The acceleration of a car, moving along a straight line, that increases its velocity from 0 to 15 is A=1.5m/s.
Acceleration = Velocity Change / Time Interval
So,
A=15/10
A=1.5m/s .
A slowing car slows down. It also accelerates because the speed changes. Imagine a car driving down the street. As speed increases, the car has positive acceleration. When a car turns a corner at a constant speed it changes direction and thus accelerates. The faster the rotation the greater the acceleration.
Therefore, any change in velocity in magnitude and direction causes acceleration. The main causes of poor acceleration are air and fuel issues and sensor issues. However mechanical problems can also cause poor performance. The car is moving forward and accelerating in the positive direction so the acceleration is in the same direction as the car is moving. An object can also have positive acceleration if it slows down while moving in a negative direction.
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A rocket toy moves straight upward, starting from rest with an acceleration of 4 m/s2. It runs out of fuel at the end of 10 s and continues to coast upward, reaching a maximum height before falling back to earth
a. Find the rocket’s velocity and position at the end 10 s
B. Find the velocity before the rocket crashes on the ground
C. Find the time flight of rocke
Density of iron is 7860 Kg/m^3 what is the mass of iron sphere whose diameter is 0.5 m?
The mass of the given iron sphere of diameter 0.5 m is 510.9 kg
Density of the iron sphere = 7890 kg/m³
Diameter of the iron sphere = 0.5 m
Radius of the iron sphere = 0.25m
The formula to find the mass of the iron sphere is
ρ = m / v
where ρ is the density
m is the mass
v is the volume.
So , to find the mass , we need to know the volume of the sphere
Volume of the sphere = 4/3 πr³
Let us substitute the value of r in the above equation,
V = 4/3 x 3.14 x (0.25)³
= 4/3 x 3.14 x 0.0156
= 0.196/3
V = 0.065 m³
So , now let us find the mass of the given sphere
ρ = m / v
m = ρv
m = 7860 x 0.065
m = 510.9 kg
Therefore , the mass of the given iron sphere is 510.9 kg
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If the slope of a position versus time graph is negative, what does that imply about the motion of an object? A. The object has a negative acceleration. B. The object turned around. C. The object is moving in the negative direction. D. All of the above
ANSWER
A. The object has a negative velocity
EXPLANATION
We want to describe the slope of a position-time graph that is negative.
If the slope of a position-time graph is negative, it implies that the velocity is also negative. In other words, as time increases, the velocity also decreases.
As velocity decreases, it implies that the acceleration also decreases.
Therefore, we have that the acceleration of the object is negative.
The answer is option A.
A 320.-kg reindeer stands in the middle of the railroad tracks, frozen by the lights of an oncoming 10,000.-kg train that is traveling at 10.0 m/s. The engineer sees the reindeer but is unable to stop the train in time and the reindeer rides down the track sitting on the cowcatcher. What is the new combined velocity of the reindeer and train?
Answer:
9.69 m/s
Explanation
By the conservation of momentum, we can write the following equation:
[tex]\begin{gathered} p_i=p_f \\ m_1v_{i1}+m_2v_{i2}=(m_1+m_2)v_f \end{gathered}[/tex]Where m1 is the mass of the reindeer, m2 is the mass of the train, vi1 is the initial velocity of the reindeer, vi2 is the initial velocity of the train and vf is the final velocity of the reindeer and train.
So, replacing m1 = 320 kg, m2 = 10,000 kg, vi1 = 0 m/s and vi2 = 10 m/s, we get
[tex]\begin{gathered} 320(0)+10,000(10)=(320+10000)v_f_{} \\ 0+100,000=(10320)v_f \\ 100,000=10,320v_f \end{gathered}[/tex]Now, solving for the final velocity, we get
[tex]\begin{gathered} v_f=\frac{100,000}{10,320} \\ v_f=9.69\text{ m/s} \end{gathered}[/tex]Therefore, the new combined velocity of the reindeer and train is 9.69 m/s
It takes a continent approximately 250,000,000 years to move half-way around Earth, a distance of about 19,200 km or 1,920,000,000 cm. The speed of the continent in what per what can be found by dividing 1,920,000,000 cm by 250,000,000 years.
The speed of the continent is found to be 7.68 centimeters per hour when 1920000000cm is divided by 250000000years.
