Answer:
A. Howard Gardner
Explanation:
Howard Gardner was the educator who did not contribute to study or play.
- His parents fled Nazi Germany with their first son who died in accident before Howard was born.
-He was not allowed to play sports.
- He was an excellent pianist.
-He attended Harvard university.
-Influenced by Erickson
Among Howard Gardner, Jean Jacques Rousseau, Johann Amos Comenius, and Friedrich Froebel, Friedrich Froebel is the one most directly associated with the study of play for establishing the kindergarten system.
Explanation:The question asks who among the listed educators did not contribute to the study or theory of play. Let's examine the contributions of each figure to determine the answer:
Howard Gardner is known for his theory of multiple intelligences, which includes the idea of bodily-kinesthetic intelligence, relating to the way we control our body motions and handle objects, suggesting a recognition of play's importance.Jean Jacques Rousseau emphasized naturalism and the importance of allowing children to interact with their environment, which is in line with the principles of learning through play.Johann Amos Comenius advocated for educational reform that included playful learning, which he saw as a natural way for children to learn.Friedrich Froebel is recognized as the creator of the concept of kindergarten, where play is a fundamental educational approach.Considering the contributions of these educators, Friedrich Froebel is the one who stands out the most as being directly linked to the concept of play in education by virtue of establishing the kindergarten system.
Two identical small insulating balls are suspended by separate 0.26-m threads that are attached to a common point on the ceiling. Each ball has a mass of 7.75 10^-4 kg. Initially the balls are uncharged and hang straight down. They are then given identical positive charges and, as a result, spread apart with an angle of 35° between the threads. (a) Determine the charge on each ball. (answer in C)
The charge on each ball is approximately [tex]\( 8.08 \times 10^{-9} \) C[/tex].
Step 1
To determine the charge on each ball, let's analyze the forces acting on one of the charged balls when they are in equilibrium. The forces are:
1. Gravitational force [tex](\( F_g \))[/tex] downward.
2. Tension [tex](\( T \))[/tex] in the thread.
3. Electrostatic force [tex](\( F_e \))[/tex] between the two charges.
Step 2
Given:
Length of the thread [tex](\( L \)) = 0.26 m[/tex]
Mass of each ball [tex](\( m \)) = \( 7.75 \times 10^{-4} \) kg[/tex]
Angle between the threads = 35°, so each thread makes an angle of 17.5° with the vertical.
Step 3
First, calculate the gravitational force:
[tex]\[ F_g = mg = 7.75 \times 10^{-4} \, \text{kg} \times 9.8 \, \text{m/s}^2 = 7.595 \times 10^{-3} \, \text{N} \][/tex]
The electrostatic force must balance the horizontal component of the tension:
[tex]\[ F_e = T \sin(17.5^\circ) \][/tex]
The vertical component of the tension balances the gravitational force:
[tex]\[ T \cos(17.5^\circ) = F_g \][/tex]
[tex]Solving for \( T \):[/tex]
[tex]\[ T = \frac{F_g}{\cos(17.5^\circ)} = \frac{7.595 \times 10^{-3} \, \text{N}}{\cos(17.5^\circ)} \approx 7.99 \times 10^{-3} \, \text{N} \][/tex]
Step 4
Now calculate the horizontal component of the tension:
[tex]\[ T \sin(17.5^\circ) = 7.99 \times 10^{-3} \, \text{N} \times \sin(17.5^\circ) \approx 2.41 \times 10^{-3} \, \text{N} \][/tex]
Step 5
This is the electrostatic force:
[tex]\[ F_e = \frac{k q^2}{r^2} \][/tex]
where [tex]\( r \)[/tex] is the distance between the balls.
Step 6
To find [tex]\( r \)[/tex]:
[tex]\[ r = 2L \sin(17.5^\circ) = 2 \times 0.26 \, \text{m} \times \sin(17.5^\circ) \approx 0.156 \, \text{m} \][/tex]
Using Coulomb's law:
[tex]\[ 2.41 \times 10^{-3} \, \text{N} = \frac{(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) q^2}{(0.156 \, \text{m})^2} \][/tex]
Step 7
Solving for [tex]\( q \)[/tex]:
[tex]\[ q^2 = \frac{2.41 \times 10^{-3} \, \text{N} \times (0.156 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2} \approx 6.53 \times 10^{-17} \, \text{C}^2 \][/tex]
[tex]\[ q \approx \sqrt{6.53 \times 10^{-17}} \approx 8.08 \times 10^{-9} \, \text{C} \][/tex]
What is the relationship between the slope of the position graph of an object and its velocity?
Answer:
Slope of position time is velocity
Explanation:
The position time graph means a relation between the position of the object and the time which is represent on a graph.
The graph line shows that how the position of an object changes with respect to time.
The slope of the position time graph shows the rate of change of position of the object with respect to time.
The rate of change of position with respect to time is called velocity.
thus, the slope of position time graph gives the velocity of the object.
The de Broglie wavelength of a 0.064-kg golf ball is 3.09 x 10^−34m. What is its speed? (h = 6.63 x 10^−34 J⋅s)
Answer:
The speed is 33.5 m/s.
Explanation:
Given that,
Mass = 0.064 kg
Wavelength [tex]\lambda= 3.09\times10^{-34}\ m[/tex]
We need to calculate the speed
Using formula of he de Broglie wavelength
[tex]\lambda=\dfrac{h}{mv}[/tex]
[tex]v=\dfrac{h}{m\lambda}[/tex]
Where, h = Planck constant
m = mass
[tex]\lambda[/tex] = wavelength
Put the value into the formula
[tex]v = \dfrac{6.63\times10^{-34}}{0.064\times 3.09\times10^{-34}}[/tex]
[tex]v=33.5\ m/s[/tex]
Hence, The speed is 33.5 m/s.
