Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.9 m/s^2 while Kathy maintains an acceleration of 4.73 m/s^2. (a) Find the time at which Kathy overtakes (b) Find the distance she travels before she catches him (c) Find the speeds of both cars at the instant she overtakes him.

Answer:

acceleration is 1.59 m/s²

required % is 16.22 %

Explanation:

given data

radius = 63 m

speed = 10 m/s

to find out

acceleration of the car and acceleration due to gravity

solution

we know car is moving in circular path

so acceleration will be, a = [tex]\frac{v^2}{r}[/tex]   .........1

put here value

a =  [tex]\frac{10^2}{63}[/tex]

a = 1.59 m/s²

so acceleration is 1.59 m/s²

so acceleration due to gravity as a percentage is 9.8 m/s²

required % = [tex]\frac{1.59}{9.8}*100[/tex]

required %  = 16.22 %

Answer:

Thickness = 123.19 nm

Explanation:

Given that:

The refractive index of the glass = 1.50

The refractive index of thin layer of magnesium fluoride = 1.38

The wavelength of the light = 680 nm

The thickness 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 thin layer of magnesium fluoride = 1.38

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

So, thickness is:

[tex]Thickness=\frac {680\ nm}{4\times 1.38}[/tex]

Thickness = 123.19 nm

Answer:

a) 231 yards

b) 55.2 yards

c) 0 yards to 260 yards

d) Infinite values

Explanation:

This situation can be described as a horizontal line that begins at point [tex]P_{1}=0 yards[/tex] (Meredith's house) and ends at point [tex]P_{2}=260 yards[/tex] (Bus stop). Where [tex]x[/tex] is the varying distance from her house, which can be calculated in the following way:

x=Final Position - Initial Position

or

[tex]x=x_{f} - x_{i}[/tex]

a) For the first case Meredith is at position [tex]x_{i}=29 y[/tex] and the bus stop at position [tex]x_{f}=260 y[/tex]. So the distance Meredith is from the bus stop is:

[tex]x=260 y - 29 y=231 y[/tex]

b) For the second case the initial position is [tex]x_{i}=204.8 y[/tex] and the final position [tex]x_{f}=260 y[/tex]. Hence:

[tex]x=260 y - 204.8  y=55.2  y[/tex]

c) If we take Meredith's initial position at her house  [tex]x_{i}=0 y[/tex] and her final position at the bus stop  [tex]x_{f}=260 y[/tex], the value of  [tex]x[/tex] varies from 0 yards to 260 yards.

d) As Meredith walks from her house to the bus stop, the variable [tex]x[/tex] assumes infinite values, since there are infinite position numbers from [tex]x=0 yards[/tex] to [tex]x=260 yards[/tex]

The answers to the possible distance covered by Meredith at the various distances from her house are;

A) distance = 231 yards

A) distance = 231 yardsB) distance = 55.2 yards

A) distance = 231 yardsB) distance = 55.2 yardsC) x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260

A) We are told that meredith walks from her house to a bus stop that is 260 yards away.

After walking, she is now 29 yards from her house. This means that she has walked a total of 29 yards from her house.

Distance left to reach bus stop = 260 - 29 = 231 yards

B) We are told that Meredith is now 204.8 yards from her house. This means that she has walked a total of 204.8 yards from here house. Thus;

Distance left to reach bus stop = 260 - 204.8 = 55.2 yards.

C) This question is basically asking for all the possible values that Meredith could have walked from her house to the bus stop.

Since she starts from her house at 0m, then it means that if the bus stop is 260 m away, then if x is the possible distance, we can say that x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260

Read more at; https://brainly.com/question/13242055

Answer:

The total charge Q of the sphere is [tex]2.094\times10^{-10}\ C[/tex].

Explanation:

Given that,

Radius = 5 cm

Charge density [tex]J= 400\ nC/m^3[/tex]

We need to calculate the total charge Q of the sphere

Using formula of charge

[tex]q=\rho V[/tex]

Where, [tex]\rho[/tex] = charge density

V = volume

Put the value into the formula

[tex]q=\rho\times(\dfrac{4}{3}\pi r^3)[/tex]

Put the value into the formula

[tex]q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3[/tex]

[tex]q=2.094\times10^{-10}\ C[/tex]

Hence, The total charge Q of the sphere is [tex]2.094\times10^{-10}\ C[/tex].

