A carousel has a diameter of 6.0-m and completes one rotation every 1.7s. Find the centripetal acceleration of the traveler in m / s2.

Answer:

173.75 km/hr in the NW direction.

Explanation:

Velocity is the time rate of change in displacement of a body. Mathematically:

v = d / t

where d = displacement

t = time

Therefore, the velocity of the car is:

v = 556 / 3.2 = 173.75 km/hr

The velocity of the car is 173.75 km/hr in the NW direction.

The velocity of a car will be "173.75 km/hr".

The velocity of something like a car moving northward on something like a prominent motorway as well as the velocity of something like a rocket launching towards spacecraft both might be determined or monitored.

Displacement, d = 556 km

Time, t = 3.2 hours

We know the relation,

→ Velocity = [tex]\frac{Displacement}{Time}[/tex]

or,

→ V = [tex]\frac{d}{t}[/tex]

By substituting the values, we get

      = [tex]\frac{556}{3.2}[/tex]

      = [tex]173.75[/tex] km/hr

Thus the response above is right.

Find out more information about velocity here:

https://brainly.com/question/6504879

Answer:

Explanation:

Ex(z,t) = Eocos(kz - ω t + φ)

k = 2π/λ  , ω = 2π f

φ = +30° , E₀ = 10³ V .

z/λ = 0.25 , ft = 0.125

Ex(z,t) = Eocos(2πz/λ - 2πf t + φ)

Putting the values given above

Ex(z,t) = 10³ cos ( 2π / 4 - 2π x .125 + 30⁰ )

= 1000cos (90⁰ - 45+30)

= 1000 cos 75

=258.8  V .

Answer:

So if its acceleration is -2m/s^2 that means every second the initial velocity would be subtracted by 2. So since it took 3 seconds 2*3=6. The initial velocity was 6 m/s

Answer:

6 A

Explanation:

Parallel connected resistors needs to be calculated as one single resistor. To do that: [tex]\frac{1}{15}[/tex]+[tex]\frac{1}{15}[/tex]+[tex]\frac{1}{15}[/tex]=[tex]\frac{3}{15}[/tex]=[tex]R^{-1}[/tex]

[tex]\frac{3}{15} ^{-1}[/tex]= 5 Ω (total resistance)

U = R* I

[tex]\frac{U}{R}[/tex]=I

[tex]\frac{30}{5}[/tex]=6 A

Answer:

V = 0.0283 m³ = 28300 cm³

T₂ = 1200 K

Explanation:

The volume of the gas can be determined by using General Gas Equation:

PV = nRT

where,

P = Pressure of Gas = (72 cm of Hg)(1333.2239 Pa/cm of Hg) = 95992.12 Pa

V = Volume of Gas = ?

n = no. of moles = mass/molar mass = (35 g)/(32 g/mol) = 1.09 mol

R = General Gas Constant = 8.314 J/ mol.k

T = Temperature of Gas = 27°C + 273 = 300 k

Therefore,

(95992.12 Pa)(V) = (1.09 mol)(8.314 J/mol.k)(300 k)

V = 2718.678 J/95992.12 Pa

V = 0.0283 m³ = 28300 cm³

The Kinetic Energy of gas molecule is given as:

K.E = (3/2)(KT)

Also,

K.E = (1/2)(mv²)

Comparing both equations, we get:

(3/2)(KT) = (1/2)(mv²)

v² = 3KT/m

v = √(3KT/m)

where,

v = r.m.s velocity

K = Boltzamn Constant

T = Absolute Temperature

m = mass of gas molecule

At T₁ = 300 K, v = v₁

v₁ = √(3K*300/m)

v₁ = √(900 K/m)

Now, for v₂ = 2v₁ (double r.m.s velocity), T₂ = ?

v₂ = 2v₁ = √(3KT₂/m)

using value of v₁:

2√(900 K/m) = √(3KT₂/m)

4(900) = 3 T₂

T₂ = 1200 K

Answer: Around 0:35 Pm or 12:35 Am

Explanation:

The equation that describes the cooling of objects can be written as:

T(t) = Ta + (Ti - Ta)*e^(k*t)

Where Ta is the ambient temperature, here Ta = 25°C.

