}{4.35} \\ =4.75\text{ s} \end{gathered}[/tex]We have to find the displacement of the second object.
Let the displacement of the second object be denoted by S'.
The displacement of the second object will be
[tex]\begin{gathered} S^{\prime}=v^{\prime}_ot^{\prime}+\frac{1}{2}a^{\prime}(t^{\prime})^2 \\ =-6.9\times4.75+\frac{1}{2}\times4.35\times(4.75)^2 \\ =-32.775+49.073 \\ =16.298\text{ m} \end{gathered}[/tex](d) The initial speed of the object will be
[tex]v^{\doubleprime}_o=6.9\text{ m/s}[/tex]We have to find the distance.
Let the distance be denoted by d'.
The distance of the second object will be
[tex]\begin{gathered} d^{\prime}=v^{\doubleprime}_ot^{\prime}+\frac{1}{2}a^{\prime}(t^{\prime})^2 \\ =6.9\times4.75+\frac{1}{2}\times4.35\times(4.75)^2_{} \\ =32.775+49.073 \\ =81.848\text{ m} \end{gathered}[/tex]If the resultant force acting on a 2.0 kg object is equal to (3.0î + 4.0ĵ) N, what is the change in kinetic energy as the object moves from (7.0î – 8.0ĵ) m to (11.0î – 5.0ĵ) m?
Answers
Answer:
ΔK = 24 joules.
Explanation:
Δ[tex]K =[/tex] Work done on the object
Work is equal to the dot product of force supplied and the displacement of the object.
[tex]W = F[/tex] * Δ[tex]s[/tex]
Δ[tex]s[/tex] can be found by subtracting the vectors (7.0, -8.0) and (11.0, -5.0), which is written as Δ[tex]s[/tex] = (11.0 - 7.0, -5.0 - -8.0) which equals (4.0, 3.0).
This gives us
[tex]W = < 3, 4 >[/tex] * [tex]< 4, 3 >[/tex] = [tex](3*4)+(4*3)[/tex] = [tex]24[/tex] J
In Physics we tend to almost always usescatter graphsline graphsbar graphspie graphs
Answers
As the different equations relationships are described by using the graphical form.
The bar graph can be used to describe the energy conservation process.
But most of the equations are described like velocity in terms of time and displacement, force in terms of mass and acceleration etc by using the line graphs.
Thus, the line graphs are mostly used in the physics from the fundamental level to the advanced level.
Hence, line graph is the correct answer.
If a total of 50 J of work are done onan object, its energy...A. decreases by 50 J.B. increases by 50 J.C. doesn't change.Help / ResourcesD. is transferred to another object.
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It is important to know that the work done refers to the amount of energy (Joules) released in a certain activity. However, in this case, if you apply certain work to an object its energy will increase by the same amount.
Hence, the answer is B.Which unit of the following is smaller?a.)one kelvinb.)one degree Celsius c.)one degree Fahrenheit
Answers
c) one degree Farenheit
Explanationto solve this let's convert 1° C into kelvin and Celsiuts, then,let's compare
Step 1
a) convert 1 °C into Kelvin:
to do that, we need to use the formula
[tex]\begin{gathered} °K=C+237\text{ \degree} \\ replace \\ °K=1\text{ +237\degree} \\ °K=238°\text{ K} \end{gathered}[/tex]hence
[tex]1\text{ \degree C=238 K}[/tex]so, one degree Kelvin is smaller than aone degree celsius ( note that 1° C= 238 °K)
Step 2
now, let's convert 1° C into Farenheit
we need to use the formula:
[tex]\begin{gathered} °F=1.8\text{ C+32} \\ where\text{ C is the measure in \degree C} \end{gathered}[/tex]replace and calculate
[tex]\begin{gathered} \begin{equation*} °F=1.8\text{ C+32} \end{equation*} \\ °F=(1.8*1)+32=33.8 \\ °F=33.8\text{ \degree F} \end{gathered}[/tex]so
[tex]1\text{ \degree C=38.3 \degree F}[/tex]so, one degree Fahrenheit is smaller than aone degree celsius ( note that 1° C= 38.3 °F)
Step 3
finally, compare °F and °K
[tex]\begin{gathered} 1\text{ \degree C=238 \degree K=38.3 \degree F} \\ 238°K=38.3°F \\ F=\frac{238}{38.3}K \\ F=6.21\text{ \degree K} \end{gathered}[/tex]so, the measure in °F is six times the measure in °K
therefore,
the smaller unit is
c) one degree Fahrenheit
I hope this helps you
A soccer ball is kicked upward from a height of 6 ft with an initial velocity of 64 ft/s. How high will it go? Use - 32 ft/s2 forthe acceleration caused by gravity. Ignore air resistance.
