An airplane flies horizontally with a constant speed of 172.0 m/s at an altitude of 1390 m. A package is dropped out of the airplane. Ignore air resistance. The magnitude of the gravitational acceleration is 9.8 m/s2. Choose the RIGHT as positive x-direction. Choose UPWARD as positive y-direction Keep 2 decimal places in all answers

(a) What is the vertical component of the velocity (in m/s) just before the package hits the ground? Pay attention to the direction (the sign).
(b) What is the magnitude of the velocity (in m/s) (including both the horizontal and vertical components) of the package just before it hits the ground?

Answer:

63.57 kg

Explanation:

weight = 140 lbs

Let the mass  is m.

1 lbs = 4.45 N

The weight of an object is defined as the force with which our earth attracts the body towards its centre.

Weight is the product of mass of the body and the acceleration due to gravity of that planet.

W = m x g

On earth surface g = 9.8 m/s^2

Now convert lbs in newton

So, 140 lbs = 140 x 4.45 = 623 N

So, m x 9.8 = 623

m = 63.57 kg

Thus, the mass is 63.57 kg.

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: D) 936 K

Explanation:

Given:

Initial temperature of the gas, [tex]T = 450\ K[/tex]

Initial Pressure of the gas, [tex]P=4\ atm[/tex]

initial volume of the gas, [tex]V=10\ L[/tex]

It it given that the process is adiabatic, so for a adiabatic process we have

Let [tex]T_f \ \ and\ \ V_f [/tex] be the final temperature and volume of the gas.

[tex]T_iV_i^{\gamma -1}=T_fV_f^{\gamma -1}[/tex]

For monotomic gas [tex]\gamma=1.67[/tex]

[tex]450\times V_i^{1.67 -1} =T_f\left (\dfrac{V_i}{3} \right )^{1.67-1}\\T_f=936 K[/tex]

Hence the final temperature of the gas is 936 K. So option D is correct

Final answer:

To find the distance the cat strikes from the edge of the piano, we can break the motion into horizontal and vertical components, and solve for the time of flight and horizontal distance traveled. By substituting the given values into relevant equations and solving, we can determine that the cat will strike the floor approximately 1.71 meters from the edge of the piano.

Explanation:

To solve this problem, we can break the motion into horizontal and vertical components. The horizontal component of the initial velocity will not change, so the cat will travel a horizontal distance of 3m/s *cos(37°) * t, where t is the time of flight. To find t, we can use the equation h = v0y * t + (1/2) * g * [tex]t^2[/tex], where h is the vertical displacement, v0y is the vertical component of the initial velocity, and g is the acceleration due to gravity.

Substituting the given values into the equation, we get 1.3m = 3m/s * sin(37°) * t - (1/2) * 9.8m/[tex]s^2[/tex] * [tex]t^2[/tex]. Solving this quadratic equation, we find two possible values for t: approximately 0.364s and 0.203s. Since the cat jumps off the piano, we only consider the positive value of t. So the cat will strike the floor approximately 3m/s * cos(37°) * 0.364s = 1.71 meters from the edge of the piano.

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]

Answer:

210.3 degrees

Explanation:

The net force exerted on charge A = 59.5 N

Use the x and y coordinates of net force to get the direction

arctan (y/x)

Answer:

F = -51.357i -29.958j

abs(F) = 59.45 N

Explanation:

To solve the problem we use coulomb's law with vectorial notation, F = q1*q2/(4*pi*eo*r^2) where q1 and q2 are the charges and r is the distance between them:

Point B exerts a force on A in '-i' direction  

Point C exerts a force on A in '-j' direction  

Fba = 4*7/(4*pi*eo*0.07^2) = 51.357N

Fca = 4*3/(4*pi*eo*0.06^2) = 29.958N  

F = -51.357i -29.958j

abd(F) = 59.45 N

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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The weight of the astronaut's life support backpack on the moon would be one-sixth of its weight on Earth, which equates to 50lb.

The acceleration due to gravity on the moon is one-sixth that on Earth. Therefore, the weight of any object on the moon will be one-sixth of its weight on Earth. Since weight is a product of mass and gravity, if gravity is reduced, weight is also reduced accordingly.

To calculate the weight of an astronaut's life support backpack on the moon, you need to divide its Earth weight (300lb) by 6 (the relative gravity of the moon compared to Earth). So, on the moon, the backpack would weigh 300/6 = 50lb.

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I have to say, i love this kind of problems.

So, we got the linear acceleration, as we all know, linear the acceleration its, in dimensional units:

[tex][a]=[\frac{distance}{time^2}][/tex].

