Cindy's average speed for the entire run is 200 m per min.
What is Cindy's average speed for the entire run?
First, to measure Cindy's average speed, we need to know that the average speed of Cindy would be the total distance she had covered while running by the total amount of time it took her to cover the distance. So, we need to measure her average rate of speed over the extent of her runs.
In order to calculate average speed we will calculate the total distance in meters traveled by Cindy.
Given:
Total distance = 2 Km
Convert km to meter, since she is running at a different speed measured in meter.
1 km = 1,000 m
Total distance = 2 Km = 2000 m
Then, the total amount of time in minutes it took her to cover this distance is: 2 (for the first 250 m) + 4 (for the next 1,000 m) + 1 (next 350 m) + 3 (time used for rest) = 10 minutes
Average speed = total distance / total time
Average speed = 2000 m / 10 minutes
Therefore, her average speed = 200 m per min
In summary, Cindy, though rested for 3 minutes, maintained an average speed of about 200 m per min for the entire run.
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Which is expected to have the largest dispersion forces?
options:
C9H20
F2
CH4
PH3
C9H20 is expected to have the largest dispersion forces among the given four options.
The strength of dispersion force increases with number of electrons in the molecule.
Number of electrons in an element = Atomic number * Number of atoms
Atomic number of
C = 6H = 1O = 8F = 9P = 15Number of electrons in C9H20 = ( 6 * 9 ) + ( 1 * 2 ) + ( 8 * 1 )
Number of electrons in C9H20 = 64
Number of electrons in F2 = 9 * 2
Number of electrons in F2 = 18
Number of electrons in CH4 = ( 6 * 1 ) + ( 1 * 4 )
Number of electrons in CH4 = 10
Number of electrons in PH3 = ( 15 * 1 ) + ( 1 * 3 )
Number of electrons in PH3 = 18
Therefore, C9H20 is expected to have the largest dispersion forces.
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assume you have a cart with a mass m = 0.530 kg. the cart accelerates from rest to a final velocity of 2.1 m/s.
1. what is the net work done on the cart?
2. if the work is done over the displacement of 0.71 m, what is the average net force on the cart?
3. if there is an uncompensated frictional force f = 0.11 n, what is the work done by friction over the same displacement?
The net work done on the cart is 1.17 J.
The average net force on the cart is 1.65 N.
The work done by frictional force over the same distance is 0.078 J.
What is net work done on the cart?The net work done on the cart is calculated by applying the principle of conservation of energy as follows;
W = ΔK.E
W = ¹/₂m(v² - u²)
where;
v is the final velocity of the caru is the initial velocity of the cartm is mass of the cartW = ¹/₂(0.53)(2.1² - 0²)
W = 1.17 J
The average net force on the cart is calculated as follows;
W = Fd
where;
F is the net forced is the displacement of the cartF = W/d
F = (1.17) / (0.71)
F = 1.65 N
The work done by frictional force over the same distance is calculated as follows;
W = fd
W = 0.11 x 0.71
W = 0.078 J
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A +15μC point charge is placed 10cm from a +7µC point charge. 1. Find the force experienced by the +15μC charge.
The electric force on the charge Fc = 9.439 x 10⁹N(m/C)²
Equation :To calculate the electric force on different charges and a certain distance,
Use formula,
Fc = k (q₁q₂)/r²
Where,
Fc is electric force
k is coulomb's constant
q₁ is charge at one point
q₂ is charge at second point
r is the distance
Known values,
q₁ = 15μC
q₂ = 7μC
k = 8.99 × 10⁹ Nm² /C²
r = 10cm
Fc = ?
Now, putting the values
Fc = 8.99 × 10⁹ Nm² /C² (15μC x 7μC)/(10cm)²
Fc = 8.99 × 10⁹ N(m/C)² x 1.05
Fc = 9.439 x 10⁹N(m/C)²
Hence, the electric force is Fc = 9.439 x 10⁹N(m/C)².
