A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .
b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
Express your answer in terms of the variable v0 and appropriate constants.
c) For what value of h does the collision occur at the instant when the first ball is at its highest point?
Express your answer in terms of the variable v0 and appropriate constants.

The minimum coefficient of kinetic friction required to stop the automobile before it hits the barrier is approximately 0.2041.

To stop the automobile before it hits the barrier, the minimum coefficient of kinetic friction needed can be found using the equation F = μk Mcg. The force of friction is equal to the product of the coefficient of kinetic friction (μk), the mass of the car (Mc), and the acceleration due to gravity (g). Since the car is stopping, its acceleration is negative, equal in magnitude to the square of its velocity divided by twice the stopping distance.

Using the given information, we can solve for the coefficient of kinetic friction:

F = μk Mcg

μk Mcg = (Mc)((-v^2)/(2d))

Substituting the values, we get:

μk = ((-v^2)/(2d))/g

where v = 20 m/s and d = 50 m.

Substituting the values in the equation, we get:

μk = ((-20^2)/(2*50))/9.8 ≈ -0.2041 ≈ 0.2041

Therefore, the minimum coefficient of kinetic friction required to stop the automobile before it hits the barrier is approximately 0.2041.

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Answer:

[tex]a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}[/tex]

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

a)

[tex]a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\[/tex]

b)

[tex]a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}[/tex]

Final answer:

The car's average acceleration for part (a) is 9.5 [tex]ft/s^2[/tex], and for part (b) it is 12.67 [tex]ft/s^2[/tex], calculated using the change in velocity over the time period.

Explanation:

Car's Average Acceleration

To find the car's average acceleration in part (a), we use the formula for average acceleration, which is the change in velocity (deltaV) divided by the time (t).

Average Acceleration (a) = (deltaV) / (t)For part (a), the initial velocity (V0) is 0 ft/s, the final velocity (V) is 38 ft/s, and the time (t) is 4.0 s.

Therefore, a = (38 ft/s - 0 ft/s) / 4.0 s = 9.5 [tex]ft/s^2[/tex]

For part (b), the final velocity (V) is now 76 ft/s and the time (t) is 3.0 s (the additional time).

Average Acceleration (a) = (76 ft/s - 38 ft/s) / 3.0 s = 12.67 [tex]ft/s^2[/tex]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant [tex]\gamma =1.4[/tex]

specific heat is given as [tex]=\frac{\gamma R}{\gamma -1}[/tex]

gas constant =287 J⋅kg−1⋅K−1

[tex]Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}[/tex]

specific heat at constant volume

[tex]Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}[/tex]

change in internal energy [tex]= Cv(T_2 -T_1)[/tex]

                            [tex]  \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}[/tex]

change in enthalapy [tex]\Delta H = Cp(T_2 -T_1)[/tex]

                                 [tex] \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}[/tex]

change in entropy

[tex]\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})[/tex]

[tex]\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})[/tex]

[tex]\Delta S = 382.79 J kg^{-1} K^{-1}[/tex]

Explanation:

Given that,

Wavelength of the laser light, [tex]\lambda=1062\ nm=1062\times 10^{-9}\ m[/tex]

The laser pulse lasts for, [tex]t=34\ ps=34\times 10^{-12}\ s[/tex]

(a) Let d is the distance covered by laser in the given by, [tex]d=c\times t[/tex]

[tex]d=3\times 10^8\times 34\times 10^{-12}[/tex]

d = 0.0102 meters

Let n is the number of wavelengths found within the laser pulse. So,

[tex]n=\dfrac{d}{\lambda}[/tex]

[tex]n=\dfrac{0.0102}{1062\times 10^{-9}}[/tex]

n = 9604.51

(b) Let t is the time need to be fit only in one wavelength. So,

[tex]t=\dfrac{\lambda}{c}[/tex]

[tex]t=\dfrac{1062\times 10^{-9}}{3\times 10^8}[/tex]

[tex]t=3.54\times 10^{-15}\ s[/tex]

Hence, this is the required solution.

Answer:

1.507 h

Explanation:

We can think that they move at constant speed, so their positions will follow

X(t) = Xo + v * t

However Carl will start moving 31 minutes later, so we can adjust the equation like this

X(t) = Xo + v * (t - t1)

Where t1 is the time at which each of them will start moving, 0 for Isaac since he starts driving right away, and 31 minutes (0.5167 h) for Carl

Also, we can considre the initial position of Carl to be 0 and his speed to be 50, while Isaac will have an initial position of 155 and a speed of -70 (negative because he is driving towards the origin of coordinates).