The current density inside a long, solid, cylindrical wire of radius a = 5.0 mm is in the direction of the central axis and its magnitude varies linearly with radial distance r from the axis according to J = J0r/a, where J0 = 420 A/m2. Find the magnitude of the magnetic field at a distance (a) r=0, (b) r = 2.9 mm and (c) r=5.0 mm from the center.
Explanation:
Given that,
Radius a= 5.0 mm
Radial distance r= 0, 2.9 mm, 5.0 mm
Current density at the center of the wire is given by
[tex]J_{0}=420\ A/m^2[/tex]
Given relation between current density and radial distance
[tex]J=\dfrac{J_{0}r}{a}[/tex]
We know that,
When the current passing through the wire changes with radial distance,
then the magnetic field is induced in the wire.
The induced magnetic field is
[tex]B=\dfrac{\mu_{0}i_{ind}}{2\pi r}[/tex]...(I)
We need to calculate the induced current
Using formula of induced current
[tex]i_{ind}=\int_{0}^{r}{J(r)dA}[/tex]
[tex]i_{ind}= \int_{0}^{r}{\dfrac{J_{0}r}{a}2\pi r}[/tex]
[tex]i_{ind}={\dfrac{2\pi J_{0}}{a}}\int_{0}^{r}{r^2}[/tex]
[tex]i_{ind}={\dfrac{2\pi J_{0}}{a}}{\dfrac{r^3}{3}}[/tex]
We need to calculate the magnetic field
Put the value of induced current in equation (I)
[tex]B=\dfrac{\mu_{0}{\dfrac{2\pi J_{0}}{a}}{\dfrac{r^3}{3}}}{2\pi r}[/tex]
[tex]B=\dfrac{\mu_{0}J_{0}r^2}{3a}[/tex]
(a). The magnetic field at a distance r = 0
Put the value into the formula
[tex]B=\dfrac{4\pi\times10^{-7}\times420\times0}{3\times5.0\times10^{-3}}[/tex]
[tex]B = 0[/tex]
The magnetic field at a distance 0 is zero.
(b). The magnetic field at a distance r = 2.9 mm
[tex]B=\dfrac{4\pi\times10^{-7}\times420\times(2.9\times10^{-3})^2}{3\times5.0\times10^{-3}}[/tex]
[tex]B = 2.95\times10^{-7}\ T[/tex]
The magnetic field at a distance 2.9 mm is [tex]2.95\times10^{-7}\ T[/tex]
(c). The magnetic field at a distance r = 5.0 mm
[tex]B=\dfrac{4\pi\times10^{-7}\times420\times(5.0\times10^{-3})^2}{3\times5.0\times10^{-3}}[/tex]
[tex]B = 8.79\times10^{-7}\ T[/tex]
The magnetic field at a distance 5.0 mm is [tex]8.79\times10^{-7}\ T[/tex]
Hence, This is the required solution.
If the retina is 1.7 cm from the lens in the eye, how large is the image on the retina of a person of height 1.5 m standing 9.0 m away?
Answer:
The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign
Explanation:
u = Object distance = 9 m
v = Image distance = 1.7 cm (as the image is forming on the retina)
[tex]h_u[/tex]= Object height = 1.5 m
Magnification
[tex]m=-\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{0.017}{9}=\frac{h_v}{1.5}\\\Rightarrow h_v=-0.00283\ m[/tex]
The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign
The image formed on the retina is 1.8 cm behind the lens.
Explanation:The image formed on the retina can be calculated using the lens formula: 1/f = 1/do + 1/di, where f is the focal length of the lens, do is the object distance from the lens, and di is the image distance.
In this case, the lens-to-retina distance is given as 2.00 cm, so di = -2.00 cm. The object distance do can be calculated as the difference between the person's height and the distance at which they are standing: do = 9.0 m - 1.5 m = 7.5 m. Substituting the values into the lens formula, we get 1/2.00 cm = 1/7.5 m + 1/di. Solving for di, we find that the image is formed 1.8 cm behind the lens.
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Assume that an oil slick consists of a single layer of molecules and that each molecule occupies a cube 1.4 nm on a side. Determine the area of an oil slick formed by 1.1 m^3 of oil. Answer in units of m^2.
Answer:
The area of oil slick is calculate as [tex]785.7 m^{2}[/tex]
Solution:
Volume of oil slick, [tex]V_{o} = 1.1 m^{3}[/tex]
The thickness of one molecule on a side, w = 1.4 mm = [tex]1.4\times 10^{- 3}[/tex]
Now, in order to determine the area of oil slick, [tex]A_{o}[/tex]:
Volume, V = [tex]Area\times thickness[/tex]
Thus
[tex]Area,\ A_{o} = \frac{V_{o}}{w}[/tex]
[tex]Area,\ A_{o} = \frac{1.1}{1.4\times 10^{- 3}} = 785.7 m^{2}[/tex]
The mass of a rocket decreases as it burns through its fuel. If the rocket engine produces constant force (thrust), how does the acceleration of the rocket change over time? Answers:- it does not chage- it increases- it decreases
Answer:
it increases-
Explanation:
When the mass of a rocket decreases as it burns through its fuel and the force ( thrust) is constant then by newtons second law of motion
F= ma here F is constant this means that ma= constant
⇒ m= F /a this implies that mass is inversely proportional to acceleration.