Answer:

Explanation:

Resistance in series is given by the sum of all the resistor in series

value of Total Resistance is given by

[tex]R_{th}=R_1+R_2+R_3+R_4+..............R_n[/tex]

Where [tex]R_{th}[/tex] is the total resistance

[tex]R_1,R_2[/tex] are the resistance in series

Current in series remains same while potential drop is different for different resistor

The value of net resistor is always greater than the value of individual resistor.

If a there is a defect in a single resistor then it affects the whole circuit in series.

Answer:

(C) The frequency decrease and intensity decrease

Explanation:

The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.

if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.

The frequency is the inverse from the wavelength, so the frequency heard will increase.

The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.

Answer:

a)  1.28 *10^5 N/C

b)2.05 *10^{-14} N

c) 4.83 *10^{-17} J

Explanation:

Given Data:

Distance between the plates, d = 4.67 mm

[tex]= (4.67) *10^{-3} m[/tex]

[/tex]= 4.67 *10^{-3} m[/tex]

Potential difference, V = 600 V

Solution:

(a) The  magnitude of the electric field between the plates is,

    [tex]E = \frac{V}{d}[/tex]  

[tex]= \frac{600 V}{4.67 *10^{-3}} m[/tex]

  [tex]= 1.28 *10^5 V/m or 1.28 *10^5N/C[/tex]  

(b) Force on electron btwn the plates is,

   F = q E

 [tex]= (1.6 *10^{-19} C) (1.28 *10^5N/C[/tex]

 [tex]= 2.05 *10^{-14} N[/tex]  

(c) Work done on the electron is

   W = F * s

 [tex]= (2.05 *10^{-14} N) * (5.31 *10^{-3} m - 2.95 *10^{-3} m)[/tex]

 [tex]= 4.83 *10^{-17} J[/tex]

Answer and Explanation:

The gravitational acceleration 'g' depend directly on the mass of the object or body and inversely on the distance or radius squared:

[tex]g = \frac{Gm_{o}}{R^{2}}[/tex]

where

[tex]m_{o}[/tex] = mass of the object

G = Gravitational constt

Thus the plastic ball is lighter and have low mass as compared to the steel and golf balls.

This is the reason that a plastic ball have a low value of acceleration as compared to that of steel and golf balls with higher values of acceleration.

Final answer:

The acceleration due to gravity is a constant (g) for all objects in the same gravitational field, and any observed difference in falling rates is likely due to air resistance, not gravitational pull. Experiments have confirmed that objects fall at the same rate regardless of their mass or composition, assuming no air resistance.

Explanation:

The acceleration due to gravity should not vary with the material of the object if we ignore effects such as air resistance. This is because, according to Newton's second law, the force acting on an object is the product of its mass and the acceleration (F = ma). Here, the force due to gravity is mg, where m is the mass and g is the acceleration due to gravity. Since a = F/m, the mass cancels out, leaving a = g, which is constant for all objects in the same gravitational field.

The question assumes that the plastic ball has a lower acceleration due to gravity, which contradicts known physics principles. All objects, regardless of their mass or composition, feel the same acceleration due to gravity near the Earth's surface, assuming no other forces, like air resistance, play a significant role. Historical experiments by scientists such as Galileo and Eötvös have confirmed the equality of gravitational acceleration (g) for different substances within exceptionally high precision.

If you observe differing acceleration rates, this can be often attributed to air resistance, not a difference in gravitational pull. Objects with a larger surface area or less aerodynamic shape, like the plastic ball, may experience greater air resistance and thus appear to fall slower, even though their acceleration due to gravity is the same as denser objects like the steel or golf ball.

Answer:

1.708*10^6 tons.

Explanation:

Number of Aluminum cans used by 1 person in 1 year = 365/2=182.5 say it as 183 cans per year.

Total number of people in US= 280,000,000

Total number of cans used by americans.

[tex] = 5.12×10^10 cans[/tex]

Weight of 1 can =1/15 pounds

Weight of all cans used in 1 year

[tex]= \frac{5.12*10^10}{15} =3.41*10^9pounds.[/tex]

we know that

1ton=2000pounds.