The distance travelled by the continent is 1920000000cm.
The time taken by the continent to complete its path is 250000000.
The speed S of the body,
S = D/T
Where,
D is the distance travelled by the body,
T is the time taken by the body to cover that distance.
The speed S of the continent is,
S = D/T
S = 1920000000cm/250000000year
S = 7.68 cm/year
Hence, the speed of the continent is 7.68 centimeters per year.
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Starting from rest, a car accelerates at 2.0 m/s2 up a hill that is inclined 5.5° above the horizontal. How far (a) horizontally and (b) vertically has the car traveled in 12 s?
Given data:
* The acceleration of the car is,
[tex]a=2ms^{-2}^{}[/tex]* The angle of the inclined plane is 5.5 degree.
* The time taken by the car is 12 s.
* The initial velocity of the car is 0 m/s.
Solution:
By the kinematics equation, the final velocity of the car on the inclined plane is,
[tex]v-u=at[/tex]where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the tiem taken,
Subsituting the known values,
[tex]\begin{gathered} v-0=2\times12 \\ v=24\text{ m/s} \end{gathered}[/tex]By the kinematics equation, the distance tarveled by the car on the inclined plane is,
[tex]v^2-u^2=2aS[/tex]where S is the distance tarveled on the inclined plane,
Substituting the known values,
[tex]\begin{gathered} 24^2-0=2\times2\times S \\ 576=4\times S \\ S=\frac{576}{4} \\ S=144\text{ m} \end{gathered}[/tex]The diagrammatic representation of the car on the inclined plane is,
The distance traveled by the car in the horizontal direction or along x-axis is,
[tex]\begin{gathered} S_x=S\cos (5.5^{\circ}) \\ S_x=144\times\cos (5.5^{\circ}) \\ S_x=143.34\text{ m} \end{gathered}[/tex]Thus, the distance tarveled by the car in the horizontal direction is 143.34 meter.
The distance tarveled by the car in the vertical direction or y-axis is,
[tex]\begin{gathered} S_y=S\sin (5.5^{\circ}) \\ S_y=144\times\sin (5.5^{\circ}) \\ S_y=13.8\text{ m} \end{gathered}[/tex]Thus, the distance tarveled by the car in the vertical direction is 13.8 meter.
The final stage of reduction of pressure in a regulator is to how many PSIG
The final stage of reduction of pressure in a regulator is to 500 PSIG.
What is a gas pressure regulator?A gas pressure regulator can be defined as a mechanical device that is designed and developed to ensure that a controlled amount of gas is supplied from a gas cylinder (source).
This ultimately implies that, a gas pressure regulator is required on a gas cylinder in order to reduce very high (unusable) pressure within the gas cylinder to a safer and usable delivery pressure.
In Science, a two-stage gas pressure regulator is designed and developed to lower very high (unusable) pressure in two (2) stages.
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Which of the following numbers has the greatest number of significant figures?a.) 144.5b.) 0.009514 c.) 10.507d.) 6.948 ✕ 10
So, the number with the greatest number of significant figures is:
0.009514. [Option B]
What voltage was needed for 2 amps to flow through a 600 lamp?
Answer:
I don't kno..................................................w
The density of mercury is 13600 kgm-3. So, the relative density of mercury is A. 13.6 B. 1.36 C. 136
The density of water is,
[tex]d_w=1000kgm^{-3}[/tex]The density of mercury is,
[tex]d_m=13600kgm^{-3}[/tex]The relative density of mercury is the density of mercury concerning the density of water.
Thus, the relative density of mercury is,
[tex]\begin{gathered} \frac{d_m}{d_w}=\frac{13600}{1000}_{} \\ \frac{d_m}{d_w}=13.6 \end{gathered}[/tex]Hence, option A is the correct answer.
yo can anyone help me solve this rq i need it for tomorrow
a. 600 meters; b. 6 minutes; c. 7.5 minutes; d. 4 minutes; e. 5 minutes; f. 125m/min; g. 50m/min; h. 60m/min; i. 150m/min; j. 100m/min; k. 80m/min and l. 1m/s, 2.5m/s, 1.66m/s and 1.33m/s.