its means when the mass decreases the acceleration must increase. hence the acceleration increases
A straight wire of length L has a positive charge Q distributed along its length. Find the magnitude of the electric field due to the wire at a point located a distance d from one end of the wire along the line extending from the wire. Imagine that distance d is much greater than the length of the wire. Intuitively, what should the magnitude of the electric field at point P be in this case
Answer:
Electric Field at a distance d from one end of the wire is [tex]E=\dfrac{Q}{4\pi \epsilon_0(L+d)d}[/tex]
Electric Field when d is much grater than length of the wire =[tex]\dfrac{Q}{4\pi \epsilon_0\ d^2}[/tex]
Explanation:
Given:
Total charge over the length of the wire = QLength of the wire = LDistance from one end of wire at which electric field is needed to find=dLet dE be the Electric Field due to the small elemental charge on the wire at a distance x from the one end of the wire and let [tex]\lambda[/tex] be the charge density of the wire
[tex]E=\dfrac{dq}{4\pi \epsilon_0x^2}[/tex]
Now integrating it over the entire length varying x from x=d to x=d+L we have and replacing [tex]\lambda=\dfrac{Q}{L}[/tex] we have
[tex]E=\int\dfrac{\lambda dx}{4\pi \epsilon_0x^2}\\E=\dfrac{Q}{4\pi \epsilon_0 (L+d)(d)}[/tex]
When d is much greater than the length of the wire then we have
1+\dfrac{L]{d}≈1
So the Magnitude of the Electric Field at point P = [tex]\dfrac{Q}{4\pi \epsilon_0\ d^2}[/tex]
The Joule (J) is a unit of : kinetic energy
electric energy
potential energy
work all of the above
Answer: work all of the above
Explanation: kinetic, electric and potential are energies so their units must be energy and the Joule is.
Answer:
work all of the above
Explanation:
In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight stretch of highway at 15 m/s. At t=0 the car begins to accelerate at 2.0 m/s2 in order to pass a truck. What is the velocity of the car after 5.0 s have elapsed?
Answer:
[tex]v_{f} =25m/s[/tex]
Explanation:
Kinematics equation for constant acceleration:
[tex]v_{f} =v_{o} + at=15+2*5=25m/s[/tex]
If the gravitational force between objects of equal mass is 2.30 x 10^‐8 N when the objects are 10.0 m apart, what is the mass of each object?
Answer:
Mass of each objects, m = 185.69 kg
Explanation:
The gravitational force between two equal masses is, [tex]F=2.3\times 10^{-8}\ N[/tex]
Separation between the masses, d = 10 m
Let m is the mass of each object. The gravitational force is given by :
[tex]F=G\dfrac{m^2}{d^2}[/tex]
[tex]m=\sqrt{\dfrac{F.d^2}{G}}[/tex]
[tex]m=\sqrt{\dfrac{2.3\times 10^{-8}\times (10)^2}{6.67\times 10^{-11}}}[/tex]
m = 185.69 kg
So, the mass of each objects is 185.69 kg. Hence, this is the required solution.
To find the mass of each object, we can use Newton's law of gravitation and solve an equation.
Explanation:To find the mass of each object, we can use Newton's law of gravitation, which states that the gravitational force between two objects is given by the equation: F = G * M₁ * M₂ / R². Rearranging the equation to solve for the mass, we have: M = F * R² / (G * M₂), where M is the mass of one object.
Using the given values, the gravitational force is 2.30 x 10^‐8 N and the distance between the objects is 10.0 m. Plugging these values into the equation, we have: M = (2.30 x 10^‐8 N) * (10.0 m)² / (6.67 × 10-¹¹ Nm²/kg² * M₂).
Since the masses of the two objects are equal, we can assume M₁ = M₂. Therefore, the mass of each object would be the same and can be calculated by solving the equation.
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A ball thrown straight up climbs for 3.0 sec before falling. Neglecting air resistance, with what velocity was the ball thrown?
Answer:
Speed, u = 29.4 m/s
Explanation:
Given that, A ball thrown straight up climbs for 3.0 sec before falling, t = 3 s
Let u is speed with which the ball is thrown up. When the ball falls, v = 0
Using first equation of motion as :
v = u + at
Here, a = -g
So, u = g × t
[tex]u=9.8\times 3[/tex]
u = 29.4 m/s
So, the speed with which the ball was thrown is 29.4 m/s. Hence, this is the required solution.
The velocity at which the ball was thrown is 29.4 m/s.
To calculate the velocity at which the ball was thrown, we use the formula below.
Formula:
v = u+gt.............. Equation 1Where:
v = Final velocity of the ballu = Initial velocity of the ballg = acceleration due to gravity of the ballt = time.From the question,
Given:
v = 0 m/s (At maximum height)g = -9.8 m/st = 3.0 sSubstitute these values into equation 1
0 = u+3(-9.8)0 = u-29.4u = 29.4 m/sHence, The velocity at which the ball was thrown is 29.4 m/s.
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In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min.Hannah runs the race in a blistering 15.3 min, so fast that she actually passes Kara during the race. How many laps has Hannah run when she passes Kara?
Answer:
No. of laps of Hannah are 7 (approx).