[tex]\frac{3.41*10^9 pounds}{2000} = 1.708*10^6 tons.[/tex]

The maximum potential is [tex]6.45 \times 10^5\ V[/tex] and the maximum charge is [tex]71 \mu C[/tex].

The electric field is the gradient of the potential. So, the electric field in terms of the potential is given as:

[tex]V=E \times r[/tex]

Here, V is the potential, E is the electric field and r is the distance.

Given:

Diameter of the dome, [tex]d=43\ cm[/tex]

Maximum electric field, [tex]E=3.00 \times10^6\ V/m[/tex]

(a) The maximum potential is computed as:

[tex]V_{max}={E}{d/2}\\V_{max}= {3.00 \times 10^6} \times \frac{0.43}{2}\\V_{max}= 6.45 \times 10^5\ V[/tex]

(b) The maximum charge is computed as:

[tex]Q= \frac{Er}{k}\\Q= \frac{3.00 \times 10^6 \times 0.215}{9 \times 10^9}\\Q=71 \mu C[/tex]

Therefore, the maximum potential is [tex]6.45 \times 10^5\ V[/tex] and the maximum charge is [tex]71 \mu C[/tex].

To know more about electric potential, click here:

https://brainly.com/question/28444459

#SPJ12

The maximum potential of the Van de Graaff generator with a diameter of 43.0 cm is 645 kV, and the maximum charge it can hold, before sparks occur due to air breakdown, is 5.11 μC.

The maximum potential of a Van de Graaff generator can be calculated using the formula V = (kQ)/r, where V is the potential, k is Coulomb's constant, Q is the charge, and r is the radius of the sphere. Given that the Van de Graaff generator has a diameter of 43.0 cm (which means the radius r is 21.5 cm or 0.215 m) and it is surrounded by air with a breakdown electric field of 3.00 x 106 V/m, the maximum surface potential (V) equals the breakdown electric field (E) times the radius (r), which gives V = E × r = 3.00 x 106 V/m × 0.215 m = 645,000 V or 645 kV.

The maximum charge on the dome can be calculated using the relationship between the electric field and charge on a sphere, which is E = (kQ)/(r2). Re-arranging this for Q gives Q = Er2/k. Inserting the known values, we get Q = (3.00 x 106 V/m) × (0.215 m)2 / (8.988 x 109 N·m2/C2) = 5.11 x 10−5 C or 5.11 μC, where k is Coulomb's constant.

Answer:

1.  1.568 x 10^6 N m^2 / C

2. -  1.568 x 10^6 N m^2 / C

Explanation:

E = 4.9 x 10^4 N/C

Side of square, a = 8 m

Area, A = side x side = 8 x 8 = 64 m^2

Angle between lane of sheet and electric field = 30°

Angle between the normal of plane of sheet and electric field,

θ = 90°- 30° = 60°

The formula for the electric flux is given by

[tex]\phi = E A Cos\theta[/tex]

(1) [tex]\phi = E A Cos\theta[/tex]

By substituting the values, we get

Ф = 4.9 x 10^4 x 64 x Cos 60 = 1.568 x 10^6 N m^2 / C

(2) [tex]\phi = E A Cos\theta[/tex]

By substituting the values, we get

Ф = - 4.9 x 10^4 x 64 x Cos 60 = - 1.568 x 10^6 N m^2 / C

The problem requires vector operations in two dimensions. Displacement is broken down into x and y-components which follow the east-west and north-south directions respectively. Total displacement being zero means the sum of the x and y components of displacement will also be zero. The fourth displacement is determined by negating the total x and y components of the first three displacements. The magnitude and direction are then obtained using Pythagorean theorem and arctan function respectively.

To solve this problem, we need to deal with the changes in displacement in terms of vector operations. Given the different directions, we need to break down the vectors into their x and y components, where x represents east-west direction, and y north-south direction. Since our spelunker starts and ends in the same place, the sum of the displacements in each dimension will also be zero.