As per the graph,
The motion of Rachel and ben starts from t=0 and ends at t = 13minutes after covering distance of 600 meters.
a. As there motion is ending at 600 meters. They travel 600 meters to go to school.
b. Ben travels for 4 minutes first, after that he stops and then again starts from 8 minutes to 10 minutes. So, it takes 6 minutes for him to go to school.
c. Rachel travels for 5 minutes first, after that she stops and then again starts from 10 minutes to 12.5 minutes. So, it takes 7.5 minutes for her to go to school.
d. Ben stops from 4 to 8 minutes on his way to school. After that he stops only at school so it would not be counted. So he stops for 4 minutes.
e. Rachel stop from 5 minutes to 10 minutes on her way. So, she stops for 5 minutes in her way to school.
f. Ben's speed in first half,
Speed = distance in first half/time taken in first half.
Ben's Speed in first half = 500/4
Ben's Speed in first half = 125 m/min.
g. Ben's speed in last part,
Speed = distance in last part/time taken in last part.
Ben's Speed in last part = 100/2
Ben's Speed in last part = 50m/min.
h. Rachel's speed in first half,
Speed = distance in first half/time taken in first half.
Rachel's Speed in first half = 300/5
Rachel's Speed in first half = 60 m/min.
i. Rachel's speed in last part,
Speed = distance in last part/time taken in last part
Rachel's Speed in last part = 300/2
Rachel's Speed in last part = 150m/min.
j. Ben's average speed = total distance/total time
Total distance = 600
Total time = 6 minutes
Average speed of Ben = 600/6 = 100 m/min.
k. Rachel's average speed = total distance/total time
Total distance = 600
Total time = 7.5 minutes
Average speed of Rachel = 600/7.5 = 80m/min.
l. Converting m/min to m/s by dividing by 60.
Converting speed in h = 60/60 = 1m/s.
Converting speed in i = 150/60 = 2.5m/s.
Converting speed in j = 100/60 = 1.66m/s.
Converting speed in k = 80/60 = 1.33m/s.
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Alex tickles his brother by stroking adjacent ________ spots on his skin.
Answer:
adjacent flat spots on his skin
Answer:
You didn't provide answer choices but I'd assume it's pressure.
When you drop a 0.38 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s2 to- ward the earth’s surface. According to New-
Answer:
At the surface it is defined by 9.80665 m/s2.
1.The water in the plumbing in a house is at a gauge pressure of 74,000 Pa. What force does this cause on the top of the tank inside a water heater if the area of the top is 0.38 m2?
Given:
The pressure inside the tank, P=74000 Pa
The area of the top of the tank, A=0.38 m²
To find:
The force exerted on the top of the tank.
Explanation:
The pressure can be described as the force exerted per unit area.
Thus the pressure inside the tank is given by,
[tex]P=\frac{F}{A}[/tex]Where F is the force exerted on the top of the tank.
On substituting the known values,
[tex]\begin{gathered} 74000=\frac{F}{0.38} \\ \implies F=74000\times0.38 \\ =28120\text{ N} \end{gathered}[/tex]Final answer:
Thus the force exerted on the top of the tank is 28120 N
What if you had a leak in the refrigerant line. Explain in terms of energy and pressure and temperature.
The most frequent HVAC crises are gas leaks, clogged drain lines, electrical problems, refrigerant leaks, and problems with ventilation.
What happens if you have a refrigerant leak?Your system could not cool as effectively as it should due to refrigerant leakage. Your system can have trouble producing enough air cooling, which would keep your house from reaching the target temperature. This can lead to an increase in your monthly energy costs and makes your system work harder.Your system could not cool as effectively as it should due to refrigerant leakage. Your system can have trouble producing enough air cooling, which would keep your house from reaching the target temperature. This can lead to an increase in your monthly energy costs and makes your system work harder.The most frequent HVAC crises are gas leaks, clogged drain lines, electrical problems, refrigerant leaks, and problems with ventilation.To learn more about : Refrigerant
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Question 11
5 pts
Gas Laws: A balloon is carefully filled with one breath of air (about 500ml) underwater at
a depth of 100ft, then tied shut. At this depth, the balloon is under 4 atmospheres of
pressure. Assuming negligible compression by the rubber of the balloon, what is the
volume of the air (in m³) once the balloon rises to the surface of the water and
experiences 1 atm of pressure? Modify the Ideal Gas Law to solve this conservation
problem.
