Answers
From the graph of the motion of the car, the deceleration of the car is -1.5 m/s².
What is motion?Motions is defined as the change in the position of a body.
There are several forms of motion that a body can undergo, they include:
translational motion - this is motion along a straight lineCircular or rotational motion - this is motion of a body in a circular pathOscillatory motion - this is a periodically repeating motion of a bodyRandom motion - this is the haphazard motion of a body.Considering the motion graph of the car as shown, the car undergoes a translational motion in a straight path. The car changes its speed over time.
Between O and A, the speed of the car increases steadily, and this part is known as acceleration.
Between A and, the car moves with constant speed.
Between B and C, the speed of the car steadily decreases which is known as deceleration.
deceleration= change in speed/ change in timeThe deceleration of the car = (0 - 30) / (50 - 30)
The deceleration of the car =-1.5 m/s²
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Complete question:
A graph of a car's motion is shown below. What is the deceleration of the car from B to C?
Related Questions
Do gas giant planets have geology? Why or why not?
Answers
Geologists research the Earth's components, processes, outcomes, physical characteristics, and past.
Why do people decide to study geology?Geology examines some of the most crucial challenges facing modern society, such as the use of renewable energy sources, environmental sustainability, climate change.
How do gas giant planets fare?At higher densities, the atmosphere turns into a dense mixture of steam, hydrogen, and helium when the planet's ocean boils. A planet's atmosphere begins to increase quickly when its mass reaches a few times that of Earth. This atmosphere will eventually create a gas giant planet like Jupiter because it will grow faster than the planet's solid core.
How are gas giants made up?A massive planet that is primarily made of helium or hydrogen is known as a gas giant. In contrast to our solar system's Jupiter and Saturn, these planets feature swirling gases atop a solid core rather than hard surfaces.
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the product of mass and velocity is also known as linear momentum.1. true2. false
Answers
Given the statement:
The product of mass and velocity is also known as linear momentum.
Let's determine if the statement is true or false.
Linear momentum can be defined as the product of a body's mass times the velocity.
Therefore, we can say the statement is true.
ANSWER:
1. True
As Rain passes a classroom, they hear the instructor say, "An early school of thought in psychological science that was influenced by Charles Darwin's work was that of . . . ." Which school of thought was the instructor MOST likely describing?
Please choose the correct answer from the following choices, and then select the submit answer button.
Answer choices
A: functionalism
B: structuralist
C: humanist
D: psychoanalytic
Answers
The school of thought in which the instructor is most likely describing is functionalism and is denoted as option A.
What is Functionalism?This is referred to as a theory which is based on the doctrine of all aspects of a society and touches the mental states.
This type of theory also believes that institutions perform positive functions and so on. Darwin was a scientist who was involved in putting in positive efforts in the science field due to the skills which were developed and was solved many problems which is therefore the reason why Functionalism was chosen as the correct choice.
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How human memory is complicated and can even sometimes be enhanced?
Answers
Human memory is located not in one particular place in the brain but is instead a brain-wide process in which several different areas of the brain
How human memory is complicated sometimes to be enhanced?Human memory is complex, and genetics are still trying to bare the mechanisms that lead to memories being formed. It seems that our human memory is discovered not in one particular place in the brain but is instead a brain-wide process in which several different areas of the brain or memory is a skill, and just like other skills, it can be improved with practice and healthy overall habits. You can start small. For example, a human memory picks a new challenging activity to learn, incorporates a few minutes of exercise into your day, maintains a sleep schedule, and eats a few more green vegetables, fish, and nuts. Procedural human memories (motor skills) are the last ability to be destroyed.
so we can conclude that human memory is located not in one particular place in the brain but is instead a brain-wide process in which several different areas of the brain are.
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Which change to this process would shift the equilibrium to produce the
maximum possible amount of SO3?
2SO2 +022SO3 + energy
A. Removing the O₂ as it forms
B. Removing the SO3 as it forms
C. Increasing the temperature
D. Decreasing the pressure
B
Answers
The change to this process would shift the equilibrium to produce the maximum possible amount of SO3 is Decreasing the pressure. Option D.
Increasing the temperature shifts the equilibrium to the right with higher vapor concentrations, but keeping the system at this elevated temperature restores equilibrium. You can predict how given stress or change in conditions will affect equilibrium.