Solution:
According to the question:
The total distance to be covered, D = 5000 m
The distance for each lap, x = 400 m
Time taken by Kara, [tex]t_{K} = 17.9 min = 17.9\times 60 = 1074 s[/tex]
Time taken by Hannah, [tex]t_{H} = 15.3 min = 15.3\times 60 = 918 s[/tex]
Now, the speed of Kara and Hannah can be calculated respectively as:
[tex]v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s[/tex]
[tex]v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s[/tex]
Time taken in each lap is given by:
[tex](v_{H} - v_{K})t = x[/tex]
[tex](5.45 - 4.65)\times t = 400[/tex]
[tex]t = \frac{400}{0.8}[/tex]
t = 500 s
So, Distance covered by Hannah in 't' sec is given by:
[tex]d_{H} = v_{H}\times t[/tex]
[tex]d_{H} = 5.45\times 500 = 2725 m[/tex]
No. of laps taken by Hannah when she passes Kara:
[tex]n_{H} = \frac{d_{H}}{x}[/tex]
[tex]n_{H} = \frac{2725}{400} = 6.8[/tex] ≈ 7 laps
The number of laps that Hannah has run when she passes Kara is 7 laps.
Speed of each athlete
The speed of each athlete is calculated as follows;
Kara = (5000) / (17.9 x 60) = 4.66 m/s
Hannah = (5000) / (15.3 x 60) = 5.47 m/s
Time taken in each lapThe time taken in each lap if Hannah passes kara is calculated as follow;
(5.47 - 4.66)t = 400
0.81t = 400
t = 493.83 s
Distance covered by Hannah when she passes kara;
d = 493.83 x 5.47 = 2,701.25 m
Number of lapsn = 2,701.25/400
n = 6.8 ≈ 7 laps.
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A long, nonconducting cylinder (radius = 6.0 mm) has a nonuniform volume charge density given by αr2, where α = 6.2 mC/m5 and r is the distance from the axis of the cylinder. What is the magnitude of the electric field at a point 2.0 mm from the axis?
Answer: 2.80 N/C
Explanation: In order to calculate the electric firld inside the solid cylinder
non conductor we have to use the Gaussian law,
∫E.ds=Q inside/ε0
E*2πrL=ρ Volume of the Gaussian surface/ε0
E*2πrL= a*r^2 π* r^2* L/ε0
E=a*r^3/(2*ε0)
E=6.2 * (0.002)^3/ (2*8.85*10^-12)= 2.80 N/C
The car is traveling at a constant speed v0 = 80 km/hr on the level portion of the road. When the 4-percent (tan q = 4/100) incline is encountered, the driver does not change the throttle setting and consequently the car decelerates at the constant rate g sin q. Determine the speed of the car: (a) 28 seconds after passing point A ,and (b) when s = 620 m.
Answer:
11.27 m /s
2.98 m / s.
Explanation:
80 km / h = 22.22 m /s
Tanq = 4 / 100
Sinq = .0399
Deceleration acting on inclined plane = g sinq
= 9.8 x .0399
= .3910
Initial speed u = 22.22 m/s
acceleration = - .3910 ms⁻²
v = u - a t
= 22.22 - .3910 x 28
= 22.22 - 10.95
= 11.27 m /s
b ) v² = u² - 2 a s
v² = ( 22.22) ² - 2 x .3910 x 620
= 493.7284 - 484.84
= 8.8884
v = 2.98 m / s.
A step index fiber has a numerical aperture of NA = 0.1. The refractive index of its cladding is 1.465. What is the largest core diameter for which the fiber remains single-moded at the wavelength of 1.3 μm?
Answer:
diameter = 9.951 × [tex]10^{-6}[/tex] m
Explanation:
given data
NA = 0.1
refractive index = 1.465
wavelength = 1.3 μm
to find out
What is the largest core diameter for which the fiber remains single-mode
solution
we know that for single mode v number is
V ≤ 2.405
and v = [tex]\frac{2*\pi *r}{ wavelength} NA[/tex]
here r is radius
so we can say
[tex]\frac{2*\pi *r}{ wavelength} NA[/tex] = 2.405
put here value
[tex]\frac{2*\pi *r}{1.3*10^{-6}} 0.1[/tex] = 2.405
solve it we get r
r = 4.975979 × [tex]x^{-6}[/tex] m
so diameter is = 2 × 4.975979 × [tex]10^{-6}[/tex] m
diameter = 9.951 × [tex]10^{-6}[/tex] m
A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance, the Coulomb force between the two is to be maximized?
Answer:
Explanation:
The two charges are q and Q - q. Let the distance between them is r
Use the formula for coulomb's law for the force between the two charges
[tex]F = \frac{Kq_{1}q_{2}}{r^{2}}[/tex]
So, the force between the charges q and Q - q is given by
[tex]F = \frac{K\left ( Q-q \right )q}}{r^{2}}[/tex]
For maxima and minima, differentiate the force with respect to q.
[tex]\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )[/tex]
For maxima and minima, the value of dF/dq = 0
So, we get
q = Q /2
Now [tex]\frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}[/tex]
the double derivate is negative, so the force is maxima when q = Q / 2 .
A car with an initial speed of 6.64 m/s accelerates at a uniform rate of 0.85 m/s^2 for 3.7s. The final speed of the car is 9.8 m/s. What is the cars displacement after that time? answer in km.
