Based on the given figure, you have that the components of the forces on the direction of the incline are:
applied force:
Fa = 500N*cos(-30°) = 433.01N
friction force:
Fr = µk*N = µk*m*g*cos(30°) = (0.42)(40kg)(9.8 m/s^2)cos(30°) = 142.58N
gravitational force:
Fg = m*g*sin(30°) = (40kg)(9.8)sin(30°) = 196.0N
Next, consider that the work done by a force is equal to the product of the force and the distance at which the force is applied. In this case, such distance is 8m.
Then, you have:
Work done by the applied force:
WFa = (433.01N)(8.0m) = 3464.08 J
Work done by the friction force:
WFr = (142.58N)(8.0m) = 1140.64 J
Work done by the gravitational force:
WFg = (196.0N)(8.0m) = 1568.0 J
Finally, the net work, based on the direction of the forces, is:
Net work = WFa - WFr - WFg = 3464.08J - 1140.64J - 1568.0J = 755.44J
Hence, the net work is 755.44 J
Answer: Wfr = -1883.25 J
Explanation:
1) In the x direction (direction of motion of the crate):
Fcos(30^o) - f_fr - mgsin(30^o ) = 0 (because a_x = 0)
f_fr = Fcos(30^o ) - mgsin(30^o)
f_fr = (498)(√3/2) - (40)(9.81) (1/2) = 235.28 N
⇒ W_fr = f_fr.s.cos(180^o) = -(235.28)(8.0) = -1882.25 J (1)
2) In the y direction (perpendicular to the motion of the crate):
n - Fsin(30^o) - mgcos(30^o) = 0 (because a_y = 0)
⇒ n = Fsin(30^o) + mgcos(30^o) = (498)(1/2) + (40)(9.81)(√3/2) = 588.83 N
⇒ f_fr = μ_k.n = (0.40)(588.83) = 235.53 N
⇒ W_fr = f_fr.s.cos(180^o ) = -(235.53)(8.0) = -1884.24 J (2)
Average of calculations (1) & (2): W_fr = -1883.25 J
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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The volume of asamplea saof nitrogen gas is 88mLat 25°C and 1.0 atm. Whatis its volume in LatSTP?V₁=88mL T₁=25°CV₂=? T₂=23°2
Given:
V1= 88 ml
P1= 1 atm
T1= 25°C= 298.15 K
STP conditions:
P2= 1atm
T2= 273.15 K
Apply :
[tex]\frac{P1V1}{T1}=\frac{P2V2}{T2}[/tex]Replacing:
[tex]\frac{1(88)}{298.15}=\frac{1(V2)}{273.15}[/tex][tex]88(273.15)=298.15V2[/tex][tex]\frac{88(273.15)}{298.15}=V2[/tex]V2= 80.62 ml
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.
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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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.
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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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.
A passenger on a boat moving at 2.50 m/s on a still lake walks up a flight of stairs at a speed of 0.60 m/s (Figure 1). The stairs are angled at 45∘ pointing in the direction of motion as shown. What is the magnitude of the velocity of the passenger relative to the water?
The magnitude of the velocity of the passenger on a boat moving at 2.50 m/s on a still lake relative to the water is 2.95 m / s
Velocity of man = 0.6 m / s
Velocity of ship = 2.5 m / s
Resolving the velocity of man walking at an angle of 45° into its X and Y components.
X-components of velocity of man = 0.6 * cos 45°
X-components of velocity of man = 0.42 m / s
Y-components of velocity of man = 0.6 * sin 45°
Y-components of velocity of man = 0.42 m / s
[tex]V_{x}[/tex] = 0.42 + 2.5
[tex]V_{x}[/tex] = 2.92 m / s
[tex]V_{y}[/tex] = 0.42 m / s
V² = [tex]V_{x}[/tex]² + [tex]V_{y}[/tex]²
V² = 2.92² + 0.42²
V² = 8.53 + 0.18
V = √ 8.71
V = 2.95 m / s
Therefore, the magnitude of the velocity of the passenger relative to the water is 2.95 m / s
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A vertical wire, suspended from one end is stretched by attaching a weight of 20N to the lower end. If the weight extends the wire by Imon. Calculate the energy gained
A vertical wire suspended from one stop is tensioned by attaching a 20N weight to the lower stop. If you have weight, use Imon to stretch the wire. Energy source: `W=(1)/(2) f xx l=(1)/(2) xx 20 xx 10^(-3)` `= 0.01J`.
In physics, power is a quantitative property that is transferred to an identifiable frame or physical device in the form of labor productivity and warmth and tenderness. Electricity is a stored quantity. Energy conservation regulations state that energy can be transformed into form, but cannot be created or destroyed.
Energy exists in various characteristic documents. Examples include light energy, thermal energy, mechanical energy, gravity, electrical energy, sound electricity, chemical power, and nuclear or nuclear energy. Each form can be transformed or transformed into a different bureaucracy. Power is in everything we consume, eat, and use. Power supplies and regulates the body's natural internal abilities. It is used to support cells and tissues in the body, build muscle, and is important for maintaining homeostasis. The harsher the environment, the more energy it takes to maintain.
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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
Suppose you apply a flame and heat one liter of water, raising its temperature 10 degrees celsius. If you transfer the same heat energy to two liters, how much will the temperature rise?
Transferring the same heat energy to two litres the temperature will rise by 5 degree celsius.
What is temperature and how the temperature rise to be 5 degree celsius?Temperature is the measure of degree of hotness and coldness , to either measure the temperature of room or weather.Here in this question we are applying a flame and heat one litre of water by which the temperature is raised 10 degree celsius.Now we are transferring the same heat energy to two litres , then the temperature rise is asked.The d , density of water =1000 kg/m^3 , using the equation Q= mCdelT , where C = 4.18 KJ.The heat we got is 41.8 KJ, the change is temperature comes out to be 5 degree celsius.To know more about temperature visit:
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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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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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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
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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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).
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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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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.
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.
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
How do I solve this?
An Alaskan rescue plane drops a package of emergency rations to a
stranded party of explorers. If the plane is traveling horizontally at 40.0
m/s at a height of 100 m off the ground, where does the package strike
relative to where it is released? (Hint. Use the plane as your origin (x=0,
y=0 and remember to use our 6 step process)
The horizontal distance travelled by the package is 180.8 m.
What is the time of motion of the package?
The time of motion of the package is the time taken for the package to travel from its initial position to the given height.
The time taken for the package to travel through the given distance is calculated as follows;
h = vt + ¹/₂gt²
where;
v is the vertical velocity of the package = 0t is time of motionh is the height through which the package travelledg is acceleration due to gravityh = ¹/₂gt²
2h = gt²
t = 2h/g
t = √(2h/g)
t = 4.52 seconds
The horizontal distance travelled by the package is calculated as follows;
X = Vxt
where;
Vx is the horizontal velocity of the packaget is the time of motionX = (40 m/s) x (4.52 s)
X = 180.8 m
Thus, the distance the package strike relative to where it was released is the horizontal displacement of the package.
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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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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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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.)
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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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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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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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
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.
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
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