Sulfur trioxide formation is very slow at low temperatures. Increasing the temperature increases the speed at which equilibrium is reached. Decreasing the concentration of the reactants shifts the equilibrium to the left and produces less product. As the product concentration decreases the equilibrium shifts to the right producing more products.
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Answer: B
Explanation:
Using the vector arrows in the virtual lab, depict the following path: You drive 14 blocks East, 7 blocks North, and 2 blocks West. Use the vertex of the graph (coordinates 0,0) as you lay out the vectors. What is the displacement between your starting and end points? What is the angle of displacement?
Answers
The displacement between starting and endpoints is 13.89 units. Northeast.
A vector is an item that has both importance and direction. Geometrically, we can photo a vector as a directed line segment, whose duration is the importance of the vector, and with an arrow indicating the direction. The course of the vector is from its tail to its head. A displacement is a vector whose length is the shortest distance from the initial to the final function of a factor.
let 1 block = 1 unit
so, 14 block east means = 14î
7 block north = 7 j
2 block west = -2î
Net displacement in x- direction = 14î - 2î
= 12î
displacement towards y-direction = 7 j
The resultant R = [tex]\sqrt{12^{2} +7^{2} }[/tex]
= [tex]\sqrt{193}[/tex]
= 13.89
The angle of displacement = 13.89/ 14î and 13.89/7 j
or
The angle of displacement = 12 tans 7/12
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why is this situation seem impossible?
Answers
Answer:
The situation seems impossible because in order for the liquid to be tilted the glasses would also have to be tilted.
Explanation:
All of the glasses are in line and perfectly straight. There is no way that the liquid in the glasses would be tilted like how they are in the photo.
Any physics help out there?please
A baseball weighing 1.42 N has been dropped from a balloon and is falling with an acceleration of 4.9 m/s². What air resistance
force acts on it?
Ο 0Ν
O 0.71 N
O 1.42 N
O 9.8 N
Answers
Answer:
f = 0.71 N
Explanation:
Apply the 2º Newton's Law to baseball:
∑F = m.a
So:
W - f = m * a
Where:
W: weight;
f: resistance force;
m: mass;
a: resulting acceleration.
But:
W = m*g
1.42 = m * 9.8
m = 1.42/9.8
Thus:
W - f = m * a
m * g - f = m * a
1.42 - f = (1.42/9.8) * 4.9
f = 1.42 - (1.42/9.8) * 4.9
f = 1.42 - 0.71
f = 0.71 N
In the first movement of Beethoven's piano sonata No. 12, the theme is heard first followed by 2 different variations of it.
A. True
B. False, the theme is heard first followed by three different variations of it.
C. False, the theme is heard first followed by four different variations of it.
Answers
All choices offered are false.
The theme is heard first, followed by FIVE different variations of it.
===> 5
Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s. B) Find the minimum static friction coefficient between the tires and the road.
Answers
Answer:
At least [tex]0.128[/tex] (rounded up, to three significant figures as in speed,) assuming that the road is level and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
When an object is in a circular motion, the magnitude of the net force can be found in terms of the mass and speed of that object, as well as the radius of the circular path:
[tex]\begin{aligned} (\text{net force}) &= \frac{(\text{mass}) \, (\text{speed})^{2}}{(\text{radius})} \end{aligned}[/tex].
Let [tex]m[/tex] denote the mass of this vehicle. Let [tex]v[/tex] denote the speed of this vehicle. Let [tex]r[/tex] denote the radius of this circular path. While in this circular motion, the magnitude of the net force will be on this vehicle will be:
[tex]\begin{aligned} (\text{net force}) &= \frac{(\text{mass}) \, (\text{speed})^{2}}{(\text{radius})} = \frac{m\, v^{2}}{r} \end{aligned}[/tex].
Forces on this vehicle include:
Gravitational attraction from the earth (weight) of magnitude [tex]m\, g[/tex], pointing downwards,Normal force from the road, pointing upwards, andStatic friction from the road, pointing towards the center of the curve.If this road is level, the normal force from the road will balance the weight of this vehicle. The magnitude of the normal force will be equal to that of the weight of this vehicle, [tex]m\, g[/tex].
Let [tex]\mu_{\text{s}}[/tex] denote the static friction coefficient between the tire and the road. The static friction between the vehicle and the road cannot exceed [tex](\text{static friction coefficient}) \times (\text{normal force})[/tex].