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A polar bear initially running at 4.0 m/s accerlerates uniformly for 18 s. If the bear travels 135 m in this time what is it's maximum speed
Answer:
Explanation:4.0-135m/ 18s = 131/18s = -7.2 m/
Astrologers claim that your personality traits are determined by the positionsof the planets in relation to you at birth. Scientists argue that these gravitationaleffects are so small that they are totally insignificant. Compare the gravitationalattraction between you and Mars to the gravitational attraction between youand your 70.0 kg doctor at the moment of your birth, if the doctor stands0.500 m away (Note: MM = 6.42 x 10^23 ky, de to M = 7.83 x 10^10 m. This is theaverage distance between Earth and Mars. This distance varies as the twoplanets orbit the sun.)
ANSWER:
The force exerted by the doctor is greater
STEP-BY-STEP EXPLANATION:
Given:
Mass of Mars = 6.42 x 10^23 kg
Mass of Doctor = 70 kg
Mass at birth = m
Distance between Earth and Mars = 7.83 x 10^10 m
Distancia Doctor = 0.5 m
G = 6.67 x 10^- 11 N*m^2/kg^2
We calculated the force in each based on the mass at birth.
[tex]\begin{gathered} F_{mars}=\frac{G\cdot M_{mars}\cdot m}{d_{EM}^2}=\frac{6.67\cdot10^{-11}\cdot6.42\cdot10^{23}}{(7.83\cdot10^{10})^2}m=6.98\cdot10^{-9}m \\ F_{doc}=\frac{G\cdot M_{doc}\cdot m}{d_2^2}=\frac{6.67\cdot10^{-11}\cdot70}{\left(0.5\right)^2}\:m=1.87\cdot10^{-8}m \end{gathered}[/tex]Which means that the force exerted by the doctor is greater
The light from the sun has higher frequencies from one side of the sun than from the other side. What does that tell you about the sun?
If the light from the sun has higher frequencies from one side of the sun than from the other side, it is proof that the Sun is rotating.
Doppler effect states that, if a person is standing still and a source ( sound / light ) is moving towards him, the frequency of the wave emitted from the object will increase and if the source ( sound / light ) is away from him, the frequency of the wave emitted from the object will decrease.
So, if the light from the sun has higher frequencies from one side of the sun than from the other side, it means that the Sun is rotating. The higher frequencies points are the points that rotating towards Earth and lower frequencies points are the points that rotating away from Earth.
Therefore, if the light from the sun has higher frequencies from one side of the sun than from the other side, it is proof that the Sun is rotating.
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5. Three skiers were racing across flat, smooth snow. The skiers in yellow and purple sweaters have the same mass,
which is more mass than the skier in the red sweater. Each skier was bumped by another skier, but not from the same
direction. Which skier(s) experienced the strongest force when bumped? How do you know?
The skier(s) who will experienced the strongest force when bumped are the yellow and purple skiers.
What is force?
A force is an effect which has the power to change any object's motion. When a mass-containing object increases its velocity, such as when it travels away from rest, a force can cause it to accelerate. A push or a pull is a straightforward method to explain force. Since a force has both magnitude and direction, it is a vector quantity. It is computed using the SI unit of newton (N). F is used to represent force (formerly P).
The skier(s) who will experienced the strongest force when bumped are the yellow and purple skiers. We can come to these conclusion because they have more mass than the red skier, but they changed speed by the same amount as the red skier did.
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In a laboratory experiment, a student was exploring the stretch of a spring based on the pull of the force. The student used weights of 1 N, 2 N, 3 N, 4 N, and 5 N. The points were plotted, and the relationship was analyzed. The graph is shown.
A graph titled stretch versus force is shown with stretch on the vertical axis and force on the horizontal axis. The graph is a straight diagonal line from bottom left to top right going through points 0 comma 0, 1 comma 1.0, 2 comma 3.6, 3 comma 5.5, 4 comma 7.2, and 5 comma 9.1.
Which of the following values would be considered an extrapolation?