Then we have these two equations:

x(t) = 0 + 50 * (t - 0.5167)

x(t) = 155 - 70 * t

These are equations of lines, the point where they intersect determines the place and time they meet.

50 * (t -0.5167) = 155 - 70 * t

50 * t - 25.83 = 155 - 70 * t

120 * t = 180.83

t = 180.83 / 120 = 1.507 h

This is the time Isaac will be driving

Issac will drive for [tex]\boxed{1.507\text{ hr}}[/tex] before he meets carl.

Further Explanation:

The acceleration of the driving is ignored. It means that the motion of the cars is purely linear on the highways.

Given:

The distance between Carl and Issac is [tex]155\text{ miles}[/tex].

The speed of Carl is [tex]50\text{ mph}[/tex].

The speed of Issac is [tex]70\text{ mph}[/tex].

The time taken by Carl to finish lunch before starting to drive is [tex]31\text{ min}[/tex].

Concept:

Since Carl is eating lunch and he will start after 31 minutes, Issac will cover some distance within this time.

The distance covered by Issac before Carl starts to drive is,

[tex]\begin{aligned}d_1&=\text{speed}\times\text{time}\\&=50\text{ mph}\times\left(\dfrac{31}{60}\right)\text{hr}\\&=36.16\text{ mile}\end{aligned}[/tex]

Now after Issac has covered [tex]36.16\text{ miles}[/tex], Carl starts to drive. So, the distance left between Carl and Issac to be covered after carl starts to drive is,

[tex]\begin{aligned}d&=155\text{ miles}-36.16\text{ miles}\\&=118.84\text{ miles}\end{aligned}[/tex]

So, now the sum of the distances covered by Issac and Carl in time [tex]t[/tex] should be equal to the [tex]118.84\text{ miles}[/tex].

[tex]\begin{aligned}D_{\text{Issac }}+D_{\text{Carl}}&=118.84\text{ miles}\\(V_{Issac}\times t)+(V_{Carl}\times t)&=118.84\text{ miles}\end{aligned}[/tex]

Substituting the values of speed of Issac and Carl,

[tex]\begin{aligned}(70\times t)+(50\times t)&=118.84\text{ miles}\\120t&=118.84\text{ miles}\\t&=\frac{118.84}{120}\text{ hr}\\t&=0.990\text{ hr}\end{aligned}[/tex]

So, Issac and Carl meet after [tex]0.990\text{ hr}[/tex] after Carl starts driving.

So, the time for which Issac has been driving is,

[tex]\begin{aligned}T_{\text{ Issac}}&=0.990\text{ hr}+\dfrac{31}{60}\text{ hr}\\&=(0.990+0.516)\text{ hr}\\&=1.507\text{ hr}\end{aligned}[/tex]

Thus, Issac will drive for [tex]\boxed{1.507\text{ hr}}[/tex] before he meets carl.

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Answer Details:

Grade: College

Subject: Physics

Chapter: Linear motion in one dimension

Keywords:

Carl, Issac, Speed, meet, 155 miles, eating lunch, 70 mph, 50 mph, 31 min, between two locations, highway forms a straight line.

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

[tex]OB=0.40 km[/tex]

And the Max distance towards bookstore is,

[tex]OA=3.65 km[/tex]

Now the distance between the Joy and Max will be,

By applying pythagorus theorem,

[tex]AB=\sqrt{OB^{2}+OA^{2}}[/tex]

Substitute 0.40 km for OB and 3.65 km for OA in the above equation.

[tex]AB=\sqrt{0.40^{2}+3.65^{2}}\\AB=\sqrt{13.4825} \\AB=3.67 km[/tex]

Therefore the distance between there destination is 3.67 km.

Answer:

The car will reach 20 m/s in 4.76 seconds

Explanation:

The car moves following an uniformly accelerated motion, therefore we know that [tex]v=v_0+ a(t-t_0)[/tex]

Where [tex]v_0 = 0[/tex] are [tex]t_0 =0[/tex]

Then [tex]t= \frac{v}{a}=\frac{20 m/s}{4.2 m/s^{2}}=4.76s[/tex]

The distances traveled by the child on the toboggan are [tex]0.9, 3.6, and\ 8.1[/tex] respectively.