Answer:
So the car displacement after 3.7 sec is 0.030 km
Explanation:
We have given initial velocity u = 6.64 m/sec
Acceleration [tex]a=0.85m/sec^2[/tex]
Time t = 3.7 sec
Final velocity v = 9.8 m/sec
We have to find the displacement after that time
From second equation of motion we know that [tex]s=ut+\frac{1}{2}at^2[/tex], here s is displacement, u is initial velocity, t is time , and a is acceleration
So displacement [tex]s=ut+\frac{1}{2}at^2=6.64\times 3.7+\frac{1}{2}\times 0.85\times 3.7^2=30.386m[/tex]
We know that 1 km = 1000 m
So 30.386 m = 0.030 km
Answer:
The displacement of car after that time is 30.56 m.
Explanation:
Given that,
Initial velocity = 6.64 m/s
Acceleration = 0.85 m/s²
Time = 3.7 s
Final velocity = 9.8 m/s
We need to calculate the displacement
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
Put the value into the formula
[tex]s=\dfrac{9.8^2-6.64^2}{2\times0.85}[/tex]
[tex]s =30.56\ m[/tex]
Hence, The displacement of car after that time is 30.56 m.
A differential manometer is used to measure the drop in pressure across a filter at a water (rho = 1.00 g/cm3) processing plant. The density of the manometer fluid is 8.75 g/cm3. The height of the manometer fluid on side 1 is 275 mm. The height of the manometer fluid on side 2 is 293 mm. What is the pressure difference between points 1 and 2 in psi?
Answer:
0.224 psi
Explanation:
The pressure using a differential manometer is calculated with the delta H.
Delta H = 293 - 275 = 18 mm
The formula for the pressure is:
P = rho * g * h,
where rho : density of the fluid inside the manometer
g : gravitational acceleration
h : delta H inside the manometer.
It is importar the use of units.
8.75 g/cm3 = 8750 kg/m3
g = 9.8 m/s2
h = 18 mm = 0.018 m
P = 1543,5 Pa ; 1 psi = 6894.8 Pa
P = 1543,5/6894,8 = 0.224 psi
Specific heat depends on several factors. Pick the factor below that you suspect will not affect specific heat. composition
state of matter
mass
Answer:
COMPOSITION
Explanation:
the correct answer is COMPOSITION.
specific heat of the body can be defined as the heat required to raise the temperature of a unit mass of a body by 1 °C.
so, A composition of the material is not affected by the specific heat.
The unit of specific heat is Joule per kelvin (J / K ).
A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true anomaly of 150. If the speed of the meteoroid at that time is 2.23 km/s, calculate
(a) the eccentricity of the trajectory;
(b) the altitude at closest approach; and
(c) the speed at the closest approach.
Explanation:
Given that,
Distance = 402000 km
Speed = 2.23 m/s
Angle = 150
(a). We need to calculate the eccentricity of the trajectory
Using formula of eccentricity
[tex]\epsilon=\dfrac{v^2}{2}-\dfrac{\mu}{r}[/tex]
Put the value into the formula
[tex]\epsilon=\dfrac{2.23^2}{2}-\dfrac{398600}{402000}[/tex]
[tex]\epsilon=1.4949\ km^2/s^2[/tex]
We need to calculate the angular momentum
Using formula of the angular momentum
[tex]h^2=-\dfrac{1}{2}\dfrac{\mu^2}{\epsilon}(1-e^2)[/tex]
[tex]h^2=-\dfrac{1}{2}\dfrac{(398600)^2}{1.4949}(1-e^2)[/tex]
[tex]h^2=-5.3141\times10^{10}(1-e^2)[/tex]...(I)
The orbit equation is
[tex]h^2=\mu r(1+e\cos\theta)[/tex]
[tex]h^2=398600\times402000(1-+\cos150)[/tex]
[tex]h^2=16.02372\times10^{10}(1-e0.8660)[/tex]
[tex]h^2=16.02372\times10^{10}-e13.877\times10^{10}[/tex]....(II)
Equating the value of h²
[tex]-5.3141\times10^{10}(1-e^2)=16.02372\times10^{10}-e13.877\times10^{10}[/tex]
[tex]-5.3141+5.3141e^2=13.877e+16.02372[/tex]
[tex]5.3141e^2+13.877e-21.33782=0[/tex]
[tex]e = 0, 1.086[/tex]
(b). We need to calculate the altitude at closest approach
Put the value of e in equation (I)
[tex]h^2=16.02372\times10^{10}-1.086\times13.877\times10^{10}[/tex]
[tex]h^2=9.53298\times10^{9}\ km^4/s^2[/tex]
Now, using the formula of the altitude at closest
[tex]r_{perigee}=\dfrac{h^2}{\mu}\dfrac{1}{1+e}[/tex]
[tex]r_{perigee}=\dfrac{9.53298\times10^{9}}{398600}\dfrac{1}{1+1.086}[/tex]
[tex]r_{perigee}=11465\ km[/tex]
So, The altitude is
[tex]z_{perigee}=r_{perigee}-r_{earth}[/tex]
[tex]z_{perigee}=11465-6378[/tex]
[tex]z_{perigee}=5087\ km[/tex]
(c). We need to calculate the speed at the closest approach.
Using formula of speed
[tex]v_{perigee}=\dfrac{h}{r_{perigee}}[/tex]
[tex]v_{perigee}=\dfrac{\sqrt{9.53298\times10^{9}}}{11465}[/tex]
[tex]v_{perigee}=8.516\ km/s[/tex]
Hence, This is the required solution.
(a) The eccentricity of the meteoroid's trajectory is approximately -0.226. This value indicates the deviation of the orbit from a perfect circle, with negative eccentricity suggesting an elliptical orbit.
(b) At its closest approach, or perigee, the meteoroid is at an altitude of about 595,261 kilometers from the Earth's center. This is the point in its trajectory where it is closest to Earth.
