Two cars drive on a straight highway. At time t=0, car 1 passes mile marker 0 traveling due east with a speed of 20.0 m/s . At the same time, car 2 is 0.80 km east of mile marker 0 traveling at 26.0 m/s due west. Car 1 is speeding up with an acceleration of magnitude 0.40 m/s^2 , and car 2 is slowing down with an acceleration of magnitude 0.80 m/s^2 . a.) Write x-versus-t equations of motion for both cars, taking east as the positive direction.
b.) At what time do the cars pass next to one another?

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

[tex]D = \sqrt{(d_{m})^2+(d_{w})^2}[/tex]

Where, [tex]d_{m}[/tex] =distance of Machmer Hall

[tex]d_{w}[/tex] =distance of Witless

Put the value into the formula

[tex]D = \sqrt{(400)^2+(180)^2}[/tex]

[tex]D=438.63\ m[/tex]

Hence, The distance from Witless to Machmer is 438.63 m.

Answer:

a) 4.325*10^10 V/m

b) 1.689*10^10 V/m

c) 0

d) 6.384 C/m^3

Explanation:

Hello!

The electric field of a sphere of uniform charge is a piecewise function, let a be the raius of the sphere

For r<a:

  [tex]E = kQ\frac{r}{a^{3}}[/tex]

For r>a:

[tex]E=kQ/r^{2}[/tex]

Since a=0.26m and k= 8.987×10⁹ N·m²/C²

 a)  

[tex]E= 8.987\times10^{9}\frac{0.47C \times0.18m}{(0.26m)^{3}}[/tex]

     E=4.325*10^10 V/m

b)

[tex]E= 8.987\times10^{9}\frac{0.47C}{(0.5m)^{2}}[/tex]

   E=1.689*10^10 V/m

c)

Since r --> ∞         1/r^2 --> 0

 E(∞)=0

d)

The charge density may be obtained dividing the charge by the volume of the sphere:

[tex]\rho = \frac{Q}{V} =\frac{0.47C}{\frac{4}{3} \pi (0.26m)^{3}}=6.384 C/m^{3}[/tex]

Answer:

Explanation:

The motion of the pendulum will be be SHM with time period equal to

T = [tex]2\pi\sqrt{\frac{l}{g} }[/tex]

l = .75 m , g = 9.8

T = [tex]2\pi\sqrt{\frac{.75}{9.8} }[/tex]

T = 1.73 s .

Time to reach the point of maximum velocity or maximum kinetic energy

= T /4

= 1.73 /4

= 0.43 s

For notes on Guitar , The formula is

n = [tex]\frac{1}{2l} \sqrt{\frac{T}{m} }[/tex]

n is fequency of the note , T is tension of string , m is mass per unit length of string , l is length of string.

For fundamental note , l = .648 m and f = 147 Hz

147 = [tex]\frac{1}{2\times.648} \sqrt{\frac{T}{m} }[/tex]

For G note

110 = [tex]\frac{1}{2\times l} \sqrt{\frac{T}{m} }[/tex]

[tex]\frac{147}{110} =\frac{l}{.648}[/tex]

l = 866 mm

Distance from the end of string

866 - 648 = 218 mm or 245 mm

Option c ) is correct .

Answer:

Explanation:

The electric Force and gravitational Force are the same:

[tex]m*g=E*q[/tex]

[tex]E=m*g/q=1.0*10^{-2}*9.81/(2.9*10^{-6})=3.38*10^{4}N/C[/tex]

If the Electric field is doubled, the Electric Force is doubled, so its new value is twice the weight of the object. If we add this new electric force (upwards), with the weight (downwards), we have a resulting force upwards = one time the weight. in conclusion the acceleration will have a value equal to gravity but pointing upwards.

Answer:

Option b

Explanation:

An object is said to fall freely when there is no force acting on the object other than the gravitational force. Thus the acceleration of the object is solely due to gravity and no other acceleration acts on the body.

Also the initial velocity of the body in free fall is zero and hence less than the final velocity.