Since [tex](\text{normal force}) = m\, g[/tex], the maximum possible value of static friction will be:
[tex]\begin{aligned} & (\text{static friction coefficient}) \times (\text{normal force}) \\ =\; & (\mu_{\text{s}})\, (m\, g) \\ =& \mu_{\text{s}}\, m\, g\end{aligned}[/tex].
Under the assumptions, the weight and normal force on this vehicle will be balanced. As a result, the net force on this vehicle will be equal to static friction and should also be no greater than [tex](\text{static friction coefficient}) \times (\text{normal force}) = \mu_{\text{s}}\, m\, g[/tex].
In other words:
[tex]\begin{aligned}(\text{net force}) &= (\text{static friction}) \\ &\le (\text{maximum static friction}) = \mu_{\text{s}}\, m\, g \end{aligned}[/tex].
Additionally, from circular motion:
[tex]\begin{aligned} (\text{net force}) &= \frac{m\, v^{2}}{r} \end{aligned}[/tex].
Therefore:
[tex]\begin{aligned}\frac{m\, v^{2}}{r} \le \mu_{\text{s}}\, m\, g \end{aligned}[/tex].
Rearrange this inequality to separate the coefficient of static friction, [tex]\mu_{\text{s}}[/tex]:
[tex]\begin{aligned} \mu_{\text{s}} \ge \frac{v^{2}}{r\, g}\end{aligned}[/tex].
(Note that [tex]m[/tex] and [tex]g[/tex] are both greater than [tex]0[/tex].)
Substitute in [tex]v = 25.0\; {\rm m\cdot s^{-1}}[/tex], [tex]r = 500\; {\rm m}[/tex], and [tex]g \approx 9.81\; {\rm m\cdot s^{-2}}[/tex]:
[tex]\begin{aligned} \mu_{\text{s}} &\ge \frac{v^{2}}{r\, g} \\ & \approx \frac{(25.0\; {\rm m\cdot s^{-1}})^{2}}{(500\; {\rm m})\, (9.81\; {\rm m\cdot s^{-2}})} \\ &\approx \frac{25.0^{2}}{500\times 9.81} \\ &\approx 0.128\end{aligned}[/tex].
(Rounded up.)
As a sample of water is heated, the energy is used to overcome the attractions between the molecules so that they begin to move past one another. Then, the kinetic energy of its molecules increases. Which sections of the heating curve illustrate this process? A.A followed by CB.C followed by BC.B followed by CD.A followed by B
Answers
Given that as the sample of water is heated, the kinetic energy of its molecules increases.
Let's determine the sections of the heating curve which illustrates this process.
At point A, we can see the temperature of the water at 0 degrees, then as the temperature increases, the water molecules absorb more energy.
But as the phase changes, the temperature remains constant.
Since section A shows constant temerature, while section B also indicates constant temperature, we can sat section A followed by section B shows this process.
The sections which show as the energy overcome the attractions between molecules so that they begin to mover past one another can be said to be Section A followed by Section B
ANSWER:
D. A followed by B.
The January 1990 issue of Arizona Trend contains a supplement describing the 12 “best” golf courses in the state. The yardages (lengths) of these courses are as follows: 6981, 7099, 6930, 6992, 7518, 7100, 6935, 7518, 7013, 6800, 7041, and 6890. a/ Calculate the sample mean and sample standard deviation. b/ Find median.
Answers
The standard deviation of the sample means X, calculated earlier is the standard deviation of the population divided by the square root of the sample size.
Formula :
Mean : Mean = Sum of X values / N(Number of values)
We find mean=7068.08333
and standard deviation = 226.50003.
The standard deviation of the sample means known as the standard error of the mean is less than the population standard deviation and equal to the population standard deviation divided by the square root of the sample size. the sample mean is the average value found in the sample.
Samples are just a fraction of the total. For example, if you work for a polling company and want to know how much people spend on food each year, you don't need to survey over 300 million people. I also found that the sample mean is the arithmetic mean of all the values in the sample. The sample variance measures how the data are distributed and the sample standard deviation is the square root of the variance.
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Two identical 6.8kg balls are in contact with one another. The gravitational attraction between the balls is 6.2E-08 N A. What is the radius of one of these balls?
Answers
Given that the mass of each ball is m = 6.8 kg
The distance between them is d = 2r
Here, r is the radius of the ball.