Force = 1 N, stretch = 1.9 cm
Force = 3 N, stretch = 5.5 cm
Force = 4.6 N, stretch = 8.3 cm
Force = 5.4 N, stretch = 9.6 cm
An extrapolation from the graph given the points on the line of best fit is Force = 5.4 N, stretch = 9.6 cm.
What is Hooke's law?The Hooke's law states that as long as the elastic limit is not exceeded, the extension of the object is proportional to the magnitude of the force that have been applied. This would help us answer the question
Now we can see that in the plot of the Hooke's law, the force have been plotted on the horizontal axis and the extension or the stretch have been plotted on the vertical axis. The areas that runs through the line of best fit have been shown ad from this line of best fit we could be able to obtain an extrapolation.
The point on the graph that can be regarded as an extrapolation is the point that have been marked Force = 5.4 N, stretch = 9.6 cm.
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Answer:
the answer is D
Explanation:
force+5.4N, stretch=9.6
A motorcycle travels at 55m / s, taking 5s to stop and relax Relax. Calculate the acceleration of that motor.
The acceleration of the motorcycle = -11m/s²
Explanation:The initial speed of the motorcycle, u = 55 m/s
The motorcycle takes 5 seconds to stop and relax
That is, the final speed, v = 0 m/s
Time, t = 5 seconds
Acceleration = (Change in speed)/Time
[tex]\begin{gathered} a=\frac{v-u}{t} \\ a=\frac{0-55}{5} \\ a=\frac{-55}{5} \\ a=-11m/s^2 \end{gathered}[/tex]The acceleration of the motorcycle = -11m/s²
PLEASE HELP! 20 points!!
The infamous chicken is dashing toward home plate with a speed of 5.8m/s when he decided to hit the dirt. The chicken slides for 1.1s. Just reaching hte plate as he stops(safe, of course). (A)What is the magnitude and direction of the chickens acceleration? (B) How far did the chicken slide?
please show all work!
The chicken needs to have an acceleration such that after time t his Velocity is 0, shown in the equation:
0 = V + at
-5.8 = 1.1 x a
a = -5.8 / 1.1
a = -5.2727 m/s/s.
The distance traveled in a time t with velocity V and acceleration is given by:
D = Vot + 1/2 at2
D = 5.8 x 1.1 - 1/2 x 5.2727 x 1.12
D = 3.19m.
Acceleration is the name we give to the process of changing velocity. Velocity is both speed and direction, so there are only two ways to accelerate. Either change speed, change direction, or change both. There are three main types of accelerated motion uniform acceleration, unequal acceleration, and average acceleration.
The term uniform acceleration refers to the motion of a body moving in a straight line with an increasing velocity at equal time intervals. Velocity is the rate of change of displacement. Acceleration is the rate of change of velocity. Velocity is a vector quantity because it consists of both magnitude and direction. Since acceleration is the rate of change of velocity it is also a vector quantity. Acceleration occurs whenever an object increases or decreases its speed or changes direction.
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Which of the following equations
best represents "y is directly related to the square of x"?
A) y= kx
B) y = kx2
C) y = k(1/x)
D) y = K(1/x2)
"Y is directly related to the square of x" is represented by the polynomial equation y = kx². A polynomial equation is one that has the polynomial factored down to zero.
What is meant by polynomial equation?Polynomial equations would be those created using exponent, variables, and factors. The greater of its potential exponents is referred to as the equation's degree. By multiplying polynomials as according their degree and the variables in the equation, we can find the solutions.
What is a polynomial equation's degree?The highest or greatest power of a variable in a polynomial equation is referred to as the degree of the polynomial. The degree denotes the polynomial's highest exponential power (ignoring the coefficients).
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Help! I have tried several times to figure out this question on coordinates and turning counter clockwise from the x axis. My professors notes indicate I’m way off!