For displacement 1, moving 180m west, the x component would be -180, and y component would be 0. For displacement 2, moving 230m in a direction 45° east of south, the x would be -230×sin(45) and y would be -230×cos(45). For displacement 3, moving 280m at 30° east of north, the x would be 280×cos(30) and y would be 280×sin(30).

To determine the fourth displacement, we sum up the x and y components for displacement 1,2 and 3 and then negate them to get the x and y component of the fourth displacement. We then use the Pythagorean theorem to calculate the magnitude of the 4th displacement which is square root of (sum x² + sum y²). The direction can be obtained by calculating the arctan of the total y component / total x component.

https://brainly.com/question/20047824

#SPJ12

Answer:

16.17 m/s

Explanation:

h = 3.2 m

u = 18.1 m/s

Angle of projection, θ = 49°

Let H be the maximum height reached by the ball.

The formula for the maximum height is given by

[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]

[tex]H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m[/tex]

The vertical distance fall down by the ball, h'  H - h = 9.52 - 3.2 = 6.32 m

Let v be the velocity of ball with which it strikes the ground.

Use third equation of motion for vertical direction

[tex]v_{y}^{2}=u_{y}^{2}+2gh'[/tex]

here, uy = 0

So,

[tex]v_{y}^{2}=2\times 9.8 \times 6.32[/tex]

vy = 11.13 m/s

vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s

The resultant velocity is given by

[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]

[tex]v=\sqrt{11.87^{2}+11.13^{2}}[/tex]

v = 16.27 m/s

Answer:

the critical angle of the flint glass is 37.04⁰

Explanation:

to calculate the critical angle for total internal reflection.

given,

wavelength of the flint glass =  λ  = 580.0 nm

                                                       = 580 × 10⁻⁹ m        

critical angle  = sin^{-1}(\dfrac{\mu_a}{\mu_g})

at the wavelength of 580.0 nm the refractive index of the glass is 1.66

refractive index of air = 1                        

critical angle  = sin^{-1}(\dfrac{1}{1.66})

                      = 37.04⁰              

hence, the critical angle of the flint glass is 37.04⁰

Answer: The average speed is 27,24 mph (exactly 1008/37 mph)

Explanation:

This is solved using a three rule: We know the speeds and the distances, what we can obtain from it is the time used. It is done like this:

1h--->18mi

X ---->20 mi, then X=20mi*1h/18mi= 10/9 h=1,111 h

1h--->56mi

X ---->20 mi, then X=20mi*1h/56mi= 5/14 h=0,35714 h

Then the average speed is calculated by taking into account that it was traveled 40mi and the time used was 185/126 h=1,468 h and since speed is distance over time we get the answer. Average speed= 40mi/(185/126 h)=1008/37 mph=27,24 mph.

Answer:

The flux through one face of a cube is  [tex]- 5.2730\times 10^{4} Wb[/tex]

Given:

Charge, Q = [tex]- 2.80\micro C = -2.80\times 10^{- 6} C[/tex]

Solution:

The electric flux, [tex]\phi_{E} = \frac{Q}{\epsilon_{o}}[/tex]

Since, there are 6 faces in a cube, the flux through one face:

[tex]\phi_{E} = \frac{Q}{6\epsilon_{o}}[/tex]

[tex]\phi_{E} = \frac{-2.80\times 10^{- 6}}{6\epsilon_{o}}[/tex]

[tex]\phi_{E} = - 5.2730\times 10^{4} Wb[/tex]

Answer:

a) t = 3.01s

b) 15th floor

Explanation:

First we need to know the distance the elevator has descended before the bolt fell.

[tex]\Delta Y_{e} = -V_{e}*t = -5.7 * 4.95 = -28.215m[/tex]

Now we can calculate the time that passed before both elevator and bolt had the same position:

[tex]Y_{b}=Y_{e}[/tex]

[tex]Y_{ob}+V_{ob}*t-g*\frac{t^{2}}{2} = Y_{oe} - V_{e}*t[/tex]

[tex]0+0-5*t^{2} = -28.215 - 5.7*t[/tex]   Solving for t:

t1 = -1.87s    t2 = 3.01s

In order to know how the amount of floors, we need the distance the bolt has fallen:

[tex]Y_{b}=-g*\frac{t^{2}}{2}=-45.3m[/tex]  Since every floor is 3m:

Floors = Yb / 3 = 15 floors.