The gravitational force of attraction is
[tex]F=\text{ 6.2}\times10^{-8}\text{ N}[/tex]We have to find the radius of the ball.
The gravitational force formula is
[tex]F=\frac{Gmm}{(2r)^2}[/tex]Here, the universal gravitational constant is
[tex]G=\text{ 6.67}\times10^{-11}Nm^2kg^{-2}[/tex]The radius will be
[tex]\begin{gathered} r=\sqrt[]{\frac{Gmm}{4F}} \\ =\sqrt[]{\frac{6.67\times10^{-11}\times6.8\times6.8}{4\times6.2\times10^{-8}}} \\ =\text{ 0.114 m} \end{gathered}[/tex]Thus, the radius of one of these balls is 0.114 m
In December, you measure the location of stars A, B, and C in the night sky.6 months later, you measure the location of starts A, B, and C again.By doing this, you calculate the parallax.Here is your data:Star A: .67Star B: 1.2Star C: 35According to your parallax data, put the stars in order from CLOSEST to FURTHESTaway.A. A, B, CB. B, A, CC. C, A, B
Answers
iven:
he aangles of the stars;
Star A: p_A=0.67 arcsec
Star B: p_B=1.2 arcsec
Star C: p_C=0.35 arcsec
To find:xoplanation:
Order of the stars according to their distance.
he distacnce of a star in parsec is calculated using the formula,
[tex]D=\frac{1}{p}[/tex]Where p is the angles made by the stars in arcseconds.
On substituting the known values:
Star A,
[tex]\begin{gathered} D_A=\frac{1}{p_A} \\ =\frac{1}{0.67} \\ =1.5\text{ parsec} \end{gathered}[/tex]Star B,
[tex]\begin{gathered} D_B=\frac{1}{p_B} \\ =\frac{1}{1.2} \\ =0.83\text{ parsec} \end{gathered}[/tex]Star C,
[tex]\begin{gathered} D_C=\frac{1}{p_C} \\ =\frac{1}{0.35} \\ =2.9\text{ parsec} \end{gathered}[/tex]inal answer:
Thus the closest star is star B and the farthest star is star C.
Thus the correct answer is option B.
A 1400 kg car drives at 27 m/s over a circular hill that has a radius of 460 m as shown in (Figure 1). At the point shown in (Figure 1), what is the normal force on the car?
Answers
The normal force acting on the car that weighs 1400 kg and drives at 27 m / s is 8390.82 N
The car is not horizontal. So the force mg is resolved into its horizontal and vertical component.
cos θ = Adjacent side / Hypotenuse
cos 30° = r / mg
r = m g cos 30°
Since the path is a circular, the net force acting on the car is equal to the centripetal force,
m g cos 30° - N = m v² / r
m = Mass
g = Acceleration due to gravity
N = Normal force
v = Velocity
r = Radius
N = m g cos 30° - m v² / r
N = ( 1400 * 9.8 * 0.87 ) - [ ( 1400 * 33² ) / 430 ]
N = 11936.4 - 3545.58
N = 8390.82 N
Therefore, the normal force acting on the car is 8390.82 N
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Why are the accelerations due to gravitational force on the moon and the Earth different? Do you think you could shield a gravitational field in a vacuum?
Answers
The acceleration due to gravitational force on the moon and the Earth are different because the masses of the moon and the sun are different. Owing to the fact that gravity is a universal force, it is not possible to shield the gravitational field in a vacuum.
What is acceleration due to gravity?We know that there is a force that acts on any object that is found on the earth's surface. Now this force tends to act towards the center of the earth. We must realize that force is a vector and a such there must be the direction of a force at any given time that we are trying to observe the force.
We can then say that this force that acts on an object that has its direction as the center of the earth is what we call the gravitational force. This force does not only act on the earth but in the moon as well as act on other planets.
Having said this, let us bear in mind that the value of the acceleration due to gravity depends on the mass of the body. We know that the earth is larger than the moon hence the acceleration due to gravity is larger on the earth than on the moon.
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A puck of mass m = 0.095 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless string of length L = 0.69 m. The puck makes one revolution every t = 0.45 s.
Part (b) The string breaks suddenly. How fast, in meters per second, does the puck move away?
Answers
The speed in meters per second if the string breaks suddenly is 9.6 m/s.
