Which of the following are examples of 1-D motion?
a) A person pacing back and forth down a hallway in a straight line
b) An airplane traveling from Boise to Seattle
c) A student walking from Boone to KAIC
d) A car driving on a straight road

Answer:

[tex]V_{syrup}= 60ml[/tex] and [tex]V_{elixir}= 40ml[/tex]

Explanation:

Well, we need 100ml of solution, with a concentration of 2mg/ml. So the total amount of drug we need is :

[tex]M_{drug} =2\frac{mg}{ml}\times100ml = 200mg[/tex]

But the drug concentration is  5mg/ml, so the amount of elixir needed to get 200mg of drug is:

[tex]V_{elixir}= \frac{200mg}{5mg/ml}=40ml[/tex]

And so the amount of cherry flavoured syrup needed is:

[tex]V_{syrup}= 100-40 = 60ml[/tex]

Hope my answer helps.

Have a nice day!

Answer:

a) It moved to a lower potential

b) ΔФ = - 86.28 Volt

Explanation:

The energy of an charged particle and the electric potential are related by the potential al kinetic energies:

[tex]\frac{\text{mv}^2}{2} = e\phi[/tex]

If we consider an electron:

m = 9.10938*10^-31 Kilogram

e = 1.60218*10^-19 Coulomb

And the potential diference may be calculed by:

[tex]\Delta \phi =\frac{m \left(v_f^2-v_i^2\right)}{2 e}[/tex]

Replacing all the values we get:

ΔФ = - 86.28 (Kilogram Meter^2)/(Coulomb Second^2) = -86.28 Volts

Final answer:

When an electron's speed decreases, it moves to a higher potential due to the work done by the electric force. The potential difference across which the electron traveled can be calculated using the given formula.

Explanation:

a) It moved to a lower potential

b) ΔФ = - 86.28 Volt

Explanation:

The energy of an charged particle and the electric potential are related by the potential al kinetic energies:

[tex]\frac{\text{mv}^2}{2} = e\phi[/tex]

If we consider an electron:

m = 9.10938 x 10⁻³¹ Kilogram

e = 1.60218 x 10⁻¹⁹ Coulomb

And the potential diference may be calculed by:

[tex]\Delta \phi =(m \left(v_f^2-v_i^2\right))/(2 e)[/tex]

Replacing all the values we get:

ΔФ = - 86.28 (Km²)/(Cs²) = -86.28 Volts

Answer:

[tex]V=1.33m/s[/tex]   to the right

Explanation:

The balls collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V.

We need to use the conservation of momentum Law, the total momentum is the same before and after the collision.

In the axis X:

[tex]m_{1}*v_{o1}-m_{2}*v_{o2}=(m_{1}+m_{2})V[/tex]     (1)

[tex]V=(m_{1}*v_{o1}-m_{2}*v_{o2})/(m_{1}+m_{2})=(2*3-1*2)/(2+1)=1.33m/s[/tex]

Answer:

(a) 5.8s

(b) 41.36m/s

(c) 99.52m

    5.21s

Explanation:

(a) This is the total time it takes the stone to reach its maximum height above the cliff and strike the ground at the base of the cliff after projection.

let the height attained by the stone above the cliff be [tex]h_1[/tex] and the time taken to attain this height be [tex]t_1[/tex]. We can safely assume acceleration due to gravity to be taken as [tex]g=9.8m/s^2[/tex].

We use the first equation of motion under free fall to obtain [tex]t_1[/tex] as follows;

[tex]v=u-gt_1............(1)[/tex]

given: u = 15.5m/s

Where [tex]v[/tex] is the final velocity and u is the initial velocity. The negative sign in the equation indicates the fact that the stone is moving upwards against gravitational pull. The final velocity [tex]v=0[/tex] at height [tex]h_1[/tex] because the stone will momentarily at the maximum height come to rest before it begins to fall back downwards.