Given,
The coordinates (-5,-1.3)
To find
The polar coordinates,
Explanation
The polar coordinates are written in the form
[tex](r,\theta)[/tex]Thus,
[tex]r=\sqrt{(-5)^2-(-1.3)^2}=5.16[/tex][tex]\begin{gathered} \theta=tan^{-1}(-\frac{-1.3}{-5}) \\ \Rightarrow\theta=-14.57^o \\ \Rightarrow\theta=360-14.57=345.43^o \end{gathered}[/tex]Conclusion
The polar coordinates are
[tex](5.16,345.43)[/tex]module 2 question 10
A ship sets sail from Rotterdam, The Netherlands, heading due north at 9.00 m/s relative to the water. The local ocean current is 1.52 m/s in a direction 40° north of east. What is the velocity of the ship relative to the earth?
m/s
° N of E
The velocity of the ship heading due north at 9.00 m/s relative to the water has a velocity of 10 m / s relative to the earth.
Let S represent ship, W represent water and E represent Earth.
[tex]V_{SW}[/tex] = 9 m / s N
[tex]V_{WE}[/tex] = 1.52 m / s NE
Resolving [tex]V_{OE}[/tex] into its x and y components.
cos θ = Adjacent side / Hypotenuse
cos 40° = [tex]V_{OE}_{x}[/tex] / [tex]V_{OE}[/tex]
[tex]V_{WE}_{x}[/tex] = 0.77 * 1.52
[tex]V_{WE}_{x}[/tex] = 1.17 m / s
sin θ = Opposite side / Hypotenuse
sin 40° = [tex]V_{OE}_{y}[/tex] / [tex]V_{OE}[/tex]
[tex]V_{WE}_{y}[/tex] = 0.64 * 1.52
[tex]V_{WE}_{y}[/tex] = 0.97 m / s
[tex]V_{SE}[/tex] = [tex]V_{SW}[/tex] + [tex]V_{WE}[/tex]
[tex]V_{X}[/tex] = [tex]V_{SW}_{x}[/tex] + [tex]V_{WE}_{x}[/tex]
[tex]V_{X}[/tex] = 0 + 1.17
[tex]V_{X}[/tex] = 1.17 m / s
[tex]V_{Y}[/tex] = [tex]V_{SW}_{y}[/tex] + [tex]V_{WE}_{y}[/tex]
[tex]V_{Y}[/tex] = 9 + 0.97
[tex]V_{Y}[/tex] = 9.97 m / s
R = √ [tex]V_{X}[/tex]² + [tex]V_{Y}[/tex]²
R = √ 1.17² + 9.97²
R = √ 1.37 + 99.4
R = √ 100.77
R = 10 m / s
Therefore, the velocity of the ship relative to the earth is 10 m / s
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A graph of a relationship is shown.
A graph titled force versus distance is shown with force on the vertical axis and distance on the horizontal axis. The graph is a downward curve from top left to bottom right but sits very close the axes.
Which of the following best describes the relationship between the two variables?
Distance is directly related to force.
Distance is inversely related to force.
Force is directly related to the square of distance.
Force is inversely related the square of distance.
From the direction of the line of best fit, we can say that; distance is inversely related to force.
What is a graph?The term graph has to do with the relationship between two variables that have been shown by the use of the cartesian axes. We know that the cartesian axis are vertical and horizontal. The nature of the line of best fit gives us an idea of the relationship between the variables. This also helps us to understand the slope of the graph
In this case, we can see that a graph titled force versus distance is shown with force on the vertical axis and distance on the horizontal axis. The graph is a downward curve from top left to bottom right but sits very close the axes.
Now we can see that the slope of the graph must be a negative slope. The implication of this is that distance is inversely related to force.
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A ferry approaches shore, moving north with a speed of 6.2 m/s relative to the dock. A person on the ferry walks from one side of the ferry to the other, moving east with a speed of 1.4 m/s relative to the ferry.
What is the speed of the person relative to the dock?