The distance traveled by the child on the toboggan can be calculated using the equations of motion for uniformly accelerated motion. The equation that relates distance (d), initial velocity (u), acceleration (a), and time (t) is given by:

[tex]d = ut + (1/2)at^2[/tex]

In this case, the child starts at rest [tex](u = 0)[/tex] and has an acceleration of a[tex]= 1.8 m/s^2[/tex].

Calculate the distances for the given times:

(a) For t = 1.0 s:

[tex]d = (0) \times (1.0 ) + (1/2) \times (1.8 ) \times (1.0 )^2\\d = 0 + 0.9 \\d = 0.9\ m[/tex]

(b) For t = 2.0 s:

[tex]d = (0) \times (2.0 ) + (1/2) \times (1.8 ) \times (2.0 )^2\\d = 0 + 3.6\\d = 3.6\ m[/tex]

(c) For t = 3.0 s:

[tex]d = (0) \times (3.0 ) + (1/2) \times (1.8 ) \times (3.0)^2\\d = 0 + 8.1\\d = 8.1\ m[/tex]

So, the distances traveled by the child on the toboggan are [tex]0.9, 3.6, and\ 8.1[/tex] respectively.

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Final answer:

Using the kinematic equation for motion with constant acceleration, the distance traveled by the child on the toboggan is 0.9 m after 1.0 s, 3.6 m after 2.0 s, and 8.1 m after 3.0 s.

Explanation:

To solve for the distance traveled by the child on the toboggan in each case, we'll use the kinematic equation for motion with constant acceleration: d = v_i * t + \frac{1}{2} * a * t^2, where d is the distance traveled, v_i is the initial velocity, t is the time, and a is the acceleration.

Since the child starts at rest, the initial velocity v_i is 0 m/s. The acceleration a is given as 1.8 m/s^2.

Answer:7 cm

Explanation:

Given

one sphere has a radius(r) of 4.85 cm

Also another sphere has a mass 3 times the previous one

since the density remains same therefore

second sphere has a volume 3 times of former.

Let R be the radius of bigger Sphere

[tex]\frac{4\pi R^3}{3}=3\times \frac{4\pi r^3}{3}[/tex]

[tex]R=\left ( 3\right )^\frac{1}{3}r=1.44\times 4.85=6.99 \approx 7 cm[/tex]

Answer:

Car will take 5 sec to reach from 0 m/sec to 25 m/sec

Explanation:

We have given initial velocity u = 0 m/sec

And final velocity v = 25 m/sec

Acceleration [tex]a=5m/Sec^2[/tex]

From first equation of motion we know that [tex]v=u+at[/tex], here v is final velocity, u is initial velocity, a is acceleration and t is time

So [tex]25=0+5\times t[/tex]

[tex]t=4 sec[/tex]

So car will take 5 sec to reach from 0 m/sec to 25 m/sec

Answer:

[tex]t=4.06s[/tex]

Explanation:

From the exercise we know

[tex]y_{o}=1.2m\\v_{o}=20m/s\\y=1.6m\\g=-9.8m/s^2[/tex]

To find how long does it takes the marker to reach the highest point we need to use the equation of position:

[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]

[tex]1.6m=1.2m+(20m/s)t-\frac{1}{2}(9.8m/s^2)t^2[/tex]

[tex]0=-0.4+20t-4.9t^2[/tex]

Now, we need to use the quadratic formula:

[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]

[tex]a=-4.9\\b=20\\c=-0.4[/tex]

Solving for t

[tex]t=0.020s[/tex] or [tex]t=4.06s[/tex]

So, the answer is t=4.06s because the other option is almost 0 and doesn't make any sense for the motion of the marker

Final answer:

To calculate the distance the firefighters should position their water cannon from the building, analyze the vertical motion of the water stream. Therefore, the firefighters should position their water cannon approximately 11.40 meters from the building.

Explanation:

To find the horizontal distance from the building where the firefighters should position their water cannon, we can use the projectile motion equations.