What is the angular speed?The angular speed is the speed of an object that is moving along a circular path. In this case, we have a puck of mass m = 0.095 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless string of length L = 0.69 m. The puck makes one revolution every t = 0.45 s.
Let us recall that in order to find the angular speed as is required by the question "how fast in meters per second, does the puck move away?" we have to get the angle turned as well as the time taken.
We are told here that the angle turned is one revolution or 2π radians and the time from the question is 0.45 s. Thus the speed is gotten from;
ω = θ/t
ω = 2π radians/0.45 s
ω = 13.96 rad/s
But v = rω
v = 13.96 rad/s * 0.69 m
v = 9.6 m/s
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module 1 question 13
At the end of a race a runner decelerates from a velocity of 9.00 m/s at a rate of 0.400 m/s2.
(a) How far does she travel in the next 14.0 s?
m
(b) What is her final velocity?
m/s
Answers
The distance travelled by the runner is 86.8 meters and the final velocity is 3.4 m/s.
Using kinematic equation we can find the distance travelled.
S = ut + 1/2 at^2
Where S = distance, u = initial velocity, t = time and a = acceleration.
The runner decelerates from a velocity of 9.00 m/s at a rate of 0.400 m/s^2. Initial velocity is 9.00 m/s and acceleration is -0.4 m/s^2 here. The distance travelled by the runner in the next 14 seconds is,
S = 9.00*14.0 + 1/2*(-0.4)*196
S = 86.8 m
The final velocity of the runner is using the formula v = u + at
v = 9.00+(-0.4)*14
v = 3.4 m/s
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9. What is mass of a piece of steel if after being heated with 2500J its temperature increased by 4oC? 9a. A 3 kg of some metal is heated with 801J. Its temperature increased by 0.3oC. What material is it?
Answers
We are given the following information
Energy = 2500J
Change in temperature = 40 °C
We are asked to find the mass of steel.
The mass of the steel can be found using the equation below
[tex]Q=m\cdot c\cdot\Delta T[/tex]Where c is the specific heat of the steel that is 420 J/(kg°C)
Re-arranging the equation for m and substituting the given values
[tex]undefined[/tex]Kieran is driving his Chevrolet Spark around a level turn with a radius of curvature of 70.9 m at a speed of 22.4 m/s. Calculate the acceleration of the car.
Question 18 options:
3.2 m/s/s
224.4 m/s/s
0.3 m/s/s
7.1 m/s/s
Answers
The acceleration of the Chevrolet Spark around a level turn with a radius of curvature of 70.9 m at a speed of 22.4 m/s is 7.1 m / s / s
a = v² / R
a = Centripetal acceleration
v = Linear velocity
R = Radius
v = 22.4 m / s
R = 70.9 m
a = 22.4² / 70.9
a = 7.1 m / s / s
The acceleration of a body moving around in a circle is called as centripetal acceleration. It is a vector quantity. It can be denoted as m / s²
Therefore, the acceleration of the car is 7.1 m / s / s
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Each circled letter in
the circuit diagram represents a meter that is used to
measure a quantity in the circuit. Which meters show the amount of charge
passing a point each second?
A. A and C
B. A and B
C. B and D
D. C and D
Answers
The meters in the circuit that shows the amount of charge passing a point each second are A and C ( A )
In the circuit, the meter A and C are connected in series to a resistor and the meters B and D are connected parallel to a resistor. Ammeters should be connected in series and voltmeters should be connected in parallel because the resistance in ammeter is zero and the resistance in a voltmeter is infinite.
Ammeter is used to measure the current flowing through a circuit. It can also be said as amount of charge passing through the circuit. Voltmeter is used to measure the voltage across two points in a circuit.
Therefore, the meters in the circuit that shows the amount of charge passing a point each second are A and C ( A )
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write any two laws of friction
Answers
Answer:
The friction of the moving object is proportional and perpendicular to the normal force.
The friction experienced by the object is dependent on the nature of the surface it is in contact with.
Explanation: two laws of friction
Answer: The friction of the moving object is proportional and perpendicular to the normal force.
The friction experienced by the object is dependent on the nature of the surface it is in contact with.
A box weighing 150 N is being pushed with constant velocity across a horizontal surface. If the force of friction between the box and the surface is 135 N, what is the coefficient of friction? Explain all of your work!!
brainliest answer for the best reply.