1 Liter 0.001 Cubic Meter
1 atm = 1.01x 105 Pascals
The volume of the gas once it reaches the surface of water is 2 liters.
The volume of the air in balloon at depth of 100ft (30m) is 500ml.
The pressure at this point is 4 atm.
Assuming that the balloon have no compression by rubber of balloon the volume of air at the water surface is V.
The pressure at the surface of water is 1 atms.
As we know, from the ideal gas equation,
PV = nRT
Where,
P is the Pressure of gas,
V is the volume of the gas,
n is the number of moles,
R is the gas constant whose value is 0.082057 L atm mol-1 K-1,
T is the temperature.
Assuming that the temperature is constant,
We know,
PV = nRT
All quantities on the right side are constants,
So, we can write,
P₁V₁ = P₂V₂
Putting all the the values,
4(0.5) = 1V₂
V₂ = 2 Liters.
The volume of the air at the surface is 2 liters.
1 liter = 0.001 m
Hence,
2 liters = 0.002 m³
So the volume of air at the surface is 0.002m³.
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Hello! So I need some help with this question. I don’t quite understand it and I did have another tutor figure it out and the answer was wrong. It’s not his answer of 58.04. Can you help me?
ANSWER:
58 m/s
STEP-BY-STEP EXPLANATION:
Given:
Initial velocity (u) = 13 m/s
The correct answer is 58 m/sDistance (d) = 400 m
Acceleration (a) = 4 m/s²
We apply the following formula to determine the final velocity:
[tex]\begin{gathered} v^2=u^2+2as \\ \\ \text{ We replacing:} \\ \\ v^2=13^2+2(4)(400) \\ \\ v^2=169+3200 \\ \\ v=\sqrt{3369} \\ \\ v=58.04\cong58\text{ m/s} \end{gathered}[/tex]The correct answer is 58 m/s
A string 0.50 m long is stretched under a tension of 2.0 x 102 N and its fundamental frequency is 400 Hz. If the length if the string is shortened to 0.35 m and the tension is increased to 4.0 x 102 N, what is the new fundamental frequency?
Given,
The initial length of the string, L₁=0.50 m
The tension on the string, T₁=2.0×10² N
The initial fundamental frequency of the string, f₁=400 Hz
The length of the string after it was shortened, L₂=0.35 m
The increased tension on the string, T₂=4.0×10² N
The fundamental frequency of the string before it was shortened is given by,
[tex]f_1=\frac{\sqrt[]{\frac{T_1}{\mu}}}{2L_1}[/tex]Where μ is the mass per unit length of the string.
On rearranging the above equation,
[tex]\begin{gathered} 4L^2_1f^2_1=\frac{T_1}{\mu} \\ \Rightarrow\mu=\frac{T_1}{4L^2_1f^2_1} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} \mu=\frac{2.0\times10^2}{4\times0.50^2\times400^2} \\ =1.25\times10^{-3}\text{ kg/m} \end{gathered}[/tex]The fundamental frequency after the string was shortened is given by,
[tex]f_2=\frac{\sqrt[]{\frac{T_2}{\mu}}}{2L_2}[/tex]On substituting the known values,
[tex]\begin{gathered} f_2=\frac{\sqrt[]{\frac{4\times10^2}{1.25\times10^{-3}}}}{2\times0.35} \\ =808.1\text{ Hz} \end{gathered}[/tex]Thus the fundamental frequency after the string is shortened and the tension is increased is 808.1 Hz
Which planets in our solar system revolve the fastest? and Why?
A. The gas giant planets
B.Planets closest to the Sun
C. Planets farthest from the Sun
D. All planets revolve at the same speed
On a plot of angular position versus time t for the rotation of a disk, what corresponds to the disk’s angular velocity at any given instant t1?
a. inverse of the slope at the point on the plot corresponding to t1
b. intercept on the vertical axis
c. slope at the point on the plot corresponding to t1
d. intercept on the horizontal axis
Option d i.e; Slope at the point on the plot corresponding to t1 corresponds to disk's angular velocity at any given instant t1.
The angular velocity (w), a vector quantity in uniform circular motion, is determined by dividing the angular displacement (Δ), also a vector quantity, by the change in time (Δt). Speed is equal to |w|R and is calculated by dividing the arc length travelled (S) by the change in time (Δt).