(c) The speed of the meteoroid at its closest approach is approximately 3.69 kilometers per second. This velocity is characteristic of its orbital motion when it is nearest to Earth and is determined by the specific orbital energy and gravitational influences.
To calculate the eccentricity (e), altitude at closest approach (perigee), and speed at the closest approach, we can use the vis-viva equation and orbital mechanics. The vis-viva equation relates the specific orbital energy (ε) to the semi-major axis (a), eccentricity (e), and velocity (v) of an object in orbit:
ε = v²/2 - μ/a
Where:
ε = specific orbital energy
v = velocity of the object
μ = standard gravitational parameter of Earth (approximately 3.986 x 10^5 km³/s²)
a = semi-major axis of the trajectory
We are given the speed of the meteoroid at a distance of 402,000 km from the Earth's center, which we'll use as v. We'll also use μ for Earth's gravitational parameter.
First, let's find the semi-major axis (a):
ε = v²/2 - μ/a
Rearrange to solve for a:
a = μ / (2μ/v² - 1/v²)
a = (3.986 x 10^5 km³/s²) / (2 * (3.986 x 10^5 km³/s²) / (2.23 km/s)² - (1 / (2.23 km/s)²))
a ≈ 491,453 km
(a) The semi-major axis (a) is approximately 491,453 km.
Next, let's calculate the eccentricity (e) using the true anomaly (ν) and the semi-major axis:
e = cos(ν) - r/a
e = cos(150°) - (402,000 km) / (491,453 km)
e ≈ -0.226
(b) The eccentricity (e) is approximately -0.226.
Now, to find the altitude at closest approach (perigee), we need to calculate the perigee distance (rp) using the semi-major axis and eccentricity:
rp = a(1 - e)
rp = (491,453 km) * (1 - (-0.226))
rp ≈ 595,261 km
(c) The perigee distance is approximately 595,261 km.
To find the speed at closest approach (closest approach velocity), we can use the vis-viva equation again:
v = sqrt(μ * (2/r - 1/a))
Where r is the distance from the center of the Earth (rp at perigee).
v = sqrt((3.986 x 10^5 km³/s²) * (2 / (595,261 km) - 1 / (491,453 km)))
v ≈ 3.69 km/s
(c) The speed at closest approach is approximately 3.69 km/s.
Learn more about meteoroid's here:
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate the time required after impact for a puck to lose 10% of its initial speed. Assume air is at 15o C and has a dynamic viscosity of 1.75´10-5 N×s/m2 .
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 ×[tex]10^{-4}[/tex] m
dynamic viscosity = 1.75 ×[tex]10^{-5}[/tex] Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ [tex]\frac{du}{dy}[/tex]
so
= µ [tex]\frac{v}{h}[/tex] ............1
put here value
= 1.75×[tex]10^{-5}[/tex] × [tex]\frac{v}{10^{-4}}[/tex]
= 0.175 v
and
area between air and puck is given by
Area = [tex]\frac{\pi }{4} d^{2}[/tex]
area = [tex]\frac{\pi }{4} 0.1^{2}[/tex]
area = 7.85 × [tex]\frac{v}{10^{-3}}[/tex] m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 × [tex]10^{-3}[/tex]
force = 1.374 × [tex]10^{-3}[/tex] v
and now apply newton second law
force = mass × acceleration
- force = [tex]mass \frac{dv}{dt}[/tex]
- 1.374 × [tex]10^{-3}[/tex] v = [tex]0.03 \frac{0.9v - v }{t}[/tex]
t = [tex] \frac{0.1 v * 0.03}{1.37*10^{-3} v}[/tex]
time = 2.18
so time required after impact for a puck is 2.18 seconds
To calculate the time required for the puck to lose 10% of its initial speed, you can use the equation for deceleration. First, find the final velocity of the puck using the given 10% decrease. Then, calculate the acceleration of the puck using the equation for acceleration. Finally, substitute the acceleration back into the equation for time to find the answer.
Explanation:To calculate the time required for the puck to lose 10% of its initial speed, we need to find the deceleration of the puck. We can use the equation for deceleration, which is a = (v_f - v_i) / t, where a is the acceleration, v_f is the final velocity, v_i is the initial velocity, and t is the time. In this case, since the puck is losing speed, we can use the negative value of the acceleration.
Given that the initial speed of the puck is v_i = (2 * distance) / t, we can calculate the final velocity v_f = 0.9 * v_i, where 0.9 represents the 10% decrease. Substituting these values into the deceleration equation, we can solve for t as follows:
t = (v_f - v_i) / a = (0.9 * v_i - v_i) / (-a) = 0.1 * v_i / a.
Now we can find the acceleration by using the equation a = (6 * π * η * r) / (m * v_i), where η is the dynamic viscosity of air, r is the radius of the puck (which is half the diameter), m is the mass of the puck, and v_i is the initial velocity. Substituting the given values, we can calculate the acceleration. Finally, substituting the acceleration back into the equation for t, we can calculate the time required for the puck to lose 10% of its initial speed.
A charge of -3.35 nC is placed at the origin of an xy-coordinate system, and a charge of 1.75 nC is placed on the y axis at y = 3.90 cm . A. If a third charge, of 5.00 nC , is now placed at the point x = 2.70 cm , y = 3.90 cm find the x and y components of the total force exerted on this charge by the other two charges.