Given:

[tex]- Initial \ velocity \ of \ the \water \ cannon, \( v_0 = 25.0 \, \text{m/s} \)\\- Launch angle, \( \theta = 52.0^\circ \)\\- Vertical displacement, \( y = 12.0 \, \text{m} \)\\- Vertical acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)\\[/tex]

First, we'll find the time it takes for the water to reach a height of 12.0 m. We'll use the kinematic equation for vertical motion:

[tex]\[ y = v_{0y}t + \frac{1}{2}gt^2 \][/tex]

Where:

[tex]- \( v_{0y} \) is the initial vertical component of the velocity\\- \( t \) is the time[/tex]

Since the initial velocity is at an angle, we need to find its vertical component:

[tex]\[ v_{0y} = v_0 \sin(\theta) \][/tex]

Substitute the given values:

[tex]\[ v_{0y} = 25.0 \, \text{m/s} \times \sin(52.0^\circ) \]\[ v_{0y} \approx 25.0 \, \text{m/s} \times 0.788 \]\[ v_{0y} \approx 19.7 \, \text{m/s} \][/tex]

Now, let's use the equation for vertical motion to solve for \( t \):

[tex]\[ 12.0 \, \text{m} = (19.7 \, \text{m/s})t - \frac{1}{2}(9.81 \, \text{m/s}^2)t^2 \][/tex]

This is a quadratic equation, we can solve it to find \( t \). Let's denote [tex]\( a = -4.905 \, \text{m/s}^2 \) and \( b = 19.7 \, \text{m/s} \):[/tex]

[tex]\[ -4.905t^2 + 19.7t - 12.0 = 0 \][/tex]

Now, we can use the quadratic formula to solve for \( t \):[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Let's calculate \( t \).

We have the quadratic equation:

[tex]\[ -4.905t^2 + 19.7t - 12.0 = 0 \][/tex]

Using the quadratic formula:

[tex]\[ t = \frac{-19.7 \pm \sqrt{(19.7)^2 - 4(-4.905)(-12.0)}}{2(-4.905)} \]\[ t = \frac{-19.7 \pm \sqrt{389.21 - 235.2}}{-9.81} \]\[ t = \frac{-19.7 \pm \sqrt{154.01}}{-9.81} \]\[ t = \frac{-19.7 \pm 12.41}{-9.81} \][/tex]

Now, we have two possible values for \( t \):

[tex]1. \( t_1 = \frac{-19.7 + 12.41}{-9.81} \)\\2. \( t_2 = \frac{-19.7 - 12.41}{-9.81} \)[/tex]

Calculating each value:

[tex]1. \( t_1 = \frac{-7.29}{-9.81} \) \( t_1 \approx 0.742 \) seconds\\2. \( t_2 = \frac{-32.11}{-9.81} \) \( t_2 \approx 3.27 \) seconds[/tex]

Since the time \( t \) cannot be negative, we'll take [tex]\( t = t_1 \approx 0.742 \) seconds[/tex].

Now, to find the horizontal distance \( x \) from the building, we can use the equation for horizontal motion:

[tex]\[ x = v_{0x} \cdot t \][/tex]

where \( v_{0x} \) is the initial horizontal component of the velocity, given by:

[tex]\[ v_{0x} = v_0 \cdot \cos(\theta) \][/tex]

Substituting the given values:

[tex]\[ v_{0x} = 25.0 \, \text{m/s} \cdot \cos(52.0^\circ) \]\[ v_{0x} \approx 25.0 \, \text{m/s} \cdot 0.615 \]\[ v_{0x} \approx 15.375 \, \text{m/s} \][/tex]

Now, we can find \( x \):

[tex]\[ x = 15.375 \, \text{m/s} \cdot 0.742 \, \text{s} \]\[ x \approx 11.40 \, \text{m} \][/tex]

Therefore, the firefighters should position their water cannon approximately 11.40 meters from the building.

Answer:

W = 290.7 dynes*cm

Explanation:

d = 1/5 cm = 0.2 cm

The force is in function of the depth x:

F(x) = 1000 * (1 + 2*x)^2

We can expand that as:

F(x) = 1000 * (1 + 4*x + 4x^2)

F(x) = 1000 + 4000*x + 4000*x^2

Work is defined as

W = F * d

Since we have non constant force we integrate

[tex]W = \int\limits^{0.2}_{0} {(1000 + 4000*x + 4000*x^2)} \, dx[/tex]

W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2

W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3

W = 200 + 80 + 10.7 = 290.7 dynes*cm

In this exercise, we have to use work knowledge to calculate the total, in this way:

[tex]W = 290.7[/tex]

Given some information about the exercise we find that:

Using the working definition we find that:

[tex]W = F * d\\W = [1000*x + 2000*x^2 + 1333*X^3]\\W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3\\W = 200 + 80 + 10.7 = 290.7[/tex]

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Answer:

Voltage, V = 0.0524 volts

Explanation:

Thickness of the membrane, [tex]d=7.95\ nm=7.95\times 10^{-9}\ m[/tex]

Electric field strength, [tex]E=6.6\ MV/m=6.6\times 10^6\ V/m[/tex]

We need to find the voltage across it. The relationship between the voltage, electric field and the distance is given by :

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

[tex]V=6.6\times 10^6\ V/m\times 7.95\times 10^{-9}[/tex]

V = 0.0524 volts

So, the voltage across the thick membrane is 0.0524 volts. Hence, this is the required solution.