Ti is the initial temperature of the body, we have Ti = 37°C.

t is the time.

k is a constant.

So our equation is:

T(t) = 25°C +12°C*e^(k*t)

at 2pm, the temperature was 33°C

at 3pm, the temperature was 31°C.

we want to find the hour where we have our t = 0, suppose this hour is X.

then we can write our times as:

2pm ---> 2 - X

3pm ----> 3 - X

and our equations are:

33°C = 25°C + 12°C*e^(k2 - k*X)

31° = 25°C + 12°C*e^(k3 - k*X)

So we have two equations and two variables, let's solve the system.

first, simplify it a bit, for the first eq:

33 - 25 = 12*e^(k2 - k*X)

8/12 = e^(k2 - k*X)

ln(8/12) = k*2 - k*X

for the second equation we have:

31 - 25 = 12*e^(k3 - k*X)

6/12 = e^(k3 - k*X)

ln(6/12) = k*3 - k*X

So our equations are:

1) ln(2/3) = 2*k - X*k

2) ln(1/2) = 3*k - X*k

First, let's isolate one of the variables in one of the equations. let's isolate k in the first equation.

ln(2/3)/(2-X) = k

now we can replace it in the second equation:

ln(1/2) = 3*ln(2/3)/(2 - X) - X*ln(2/3)/(2-X)

now let's solve it for X, i will take a = ln(1/2) and b = ln(2/3) so it is easier to read.

a = 3*b/(2 - X) - X*b/(2 - X)

a*(2 - X) = 3*b - X*b

2a - aX = 3b - Xb

X(a - b) = 2a - 3b

X = (2*ln(1/2) - 3*ln(2/3))/(ln(1/2) - ln(2/3)) = 0.590

now, knowing that one hour has 60 minutes, then this is:

0.59*60m = 35 minutes

So the hour of death is 0:35 Pm or 12:35 Am

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W'  the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

Answer :the resultant of two vectors can be found using either the parallelogram method or the triangle method. don't know if this was helpful ?

Explanation:

Answer:

Analytical method.

Answer:

Tension of the wire(T) = 169 N

Explanation:

Given:

f = 65Hz

Length of the piano wire (L) = 2 m

Mass density = 5.0 g/m² = 0.005 kg/m²

Find:

Tension of the wire(T)

Computation:

f = v / λ

65 = v / 2L

65 = v /(2)(2)

v = 260 m/s

T = v² (m/l)

T = (260)²(0.005/2)

T = 169 N

Tension of the wire(T) = 169 N

Answer:

e.) power you supply to the crate

Explanation:

According to given data, we have:

F = Force exerted on the crate

M = Mass of the crate

v = Speed of motion of the crate

d = Distance traveled by the crate across the floor

t = Time interval passed

Now, we try to analyze the given quantity:

=> F d/t

=> (Force)(Displacement)/(Time)

but, (Force)(Displacement) = Work Done

Therefore,

=> Work Done/Time

but, Work Done/Time = Power

Therefore,

=> Power

Hence, the quantity F d/t is:

e.) power you supply to the crate

Answer:

  B/4

Explanation:

The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:

The field at twice the distance is B/4.

Answer:

The rms voltage (in V) measured across the secondary coil is 459.62 V

Explanation:

Given;

number of turns in the primary coil, Np = 375 turns

number of turns in the secondary coil, Ns = 1875 turns

peak voltage across the primary coil, Ep = 130 V

peak voltage across the secondary coil, Es = ?