Now, we got the radius

[tex][r] = [distance][/tex]

the angular frequency

[tex][\omega] = \frac{1}{s}[/tex]

and the mass

[tex][m]=[mass][/tex].

Now, the acceleration doesn't have units of mass, so it can't depend on the mass of the particle.

The distance in the acceleration has exponent 1, and so does in the radius. As the radius its the only parameter that has units of distance, this means that the radius must appear with exponent 1. Lets write

[tex]a \propto r[/tex].

The time in the acceleration has exponent -2 As the angular frequency its the only parameter that has units of time, this means that the angular frequency must appear, but, the angular frequency has an exponent of -1, this means it must be squared

[tex]a \propto r \omega^2[/tex].

We are almost there. If this were any other problem, we would write:

[tex]a = A r \omega^2[/tex]

where A its an dimensionless constant. Its common for this constants to appears if we need an conversion factor. If we wanted the acceleration in cm/s^2, for example. Luckily for us, the problem states that there is no dimensionless constant involved, so:

[tex]a = r \omega^2[/tex]

Final answer:

The linear acceleration a of a particle in circular motion depends on the radius r of the circle and the square of the angular frequency ω, which is expressed by the formula a = rω².

Explanation:

To determine how the linear acceleration a in m/s² of a particle traveling in a circle depends on the properties such as the radius r of the circle, the angular frequency ω in s⁻¹, and the mass m of the particle, we can use dimensional analysis and known relationships from physics.

Firstly, we recognize that mass m does not directly affect a, because a depends on the force per unit mass. By considering the relationship between linear and angular variables, we know that linear velocity v is related to ω by v = rω, and the centripetal acceleration of an object moving in a circle is ac = v²/r. Substituting v = rω into the expression for centripetal acceleration, we get ac = (rω)²/r = rω².

Therefore, the linear acceleration a of a particle undergoing circular motion is directly proportional to the radius r and the square of the angular frequency ω. This relationship can be expressed as a = rω², where a is the linear acceleration, r is the radius of the circle, and ω is the angular frequency.

Answer:

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

average velocity must be zero

Explanation:

As we know that the distance moved by the boat is given as

[tex]d = 48 km[/tex]

now the time taken by the boat to move to and fro is given as

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

[tex]t = \frac{48 + 48}{48}[/tex]

[tex]t = 2 hrs[/tex]

Time taken by Boat B to cover the distance

[tex]t = \frac{48}{24} + \frac{48}{72}[/tex]

[tex]t = 2.66 h[/tex]

Part 1)

Boat A will win the race

Part 2)

Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line

Part 3)

Since the displacement of Boat A is zero

so average velocity must be zero

Final answer:

Boat A wins the race since it takes a total of 2 hours for the round trip, which is faster than Boat B's 2.6667 hours. Boat A wins by approximately 48 km when Boat B finishes. The average velocity of the winning boat (Boat A) is 48 km/h.

Explanation:

To determine which boat wins and by how much, we need to calculate the time each boat takes to make the round trip across the 48-km-wide lake. For Boat A, since it travels at 48 km/h both ways, the time to go across and back is simply the total distance (48 km there and 48 km back, for a total of 96 km) divided by the speed (48 km/h):

Time for Boat A = Total Distance / Speed = 96 km / 48 km/h = 2 hours.
For Boat B, the time to go across at 24 km/h and come back at 72 km/h must be calculated separately and then summed:

Time to go across for Boat B = Distance / Speed = 48 km / 24 km/h = 2 hours.
Time to come back for Boat B = Distance / Speed = 48 km / 72 km/h = 0.6667 hours.
Total time for Boat B = 2 hours + 0.6667 hours = 2.6667 hours.
Boat A wins since it takes less time to complete the round trip.

To find out by how much Boat A wins, we need to convert the time difference into distance. Since Boat A has already finished when Boat B has 0.6667 hours to go, we calculate how far Boat B would travel in that time at their average speed on the way back (72 km/h):

Difference in distance = Time difference x Speed of Boat B on return = 0.6667 hours x 72 km/h = 48 km.
For the average velocity of the winning boat (Boat A), considering the round trip without any rest or waiting time, it remains constant at 48 km/h because it travels at that speed in both the outward and return journeys.

Answer:

a) 12.56 rad per minute

b)62.8 inches per minute

Explanation:

Given:

a)

Let [tex]\omega[/tex] be the angular velocity of the blade which is given by

[tex]\omega =2\pi f\\\omega=2\times 3.14 \times 2\\\omega=12.56\ \rm rad/min[/tex]

b)

Let v be the linear speed of the tip of the blade which is given by

[tex]v=\omega \times R\\v=12.56\times 5\\v=62.8\ \rm inches/min[/tex]

Hence we have calculated the angular velocity and linear speed.