Answers
Answer:
uK=.9
Explanation:
friction=Fn(uK)
friction=135
Fn=150
135/150=.9
You need to move a 20 kg box along a horizontal floor at an acceleration of 1.4m/s? by pulling
on a rope attached to the box. How much force do you need if you
a) Apply the force horizontally?
b) Apply the force at 30° above the horizontal?
c)
Apply the force at 30° below the horizontal?
Answers
The magnitude of the force in each case is;
a) 28 N
b) 14 N
c) -14 N
What is force ?In physics, we know that force is a vector quantity. The force is a push or a pull and it must have direction thus we would see that the magnitude of the force would tend to differ in any direction that we would have to consider in the problem above. Hence, we shall now try to see the work that is done in each of the cases that involves the box as have been shown in the question that we can see above.
a) If the force have been applied horizontally, then the work that is done is the product of the mass, the acceleration of the body
F = ma = 20 Kg * 1.4m/s^2 = 28 N
b) If the force is 30° above the horizontal then it has a vertical direction we have;
F = ma sin 30°
F = 20 Kg * 1.4m/s^2 * sin 30° = 14 N
C) Now we have the negative vertical direction thus;
F = 20 Kg * 1.4m/s^2 * (-sin 30° ) = -14 N
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Can someone help me please
Answers
Given,
The distance of the race, d=100 m
The extended distance of the race, s=200 m
The velocity of the red car throughout the race, u₁=10 m/s
The initial velocity of the blue car, u₂=0 m/s
The blue car gains a velocity of 2 m/s every second.The constant acceleration of the blue car is
hus t
[tex]\begin{gathered} a=\frac{2}{1} \\ =2\text{ m/s}^2 \end{gathered}[/tex]As the red car maintains the same velocity, the speed when it reaches the finish line will be 10 m/s.The time it takes for the red car to reach the finish line when the race was 100 m
[tex](t_1)_{100}=\frac{d}{u_1}[/tex]On substituting the known values,
[tex]\begin{gathered} (t_1)_{100}=\frac{100}{10} \\ =10\text{ s} \end{gathered}[/tex]The time it takes for the red car to reach the finish line after the race is extended,
[tex](t_1)_{200}=\frac{s}{u_1}[/tex]On substituting the known values,
[tex]\begin{gathered} (t_1)_{200}=\frac{200}{10} \\ =20\text{ s} \end{gathered}[/tex]From the equation of motion,
The final velocity of the blue car, when it reaches the finish line of 100 m race is given by the equation of motion,
[tex]v^2_{100}=u^2_2+2ad[/tex]Where v₁₀O is the final velocity of the blue car at the end of the 100 m race.n substituting the known values,₀
[tex]\begin{gathered} v^2_{^{}100}=0+2\times2\times100 \\ =400 \\ v=\sqrt[]{400} \\ =20\text{ m/s} \end{gathered}[/tex]hus the speed of the blue car when it reaches the finish line of 100 m race is 20 m/s
he time it takes for the blue car to reach the end of the 100 m race is given by,
[tex]v_{100}=u_2+a(t_2)_{100}[/tex]Where (t₂)₁₀₀ is the time it takes for the blue car to reach the end of the 100 m race.
On substituting the known values in the above equation,
[tex]\begin{gathered} 20=0+2(t_2)_{100} \\ \Rightarrow(t_2)_{100}=\frac{20}{2} \\ =10\text{ s} \end{gathered}[/tex]hus both cars take 10 s to reach the end of the 100 m race. Thus they both reach the finitsh line together.
The time it takes for the blue car to reach the end of the 200 m race can be calculated using the equation,
[tex]s=u_2(t_2)_{200}+\frac{1}{2}a\lbrack(t_2)_{200}\rbrack^2[/tex]Where (t₂)₂₀₀ is the time it takes for the blue car to reach the end of the 200 m race.
On substituting the known values,
[tex]\begin{gathered} 200=0+\frac{1}{2}\times2\times\lbrack(t_2)_{200}\rbrack^2 \\ \lbrack(t_2)_{200}\rbrack^2=200 \\ \Rightarrow(t_2)_{200}=\sqrt[]{200} \\ =14.14\text{ s} \end{gathered}[/tex]hus while rthe ed car takes 200 s to reach the finish line of the 200 m race, the blue car takes 14.14 s.
Therefore, if the race was extended, the blue car will reach the finish line first.