As the body falls down closer to earth, it experiences more gravitational pull and the velocity increases as it falls down and the moment it touches the ground the period of free fall ends at that instant.

Thus the final velocity of an object in free fall is not zero because the final velocity is the velocity before coming in contact with the ground.

Answer:

Thickness = 103.38 nm

Explanation:

Given that:

The refractive index of the thin soap bubble = 1.33

The wavelength of the light = 550 nm

The minimum thickness that produces a bright fringe can be calculated by using the formula shown below as:

[tex]Thickness=\frac {\lambda}{4\times n}[/tex]

Where, n is the refractive index of the thin soap bubble = 1.33

[tex]{\lambda}[/tex] is the wavelength

So, thickness is:

[tex]Thickness=\frac {550\ nm}{4\times 1.33}[/tex]

Thickness = 103.38 nm

Answer:

u(intial velocty)= 10.688 m/s apprx

Explanation:

given data:

height of apartment = 16 m

time of hiitng = 1.02 s

Using equation of motion we have

h = ut + 0.5gt2 putting all value to get inital velocity value

16 = u(1.02)+ 0.5(9.8)(1.02)^2

SOLVING FOR u

16 - 5.09 = U1.02

u(intial velocty)= 10.688 m/s apprx

Direction will be towards ground

To find the ball's initial velocity, use the equation of motion for free fall and solve for v0. The initial velocity of the ball is approximately 10.78 m/s.

To find the ball's initial velocity, we can use the equation of motion for free fall: d = v0t + 0.5gt2, where d is the distance, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity. In this case, d = 16.0 m, t = 1.02 s, and g = 9.8 m/s2. Plugging in these values, we can solve for v0.

16.0 = v0(1.02) + 0.5(9.8)(1.02)2

Simplifying the equation gives us:

16.0 = 1.02v0 + 5.01

Subtracting 5.01 from both sides gives:

10.99 = 1.02v0

Finally, dividing both sides by 1.02 gives:

v0 ≈ 10.78 m/s

Answer:

8.953 s

Explanation:

Here a time of 8953 ms is given.

Some of the prefixes of the SI units are

mili = 10⁻³

micro = 10⁻⁶

nano = 10⁻⁹

kilo = 10³

The number is 8953.0

Here, the only solution where the number of significant figures is mili

1 milisecond = 1000 second

[tex]1\ second=\frac{1}{1000}\ milisecond[/tex]

[tex]\\\Rightarrow 8953\ milisecond=\frac{8953}{1000}\ second\\ =8.953\ second[/tex]

So 8953 ms = 8.953 s

Answer:

E   = 5291.00 N/C

Explanation:

Expression for capacitance is

[tex]C = \frac{\epsilon  A}{d}[/tex]

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

[tex]C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}[/tex]

[tex]C = 24.06 \epsilon[/tex]

[tex]C = 24.06\times 8.854\times 10^{-12} F[/tex]

[tex]C =2.1\times 10^{-10} F[/tex]

We know that capacitrnce and charge is related as

[tex]V = \frac{Q}{C}[/tex]

 [tex]= \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}[/tex]

v = 9.523 V

Electric field is given as

[tex]E = \frac{V}{d}[/tex]

   = [tex]\frac{9.52}{1.8*10^{-3}}[/tex]

E   = 5291.00 V/m

E   = 5291.00 N/C

Answer:

Part a)

[tex]v = 21.9 m/s[/tex]

Part b)

[tex]y = 4.17 m[/tex]

So he will not able to block the goal

Part c)

[tex]y = 0.98 m[/tex]

yes he can stop the goal

Explanation:

As we know by the equation of trajectory of the ball

[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

[tex]y = 3 m[/tex]

[tex]x = 45.7 m[/tex]

[tex]\theta = 45 degree[/tex]

now from above equation we have

[tex]y = 45.7 tan 45 - \frac{9.81(45.7)^2}{2v^2cos^245}[/tex]

[tex]3 = 45.7 - \frac{20488}{v^2}[/tex]

[tex]v^2 = 479.81[/tex]