The voltage across a 7.95 nm thick membrane with an electric field strength of 6.60 MV/m, is found to be 52.47 mV.

To find the voltage across a membrane, we can use the relationship between electric field strength (E) and voltage (V) given by the formula:

V = E * d

where:

V is the voltage

E is the electric field strength

d is the thickness of the membrane

In this case, the electric field strength (E) is given as 6.60 MV/m and the thickness of the membrane (d) is 7.95 nm.

First, convert the thickness from nanometers to meters:

d = 7.95 nm = 7.95 * 10⁻⁹ m

Now use the formula to calculate the voltage:

V = 6.60 * 10⁶ V/m * 7.95 * 10⁻⁹ m

V = 5.247 * 10⁻² V = 52.47 mV

Therefore, the voltage across the 7.95 nm thick membrane is 52.47 mV.

Answer:

T = 450 C

Explanation:

given,

pressure of rigid tank containing water = 200 kPa

water start condensing at 100 °C

vapor saturation pressure at 100 °C = 101421 Pa (from saturation table)

specific volume  = 1.6718 m³/kg

From super heated steam table

steam table is two set of table which is of energy transfer properties of water and steam.

at v = 1.6718 m³/kg and pressure = 200 kPa

temperature is equal to T = 450 C

Answer:

Frequency will be 4.30 kHz

Explanation:

We have inductance [tex]L=37\mu H=37\times 10^{-6}H[/tex]

And capacitance [tex]C=37\mu H=37\times 10^{-6}F[/tex]

Inductive reactance is given by [tex]X_L=\omega L[/tex]

And capacitive reactance [tex]X_C=\frac{1}{\omega C}[/tex]

As in question it is given that inductive reactance and capacitive reactance is same

So [tex]X_L=X_C[/tex]

[tex]\omega L=\frac{1}{\omega C}[/tex]

[tex]\omega ^2=\sqrt{\frac{1}{LC}}=\frac{1}{\sqrt{37\times 10^{-6}\times 37\times 10^{-6}}}=27027.027rad/sec[/tex]

We know that angular frequency [tex]\omega =2\pi f[/tex]

[tex]2\times 3.14\times f=27027.027[/tex]

f = 4303.66 Hz =4.30 kHz

Answer:

Explanation:

The rock will decelerate at the rate of g or 9.8 m s⁻²while on its way up because of gravitational attraction towards the earth.  . In other words , it will accelerate with the magnitude - 9.8 m s⁻² while on its way up.

On its way down , it will accelerate due to gravitational attraction. In this case acceleration will be positive and equal to 9.8 m s⁻².

At the highest point in its trajectory , rock  becomes stationary or comes to  rest momentarily. So at the peak position , velocity is zero.

At the highest point , as the gravitational force continues to act on the body undiminished, its acceleration will remain the same ie 9.8 m s⁻² at the highest point as well. It will act in the downward direction.

Answer:

213 ft/s

Explanation:

Given:

Δx = 330 ft

v₀ = 0 ft/s

t = 3.1 s

Find: Δv

x = x₀ + ½ (v + v₀)t

330 ft = ½ (v + 0 ft/s) (3.1 s)

v = 213 ft/s

The change in velocity is:

Δv = 213 ft/s − 0 ft/s

Δv = 213 ft/s

Answer:

2.551 m/s

Explanation:

Given:

Boulder is dropped from the cliff

thus,

Initial speed of the boulder, u = 0 m/s

Final speed to be obtained, v = 90 km/h =[tex]90\times\frac{5}{18}[/tex] =25 m/s

also, in the case of free fall the acceleration of the boulder will be equal to the acceleration due to the gravity i.e g = 9.8 m/s²

Now, from the Newton's equation of motion

v = u + at

where, a is the acceleration = g = 9.8 m/s²

t is the time

on substituting the respective values, we get

25 = 0 + 9.8 × t

or

t = [tex]\frac{\textup{25}}{\textup{9.8}}[/tex]

or

t = 2.551 m/s

The statement that heuristics allow a priori theoretical concepts is true. Heuristics are mental shortcuts that can definitely be informed by a priori theoretical concepts, while empiricism emphasizes the role of experience in the formation of ideas.