[tex]\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V[/tex]

The rms voltage (in V) measured across the secondary coil is calculated as;

[tex]V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V[/tex]

Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V

Answer:

a) v = -30 - 32 t ,  s (t) = 600 - 30 t -16 t² , b) v = -32 ft / s

c) v (1) = -62 ft / s,  v (3) = -126 ft / s , d) t = 7.13 s , e)  v = -258.16 ft / s

Explanation:

a) For this exercise they give us the function of the position of the ball

          s (t) = s (o) + v_o t - 16 t²

notice that you forgot to write the super index

indicate the initial position of the ball

        s (o) = 600 ft

also indicates initial speed

        v_o = - 30 ft / s

let's substitute in the equation

        s (t) = 600 - 30 t -16 t²

to find the speed we use

       v = ds / dt

       v = v_o - 32 t

       v = -30 - 32 t

b) To find the average speed, look for the speed at the beginning and end of the time interval

t = 1 s

     v (1) = -30 -32 1

     v (1) = - 62 ft / s

t = 3 s

     v (3) = -30 -32 3

     v (3) = -126 ft / s

the average speed is

    v = (v (3) -v (1)) / (3-1)

    v = (-126 +62) / 2

    v = -32 ft / s

c) instantaneous speeds, we already calculated them

    v (1) = -62 ft / s

    v (3) = -126 ft / s

d) the time to reach the ground

in this case s = 0

    0 = 600 - 30 t -16 t²

     t² + 1,875 t - 37.5 = 0

we solve the quadratic equation

     t = [-1,875 ±√ (1,875² + 4 37.5)] / 2

     t = [1,875 ± 12.39] / 2

     t₁ = 7.13 s

     t₂ = negative

Since the time must be positive, the correct answer is t = 7.13 s

e) the speed of the ball on reaching the ground

     v = -30 - 32 t

     v = -30 - 32 7.13

      v = -258.16 ft / s

Answer:

The answer is D)Theory

Explanation:

This is due because a theory is a scientific term and is a testable model that scientists seek to explain a phenomenon. You can also find out the answer by the process of elimination it can't be data because that would be something they already know and something they use to prove not explain. It can't be law because it isn't testable but can be used to explain. So that leaves you with two answers hypothesis and theory which are very similar but it isn't hypothesis because it isn't used to explain it to help the scientists come up with a theory and accumulate what might happen.

Answer:

D. Velocity of the car

Explanation:

The centripetal force acting on the car is given by the following formula:

[tex]F_c=ma_c=m\frac{v^2}{r}[/tex]    (1)

m: mass of the car  = 888 kg

v: tangential speed of the car = 7 m/s

r: radius of the flat circular track = 59 m

By the form of the equation (1) you can notice that the greatest change in the centripetal force is obtained when the velocity v is twice. In fact, you have:

[tex]F_c=m\frac{(2v)^2}{r}=4m\frac{v^2}{r}=4F_c[/tex]

Then, the greatest values of the centripetal force is:

[tex]F_c=4(888kg)\frac{(7m/s)^2}{59m}=2949.96N[/tex]

The greatest change in Fc is obtained by changing the value of the speed

answer

D. Velocity of the car

Answer:

a.  4 V

b. 0.697 A

Explanation:

Magnetic field strength B =  0.732 T

length of rod l = 0.362 m

velocity of rod v = 15.1 m/s

a.  EMF can be calculated as

E = Blv = 0.732 x 0.362 x 15.1 = 4 V

b. If the rod is connected to a conducting rail, with resistance R = 5.74Ω

current I = V/R = 4/5.74 = 0.697 A

the current flows in a clockwise direction

Explanation:

Draw a free body diagram of the pendulum (the combination of the sphere and the massless rod).  There are three forces on the pendulum:

Weight force mg at the center of the sphere,

Reaction force in the x direction at the pivot,

Reaction force in the y direction at the pivot.

Sum the torques about the pivot O.

∑τ = I d²θ/dt²

mg (L sin θ) = I d²θ/dt²

For small θ, sin θ ≈ θ.

mg L θ = I d²θ/dt²

Since d²θ/dt² is directly proportional to θ, this fits the definition of simple harmonic motion.