Express your answer using two significant figures and include the appropriate units
So, we have that the speed of the person relative to the dock is 6.36 m/s
How to calculate the speed of the person relative to the dock?Let
v = velocity of person relative to ferry, V = velocity of ferry relative to dock and V' = velocity of person relative to dockwhere their speeds are the magnitudes of their velocities
Given that a ferry approaches shore, moving north with a speed of 6.2 m/s relative to the dock and a person on the ferry walks from one side of the ferry to the other, moving east with a speed of 1.4 m/s relative to the ferry.
We have that their velocities are
v = (1.4 m/s)iV = (6.2 m/s)jSo, V' = v + V
So, V' = (1.4 m/s)i + (6.2 m/s)j
So, the speed of the person relative to the dock is the magnitude of V which is
V' = √(v² + V²)
So, substituting the value of the variables into the equation, we have
V' = √(v² + V²)
V' = √[(1.4 m/s)² + (6.2 m/s)²]
V' = √[1.96 m²/s² + 38.44 m²/s²]
V' = √[40.4 m²/s²]
V' = 6.36 m/s
So, the speed of the person relative to the dock is 6.36 m/s
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What is the mass of a 1.2 N book?
Answer:
w= MG
m=g/w
m=g/1.2 (g Constant= 10mls)
m= 8.33Kg
In your own words, answer the following 3 questions about an object experiencing blue shift? 1. Describe the motion of an object experiencing blue shift? 2. What is happening to its wavelength and frequency? 3. Why?
1. when an object is moving towards us the light from the object is known as blueshift.
2. As a wavelength increases in size, its frequency, and energy E decrease.
3. The frequency of a wave increases, and its wavelength decreases.
When an object is moving away from us the light from the object is known as redshift, and when an object is moving towards us the light from the object is known as blueshift. Astronomers use redshift and blueshift to deduce how far an object is away from Earth, the concept is key to charting the universe's expansion.
As the frequency of a wave decreases the wavelength increases as long as the wave's velocity remains constant. If the velocity of a wave does not change when frequency decreases, it means that a smaller number of crests or troughs pass a set point in a given amount of time.
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In the particle model, the object in the motion diagram is replaced by
Answer:
Single Points
Explanation:
In the particle model, the object in the motion diagram is replaced by ____.
a. an arrow showing direction
b. a series of single points
c. a large dot
d. a scalar-colored green
Hope this helps.
An object starts at the -4 meter position; when does it get to a position of 55 meters?
The original velocity was 8 m/s and the acceleration is 3 m/s². [1.5p point]
The object at -4-meter position, gets to a position of 55 meters after a time of 2 secs if the original velocity was 8 m/s and the acceleration is 3 m/s².
Slope of a location graphThe velocity of an object is represented by the slope of a location graph. Thus, the slope’s value at a given instant corresponds to the object’s velocity at that precise moment.
What is the position-time graph of an object’s slope?The idea behind this is that an object’s velocity is equal to the slope of the line on a position-time graph. The slope of the line will be +4 m/s if the object is travelling at a velocity of +4 m/s.
What does the slope indicate about the velocity of an object?The object’s acceleration is shown by the slope of a velocity graph. Therefore, the slope’s value at a given instant corresponds to the object’s current rate of acceleration.
The time can be calculated as
disp = 55 + 4
= 59m
u = 8 m/s
a = 3 m/s².
t =?
s = ut + 1/2 at²
59 = 8(t) + 1/2 x 3 t²
⇒ 3/2 t² + 8t - 59 = 0
⇒ t = [tex]\frac{-8\sqrt{64+354} }{6}[/tex]
= 12/6
= 2sec
Thus, the time taken is 2 sec.
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You are holding a 200-g apple.
What is the force that you exert on the apple?
What is the force that the apple exerts on you? Support your answer with free
body diagrams.
On the apple, we apply "antigravitational force."
The apple exerts "gravitational force" on us.
The gravitational attraction, an attractive force whose magnitude is directly proportional to the masses of the two objects and inversely proportional to the square of their distance from one another, draws all mass-containing things together.
You can never decrease the gravitational pull produced by adding mass. It is possible for gravitational forces acting in opposition to one another to cancel one another out and leave no net force. It would be conceivable for gravity to push objects instead of always dragging them if it weren't always additive.