Regarding the question on the statements about heuristics and empiricism, the best choice would be 'A'. Heuristics allow a priori theoretical concepts'. Heuristics are mental shortcuts or rules of thumb that simplify decisions, particularly under conditions of uncertainty. They are not necessarily based on theoretical concepts but rather on practical, personal experience or common sense. However, they can certainly be informed by a priori theoretical concepts in the sense that our theoretical understanding can shape the rules of thumb we use.

On the other hand, empiricism is a philosophical system that emphasizes the role of experience, especially sensory perception, in the formation of ideas, while discounting a priori reasoning, intuitiveness, and innate ideas. Hence, options B, C, and D are incorrect. Option E is also incorrect as heuristics can definitely be based on a trial and error approach.

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Answer:

(A) 476 kPa

Explanation:

If the volume remains constant, the ideal gas law says:

P/T=constant

so: P1/T1=P2/T2

P2=P1*T2/T1=460*298/288=476KPa

Answer:

A. The final pressure at the tire would be 476 kPa

Explanation:

Since the volume is constant the ideal gas equation would be used to obtain the final pressure at the tire.

Given

the initial temperature [tex]T_{1}[/tex] = 288 K

the initial pressure [tex]P_{1}[/tex] = 460 k Pa

the final temperature [tex]T_{2}[/tex] = 298 K

the final pressure [tex]P_{2}[/tex] = ?

Using the ideal gas equation;

PV = nRT

[tex]P_{1}[/tex] / [tex]T_{1}[/tex]  = [tex]P_{2}[/tex] / [tex]T_{2}[/tex]

Making [tex]P_{2}[/tex] the subject formula

[tex]P_{2}[/tex]  = ([tex]P_{1}[/tex] x [tex]T_{2}[/tex] ) / [tex]T_{1}[/tex]

[tex]P_{2}[/tex]  = (460 x 298) / 288

[tex]P_{2}[/tex]  = 475.972 kPa

[tex]P_{2}[/tex]  ≈ 476 kPa

Therefore the final pressure at the tire would be 476 kPa

Answer:

[tex]V=\frac{4}{3}\pi R^{3}[/tex]

Explanation:

The volume of a body is defined as the capacity of the object. The amount of matter that object contains is called its volume.

All the three dimensional objects have volume.

The SI unit of volume is m^3. The volume of liquids is measured by teh unit litre or milli litre.

The volume of the sphere is given by

[tex]V=\frac{4}{3}\pi R^{3}[/tex]

where, R is the radius of the sphere.

Final answer:

To achieve an overall magnification of 200 with the given measurements, the eyepiece lens would need to have a focal length of approximately 5.68 cm.

Explanation:

To calculate the eyepiece focal length that will give an overall angular magnification of 200 in a microscope, we can use the formula of magnification for a compound microscope. Given that the distance between the objective and eyepiece lenses is 19 cm and the objective lens has a focal length of 5.5 mm, we can set up the equation as follows:

M = (D/fo) × (25/fe)

Where:

In the given problem, M = 200 and fo = 5.5 mm (which we should convert to centimeters to keep consistent units, thus fo = 0.55 cm). We are solving for fe.

Using the formula, we substitute the values:

200 = (25/0.55) × (25/fe)

Solving for fe, we find:

fe = (25/0.55) × (25/200)

fe = (45.45) × (0.125)

fe = 5.68 cm

Therefore, the eyepiece lens would need to have a focal length of approximately 5.68 cm to achieve an overall magnification of 200.