If you wish, you can use parallel axis theorem to find the moment of inertia about O:

I = Icm + md²

I = ⅖ mr² + mL²

mg L θ = (⅖ mr² + mL²) d²θ/dt²

gL θ = (⅖ r² + L²) d²θ/dt²

Answer:

A) 580m

B) 0 m/s

C) 9.8m/s^2

D) downward

E) 10.87s

F) 106.62 m/s

Explanation:

A) The distance traveled by the rocket is calculated by using the following expression:

[tex]y=\frac{1}{2}at^2[/tex]

a: acceleration of the rocket = 2.90 m/s^2

t: time of the flight = 20.0 s

[tex]y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m[/tex]

B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.

C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2

D) The acceleration points downward

E) The time the rocket takes to return to the ground is given by:

[tex]t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s[/tex]

10.87 seconds

F) The velocity just before the rocket arrives to the ground is:

[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}[/tex]

Answer:

The distance from the cornea vertex to the retina is 2.36 cm

Explanation:

The question is incomplete.

The complete question is as follows;

A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.

If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?

Solution.

We use image-object reaction to calculate the distance from the cornea vertex to the retina.

Mathematically;

n1/s + n2/s’ = n2-n1/R

From the question, we identify the following;

n1 ; Refractive index of air = 1

n2 ; Refractive index of lens = 1.4

S ; Object Distance = 36 cm

S’ = ?

R ; Radius of curvature of the cornea = 0.65

Substituting these values into the equation above;

1/36 + 1.4/S’ = (1.4-1)/0.65

{S’+ 36(1.4)}/36S’ = 0.4/0.65

{S’ + 50.4}/36S’ = 0.62

S’ + 50.4 = 22.32S’

50.4 = 22.32S’ -S’

21.32S’ = 50.4

S’ = 50.4/21.32

S’ = 2.36 cm

Answer:

Vbg = 20.5 m/s

your velocity with respect to the ground Vbg = 20.5 m/s

Explanation:

Relative velocity with respect to the ground is;

Vbg = velocity of train with respect to the ground + your velocity with respect to the train

Vbg = Vtg + Vbt ......1

Given;

velocity of train with respect to the ground;

Vtg = 19 m/s

your velocity with respect to the train;

Vbt = 1.5 m/s

Substituting the given values into the equation 1;

Vbg = 19 m/s + 1.5 m/s

Vbg = 20.5 m/s

your velocity with respect to the ground Vbg = 20.5 m/s

Answer:

The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].

Explanation:

As CD is acceleration uniformly, the following equation of motion can be used to determine the angular acceleration:

[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot \Delta n[/tex]

Where:

[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.

[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.

[tex]\ddot n[/tex] - Angular acceleration, measured in revolution per square minute.

[tex]\Delta n[/tex] - Change in angular position, measured in revolutions.

The angular acceleration is cleared and calculated:

[tex]\ddot n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \Delta n}[/tex]

Given that [tex]\dot n_{o} = 0\,\frac{rev}{min}[/tex], [tex]\dot n = 430\,\frac{rev}{min}[/tex] and [tex]\Delta n = 4\, rev[/tex], the angular acceleration is:

[tex]\ddot n = \frac{\left(430\,\frac{rev}{min} \right)^{2}-\left(0\,\frac{rev}{min} \right)^{2}}{2\cdot (4\,rev)}[/tex]

[tex]\ddot n = 23112.5\,\frac{rev}{min^{2}}[/tex]

The angular accelaration measured in radians per square second is:

[tex]\alpha = \left(23112.5\,\frac{rev}{min^{2}} \right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}} \right)[/tex]

[tex]\alpha \approx 40.339\,\frac{rad}{s^{2}}[/tex]

Net torque experimented by the CD during its accleration is equal to the product of its moment of inertia with respect to its axis of rotation and angular acceleration:

[tex]\tau = I \cdot \alpha[/tex]

Where:

[tex]I[/tex] - Moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].

[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.

In addition, a CD has a form of a uniform disk, whose moment of inertia is:

[tex]I = \frac{1}{2}\cdot m \cdot r^{2}[/tex]

Where:

[tex]m[/tex] - Mass of the CD, measured in kilograms.