[tex]v = 21.9 m/s[/tex]

Part b)

If lineman is 4.6 m from the football

[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

[tex]y = 4.6tan45 - \frac{9.81(4.6^2)}{2(21.9^2)cos^245}[/tex]

[tex]y = 4.17 m[/tex]

So he will not able to block the goal

Part c)

If lineman is 1 m from the football

[tex]y = x tan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

[tex]y = 1tan45 - \frac{9.81(1^2)}{2(21.9^2)cos^245}[/tex]

[tex]y = 0.98 m[/tex]

yes he can stop the goal

Answer:

The tension is 14 N

Explanation:

For this problem we have to use newton's law, so:

[tex]F=m*a[/tex]

The second string is connected to the mass of 8 kg, but the mass of 8 kg is connected to the mass of 20kg, so we can say that the second string is handling the two masses. so:

[tex]F=28kg*0.5\frac{m}{s^2}=14N[/tex]

Answer:

Average force, F = 286.72 N

Explanation:

Given that,

Mass of the baseball, m = 140 g = 0.14 kg

Speed of the ball, v = 32 m/s

Distance, h = 25 cm = 0.25 m

We need to find the average  force exerted by the ball on the glove. It is solved using the conservation of energy as :

[tex]\dfrac{1}{2}mv^2=mgh[/tex]

F = mg

[tex]\dfrac{1}{2}mv^2=Fh[/tex]

[tex]F=\dfrac{mv^2}{2h}[/tex]

[tex]F=\dfrac{0.14\times (32)^2}{2\times 0.25}[/tex]

F = 286.72 N

So, the average force exerted by the ball on the glove is 286.72 N. Hence, this is the required solution.

Answer: Your code returns a number of 99.123456789 +0.00455679

Ok, you must see where the error starts to affect your number.

In this case, is in the third decimal.

So you will write 99.123 +- 0.004 da da da.

But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.

in the error, after the 4 comes a 5, so it rounds up.

So the final presentation will be 99.123 +- 0.005

you are discarding all the other decimals because the error "domains" them.

Answer:

the expected distance is 4.32 m

Explanation:

given data

half life time = 1.8 × [tex]10^{-8}[/tex] s

speed = 0.8 c = 0.8 × 3 × [tex]10^{8}[/tex]

to find out

expected distance over

solution

we know c is speed of light in air is 3 × [tex]10^{8}[/tex] m/s

we calculate expected distance by given formula that is

expected distance = half life time × speed   .........1

put here all these value

expected distance = half life time × speed

expected distance = 1.8 × [tex]10^{-8}[/tex] ×  0.8 × 3 × [tex]10^{8}[/tex]

expected distance = 4.32

so the expected distance is 4.32 m

Answer:

The rock thrown from the top of the cliff.

Explanation:

This is more of a conceptual question. The rock thrown from the top of the cliff will be accelerated downwards, that means, accelerated towards the bird nest, will the rock thrown from the bottom will be accelerated downwards too, but, in this case, this means that it will be accelerated against the direction of the bird nest.

The rock thrown downwards from the top of the cliff will hit the bird nest located at H/2 first, as it is continuously accelerated by gravity, unlike the upward-thrown rock which initially decelerates.

Which Rock Hits the Nest First?

When considering the rocks thrown by the two children - one upwards from the ground and one downwards from the top of the cliff - the rock that hits the bird nest located at H/2 first would be the one thrown downwards. This is based on the principles of kinematics in physics, which describe the motion of objects without considering the forces that cause the motion. The initial velocities of both rocks are the same in magnitude but opposite in direction; however, gravity only decelerates the upward-thrown rock and accelerates the downward-thrown rock, resulting in the latter reaching the nest quicker.

The rock thrown upwards will slow down as it reaches a height of H/2 and has to combat gravity's pull, which is not the case for the downward-throwing rock. Although both rocks are subjected to the same acceleration due to gravity, the downward-thrown rock will cover the distance to the nest faster due to its uninterrupted acceleration. Ignoring air resistance, we don't have to consider any opposing forces which might otherwise affect the time it takes for each rock to reach the nest.