Answer: Kilogram- newton is wrong unit for Torque

Idem pound-inch is also wrong unit for Torque

Explanation: As it is well known the torque is defined as :

Τorque= F x R

so its UNITS are Newton*meter (SI)

or in the Imperial System is often use  Pound-foot

Answer:

The snail travel at the end of 5 s with a velocity of 12 m/s and the distance of the snail is 22.5 m.

Explanation:

Given that, the initial velocity of the snail is,

[tex]u=2m/s[/tex]

And the acceleration of the snail is,

[tex]a=1m/s^{2}[/tex]

And the time taken by the snail is,

[tex]t=5 sec[/tex]

Now according to first equation of motion,

[tex]v=u+at[/tex]

Here, u is the initial velocity, t is the time, v is the final velocity and a is the acceleration.

Now substitute all the variables

[tex]v=2m/s+ 1 \times 5 sec\\v=7m/s[/tex]

Therefore, the snail travel at the end of 5 s with a velocity of 7 m/s.

Now according to third equation of motion.

[tex]v^{2}- u^{2}=2as\\ s=\frac{v^{2}- u^{2}}{2a} \\[/tex]

Here, u is the initial velocity, a is the acceleration, s is the displacement, v is the final velocity.

Substitute all the variables in above equation.

[tex]s=\dfrac{7^{2}- 2^{2}}{2(1)}\\s=\dfrac{45}{2}\\ s=22.5m[/tex]

Therefore the distance of the snail is 22.5 m.

Answer:

[tex]t=2.8h[/tex]

[tex]t=10080s[/tex]

[tex]t=168 min[/tex]

Explanation:

From this exercise we have velocity and distance. Using the following formula, we can calculate time:

[tex]v=\frac{d}{t}[/tex]

Solving for t

[tex]t=\frac{d}{v}=\frac{26.22mi}{9.39mi/h} =2.8h[/tex]

[tex]t=2.8h*\frac{3600s}{1h} =10080s[/tex]

[tex]t=2.8h*\frac{60min}{1h} =168min[/tex]

Answer:

width of fringes are increased

Explanation:

The width of central maxima is given by the following expression

Width = 2 x Dλ / d

Due to increased width ,  position of a fringe  moves away from the centre.

Answer:

plane speed: 525mph, jetstream speed=75mph, in explanation it is solved with a linear equations system

Explanation:

First lets name each speed

vs:=speed of the jetstream

vp:=speed of the plane

Now when in the jetstream direction the speeds are added and on the opposite direction are subtracted, then we get these equations, that are linear.

1800 mi=(vp+vs)*3h

1800 mi=(vp-vs)*4h

which is a linear equation system equivalent to:

600 mph=vp+vs (1)

450 mph=vp-vs  (2)

Now from (2) vp= 450mph+vs (3), replacing this in (1) we get:

600mph=(450mph+vs)+vs=450mph+2*vs, then 2*vs=150mph or vs=*75mph, this is the jetstream speed, replacing this in (3) we get the plane speed too vp=450 mph +75mph = 525 mph

Explanation:

Given that,

Initial sped of the runner, u = 0

Final speed of the runner, v = 30 ft/s

Acceleration of the runner, [tex]a=8\ ft/s^2[/tex]

Let t is the time taken by the runner. It can be calculated using first equation of motion as :

[tex]t=\dfrac{v-u}{a}[/tex]

[tex]t=\dfrac{30-0}{8}[/tex]

t = 3.75 seconds

Let s is the distance covered by the runner. Using the second equation of motion as :

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

[tex]s=\dfrac{1}{2}\times 8\times (3.75)^2[/tex]

s = 56.25 feet

Hence, this is the required solution.

Final answer:

To reach a maximum speed of 30 f/s from rest with an acceleration of 8 f/s², it takes 3.75 seconds. During this time, the runner travels a distance of 56.25 feet.

Explanation:

The question involves calculating the time it takes for a runner to reach a maximum speed of 30 feet per second (f/s) from rest with an acceleration of 8 feet per second squared (f/s²), and then finding the distance traveled during this time. This can be solved using the basic kinematics equations.