[tex]r[/tex] - Radius of the CD, measured in meters.

If [tex]m = 0.017\,kg[/tex] and [tex]r = 0.07\,m[/tex], then:

[tex]I = \frac{1}{2}\cdot (0.017\,kg)\cdot (0.07\,m)^{2}[/tex]

[tex]I = 4.165\times 10^{-5}\,kg\cdot m^{2}[/tex]

Now, the net torque exerted on CD is:

[tex]\tau = (4.165\times 10^{-5}\,kg\cdot m^{2})\cdot \left(40.339\,\frac{rad}{s^{2}} \right)[/tex]

[tex]\tau = 1.680\times 10^{-3}\,N\cdot m[/tex]

The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].

Answer:

Explanation:

moment of inertia of the rod = 1/3 mL² , m is mass and L is length of rod.

1/3 x .154 x .35²

= .00629

moment of inertia  of putty about the axis of rotation

= m d² , m is mass of putty and d is distance fro axis

= .011 x( .35 / 3 )²

= .00015

Total moment of inertia I = .00644 kgm²

angular momentum of putty about the axis of rotation

= mvRsinθ

m is mass , v is velocity , R is distance where it strikes the rod and θ is angle  with the rod at which putty strikes

= .011 x 9 x .35 / 3 x sin 29

= .0056

Applying conservation of angular momentum

angular momentum of putty = angular momentum of system after of collision

.0056 =  .00644 ω where ω is angular velocity of the rod after collision

ω = .87 rad /s .

Rotational energy

= 1/2 I ω²

I is total moment of inertia

=  .5 x .00644 x .87²

= 2.44 x 10⁻³ J .

Answer:

2500 N/m²

Explanation:

Pressure: This can be defined as the force acting normally on a surface per unit area.

The expression for pressure is give as

P = F/A...................... Equation 1

Where P = pressure (N/m²), F = force (N), A = Contact area (m²)

Given: F = 500 N, A = 2000 cm² = (2000/10000) m = 0.2 m.

Substitute into equation  1

P = 500/0.2

P = 2500 N/m²

Hence the pressure exerted to the surface is 2500 N/m²

Electrical current is defined as the flow of electric charge per unit time.

Answer:

It will take 7.75 seconds for the car to go 155m

Explanation:

From the question, we can understand that the distance covered by the moving car is got by a product of its speed and the time it travels.

i.e distance = speed X time.

However, in this case, we have the distance travelled and the speed of the car, and we are looking for the time of travel

TO solve this, we will simply make the travel time the subject of the formula in the equation above.

i.e time = distance / speed

time = 155/20= 7.75 seconds.

Hence, it will take 7.75 seconds for the car to go 155m

Answer:

B. 9.4 s

Explanation:

In order to calculate the total time taken by the ball to hit the ground, we first analyze the upward motion. We will use subscript 1 for upward motion. Now, using 1st equation of motion:

Vf₁ = Vi₁ + gt₁

where,

Vf, = Final Velocity in upward motion = 0 m/s (ball stops at highest point)

Vi = Initial Velocity in upward motion = 36.2 m/s

g = - 9.8 m/s² (negative due to upward motion)

t₁ = Time taken in upward motion = ?

Therefore,

0 m/s = 36.2 m/s + (-9.8 m/s²)(t₁)

t₁ = (36.2 m/s)/(9.8 m/s²)

t₁ = 3.7 s

Now, using 2nd equation of motion:

h₁ = (Vi₁)(t₁) + (0.5)(g)(t₁)²

where,

h₁ = distance from top of building to highest point ball reaches = ?

Therefore,

h₁ = (36.2 m/s)(3.7 s) + (0.5)(-9.8 m/s²)(3.7 s)²

h₁ = 133.58 - 66.86 m

h₁ = 66.72 m

No, considering downward motion and using subscript 2, for it.