B. Find the magnitude of this force.
C. Find the direction of this force. ( ° below the +x axis )
Answer:
Explanation:
Force due to charges 1.75 and 5 nC is given below
F =K Q₁Q₂ / d²
F₁ = [tex]\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{(2.7\times10^{-2})^2}[/tex]
F₁ = 10.8 X 10⁻⁵ N . It will at in x direction.
Force due to other charge placed at origin
F₂ = [tex]\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{22.5\times10^{-4}}[/tex]
F₂ = 3.5 x 10⁻⁵ N.
Its x component
= F₂ Cos θ
= 3.5 x 10⁻⁵ x 3.9/ 4.74
= 2.88 x 10⁻⁵ N
Its y component
F₂ sin θ
= 3.5 x 10⁻⁵ x 2.7/4.743
= 1.99 x 10⁻⁵ N
Total x component
= 10.8 X 10⁻⁵ +2.88 x 10⁻⁵
= 13.68 x 10⁻⁵ N.
Magnitude of total force F
F² = (13.68 x 10⁻⁵)² + (1.99 x 10⁻⁵ )²
F = 13.82 X 10⁻⁵ N
Direction θ with x axis .
Tanθ = 1.99/ 13.68
θ = 8 °
Astone is thrown directly upward with an initial speed of 9.6 m/s from a height of 12.8 m. After what time interval (in s) does the stone strike the ground? Use g 9.8 m/s^2 Enter a number with 2 digits behind the decimal point.
Answer:
1.89 seconds
Explanation:
t = Time taken
u = Initial velocity = 9.6 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
[tex]v=u+at\\\Rightarrow 0=9.6-9.8\times t\\\Rightarrow \frac{-9.6}{-9.8}=t\\\Rightarrow t=0.97 \s[/tex]
Time taken to reach maximum height is 0.97 seconds
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=9.6\times 0.97+\frac{1}{2}\times -9.8\times 0.97^2\\\Rightarrow s=4.7\ m[/tex]
So, the stone would travel 4.7 m up
So, total height ball would fall is 4.7+12.8 = 17.5 m
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 17.5=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{17.5\times 2}{9.8}}\\\Rightarrow t=1.89\ s[/tex]
Time taken by the stone to travel 17.5 m is 1.89 seconds
Light travels faster in warmer air. On a sunny day, the sun can heat a road and create a layer of hot air above it. Let's model this layer as a uniform one with a sharp boundary separating it from the cooler air above. Use this model to explain the formation of a mirage appearing like the shiny surface of a pool of water.
Answer:
Explanation:
If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.
n = c / v
If light travels faster in warmer air, it will have a lower refractive index
nh < nc
Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:
n1 * sin(θ1) = n2 * sin(θ2)
sin(θ2) = sin(θ1) * n1/n2
If blue light from the sky passing through the hot air will cross to the cold air, then
n1 = nh
n2 = nc
Then:
n1 < n2
So:
n1/n2 < 1
The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.
An infinitely long line charge of uniform linear charge density λ = -3.00 µC/m lies parallel to the y axis at x = -3.00 m. A point charge of 3.00 µC is located at x = 1.00 m, y = 2.00 m. Find the electric field at x = 2.00 m, y = 1.50 m.
Answer:
E=[8.1X-9.63Y]*10^{3}N/m
Explanation:
Field in the point is the sum of the point charge electric field and the field of the infinite line.
First, we calculate the point charge field:
[tex]E_{Charge}=\frac{1}{4\pi \epsilon_0} *\frac{Q}{||r_p -r||^2} *Unitary vector\\||r_p -r||^2=(x_p-x)^2+(y_p -y)^2=1.25 m^2\\Unitary Vector=\frac{(r_p -r)}{||r_p -r||}=\frac{(x_p-x)X+(y_p -y)Y}{||r_p -r||}\\=\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y \\E_{Charge}=K*\frac{3\mu C}{1.25m^2}*(\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y)[/tex]
It is vectorial, where X and Y represent unitary vectors in X and Y. we recall the Coulomb constant k=[tex]\frac{1}{4\pi \epsilon_0}[/tex] and not replace it yet. Now we compute the line field as follows:
[tex]E_{Line}=\frac{\lambda}{2\pi \epsilon_0 distance} *Unitary Vector\\Unitary Vector=X[/tex] (The field is only in the perpendicular direction to the wire, which is X)
[tex]E_{Line}=\frac{-3\mu C/m*2}{2*2\pi \epsilon_0 5*m}X=K*\frac{6\mu C/m}{ 5*m}(-X)[/tex]
We multiplied by 2/2 in order to obtain Coulomb constant and express it that way. Finally, we proceed to sum the fields.
[tex]E=K*\frac{3\mu C}{1.25m^2}*(\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y)+K*\frac{6\mu C}{ 5*m^2}(-X)\\E=K*[2.15-1.2]X-K*[1.07]Y \mu N/m\\E=K*[0.9X-1.07Y] \mu C/m^2\\E=[8.1X-9.63Y]*10^{3}N/m[/tex]
A rocket sled used in old Air Force test could accelerate a volunteer at rates of 90m/s^2. With this acceleration being constant, how long would it take for one of these rocket sleds to travel the length of a one kilometer test track if it starts from rest? A) 11.1sec B) 22.2sec C)4.71sec D) 32.2sec
Answer:
option (c) 4.71 seconds
Explanation:
Given:
Distance to be covered = 1 km = 1000 m
Acceleration, a = 90 m/s²
From Newton's equation of motion
[tex]s=ut+\frac{1}{2}at^2[/tex]
where,
s is the distance
u is the initial speed = 0 m/s ( since it starts from rest )
a is the acceleration
t is the time
on substituting the respective values, we get
[tex]1000=0\times t+\frac{1}{2}\times90\times t^2[/tex]
or
45t² = 1000
or
t² = 22.22
or
t = 4.71 seconds
Hence, the correct answer is option (c) 4.71
A baseball is hit with a speed of 47.24 m/s from a height of 0.42 meters. If the ball is in the air 5.73 seconds and lands 130 meters from the batters feet, (a) at what angle did the ball leave the bat? (b) with what velocity will the baseball hit the ground?