Hence from equation (1) we obtain the following,

[tex]0=15.5-9.8t_1\\9.8t_1=15.5\\hence\\t_1=15.5/9.8\\t_1=1.58s[/tex]

To get [tex]h_1[/tex] we use the third equation as follows;

[tex]v^2=u^2-2gh_1[/tex] ( the body is moving upward so g is negative)

[tex]0^2=15.5^2-2*9.8*h_1\\0=240.25-19.6h_1\\19.6h_1=240.25\\therefore\\h_1=240.25/19.6\\h_1=12.26m[/tex]

Next we obtain the time it takes to fall back from the maximum height downwards to the base of the cliff. Let this time be [tex]t_2[/tex]. We use the second equation of motion.

[tex]H=ut+gt_2^2/2............(2)[/tex]

( g is positive because the stone is falling downwards)

However in this case, u = 0 because the stone is falling freely from rest downwards.

[tex]H=h_1+75m=12.26+75\\H=87.26m\\[/tex]

Substituting into equation (2), we obtain;

[tex]87.26=(0*t_2)+9.8t_2^2/2[/tex]

Simplifying further we obtain;

[tex]4.9t_2^2=87.26\\t_2^2=87.26/4.9\\ =17.81\\t_2=\sqrt{17.81}=4.22s[/tex]

Hence the total time spent in air = 1.58+4.22 = 5.8s

(b) We use the third equation of motion to find the velocity with which the stone strikes the ground.

[tex]v^2=u^2+2gH....... (3)[/tex]

the stone is falling downwards in this case from height H from rest, u = 0, v is the final velocity with which is strikes the ground. Equation (3) can therefore be reduced to the following form by putting u = 0;

[tex]v=\sqrt{2gH}\\v=\sqrt{2*9.8*87.26} \\v=41.36m/s[/tex]

(c) The total distance travelled is given as follows;

[tex]H_{total}=h_1+H\\H_{total}=12.26+87.26=99.52m[/tex]

When the package was dropped from the ascending helicopter, it will be projected upwards with an initial velocity equal to that of the helicopter, attain a maximum height and then fall back downwards. The total time spent in air by the package is the sum of the time it takes to attain maximum height and the time it takes to fall to the ground from the maximum height. This solution is similar to that of part (a) of this question.

To find the time it takes to attain maximum height, we use equation (1): v = 0, u = 5.4m/s and g is negative since the package is moving upward against gravity. Hence;

[tex]0^2=5.4^2+9.8t_1\\9.8t_1=5.4\\t_1=5.4/9.8\\t_1=0.55s[/tex]

Similarly to the previous solution, we obtain the maximum height as follows;

[tex]v^2=u^2-2gh_1\\0^2=5.4^2-2*9.8*h_1\\19.6h_1=29.16\\h_1=29.16/19.6\\h_1=1.49m[/tex]

therefore maximum height is

H = 105+1.49 = 106.49m

The time taken by the package to fall from H to the ground is given by equation (2), where u = 0 since the package is falling from rest; g is positive in this case.

[tex]106.49=(0*t)+ 9.8t^2/2\\106.49=4.9t^2\\t^2=106.49/4.9=21.73\\t=\sqrt{21.73}=4.66s[/tex]

therefore the total time spent by the package before striking the ground is given by;

[tex]t_{total}=0.55s+4.66s=5.21s[/tex]

The time it takes for a stone thrown upward from a cliff to reach the bottom, and its speed just before impact, are found using kinematic equations that consider initial velocity, height of the cliff, and acceleration due to gravity.

When a stone is thrown vertically upward with an initial speed from the edge of a cliff, we can calculate how long it takes to reach the bottom by using the equations of motion under constant acceleration due to gravity. In this case, we can use the kinematic equation:

s = ut + ½at²

Where s is the displacement, u is the initial velocity, t is time, and a is the acceleration due to gravity (9.81 m/s²). Let's break down the initial question into parts (a) and (b):

Part (a):

Determine how much later the stone reaches the bottom of the cliff by solving the equation for t, considering that the stone must travel the height of the cliff plus the additional distance it ascends before starting to fall back down.