Using 2nd equation of motion:

h₂ = (Vi₂)(t₂) + (0.5)(g)(t₂)²

where,

h₂ = height of the highest point from ground = h₁ + height of building

h₂ = 66.72 m + 90 m = 156.72 m

Vi₂ = Initial Speed during downward motion = 0 m/s (ball stops for a moment at highest point)

t₂ = Time Taken in downward motion = ?

g = 9.8 m/s²

Therefore,

156.72 m = (0 m/s)(t₂) + (0.5)(9.8 m/s²)(t₂)²

t₂² = (156.72 m)/(4.9 m/s²)

t₂ = √31.98 s²

t₂ = 5.7 s

Now, the total time taken by ball to reach the ground is"

Total Time = T = t₁ + t₂

T = 3.7 s + 5.7 s

T = 9.4 s

Therefore, the correct answer is:

B. 9.4 s

Responder: 12000N

Explicación: Usando la fórmula para encontrar la eficiencia de una máquina. Eficiencia = ventaja mecánica / relación de velocidad × 100%

Dado MA = Carga / Esfuerzo

Relación de velocidad = distancia recorrida por esfuerzo (brazo de fuerza) / distancia recorrida por carga (brazo de resistencia)

MA = Carga / 3000

VR = 2 / 0.5 VR = 4

Asumiendo que la eficiencia es 100% 100% = (Carga / 3000) / 4 × 100%

1 = (Carga / 3000) / 4

4 = Carga / 3000

Carga = 4 × 3000

Carga = 12000N

Esto significa que el peso máximo que se puede levantar es 12000N

Answer:

a) v = -12t + 64

b) t = 5.33s

c) s = 362.66ft

d) t = 13.10s

e) v = 93.2ft/s

Explanation:

You have the following equation for the height of a stone thrown in Mars:

[tex]s(t)=-6t^2+64t+192[/tex]       (1)

a)  The velocity of the stone after t seconds is obtained with the derivative of s in time:

[tex]v=\frac{ds}{st}=-12t+64[/tex]   (2)

The equation for the speed of the stone is v = -12t + 64

b) The highest point is obtained when the speed of the stone is zero. Then, from the equation (2) equal to zero, you can obtain the time when the stone is at its maximum height:

[tex]-12t+64=0\\\\t=5.33s[/tex]

The time in which the stone is at the maximum height is 5.33s

c) For this time the stone is at the maximum height. Then, you replace t in the equation (1):

[tex]s(1)=-6(5.33)^2+64(5.33)+192=362.66ft[/tex]

the maximum height is 362.66 ft

d) To find the time when the stone arrive to the ground you equal the equation (1) to zero and you solve for t:

[tex]0=-6t^2+64t+192[/tex]

you use the quadratic formula:

[tex]t_{1,2}=\frac{-64\pm\sqrt{64^2-4(-6)(192)}}{2(-6)}\\\\t_{1,2}=\frac{-64\pm 93.29}{-12}\\\\t_1=13.10s\\\\t_2=-2.44s[/tex]

You use the result with positive values because is the onlyone with physical meaning.

The time for the stone hits the ground is 13.10 s

e) You replace 13.10s in the equation (2) to obtain the velocity of the stone when it strike the ground:

[tex]v=-12t+64=-12(13.10)+64=-93.2\frac{ft}{s}[/tex]

The minus sign is because the stone's direction is downward.

The speed of the stone just when it strikes the ground is 93.2ft/s

Answer:

Explanation:

Distance between fringe or fringe width =  xλ /  d

where x is location of screen and d is slit separation

Given x = 4 m

λ = 694 nm

d = .085 x 10⁻³ m

distance between fringes

= 4 x 694 x 10⁻⁹ / .085 x 10⁻³

= 4 x 694 x 10⁻⁹ / 85 x 10⁻⁶

= 32.66 x 10⁻³ m

= 32.66 mm .

3.267 cm

b )

when submerged in water , wavelength in water becomes as follows

wavelength in water = wave length / refractive index

= 694 / 1.333 nm

= 520.63 nm

new distance between fringes

3.267 / 1.333

= 2.45 cm .