Calculating Time to Reach Max Speed

To find the time, we use the equation v = at, where v is the final velocity (30 f/s), a is the acceleration (8 f/s²), and t is the time. Rearranging the equation to solve for t, we get t = v/a. Plugging in the values, t = 30 f/s / 8 f/s² = 3.75 seconds.

Calculating Distance Traveled

To find the distance traveled, we use the equation d = 0.5 * a * t², where d is the distance, a is the acceleration, and t is the time. Substituting the given values, d = 0.5 * 8 f/s² * (3.75 s)² = 56.25 feet.

Answer:[tex]3.874 m/s^2[/tex]

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

[tex]60mi/h \approx 26.8224m/s[/tex]

[tex]34mi/h \approx 15.1994 m/s[/tex]

we know acceleration is given by [tex]=\frac{velocity}{Time}[/tex]

[tex]a=\frac{15.1994-26.8224}{3}[/tex]

[tex]a=-3.874 m/s^2[/tex]

negative indicates that it is stopping the car

Distance traveled

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

[tex]\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s[/tex]

[tex]s=\frac{488.419}{2\times 3.874}[/tex]

s=63.038 m

Answer:

(a): [tex]\rm -5.627\times 10^3\ N.[/tex]

(b):  [tex]\rm 7.626\times 10^2\ N.[/tex]

Explanation:

Given:

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

[tex]\rm F=k\cdot\dfrac{q_1q_2}{r^2}[/tex]

where,

k is called the Coulomb's constant, whose value is [tex]\rm 9\times 10^9\ Nm^2/C^2.[/tex]

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

[tex]\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.[/tex]

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres, [tex]\rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.[/tex]

The new charges on both the spheres are equal and given by

[tex]\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.[/tex]

The magnitude of the force that each sphere now experiences is given by

[tex]\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.[/tex]

Answer:

a)11.6m

b)45.55s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

a)

for this problem

Vo=0

Vf=319m/min=5.3m/s

a=1.2m/s^2

we can use the ecuation number 1 to calculate the time

t=(Vf-Vo)/a

t=(5.3-0)/1.2=4.4s

then we use the ecuation number 3 to calculate the distance

X=0.5at^2

X=0.5x1.2x4.4^2=11.6m

b)second part

We know that when the elevator starts to accelerate and decelerate, it takes a distance of 11.6m and a time of 4.4s, which means that if the distance is subtracted 2 times this distance (once for acceleration and once for deceleration)

we will have the distance traveled in with constant speed.

With this information we will find the time, and then we will add it with the time it takes for the elevator to accelerate and decelerate

X=218-11.6x2=194.8m

X=VT

T=X/v

t=194.8/5.3=36.75s

Total time=36.75+2x4.4=45.55s

Answer:67.45 m

Explanation:

Given

at t=2.5 s it passes the top of a tall building and after 1.5 s it reaches maximum  height

let u is the initial velocity of stone

v=u+at

0=u-gt

[tex]u=9.81\times 4=39.24 m/s[/tex]

Let us take h be the height of building

[tex]h=ut+\frac{-1}{2}gt^2[/tex]

[tex]h=39.24\times 2.5-\frac{1}{2}\times 9.81\times 2.5^2[/tex]

h=67.45 m

Answer:

The second system must be set in motion [tex]t=0.70s[/tex] seconds later

Explanation:

The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:

[tex]T=2\pi\sqrt(\frac{m}{k} )[/tex]

One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.

If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion [tex]t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s)[/tex] seconds later

Answer:920.31 J

Explanation:

Given

Volume of water (V)[tex]=14.1 cm^3 [/tex]

mass(m)[tex]=\rho \times V=1000\times 14.1\times 10^{-6}=14.1 gm[/tex]

Temperature [tex]=8.4^{\circ} C[/tex]

Final Temperature [tex]=-7.2 ^{\circ}C[/tex]

specific heat of water(c)[tex]=4.184 J/g-^{\circ}C[/tex]

Therefore heat required to removed is

[tex]Q=mc(\Delta T)[/tex]

[tex]Q=14.1\times 4.184\times (8.4-(-7.2))[/tex]

[tex]Q=920.31 J[/tex]