Answer:
a)the ball will leave the bat at an angle of 61.3° .
b) the velocity at which it will hit the ground will be v = 27.1 m/s
Explanation:
Given,
v = 47.24 m
h = 0.42 m
t = 5.73 s
R = 130 m
a)We know that
R = v cosθ × t
cosθ = [tex]\dfrac{R}{v t } = \dfrac{130}{47.24\times 5.73 } =0.4803[/tex]
θ = 61.3°
the ball will leave the bat at an angle of 61.3° .
b)Vx = v cos(θ) = 47.24 x cos 61.3 = 22.7 m/s
v = u + at
Vy = 47.24 x sin 61.3 - 9.81 x 5.73
= -14.8 m/s
v = [tex]\sqrt{v_x^2 + v_y^2)}[/tex]
v = [tex]\sqrt{22.7^2 + -14.8^2}[/tex]
v = 27.1 m/s
the velocity at which it will hit the ground will be v = 27.1 m/s
A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The car accelerates at a uniform rate of 3.70 m/s^2 and the motorcycle at a uniform rate of 4.40 m/s^2. How much time elapses before the MC overtakes the car? How far will each have traveled during that time?
Answer:
t = 8.45 sec
car distance d = 132.09 m
bike distance d = 157.08 m
Explanation:
GIVEN :
motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t
for car
by equation of motion
[tex]d = ut + \frac{1}{2}at^2[/tex]
u = 0 starting from rest
[tex]d = \frac{1}{2}at^2[/tex]
[tex]t^2 = \frac{2d}{a}[/tex]
for bike
[tex]d+25 = 0 + \frac{1}{2}*4.40t^2[/tex]
[tex]t^2= \frac{d+25}{2.20}[/tex]
equating time of both
[tex]\frac{2d}{a} = \frac{d+25}{2.20}[/tex]
solving for d we get
d = 132 m
therefore t is[tex] = \sqrt{\frac{2d}{a}}[/tex]
[tex]t = \sqrt{\frac{2*132}{3.70}}[/tex]
t = 8.45 sec
each travelled in time 8.45 sec as
for car
[tex]d = \frac{1}{2}*3.70 *8.45^2[/tex]
d = 132.09 m
fro bike
[tex]d = \frac{1}{2}*4.40 *8.45^2[/tex]
d = 157.08 m
To find the time when the motorcycle overtakes the car, equate the distances they have traveled given their individual accelerations and solve for time. Once the time is known, calculate the distance each has traveled using the equations of motion for uniform acceleration.
Explanation:To determine when the motorcycle (MC) overtakes the car, we need to calculate the time at which both have traveled the same distance, considering the initial 25.0 m advantage of the car. We can use the equation of motion d = ut + (1/2)at2 where d is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
The car starts from rest, so its initial velocity is 0, and it accelerates at 3.70 m/s2. The motorcycle also starts from rest, with an acceleration of 4.40 m/s2, but it needs to cover an additional 25.0 m to catch up with the car.
We can set the equations equal to each other to find the time t when the distances are equal:
Distance covered by the car: Car_d = (1/2)(3.70 m/s2)t2Distance covered by the MC: MC_d = 25.0 m + (1/2)(4.40 m/s2)t2To find the time when MC overtakes the car, we equate Car_d to MC_d and solve for t:
(1/2)(3.70)t2 = 25.0 + (1/2)(4.40)t2
After solving for t, we can calculate the distance each has traveled using the original equations of motion for uniform acceleration.
Consider an equilateral triangle with sides measuring 1.0 m in length. At each point of the triangle is a +2.0 μC charge. Calculate the Coulomb force on each charge. Recall that forces are vectors and thus your answer will require a magnitude and direction for each of the three forces.
Answer:
Th magnitude of each Force will be [tex]=62.35\times10^{-3}\ \rm N[/tex]
Explanation:
Given:
Length of each side of the equilateral triangle, L=1 mMagnitude of each point charge [tex]Q=2\ \rm \mu C[/tex]Since all the charges are identical and distance between them is same so magnitude of the force between each charge is equal.
Let F be the force between the particles. According to Coulombs Law we have
[tex]F=\dfrac{kQ^2}{L^2}\\=\dfrac{9\times10^9\times (2\times10^{-6})^2}{1^2}\\F=36\times10^{-3}\ \rm N[/tex]
Now the the force on any charge by other two charges will be F and the angle between the two force is [tex]60^\circ[/tex]
Let [tex]F_{resultant}[/tex] be the force on nay charge by other two
By using vector Law of addition we have
[tex]F_{resultant}=\sqrt{(F^2+F^2+2F\times F \times cos60^\circ)}\\=\sqrt{3}F\\=\sqrt{3}\times36\times10^{-3}\ \rm N\\=62.35\times10^{-3}\ \rm N[/tex]
The angle made by the resultant vector will be
[tex]\tan\beta=\dfrac{F\sin60^\circ}{F+F\cos60^\circ}\\\beta=30^\circ[/tex]