A place or thing being free from the pull of gravity is the goal of the hypothetical phenomena known as anti-gravity. It doesn't mean balancing the force of gravity with another force, like electromagnetism or aerodynamic lift, or the lack of weight under gravity felt in free fall or orbit.
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A quality assurance engineer at a frying pan company is asked to qualify a new line of nonstick-coated frying pans. The coating needs to be 1.00 mm thick. One method to test the thickness is for the engineer to pick a percentage of the pans manufactured, strip off the coating, and measure the thickness using a micrometer. This method is a destructive testing method. Instead, the engineer decides that every frying pan will be tested using a nondestructive method. An ultrasonic transducer is used that produces sound waves with a frequency of f=25.0 kHz. The sound waves are sent through the coating and are reflected by the interface between the coating and the metal pan, and the time is recorded. The wavelength of the ultrasonic waves in the coating is 0.0760-m. What should be the time recorded if the coating is the correct thickness (1.00 mm)?
Answer:
Δt ≈ 5.3 x 10⁻⁷ s or Δt ≈ 0.5 μs.
Explanation:
Sound waves propagate at a typical speed according to the physical media.
Using the wave equation of speed:
V = λ x f
V = 0.0760 x 25 x 10³
V = 1900 m/s
Then we must use this value to find the time recorded:
V = ΔS / Δt
1900 = 10⁻³ / Δt
Δt = 10⁻³ / 1900
Δt ≈ 5.2632 x 10⁻⁷ s
Δt ≈ 5.3 x 10⁻⁷ s or Δt ≈ 0.5 μs
The wavelength of the Ultrasonic wave in the coating is 0.0760 m, then the time recorded is 0.5 μs if the coating is of correct thickness, that is of 1 mm
What is frequency?Physics refers to frequency as the quantity of waves passing a fixed location in a unit of time. Additionally, it is the quantity of vibrations or cycles that a body in periodic motion experiences in a certain period of time. One cycle or vibration is defined as the movement of a body in periodic motion across a sequence of occasions or locations before returning to its initial state. Under angular velocity, a straightforward harmonic motion is also observed.
According to the question,
Use wave equation of speed :
V = λ (wavelength) x f (frequency)
V = 0.0760 x 25 x 10³
V = 1900 m/s
The time recorded will be :
V = ΔS / Δt
1900 = 10⁻³ / Δt
Δt = 10⁻³ / 1900
Δt = 5.2632 x 10⁻⁷ s
Δt = 5.3 x 10⁻⁷ s or,
Δt = 0.5 μs
Hence, the time recorded will be 0.5 μs if the coating is of correct thickness that is 1 mm.
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Sanjay uses 225 N of force to move a wheelbarrow 18 m at a velocity of 1.8 m/s. What was the amount of power used?
Considering the definition of power, the amount of power used was 405 W.
Definition of powerPower is the amount of work (force or energy applied to a body) in a unit of time.
In this sense, mechanical power is the amount of force applied to a body in relation to the speed with which it is applied.
Mathematically, the power P is the result of multiplying the force by the execution speed:
P = F×v
where
P is the power expressed in Watts (W).F is the force expressed in newtons (N).v is the speed expressed in meters per second (m/s).Amount of power used in this caseIn this case, you know:
Force applied F = 225 N Distance moved = 18 mVelocity v = 1.8 m/sReplacing in the definition of power:
P = 225 N× 1.8 m/s
Solving
P= 405 W
Finally, the amount of power in this case is 405 W.
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A ride at Six Flags has a frequency of 2.5 Hz. What is the ride's period?
Given:
The frequency of the ride, f=2.5 Hz
To find:
The period of the ride.
Explanation:
The period of a wave is equal to the inverse of the frequency of the wave. And the frequency of the wave is equal to its period.