Answer:

a) Reversible processes

b) 1222.73 J/K and 392.144 J/K

c) Because of breaking bonds that are a stable state, while accelerating molecules in the other is less difficult.

Explanation:

Since it is equality, it should be applied to reversible processes, and even better to isothermal processes, since the Temperature remains constant. The T is the absolute temperature (measured in °K) and dQ is the heat absorbed by the system, which DEPENDS on the process.

b) The heat absorbed in the fusion process depends on the latent heat of fusion, L, of the water, which is 334 J/g. It says t kg, but I assume it was a mistake in the typing, so the change in entropy is calculated for 1 kg of water melting as follows:

[tex]\Delta S=\frac{\Delta Q}{T}=\frac{mass*L}{T}=\frac{1kg*334J/g*(1000 g/kg)}{273.16 K} \\\Delta S= 1222.73 J/K[/tex]

Now, to proceed with the change of entropy for water heated from 273.16K to 300K we use the specific heat of water, which is 4184 J/kg°K as follows:

[tex]\Delta S=m_{mass}c_{water}*ln(\frac{T_f}{T_i} )=1kg*4184\frac{J}{Kg*K}*ln(\frac{300}{273.16})\\ \Delta S=392.144 J/K[/tex]

c) In the solid state, water molecules have different bonds with other water molecules creating the crystals. In the liquid state, each molecule moves freely with less interaction between molecules. So, it required more energy to break these bonds and alter this ordered state than just accelerating the molecules in the liquid state.

Answer:

Magnitude of the resultant force (Fn₁) on q₁

Fn₁ = 0.142N (directed toward the center of the square)

Explanation:

Theory of electrical forces

Because the particle q₁ is close to three other electrically charged particles, it will experience three electrical forces and the solution of the problem is of a vector nature.

Graphic attached

The directions of the individual forces exerted by q₂, q₃ and q₄ on q₁ are shown in the attached figure.

The force (F₁₄) of q₄ on q₁ is repulsive because the charges have equal signs and the forces (F₁₂) and (F₁₃) of q₂ and q₃ on q₁ are attractive because the charges have opposite signs.

Calculation of the forces exerted on the charge q₁

To calculate the magnitudes of the forces exerted by the charges q₂, q₃, and q₄ on the charge q₁ we apply Coulomb's law:

[tex]F_{12} = \frac{k*q_1*q_2}{r_{12}^2}[/tex]: Magnitude of the electrical force of q₂ over q₁. Equation((1)

[tex]F_{13} = \frac{k*q_1*q_3}{r_{13}^2}[/tex]: Magnitude of the electrical force of q₃ over q₁. Equation (2)

[tex]F_{14} = \frac{k*q_1*q_4}{r_{14}^2}[/tex]: Magnitude of the electrical force of q₄ over q₁. Equation (3)

Equivalences

1µC= 10⁻⁶ C

Known data

q₁=q₄= 1.96 µC = 1.96*10⁻⁶C

q₂=q₃= -1.96 µC = -1.96*10⁻⁶C

r₁₂= r₁₃ = 0.47m: distance between q₁ and q₂ and q₁ and q₄

[tex]r_{14} = \sqrt{0.47^2+ 0.47^2}=0.664m[/tex]

k=8.99x10⁹N*m²/C² : Coulomb constant

F₁₂ calculation

We replace data in the equation (1):

[tex]F_{12} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}[/tex]

F₁₂ = 0.156 N Direction of the positive x axis (+x)

F₁₃ calculation

We replace data in the equation (2):

[tex]F_{13} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}[/tex]

F₁₃ = 0.156 N Direction of the negative y axis (-y)

Magnitude of the net electrostatic force between F₁₃ and F₁₂

[tex]F_{n23}= \sqrt{0.156^2+0.156^2} = 0.22N[/tex] (directed toward the center of the square)

F₁₄ calculation

We replace F₁₄ data in the equation (3):

[tex]F_{14} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.664^2}[/tex]

F₁₄ = 0.078 N (In the opposite direction to Fn₂₃)  

Calculation of the resulting force on q₁: Fn₁

Fn₁ = Fn₂₃ - F₁₄ = 0.22 - 0.078 = 0.142 N

Final answer:

The net electrostatic force on a charge can be found using Coulomb's law, considering the effects of both attraction and repulsion due to the charges' arrangement at the corners of a square. Calculations account for diagonal opposites and adjacent corners, taking advantage of the square's symmetry.