Answer:

(a) 5.45 m

(b) no change

Explanation:

refractive index of water, n = 1.333

h = 2.4 m

Let the radius of circle is r.

Here we use the concept of total internal reflection.

Let i be the angle of incidence.

By using Snell's law

Refractive index of water with respect to air is the ratio of Sin of angle of incidence to the Sin of angle of refraction.

here refraction takes place from denser medium to rarer medium

So, [tex]\frac{1}{n}=\frac{Sin i}{Sin r}[/tex]

Here, angle r is 90 for total internal reflection

[tex]\frac{1}{1.33}=\frac{Sin i}{Sin 90}[/tex]

Sin i = 0.75

i = 48.6°

According to the triangle ΔOAB

[tex]tan i = \frac{AB}{OA}[/tex]

[tex]tan 48.6 = \frac{r}{2.4}[/tex]

r = 2.73 m

Diameter = 2 r = 2 x 2.73  = 5.45 m

Thus, the diameter of the circle is 5.45 m.

(b) As the value of r depends on the angle of incidence i and angle of incidence depends on the value of refractive index. As refractive index is constant, so the value of diameter remains same.

Due to refraction, a fish sees the world outside through a circle on the water's surface. The diameter of this circle, around 3.6 meters for a fish 2.4 m deep, decreases as the fish descends due to the narrowing range of incident light angles.

The question revolves around the concept of refraction, a phenomenon that occurs when light changes the medium, therefore, also changing direction. This comes into play when light travels from water to air (or vice versa), like how a fish observes the world outside the water body.

In response to the student's question part (a): When the fish is 2.4 m below the water's surface, it sees through a circle on the surface due to refraction. Knowing that the index of refraction of water is 1.333, the apparent depth of the fish to an observer on the surface would be the real depth divided by the refractive index, or around 1.8 meters. Therefore, the light from objects outside the lake would reach the fish within a circle having a diameter of about 3.6 meters (twice the apparent depth).

For part (b) of the question: If the fish were to dive deeper, the diameter of the circle through which it could observe the world outside would diminish. This is because as the actual depth increases, the range of incident angles for light rays entering the water from air, for which total internal reflection doesn’t occur, gets narrower. Therefore, the fish's field of view on the outside world decreases.

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Answer:

[tex]W = 462.5 keV[/tex]

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

[tex]F = qE[/tex]

[tex]F = (1.6 \times 10^{-19})(1.85 \times 10^6)[/tex]

[tex]F = 2.96 \times 10^{-13} N[/tex]

now the distance moved by the electron is given as

[tex]d = 0.25 m[/tex]

so we have

[tex]W = F.d[/tex]

[tex]W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)[/tex]

[tex]W = 7.4 \times 10^{-14} J[/tex]

now we have to convert it into keV units

so we have

[tex]1 keV = 1.6 \times 10^{-16} J[/tex]

[tex]W = 462.5 keV[/tex]

Answer:

1.21 cm.

Explanation:

Given that the wavelength of light is 550 nm.

And the distance between the slits is 0.10 mm.

And the distance to the screen is 2.2 m.

As we know that the lateral separation can be defined by the formula.

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

Here, m is the order of maxima, [tex]\lambda[/tex] is the wavelength , d is the distance between the slits and D is the distance to the screen.

For first maxima m=1, and for second maxima m=2.

Difference in the lateral maxima will be,

[tex]\Delta LS=\dfrac{2\lambda D}{d}-\dfrac{1\lambda D}{d}\\\Delta LS=\dfrac{\lambda D}{d}[/tex]

Substitute all the variables in the above equation.

[tex]\Delta LS=\dfrac{550 nm\times 2.2m}{0.10mm}\\\Delta LS=\dfrac{550\times 10^{-9} nm\times 2.2m}{0.10\times 10^{-3} }\\ \Delta LS=121\times 10^{-4}m\\ \Delta LS=1.21 cm[/tex]

Therefore, the lateral separation difference  between between first order and second order maxima is 1.21 cm.

Final answer:

The weight of a 2.5-kg pumpkin is calculated by multiplying the mass by the acceleration due to gravity, resulting in 24.5 newtons (N) when rounded to one decimal place.

Explanation:

To find out the weight of a 2.5-kg pumpkin, you would need to multiply its mass by the acceleration due to gravity. In most physics problems, this is approximately 9.8 meters per second squared (m/s²). Therefore, the weight W can be calculated using the formula W = m×g, where m is the mass in kilograms and g is the acceleration due to gravity.