Thus the period of the ride is given by,
[tex]T=\frac{1}{f}[/tex]On substituting the known values,
[tex]\begin{gathered} T=\frac{1}{2.5} \\ =0.4\text{ s} \end{gathered}[/tex]Final answer:
The period of the ride is 0.4 s
what is the voltage across a 1.5 ohm resistor when a current of 4 amps flow through it
Answer:
6 v
Explanation:
V = I R
4 * 1.5 = 6V
Billie travels 3.2 km due east in 0.1 hr, then 3.2 km at 15.0 degrees eastward of due north in 0.21 hr, and finally another 3.2 km due east in 0.1 hr. What is the magnitude of his average velocity for the entire trip?
Average velocity is total distance divided by total time. The magnitude of his average velocity for the entire trip is 23.15 m/s
What is Velocity ?Velocity is the distance travel in a specific direction per time taken. It is a vector quantity. And it is measured in m/s
If Billie travels 3.2 km due east in 0.1 hr, then 3.2 km at 15.0 degrees eastward of due north in 0.21 hr, and finally another 3.2 km due east in 0.1 hr. Then,
Billie total distance S travel = 3.2 + 3.2Cos 15 + 3.2
S = 9.49 Km
Billie total time T = 0.1 + 0.21 + 0.1
T = 0.41 s
The magnitude of his average velocity for the entire trip = S/T
Average velocity = 9.49/ 0.41
Average velocity = 23.15 m/s
Therefore, the magnitude of his average velocity for the entire trip is 23.15 m/s
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A woman is sitting at a bus stop when an ambulance with a siren wailing at 440 Hz approaches at 31.5 m/s (about 70 mph). Assume the speed of sound to be 343 m/s.
(a)
What frequency (in Hz) does the woman hear?
Considering the Doppler effect, the woman sitting at the bus stop hears a frequency of 484.484 Hz.
Doppler effectThe Doppler effect is defined as the change in the apparent frequency of a wave produced by the relative motion of the source with respect to its observer.
When both the transmitter and the receiver are in motion, the frequency perceived by the receiver will increase as the receiver and transmitter increase their separation distance and will decrease whenever the separation distance between them decreases. The following expression is considered the general case of the Doppler effect:
f'=f× [(v ± vR)÷ (v ∓ vE)]
Where:
f', f: Frequency perceived by the receiver and frequency emitted by the transmitter, respectively. Its unit of measurement in the International System (S.I.) is the hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s⁻¹)v: Wave propagation speed in the medium. It is constant and depends on the characteristics of the medium. In this case, the speed of sound in air is considered to be 343 m/s.vR, vE: Receiver and transmitter speed respectively. Its unit of measure in the S.I. is the m/s±, ∓:You will use the + sign:In the numerator if the receiver approaches the senderIn the denominator if the sender moves away from the receiverYou will use the - sign:In the numerator if the receiver moves away from the senderIn the denominator if the sender approaches the receiverFrequency in this caseIn this case, you know:
f= 440 Hzv= 343 m/svR= 0vE= 31.5 m/sIn the denominator you use the - sign because the sender approaches the receiver.Reemplacing in the expression in the general case of the Doppler effect:
f'= 440 Hz× [(343 m/s ± 0)÷ (343 m/s - 31.5 m/s)]
f'= 440 Hz× [343 m/s÷ (343 m/s - 31.5 m/s)]
f'= 440 Hz× [343 m/s÷ 311.5 m/s]
f'= 440 Hz× 1.1011
f'= 484. 484 Hz
Finally, the woman hears a frequency of 484.484 Hz.
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Question 2
A toddler pushes his 7.0 kg toy box at a relatively constant velocity across the tiled floor of the family room applying a horizontal force of 35 N. Calculate the magnitude of the gravitational
force.
A. 35 N
B. 70 N
C. 135 N
D. 3N
Question 3
A toddler pushes his 7.0 kg toy box at a relatively constant velocity across the tiled floor of the family room applying a horizontal force of 35 N. Calculate the magnitude of the normal force.