Explanation:

To calculate the magnitude of the net electrostatic force on a charge in a configuration where two positive charges of 1.96 µC and two negative charges of the same magnitude are placed at the corners of a 0.47-m square, we can use Coulomb's law. Coulomb's law states that the electrostatic force (F) between two point charges is proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. It is given by F = k * |q1*q2| / r², where k is Coulomb's constant (k ≈ 8.99 x 10⁹ N·m²/C²).

Since the charges are at the corners of a square, the opposing charges will attract, and like charges will repel. Each charge experiences forces due to its interaction with the other three charges. The net force on any charge is the vector sum of the forces exerted by the other three charges. This sum can be simplified as the square has symmetry so that the forces along one diagonal cancel and the forces along the sides of the square add together to pull the charge towards the center.

The diagonal distance (r) is √2 times the side of the square (a), which is r = √2 * 0.47 m. For simplicity, let's calculate the force between one charge and its diagonal opposite, and then double it for symmetry reasons (each charge has two diagonally opposite charges).

The force due to one diagonal is: F_diagonal = (k * (1.96 x 10⁻⁶)^2) / ((√2 * 0.47)²)
Now we need to calculate this force and then double it to account for both diagonals.

The final step will be to add the forces due to charges on adjacent corners, which are of the same polarity and hence exert repulsion. Their vector sum will be directed towards the square's center due to symmetry.

Answer:

17.72° or 72.28°

Explanation:

u = 6.5 m/s

R = 2.5 m

Let the angle of projection is θ.

Use the formula for the horizontal range

[tex]R=\frac{u^{2}Sin2\theta }{g}[/tex]

[tex]2.5=\frac{6.5^{2}Sin2\theta }{9.8}[/tex]

Sin 2θ = 0.58

2θ = 35.5°

θ = 17.72°

As we know that the range is same for the two angles which are complementary to each other.

So, the other angle is 90° - 17.72° = 72.28°

Thus, the two angles of projection are 17.72° or 72.28°.

Answer:

The volume of the slab is 2527.84 cm

Explanation:

Only the first two decimals are significant digits because so are the data that the problem gives us.

Answer:

The velocity of car is 11.71 m/s.

Explanation:

Given that,

Mass of car = 2000 kg

Radius = 20 m

Coefficient of friction = 0.70

We need to calculate the velocity of car

Using relation centripetal force and frictional force

[tex]F= \dfrac{mv^2}{r}[/tex]...(I)

[tex]F=\mu mg[/tex]...(II)

From equation (I) and (II)

[tex]\dfrac{mv^2}{r}=\mu mg[/tex]

[tex]v=\sqrt{\mu\times r\times g}[/tex]

Put the value into the formula

[tex]v=\sqrt{0.70\times20\times9.8}[/tex]

[tex]v=11.71\ m/s[/tex]

Hence, The velocity of car is 11.71 m/s.

Answer:

car can move at  11 m/s without skidding.

Explanation:

given,

mass of car = 2000 kg

radius of turn = 20 m

μ = 0.7

using centripetal force

F = [tex]\dfrac{mV^2}{R}[/tex].....................(1)

and we know                                

F = μ N  = μ m g........................(2)

equating both the equation (1) and (2)                        

μ m g = [tex]\dfrac{mV^2}{R}[/tex]

v = [tex]\sqrt{\mu R g}[/tex]                                

v = [tex]\sqrt{0.7 \times 20 \times 9.81}[/tex]

v = 11.71 m/s                                            

hence, car can move at  11 m/s without skidding.