Answer:

a) Linear equation

Explanation:

Definition of acceleration

[tex]a=\frac{dv}{dt}\\[/tex]

if a=constant and we integrate the last equation

[tex]v(t)=v_{o}+a*t[/tex]

So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, [tex]d=24.44\times 2=48.88\ m[/tex]

(a) Acceleration, [tex]a=-4\ m/s^2[/tex]

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

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

[tex]s=\dfrac{-(24.44)^2}{2\times -4}[/tex]

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, [tex]a=-8\ m/s^2[/tex]

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

[tex]s=\dfrac{-(24.44)^2}{2\times -8}[/tex]

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

Answer:

55.66 m

Explanation:

While falling by 50 m , initial velocity u = 0

final velocity = v , height h = 50 , acceleration g = 9.8

v² = u² + 2gh

= 0 + 2 x 9.8 x 50

v = 31.3 m /s

After that deceleration comes into effect

In this case final velocity v = 17 m/s

initial velocity u = 31.3 m/s

acceleration a = - 61 m/s²

distance traveled h = ?

v² = u² + 2gh

(17)² = (31.3)² - 2x 61xh

h = 690.69 / 2 x 61

= 5.66 m

Total height during which he was in air

= 50 + 5.66

= 55.66 m

Answer:

The wavelength of the light is [tex]7200\ \AA[/tex].

Explanation:

Given that,

Distance between the slit centers d= 1.2 mm

Distance between constructive fringes [tex]\beta= 0.3\ cm[/tex]

Distance between fringe and screen D= 5 m

We need to calculate the wavelength

Using formula of width

[tex]\beta=\dfrac{D\lambda}{d}[/tex]

Put the value into the formula

[tex]0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}[/tex]

[tex]\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}[/tex]

[tex]\lambda=7.2\times10^{-7}\ m[/tex]

[tex]\lambda=7200\ \AA[/tex]

Hence, The wavelength of the light is [tex]7200\ \AA[/tex].

Answer:

time at which Kathy overtakes Stan is 6.70 sec

Explanation:

given data

time = 1 sec

acceleration= 3.4 m/s²

acceleration = 4.49  m/s²

to find out

the time at which Kathy overtakes Stan

solution

we consider here travel time for kathy = t1

and travel time for stan is = t2

and we know initial velocity = 0

so

t1 = 1 + t2

and distance travel equation by kinematic is

d1 = ut + [tex]\frac{1}{2}[/tex] at²

d1 = 0+  [tex]\frac{1}{2}[/tex] 4.49 (t2)²   ................1

and

d2 = 0 + [tex]\frac{1}{2}[/tex] 3.4 (1+t2)²   ..................2

and when overtake distance same so  from equation 1 and 2

[tex]\frac{1}{2}[/tex] 4.49 (t2)²  = [tex]\frac{1}{2}[/tex] 3.4 (1+t2)²

t2 = 6.703828 sec

so time at which Kathy overtakes Stan is 6.70 sec

Answer:

Answered

Explanation:

impacts of the physics of matter on aviation operations.

Thrust, drag and lift.

Thrust:

It is the force developed by airplane engines that cause it to pull forward. With the help of huge propellers of course attached to the wings.

Drag:

It is the resistive force on the plane caused by the friction between air and plane. Its magnitude depends upon surface area, speed and viscosity of the air.

Lift:

The drag produced is utilized such that one of its component acts opposite to the weight. This causes the plane to take flight and stay in air. Lift can be deduced using Bernoulli's principle.

Bernoulli's principle is equivalent to law of conservation of energy. Meaning it tries to keep the energy of a system constant. In doing so, it produces low pressure zone above the wing. Which causes a net upward force, lift.

The car lands approximately 45.67 meters away from the base of the cliff, and its impact speed is 36.1 m/s.

The situation described involves typing a projectile motion. Ignore any effect of air resistance for simplicity's sake. The stunt man's car exits the ramp at an angle, and continues its journey downwards under the force of gravity.

Firstly, we need to break down the initial velocity of the car into its horizontal and vertical components. Horizontal component (Ux) is governed by the formula Ux = U cos θ which results to 20 cos 20° = 18.79 m/s. Vertical component (Uy) is calculated by Uy = U sin θ which gives us 20 sin 20° = 6.84 m/s.

Following this, we need to calculate the total time the car is in the air (t), governed by the equation h = Uy * t - (1/2) * g * t², where g represents gravity (9.8 m/s²) and h is the height of the cliff (28m). Solving for t gives us approximately 2.43 seconds.

Finally, we find the horizontal distance (X) the car covers during the time it's in the air: X = Ux * t  = 18.79 m/s * 2.43 s ≈ 45.67 m.

To calculate the impact speed, we take the square root of the sum of the squares of the final horizontal and vertical velocities. The final vertical velocity (Vy) = Uy + g * t = 6.84 m/s + 9.8 m/s² * 2.43 s = 30.95 m/s. The impact speed then equals √((18.79 m/s)² + (30.95 m/s)²) = 36.1 m/s.

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