A. 3N
B. 35 N
C. 135 N
D. 70 N
Question 4
A toddler pushes his 7.0 kg toy box at a relatively constant velocity across the tiled floor of the family room applying a horizontal force of 35 N. Calculate the coefficient of kinetic friction between the box and the floor.
A. 0.05
B. 0.5
C. 0.7
D. 0.35
When two bodies come into contact, the friction force prevents relative motion
The box's and the floor's coefficient of friction is 1.456 102.
Reason:
Known variables include;
The toy box weighs 7.0 kg
When the toy box is pushed, its velocity is constant
The child's force is equal to 35 N.
Required:
To determine the box's and the floor's coefficient of kinetic friction
Solution:
Given that the box is moving at a constant speed, there is no net force acting on the box, which results in;
Where;
= Force of friction
Coefficient of friction, = N; Normal response, = N
Normal object behavior on a flat surface equals the weight of the object
Box weight and coefficient of friction are same.
The box's weight is W = 7.0 kg x 9.81 m/s2
what is kinetic friction?
An impeding force known as kinetic friction is present when a body or object starts to move. It should be noted that this force acts parallel to the surface and in the opposite direction to the motion of the body that is sliding with respect to the surface.
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This object is located 6.0 cm to the left of the lens, and the image forms at 28.7 cm to the right of the lens.What is the focal length of the lens?
Let's use the common formula:
1/f = 1/i + 1/o, where
f: focal length, i: image distance, o: object distance
1/f = 1/28.7 + 1/6
1/f = 0.2015
f = 4.9625 cm
In a restaurant, one cook slides a 215-g delicious pizza down the counter from left to right at 1.63 m/s. At nearly the exact same time, another cook launches a 349-g cheeseburger along the same counter from right to left at 2.10 m/s. The two items collide head‑on at the given speeds – the counter is friction‑free due to accumulated grease – and combine into a single, wonderful concoction. At what speed does the dish move?
The speed at which the dish move after combining into a single, wonderful concoction is 1.92 m / s
p = m v
p = Momentum
m = Mass
v = Velocity
According to law of conservation of momentum,
Total initial momentum = Total Final momentum
[tex]m_{1}[/tex] [tex]v_{1}[/tex] + [tex]m_{2}[/tex] [tex]v_{2}[/tex] = [tex]m_{1}[/tex] [tex]v_{1}[/tex]' + [tex]m_{2}[/tex] [tex]v_{2}[/tex]'
Since the dish is combined into a single, wonderful concoction, it will move with the same velocity,
[tex]v_{1}[/tex]' = [tex]v_{2}[/tex]' = v'
[tex]m_{1}[/tex] [tex]v_{1}[/tex]+ [tex]m_{2}[/tex] [tex]v_{2}[/tex] = [tex]m_{1}[/tex] v' + [tex]m_{2}[/tex] v'
[tex]m_{1}[/tex] [tex]v_{1}[/tex] + [tex]m_{2}[/tex] [tex]v_{2}[/tex] = ( [tex]m_{1}[/tex] + [tex]m_{2}[/tex] ) v'
v' = ( [tex]m_{1}[/tex] [tex]v_{1}[/tex] + [tex]m_{2}[/tex] [tex]v_{2}[/tex] ) / ( [tex]m_{1}[/tex] + [tex]m_{2}[/tex] )
[tex]m_{1}[/tex] = 215 g
[tex]m_{2}[/tex] = 349 g
[tex]v_{1}[/tex] = 1.63 m / s
[tex]v_{2}[/tex] = 2.10 m / s
Substituting these values in v'
v' = [ ( 215 * 1.63 ) + ( 349 * 2.10 ) ] / ( 215 + 349 )
v' = ( 350.43 + 732.9 ) / 564
v' = 1083.33 / 564
v' = 1.92 m / s
Therefore, the dish moves at 1.92 m / s
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Which of the following statements accurately describe a scientific law?
Answer:
Explanation:
Scientific laws or laws of science are statements, based on repeated experiments or observations, that describe or predict a range of natural phenomena.