So, for a 2.5-kg pumpkin, W = 2.5 kg × 9.8 m/s² = 24.5 kg·m/s². Since weight is a force, it is measured in newtons (N) in the International System of Units (SI). Therefore, the pumpkin weighs 24.5 N.

Final answer:

The weight of a 2.5-kg pumpkin is calculated by multiplying its mass by the gravitational acceleration, which results in a weight of 24.5 Newtons (N) on Earth.

Explanation:

The weight of a pumpkin is the force with which it is pulled towards the Earth due to gravity. To calculate the weight, you can use the formula:

Weight (W) = mass (m) × gravitational acceleration (g)

Where the gravitational acceleration on Earth is typically 9.8 meters per second squared (9.8 m/s²). For a 2.5-kg pumpkin, the calculation would be:

Weight (W) = 2.5 kg × 9.8 m/s²

Weight (W) = 24.5 kg·m/s²

Since weight is measured in Newtons (N), and 1 kg·m/s² is equivalent to 1 Newton, the weight of the pumpkin is 24.5 Newtons (N). When converting measurements and rounding, it's important to consider significant figures and the least precise measurement. In this calculation, both 2.5 kg and 9.8 m/s² are exact enough, so our final answer in Newtons does not need additional rounding.

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.

Answer: a) 23 m/s

Explanation: In order to solve this problem we have to consider the circular movement  where the friction force  change de direcion of the velocty to keep a circular trajectory.

By using Second Newton law, we have:

F=m*acentripeta

μ*N= m*v^2/r  where N  ( equal to mg) is the normal force and μ is the coefficient of friction. r is the redius of the trajectory.

so

the maximun speed permited to keep a circular trajectory is calculated as:

v= (μ*r*g)^1/2=(0.65*80m*9.8 m/s^2)^1/2=22.57 m/s

Answer:

Explanation:

In Both Physics and Math

y=mx+b is plotted as straight line where

m=slope of line

b=intercept on Y-axis

whereas Equation of parabola is something like this

[tex]y^2=4ax[/tex]

or

[tex]x^2=4ay[/tex]

Math is a tool to solve Physics problems so equations are same in math and physics

Answer:

The force will be zero

Explanation:

Due to the symmetric location of the +2μC charges the forces the excert over the +5μC charge will cancel each other resulting in a net force with a magnitude of zero.However in this case it would be an unstable equilibrium, very vulnerable to a kind of bucking. If the central charge is not perfectly centered on the vertical axis the forces will have components in that axis that will add together instead of canceling each other.

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:

[tex]V_{3.01}=-93.2m/s[/tex]

[tex]V_{3.005}=-93.1m/s[/tex]

[tex]V_{3.002}=-93.04m/s[/tex]

[tex]V_{3.001}=-93.02m/s[/tex]

[tex]V_{3}=-93m/s[/tex]

Explanation:

To calculate average velocity we need the position for both instants t0 and t1.

Now we will proceed to calculate all the positions we need:

[tex]Y_{3}=-99m/s[/tex]

[tex]Y_{3.01}=-99.932m/s[/tex]

[tex]Y_{3.005}=-99.4655m/s[/tex]

[tex]Y_{3.002}=-99.18608m/s[/tex]

[tex]Y_{3.001}=-99.09302m/s[/tex]

Replacing these values into the formula for average velocity:

[tex]V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s[/tex]

[tex]V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s[/tex]

[tex]V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s[/tex]

[tex]V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s[/tex]

To know the actual velocity, we derive the position and we get:

[tex]V=27-40t = -93m/s[/tex]

Answer:

5.91 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance = Speed × Time

⇒Distance = 3.1t

Distance traveled by bicycle that passes through = 3.1t

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 2.4\times (t-2)^2\\\Rightarrow s=1.2(t-2)^2[/tex]

They both travel the same distance

[tex]3.1t=1.2(t-2)^2\\\Rightarrow 31t=12\left(t-2\right)^2\\\Rightarrow 12t^2-79t+48=0[/tex]

[tex]t=\frac{79+\sqrt{3937}}{24},\:t=\frac{79-\sqrt{3937}}{24}\\\Rightarrow t=5.91, 0.67[/tex]

Hence, time taken by the bicyclist to catch the other bicyclist is 5.91 seconds

Answer:

R = 668.19 ft  

Explanation:

given,

speed of the ball thrown by the pitcher = 100 mph

to travel maximum distance θ = 45°

distance traveled by the ball = ?

using formula

1 mph = 0.44704 m/s

100 mph = 44.704 m/s

[tex]R = \dfrac{u^2sin2\theta}{g}[/tex]

[tex]R = \dfrac{44.704^2sin2\times 45}{9.81}[/tex]

R = 203.71 m

1 m  = 3.28 ft

R = 203.71 × 3.28

R = 668.19 ft  

hence, ball will go at a distance of 668.19 ft   when pitcher throw it at 100 mph.