Which object has a weight of about 22.5 N?
Answers
The object with a mass of 2.3 kg that is the rock will have a weight of 22.5 N.
W = m * g
W = Weight
m = Mass
g = Acceleration due to gravity
g = 9.8 m / s²
For the book,
m = 1.1 kg
W = 1.1 * 9.8
W = 10.78 N
For the Box,
m = 4.5 kg
W = 4.5 * 9.8
W = 44.1 N
For the Rock,
m = 2.3 kg
W = 2.3 * 9.8
W = 22.5 N
Therefore, the object that has a weight of about 22.5 N is the rock
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need help with image
Answers
Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed if there is no external force acting.
So, the correct option is :
If there is no friction, the ball will roll forever in a straight-line path.
A 0.005 kg projectile has a velocity of 255 m/s to the right. What force is required to stop this projectile in 1.45 s?
Answers
Answer:
F ≈ 0,879 N
Explanation:
First, we calculate the value of the acceleration needed to stop the projectile.
That is:
V = V₀ + a.t
0 = 255 + a . (1,45)
a = - 255 / 1,45
a = -175,86 m/s²
Using the 2° Law of Newton:
∑F = m.a
F = m . a
F = 0,005 . (175,86)
F ≈ 0,879 N
module 2 question 6
The cannon on a battleship can fire a shell a maximum distance of 48.0 km.
(a) Calculate the initial velocity of the shell.
m/s
(b) What maximum height does it reach? (At its highest, the shell is above a substantial part of the atmosphere--but air resistance is not really negligible as assumed to make this problem easier.)
m
(c) The ocean is not flat, since the earth is curved. How many meters lower will its surface be 48.0 km from the ship along a horizontal line parallel to the surface at the ship?
mDoes your answer imply that error introduced by the assumption of a flat earth in projectile motion is significant here? (Select all that apply.)
The error is significant compared to the distance of travel.
The error is insignificant compared to the distance of travel.
The error is insignificant compared to the size of a target.
The error could be significant compared to the size of a target.
Answers
( a ) R = u² sin 2θ / g
θ = 45°
R = 48 km = 48000 m
g = 9.8 m / s²
u² = ( 48000 * 9.8 ) / sin 2 ( 45 )
u² = 470400 / 1
u = 685.86 m / s
( b ) H = u² sin²θ / 2g
H = 685.85² sin²45 / 2 * 9.8
H = 12098.15 m
( c ) The radius of Earth, the horizontal distance travelled by the shell and the distance from final position of cannon and center of Earth makes a right angled triangle.
According to Pythagoras theorem,
x² = 48² + ( 6.371 * 10³)²
x² = 2304 + ( 40.589641 * [tex]10^{6}[/tex] )
x² = ( 23.04 * 10² ) + ( 405896.41 * 10² )
x² = 405919.45 * 10²
x = 6371.18
At 48 km from the ship, the distance lower form the surface the shell lands is,
d = 6371.18 - 6371
d = 0.18 km
Compared to the distance travelled of 48 km, the error of 0.18 km is insignificant
Therefore,
The initial velocity of the shell = 685.86 m / sThe maximum height reached by the shell = 12098.15 mIt will be 180 meters lower, 48 km from the ship.Riley works at the Six Flags Amusement Park on weekends and usually operates the Merry-Go-Round ride. If she stands on the ride's platform a distance of 6.63 m from its center and reaches a speed of 3.89 m/s, then what is the net inward force that would be required to keep her 50.1-kg body moving in a circle?
Question 20 options:
114.3 N
2.3 N
29.5 N
1249.5 N
Answers
The net inward force that would be required to keep her 50.1-kg body moving in a circle is 114.3 N
[tex]F_{c}[/tex] = m v² / R
[tex]F_{c}[/tex] = Centripetal force
m = Mass
v = Linear velocity
R = Radius
m = 50.1 kg
v = 3.89 m / s
R = 6.63 m
[tex]F_{c}[/tex] = 50.1 * 3.89² / 6.63
[tex]F_{c}[/tex] = 114.3 N
The net inward force acting on Riley to keep her body moving in a circle is called as Centripetal force. A centripetal force is responsible for making a body follow a curved path. The direction of centripetal force is perpendicular to the motion and towards the center of curvature.
Therefore, the net inward force that would be required to keep her 50.1-kg body moving in a circle is 114.3 N
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