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.
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.)
What is the source of the electrical energy in the brain
Answer:
What's the source of energy that powers the human brain?
Mainly sugar, glucose. Glucose is broken down, and in the process it is used to create a gradient of hydrogen ions (Oxygen is also needed, which is why we breath).
brainiest please
How long does it take for a person that is walking at 2 meters per second to travel 1200 meters
The time taken by a person that is walking at 2 meters per second to travel 1200 meters is 600 s
v = d / t
v = Velocity
d = Distance
t = Time
v = 2 m / s
d = 1200 m
t = d / v
t = 1200 / 2
t = 600 s
Velocity is the rate of change of distance with respect to time. It is denoted using V. Its unit is usually m / s.
Therefore, the time taken by thy person is 600 seconds
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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.
( 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.A curious physics student asked her teacher “how quickly is Earth moving through space?” Using the concept of relative velocity, explain why this question does not have an answer.
We know that the relative velocity of earth with respect to the object upon it is zero, but it is non-zero with respect to space.
Whenever we measure velocity, we actually are measuring relative velocity, i.e. the velocity of an object wrt another object. So say, we are saying the velocity of a car is 60 m/s, so it actually means that for us observing the car, it is moving at 60 m/s. Now say you are observing the same car while traveling on a bike at a speed of 30 m/s. Then for you, the speed of the car will now be 60- 30 m/s. This is the concept of relative velocity.
Now, when the student asks how quickly the Earth is moving through space, in order to answer that question, we need to know the relative velocity of the Earth with respect to space. But Space surrounds the earth and all other astronomical bodies, and the limits of space is considered to be infinite so there is no way to determine if it is moving at a certain velocity or not moving at all. Hence, we can't determine the velocity of Earth through space. Hence, the question doesn't have an answer.
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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?
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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Vector B~ has x, y, and z components of 4.3, 4.5, and 5.1 units, respectively. Calculate the magnitude of B~ .
The magnitude of vector B is determined as 8.05 units.
What is the magnitude of a vector?
The magnitude of a vector is the scalar value of a vector that can be used to describe the unit value of the vector.
The magnitude of a vector can also be described as the resultant of the vector because it is usually the single value of the vector that can be used to represent the vector.
The magnitude of vector B is calculated as follows;
|B| = √(x² + y² + z²)
where;
x is the x - component of the vectory is the y - component of the vectorz is the z - component of the vector|B| = √(4.3² + 4.5² + 5.1²)
|B| = 8.05 units
Thus, we can conclude that vector B with x, y, and z components of 4.3, 4.5, and 5.1 units, respectively, has a magnitude of 8.05 units.
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How human memory is complicated and can even sometimes be enhanced?
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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need help with image
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.
Do gas giant planets have geology? Why or why not?
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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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?
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
why is this situation seem impossible?
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.
write any 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.
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.
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?
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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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
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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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?
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
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
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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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?
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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An airplane flies with a constant speed of 760 km/h. How far can it travel in 1 1/2 hours?
Answer: 1140 km
Explanation: You would just add 760 to 760/2 because the airplane travels 760 km in 1 hour (760 km/h means 760 km every 1 hour) and travels 380 km in half an hour. Add those up and you would get 1140 km.
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
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.
Can someone help me please
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.
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.
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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the product of mass and velocity is also known as linear momentum.1. true2. false
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
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.
All choices offered are false.
The theme is heard first, followed by FIVE different variations of it.
===> 5
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?
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]Two students are on a balcony 19.6 m above the street. One student throws a ball vertically down ward at 14.7 m / s At the same instant, the other stu dent throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down . a. What is the difference in the time the balls spend in the air? b . What is the velocity of each ball as it strikes the ground ? c . How far apart are the balls 0.800 s after they are thrown ?
According to the given statement:
a)The difference in the time the balls spend in the air is 3s.
b)Each ball travels at a speed of 24.5 m/s as it lands.
c) The balls 0.800 s after they are thrown far apart are 23.52 m.
What exactly is meant by velocity?The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In principle, velocity is a vector quantity. It is the velocity where at distance changes. It is the migration change rate.
Briefing:A )
We have u = 14.7 m/s, acceleration = acceleration due to gravity = 9.8 m/s² , we need to find time when s = 19.6 m.
The first ball:
h = v o × t - g t² / 2
19.6 = 14.7 t + 9.8 t²/2
4.9 t² + 14.7 t - 19.6 = 0
t 1/2 = (-14.7 +/- √216.09+384.16 ) / 9.8
= 9.8/9.8 = 1 s
The second ball:
v = v o - g t
0 = 14.7 - 9.8 t
9.8 t = 14.7
t = 1.5 ( upward )
h = 19.6 + 14.7 - 9.8 × 1.5 = 30.625 m
30.625 = 9.8 × t²/2
t² = 6.25
t = 2.5 s ( downward )
1.5 s + 2.5 s = 4 s
The difference: 4 s - 1 s = 3 s.
B )
We have u = 14.7 m/s, acceleration = acceleration due to gravity = 9.8 m/s² , and t = 1 second.
The first ball:
v = v o + g t
= 14.7 + 9.8 × 1
= 24.5 m/s
The second ball:
v = g * t = 9.8 * 3 = 24.5 m/s
Each ball travels at a speed of 24.5 m/s as it lands.
C ) d 1 = 14.7 * 0.8 + 9.8 * 0.8²/2
= 11.76 + 3.136
= 14.896 m
d 2 = 14.7 * 0.8 - 9.8* 0.8²/2
= 11.76 - 3.136
= 8.624 m
d 1 + d 2 = 14.896 + 8.624 = 23.52 m
The balls 0.800 s after they are thrown far apart are 23.52 m.
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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
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
Why does the damage continue after the initial explosion of a large atomicbomb?A. Because a fission reaction has no productsB. Because atomic bombs continue going offC. Because the radiation makes people sickD. Because fusion reactions occur afterward
Large atomic bombs undergo nuclear fission.
Nuclear fission is a chain reaction.
A large amount of radiation is produced.
Thus, the damage continues after the initial explosion because the produced radiation makes people sick.
Option C is correct.
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
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
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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