Answer:[tex]\theta =32.08 ^{\circ}[/tex]

Explanation:

Given

Ball launcher is mounted on car at angle of [tex]39^{\circ}[/tex]

launching velocity is u=9.7 m/s

Speed of car =2.2 m/s

So in horizontal component of balloon speed of car will be added

Thus [tex]u_x=9.7cos39+2.2=7.53+2.2=9.73 m/s[/tex]

[tex]u_y=9.7sin39=6.10 m/s[/tex]

therefore Appeared trajectory angle is

[tex]tan\theta =\frac{6.10}{9.73}=0.627[/tex]

[tex]\theta =32.08 ^{\circ}[/tex]

Final answer:

The force that the second child must exert to keep the door from moving is 10.5 N.

Explanation:

To find the force that the second child must exert to keep the door from moving, we need to consider the torque applied by each child.

First, we calculate the torque applied by the first child:

T1 = F1 * r1 * sin(θ)

T1 = (17.5 N) * (0.600 m) * sin(90°) = 10.5 N·m

The torque applied by the second child is:

T2 = F2 * r2

T2 = F2 * (0.450 m)

Since the door is not moving, the net torque must be zero. Therefore:

T1 + T2 = 0

Substituting the values:

10.5 N·m + F2 * (0.450 m) = 0

Simplifying, we find that F2 = -10.5 N * (0.450 m) / (0.450 m) = -10.5 N

Therefore, the force that the second child must exert to keep the door from moving is 10.5 N.

Answer:

[tex] \Delta V=V_{2}-V_{1}=45.4V[/tex]

Explanation:

The energy, E, from a capacitor, with capacitance, C, and voltage V is:

[tex]E=\frac{1}{2} CV^{2}[/tex]

[tex]V=\sqrt{2E/C}[/tex]

If we increase the Voltage, the Energy increase also:

[tex]V_{1}=\sqrt{2E_{1}/C}[/tex]

[tex]V_{2}=\sqrt{2E_{2}/C}[/tex]

The voltage difference:

[tex]V_{2}-V_{1}=\sqrt{2E_{2}/C}-\sqrt{2E_{1}/C}[/tex]

[tex]V_{2}-V_{1}=\sqrt{2*55*10^{-6}/4.55*10^{-9}}-\sqrt{2*27.5*10^{-6}/4.55*10{-9}}=45.4V[/tex]

Answer:

Ball A

Explanation:

Let the initial speed of the balls be u .

Angle of projection for ball A = 20°

Angle of projection for ball B = 75°

As we know that at highest point, the ball has only horizontal speed which always remains constant throughout the motion because the acceleration in horizontal direction is zero.

Speed of ball A at highest point = u Cos 20° = 0.94 u

Speed of ball B at highest point = u Cos 75° = 0.26 u

So, the ball A has bigger speed than B.

Final answer:

To find the total increase in the length of each day over 43 centuries, we use the sum of an arithmetic series: S = n/2(2a1 + (n-1)d). With 4300 days in 43 centuries, an initial increase of 0 ms, and a daily increase of 0.002 ms, the total increase is 18471.9 milliseconds.

Explanation:

The length of the day increases because Earth's rotation is gradually slowing, primarily due to the friction of the tides. To calculate the total increase in the length of each day over 43 centuries, given that the length of a day at the end of a century is 1 ms longer than the day at the start of the century, we use the arithmetic progression formula where the total sum (S) is equal to n/2 times the first term (a1) plus the last term (an). In this case, the difference between each day (common difference, d) is constant at 0.002 ms.

The first term (a1) is 0 ms, as there is no increase on the first day of the first century, and the last term (an) after 43 centuries will be 43 ms, since we're told that each century contributes an additional 1 ms to the length of a day. So, for 43 centuries, we have n=4300 days (number of days in 43 centuries), a1=0 ms, an=43 ms, and d=0.002 ms/day.