Examples of angular velocity include a roulette ball on a wheel, a race car travelling in a circle, and a Ferris wheel. In addition, the object's angular displacement with respect to time is represented by the angular velocity of the object.
Because it is practical, angular velocity is perpendicular. Because all they are is a magnitude and a direction, a vector cannot be curved. In order to represent the rotational direction, we use the direction perpendicular to the plane.
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Calculate the amount of charge that flows past a cross-sectional area of a conductor if a current of 0.2 A flows in it for 1 minute?
Answer:
12C
Explanation:
q = It
where q is the charge in coloumbs (c)
I is the current in amperes (A)
t is the time in seconds (s)
q = 0.2× 60 (1 minute is 60seconds)
q= 12C
A stunt performer falls off a wall that is 3.1 m high and then lands on a mat.
What is his impact velocity?
Answer:7.8 M/s
Explanation:
1. A sound engineer has set up his mixing equipment a distance from an outdoor band shell.She claps her hands and hears the echo 0.250 s later. If the speed of sound was 340.0m/s, how far from the wall was she?
Given
Hears the echo at , t=0.250 s
Speed pf sound is v=340 m/s
To find
The distance from the wall
Explanation
let her distance from the wall be d.
When she hears the echo, the sound has travelled the same distance twice
So
[tex]\begin{gathered} v=\frac{2d}{t} \\ \Rightarrow340=\frac{2d}{0.250} \\ \Rightarrow d=d=42.5\text{ m} \end{gathered}[/tex]Conclusion
The distance of the wall from her is 42.5 m
A circular loop of wire of area 10 cm^2 carries a current of 25 A. At a particular instant, the loop lies in the xy-plane and is subjected to a magnetic field B =(2.0iˆ+6.0jˆ+8.0kˆ)×10−3T. As viewed from above the xy-plane, the current is circulating clockwise. (a) What is the magnetic dipole moment of the current loop? (b) At this instant, what is the magnetic torque on the loop?
The magnetic dipole moment of the current loop is 0.025 Am².
The magnetic torque on the loop is 2.5 x 10⁻⁴ Nm.
What is magnetic dipole moment?The magnetic dipole moment of an object, is the measure of the object's tendency to align with a magnetic field.
Mathematically, magnetic dipole moment is given as;
μ = NIA
where;
N is number of turns of the loopA is the area of the loopI is the current flowing in the loopμ = (1) x (25 A) x (0.001 m²)
μ = 0.025 Am²
The magnetic torque on the loop is calculated as follows;
τ = μB
where;
B is magnetic field strengthB = √(0.002² + 0.006² + 0.008²)
B = 0.01 T
τ = μB
τ = 0.025 Am² x 0.01 T
τ = 2.5 x 10⁻⁴ Nm
Thus, the magnetic dipole moment of the current loop is determined from the current and area of the loop while the magnetic torque on the loop is determined from the magnetic dipole moment.
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Which of the following statements correctly explains the relationship between FM and AM radio waves?
A.
FM radio waves carry less energy than AM radio waves. This is supported by the fact that FM radio waves have a shorter wavelength compared to AM radio waves.
B.
FM radio waves carry less energy than AM radio waves. This is supported by the fact that FM radio waves have a longer wavelength compared to AM radio waves.
C.
FM radio waves carry more energy than AM radio waves. This is supported by the fact that FM radio waves have a shorter wavelength compared to AM radio waves.
D.
FM radio waves carry more energy than AM radio waves. This is supported by the fact that FM radio waves have a longer wavelength
The relationship between the AM and the FM waves is that the wavelength of the FM waves is longer than that of the AM waves.
What are radio waves?The radio waves are one of the waves that are part of the electromagnetic spectrum and they are used in the process of communication . The reason why the radio waves could be used for the process of communication is that the radio waves do have a long wavelength this they can travel for very long miles without a significant decrease in intensity of the waves.
Thus, FM radio waves carry less energy than AM radio waves. This is supported by the fact that FM radio waves have a longer wavelength compared to AM radio waves.
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H3C- CH2-CH2-CH2-CH3name formula HC/// C-CH2-CH3name formula
1.
The number of the carbon atom is five. All the carbon atom is single-bonded.