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:

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

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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Answer:

a) 99.1 s

b) 1093.2 m

Explanation:

The equation for speed with constant acceleration is

V(t) = V0 + a * t

Mary starts accelerating with a speed of zero, so

V0 = 0

V(t) = a * t

To reach a speed of 12 m/s

t = V(t) / a

t = 12 / 0.75 = 16 s

The equation for position under constant acceleration is

X(t) = X0 + V0 * t + 1/2 * a * t^2

Mary's starting position is zero

X0 = 0

X(t) = 1/2 * a * t^2

X(16) = 1/2 * 0.75 * 16^2 = 96 m

Frank starts skating 3 seconds after Mary, and he accelerates at 1.1 m/s^2 for 9 s. He will stop accelerating at second 12 (9 + 3).

His position after accelerating will be:

X(12) = X0 + V0 * (t - 3) + 1/2 * a * (t - 3)^2

His initial position is 2000, and his initial speed is zero

x(12) = 2000 - 1/2 1.1 * (12 - 3)^2 = 1955.5 m

Shi speed will be

V(12) = -1.1 * (12 - 3) = -9.9 m/s

From there they will move at constant speed from these positions.  We can consider them as moving at constant speed starting at t0 = 16 and t0 = 12 respectively.

For Mary:

X(t) = X0 + V0 * (t - t0)

X(t) = 96 + 12 * (t - 16)

For Frank:

X(t) = 1955.5 - 9.9 * (t - 12)

Equating these two we can find the time when they meet:

96 + 12 * (t - 16) = 1955.5 - 9.9 * (t - 12)

96 + 12*t - 192 = 1955.5 - 9.9*t + 118.8

21.9*t = 2170.3

t = 2170.3 / 21.9 = 99.1 s

Replacing this time value on either equation we get the position:

X(99.1) = 96 + 12 * (99.1 - 16) = 1093.2 m

Answer:

[tex]t=d/(v_{o}*cos(\alpha ))[/tex]

[tex]y =d*tan(\alpha )- 1/2*g^{2}d^{2}/(v_{o}^{2}*(cos(\alpha ))^2)[/tex]

Explanation:

Kinematics equation in the axis X:

[tex]x=v_{o}*cos(\alpha )*t[/tex]

The projectile strikes the building at time t:

[tex]d=v_{o}*cos(\alpha )*t[/tex]

[tex]t=d/(v_{o}*cos(\alpha ))[/tex]  (1)

Kinematics equation in the axis Y:

[tex]y =v_{o}*sin(\alpha )*t - 1/2*gt^{2}[/tex]  (2)

We replace (1) in (2):

[tex]y =v_{o}*sin(\alpha )*d/(v_{o}*cos(\alpha )) - 1/2*g(d/(v_{o}*cos(\alpha )))^{2}[/tex]  

[tex]y =d*tan(\alpha )- 1/2*g^{2}d^{2}/(v_{o}^{2}*(cos(\alpha ))^2)[/tex]  (2)

Answer:

The total time to reach ground is 24.89 seconds

Explanation:

Since the height of the helicopter is given by

[tex]h(t)=3.30t^{3}[/tex] thus at time t = 2.30 seconds the height of the helicopter is

[tex]h(2.30)=3.30\times (2.30)^{3}=40.151m[/tex]

The velocity of helicopter upwards at time t = 2.30 is given by

[tex]v=\frac{dh(t)}{dt}\\\\v=\frac{d}{dt}(3.30t^{3})\\\\v(t)=9.90t^{2}\\\\\therefore v(2.30)=9.90\times (2.30)^2=120.45m/s[/tex]

Now the time after which it becomes zero can be obtained using the equations of kinematics as

1) Time taken by the mailbag to reach highest point equals

[tex]v=u+gt\\\\0=120.45-9.81\times t\\\\\therefore t_{1}=\frac{120.45}{9.81}=12.28s[/tex]

2) Time taken by the mailbag to reach ground from a height of 40.151 meters equals

[tex]s=ut+\frac{1}{2}gt^{2}\\\\40.151=120.45t+4.9t^{2}[/tex]

Solving for t we get[tex]t_{2}=0.3289secs[/tex]

Now the total time of the journey is

[tex]\\\\2\times t_{1}+t_{2}\\\\=2\times 12.28+0.3289=24.89secs[/tex]