Using the formula for the sum of an arithmetic series, S = n/2(2a1 + (n-1)d), we get the following:

S = 4300/2 [2(0) + (4300-1)(0.002 ms)]

S = 2150 [0 + 4299(0.002 ms)]

S = 2150 x 8.598 ms

S = 18471.9 ms

Therefore, over 43 centuries, the total increase in the length of each day adds up to 18471.9 milliseconds.

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.

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:

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.

Answer:

Speed= 6cm/s and velocity= 6cm/s in the negative direction

Explanation:

the change in position is from 45cm to 27 cm (moving towards the negative x direction)

[tex] \Delta x = 45 cm - 27 cm = 18 cm[/tex]

And the change in time:

[tex] \Delta t= 3 s[/tex]

Now we must define the difference between speed and velocity:

Speed is a scalar quantity, which means that it is a number. Velocity ​​is also a number but you must also indicate the direction of the movement.

Thus, the speed is:

[tex] speed= \Delta x/ \Delta t = 18cm/3s=6cm/s[/tex]

An the velocity is:

6cm/s in the negative direction

Answer:

a) 215 cm

b) 1290 mm

c) 2.81 m³

d) 265 cm

e) 1590 mm

f) 5.26 m³

Explanation:

Lower value

1 cubit = 43 cm

Length of cylinder = 5 cubits

So,

a) Length of cylinder = 5×43 = 215 cm

Diameter of cylinder = 3 cubit

1 cubit = 43 cm = 430 mm

b) Diameter of cylinder = 3×430 = 1290 mm

Radius of cylinder = 129/2 = 64.5 cm

Volume of cylinder

[tex]v=\pi r^2h\\\Rightarrow v=\pi 0.645^2\times 2.15\\\Rightarrow v=2.81 m^3[/tex]

c) Volume of cylinder = 2.81 m³

Upper value = 53 cm

Length of cylinder = 5 cubits

So,

d) Length of cylinder = 5×53 = 265 cm

Diameter of cylinder = 3 cubit

1 cubit = 53 cm = 530 mm

e) Diameter of cylinder = 3×530 = 1590 mm

Radius of cylinder = 159/2 = 79.5 cm

Volume of cylinder

[tex]v=\pi r^2h\\\Rightarrow v=\pi 0.795^2\times 2.65\\\Rightarrow v=5.26 m^3[/tex]

f) Volume of cylinder = 5.26 m³

Answer:

The maximum height that the rocket reaches is 645.5 m.

Explanation:

Given that,

Mass = 10000 kg

Acceleration = 2.25 m/s²

Distance = 525 m

We need to calculate the velocity

Using equation of motion

[tex]v^2=u^2+2as[/tex]

Put the value in the equation

[tex]v^2=0+2\time2.25\times525[/tex]

[tex]v=\sqrt{2\times2.25\times525}[/tex]

[tex]v=48.60\ m/s[/tex]

We need to calculate the maximum height with initial velocity

Using equation of motion

[tex]v^2=u^2-2gh[/tex]

[tex]h=\dfrac{v^2-u^2}{-2g}[/tex]

Put the value in the equation

[tex]h=\dfrac{0-(48.60)^2}{-2\times9.8}[/tex]

[tex]h=120.50\ m[/tex]

The total height  reached by the rocket is

[tex]h'=s+h[/tex]

[tex]h'=525+120.50[/tex]

[tex]h'=645.5\ m[/tex]

Hence, The maximum height that the rocket reaches is 645.5 m.

Answer:

the force acting on the team mate is 1.19 kN.

Explanation:

given,

mass = 196 lbm

while tackling, the deceleration is from velocity 6.7 m/s to 0 m/s

time taken for deceleration = 0.5 sec        

F = mass × acceleration

acceleration = [tex]\dfrac{0-6.7}{0.5}[/tex]              

                     = -13.4 m/s²                            

1 lbs  = 0.453 kg                      

196 lbs = 196 × 0.453  = 88.79 kg

F = 88.79 × 13.4                              

F = 1189.786 N = 1.19 kN                      

hence, the force acting on the team mate is 1.19 kN.