The formula name of the above is Pentane.
The number of the carbon atom is four. All the carbon atom is single-bonded except one triple bond at the first place.
The formula name of the above is 1-butyne.
Cory is at the park playing fetch with his dog. He throws the ball 23 m south, and his dog retrieves the ball and returns it to Cory. Cory then throws the ball 18 meters, and the dog again collects the ball and returns it to Cory. When the dog returns for the second time, what distance has the dog traveled?0 m82 m36 m23 m
In order to determine the distance traveled by the dog in the two times, consider:
In the second time the dog goes and comes back for a distance of 18 m, that is, the total distance is 18m + 18m = 36 m
In the first time the dog goes and comes back for a distance of 23 m, that is, the total distance is 23m + 23m = 46 m
Hence, for the two times the distance traveled by the dog is:
46m + 36m = 82m
A block of mass M is hung by ropes as shown. The system is in
equilibrium. The point O represents the knot, the junction of the
three ropes. The angle theta is given to be 30o
. Which of the
following statements is true concerning the magnitudes of the three
forces in equilibrium?
Their magnitude of force must be the same for the tensions to equal out.
An object at equilibrium has zero acceleration so both the magnitude and direction of the object's velocity must be constant. When the body is in equilibrium the forces are balanced. Balanced is a keyword used to describe a balanced state. Therefore the net force is zero and the acceleration is 0 m/s/s. The acceleration of a body in equilibrium must be 0 m/s/s
For a body to be in equilibrium it must not be accelerating. This means that both the net force and net torque on the object must be zero. Clearly when in equilibrium, the net force on the object is zero. According to Newton's second law of motion if the net force is zero the acceleration is also zero. When the acceleration is zero the velocity and thus the velocity is constant by definition. Option d does not apply to equilibrium in this case.
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Lesson 3
Write a descriptive essay about the basic steps of the scientific method.
Include the following essential content:
Testable hypothesis
Data collection and analysis
Peer review
Also, include where theories fit into the process of the scientific method.
A sample descriptive essay about the basic steps of the scientific method is given below:
When a person is testing a hypothesis, it is important that he collects data and then makes an analysis of it, and puts it through a controlled experiment at the lab.
Next, he would have to peer-review the results of the experiments and when this is confirmed, it becomes a theory,
What is the Scientific Method?This refers to the process of systematic observation that is used to come to a conclusion in a scientific investigation.
Hence, we can see that the descriptive essay about the scientific method is given above,.
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A spherical specimen of the mineral chalcopyrite measures 3.9 cm in diameter and has a mass of
129.67 g. What is its specific gravity?
A spherical specimen of the mineral chalcopyrite measures 3.9 cm in diameter and has a mass of 129.67 g, its specific gravity is 4.17g/cm³
What is specific gravity?The ratio of a substance's density to that of a reference substance is its specific gravity, also known as relative density.
Water, which has a density of 1.0 kg/litre at 4 °C (39.2 °F), serves as the standard of comparison for solids and liquids (62.4 pounds per cubic foot).
Dry air, which is frequently used as a comparison point for gases, has a density of 1.29 grams per litre (1.29 ounces per cubic foot) under what are known as "standard conditions" (a temperature of 0 °C and a pressure of 1 standard atmosphere).
As an illustration, the specific gravity of liquid mercury is 13.6 due to its density of 13.6 kg per litre. The density of the gas carbon dioxide, which is 1.976 grams per liter under ideal conditions, translates into a specific gravity of 1.53 (= 1.976/1.29).
The formula for specific gravity is given by
specific gravity = density of the object / density of the water
Here given that a spherical specimen of the mineral chalcopyrite measures 3.9 cm in diameter and has a mass of 129.67 g
Then its volume is
Volume of sphere = [tex]{\frac {4}{3}}\pi r^{3}[/tex]
Substituting the values we get
V = [tex]{\frac {4}{3}}\pi (\frac{3.9}{2} )^{3}[/tex]
= 31.06 cm³
Density = Mass/Volume
Density of mineral chalcopyrite = 129.67g / 31.06 cm³
= 4.17 g/cm³
Now, we can finally calculate specific gravity
specific gravity = 4.17 g/cm³ / 1g/cm³ (Density of water is 1m/1V)
= 4.17g/cm³
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