Answers
Given,
The initial height of the ball, h₁=6 ft
The initial velocity of the ball, u=64 ft/s
The acceleration due to gravity, g=-32 ft/s²
The velocity of the ball when it reaches the maximum height will be, v=0 ft/s
From the equation of motion,
[tex]v^2-u^2=2gh_2[/tex]Where h₂ is the height covered by the ball to reach the maximum height.
On substituting the known values,
[tex]\begin{gathered} 0-64^2=2\times-32\times h_2 \\ h_2=\frac{-64^2}{2\times-32} \\ =64\text{ ft} \end{gathered}[/tex]Thus the maximum height reached by the ball is
[tex]\begin{gathered} H=h_1+h_2 \\ =6+64 \\ =70\text{ ft} \end{gathered}[/tex]Thus the maximum height reached by the ball is 70 ft
A runner moving at a speed of 5.6 m/s rounds a circular track with a radius of 65 m. What is the runner’s centripetal acceleration.
Answers
The centripetal acceleration of a particle moving at constant speed v over a circular trajectory with radius r is:
[tex]a_c=\frac{v^2}{r}[/tex]Replace v=5.6 m/s and r=65 m to find the centripetal acceleration:
[tex]\begin{gathered} a_c=\frac{(5.6\frac{m}{s})^2}{(65m)} \\ =\frac{31.36\frac{m^2}{s^2}}{65m} \\ =0.48246\ldots\frac{m}{s^2} \\ \approx0.48\frac{m}{s^2} \end{gathered}[/tex]Therefore, the runner's centripetal acceleration is 0.48 m/s^2
The following table shows the time of day when tides occurred at a coastline on a certain day of a year.Coastline Tide TableDateTimeType of TideSunday, 4th June2:31 a.m.Low tide8:35 a.m.High tidePart 1: Predict the approximate time when the next high tide and the next low tide would occur at the same coastline.Part 2: Use complete sentences to explain your answer.
Answers
Given:
Coastline Tide table,
Sunday-4th June-2:31 a.m. - low tide
8:35 a.m. - high tide.
To find:
The next high tide and low tide.
Explanation:
Part 1:
Next low tide Sunday-4th June-6:47 p.m.
Next high tide: Saturday-5th June- 1:00 a.m.
Part 2:
In one lunar day, i.e., the time it takes for the moon to complete one rotation, there will be two bulges. That is in one lunar day, there will be two high tides and two low tides. One lunar day is 24 hours and 50 minutes.
Thus the time difference between a low tide to a high tide or from high tide to low tide is 6 hours and 12.5 minutes. That is for every 6 hours and 12.5 minutes a shore experiences either a low tide or a high tide.
A force of only 150 N can lift a 600 N sack of flour to a height of 0.50 m when using a lever as shown in the diagram below.
a. Find the work done on the sack of flour (in J).
b. Find the distance you must push with the 150 N force on the left side (in m).
c. Briefly explain the benefit of using a lever to lift a heavy object.
Answers
a)The work done on the sack of flour is 75 J
b)The distance must push with the 150 N force on the left side 0.5 m
c)Levers work by lowering the amount of force required to move or lift a load.
What is work done?The amount of force required to move an object a specific distance is referred to as the work done. Essentially, it is a measurement of the amount of energy that is transmitted to or from an item so that it can be moved.
Briefing:a)The effort required to lift the weight to the provided height will equal,
W = F * h
Where "h" is the height and "F" is the force. In accordance with the assertion,
F=15 N
h=0.50m
After entering the values,
W=150N×0.50 m
∴W=75 J
b)Now, if it is moved in a horizontal direction with the same amount of power and energy, we have that.
d=W/F
d=75J/150 N
d = 0.5 m
The system's ability to conserve energy supports this.
c)Levers work by lowering the amount of force required to move or lift a load. Levers achieve this by extending the force's range of motion.
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I need help with this question, this question contains Part, A, B, C, and D response.
Answers
Part A.
For trial 1 and trial 2 the velocity of the car is constant. Since the velocity is constant there is no acceleration. So, we can conclude the acceleration is equal to zero.
Part B.
Depending on the material of the floor, it affects the movement of the car. This is associated with the coefficient of friction. The higher the coefficient of friction, the greater the opposition to movement. According to this the coefficient of friction for the carpet is higher than the coefficient of friction for tile and wood.
Part C.
For the interval [8,10] in the trial 3, the velocity becomes zero, since the distance isn't changing at all. This happened because of the high coefficient of friction of the carpet, which caused the car to stop.
Part D
In order to the students have more confidence in their result, they can change the following:
1. Repeat the experiment several times on the same surface
2. Have a longer length on each surface
A low “G” note has a frequency of 98 Hz. Which frequency of note, if played with this G, would produce a dissonant sound?1) 208 Hz2) 122.5 Hz3) 131 Hz4) 147
Answers
ANSWER:
1) 208 Hz
STEP-BY-STEP EXPLANATION:
For the sound to be dissonant, the frequency must be more than double, therefore
[tex]98\cdot2=196[/tex]The only frequency greater than 196 Hz is option 1