Part (b):

The speed just before hitting can be calculated using the equation v = u + at, where v is the final velocity. For a comprehensive calculation, more elaborate explanations and use of kinematic formulas are required to solve for the time of flight, final velocity, and total distance traveled.

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Answer:

It must pass 7.1 min

Explanation:

For first order kinetics, the concentration of the reactive in function of time can be written as follows:

[A] = [A]₀e^(-kt)

where:

[A] = concentration of reactant A at time t.

[A]₀ = initial concentration of A

k = kinetic constant

t = time

We want to know at which time the concentration of A is 4 times the concentration of B ([A] = 4.00[B]). These are the equations we have:

[A] = [A]₀e^(-ka*t)

[B] = [B]₀e^(-kb*t)

[A] = 4.00[B]

[A]₀ = [B]₀

Replacing [A] for 4[B] and [A]₀ = [B]₀ in the equation [A] = [A]₀e^(-ka*t), we will get:

4[B] =[B]₀e^(-ka*t)

if we divide this equation with the equation for [B] ([B] = [B]₀e^(-kb*t)), we will get:

4 = e^(-ka*t) / e^(-kb*t)

Applying ln on both sides:

ln 4 = ln(e^(-ka*t) /  e^(-kb*t))

applying logarithmic property (log x/y = log x- log y)

ln 4 = ln (e^(-ka*t)) - ln (e^(-kb*t))

applying logarithmic property (log xⁿ = n log x)

ln 4 = -ka*t * ln e - (-kb*t) * ln e   (ln e = 1)

ln 4 = kb * t - ka *t

ln 4 = (kb -ka) t

t = ln 4 / (3.7 x 10⁻³ s⁻¹ - 4.50 x 10⁻⁴ s⁻¹) = 426. 6 s or 7.1 min

Final answer:

To reach a condition where [A] = 4.00[B], we can use the first-order rate laws to determine the time required. By substituting the given concentrations and rate constants, we can solve for time (t).

Explanation:

The given reaction is A → B. We are given that the rate constant for A is ka = 4.50e-4s^-1 and for B is kb = 3.70e-3s^-1. Since both substances decompose by first-order kinetics, we can determine the time required to reach a condition where [A] = 4.00[B].

Let's denote the initial concentration of both A and B as [A]0 = [B]0. According to the given information, if [B] = 4[B]0, then [A] must be 2[B]0 for the value of the fraction to equal 2. We can set up the following equation using the first-order rate laws:

[A] = [A]0 * e^(-ka * t)

[B] = [B]0 * e^(-kb * t)

Substituting the given values of ka, kb, and the desired concentrations, we can solve for time (t). By substituting [A] = 2[B]0 and [B] = 4[B]0 into the equations, we can solve for t:

Answer:

18.03 s

Explanation:

We have two different types of motions, the criminal moves with uniform motion while the police do it with uniformly accelerated motion. Therefore we will use the equations of these cases. We know that by the time the police reach the criminal they will have traveled the same distance.

[tex]x=vt\\x=x_{0}+v_{0}t+\frac{a}{2}t^2[/tex]

The distance between the police and the criminal when the first one starts the persecution is 0, its initial speed is also zero. So:

[tex]x=(55m/s)t\\x=\frac{6.1m/s^2}{2}t^2=(3.05m/s^2)t^2[/tex]

Equalizing these two equations and solving for t:

[tex](55m/s)t=(3.05m/s^2)t^2\\(3.05m/s^2)t^2-(55m/s)t=0\\t((3.05m/s^2)t-55m/s)=0\\t=0 \\(3.05m/s^2)t-55m/s=0\\t=\frac{55m/s}{3.05m/s^2}=18.03 s[/tex]

Answer:

Yes, there are 2 flaws

Explanation:

Electromagnetics say that, given two particles with charge, they will be attracted if they have opposite charge or repelled if they have identical charge.

Then, we can find two flaws in the Bohr model. The first one is that the electrons move in energy layers (or just layers to get the idea) far from the proton compared with the distance between the electrons themselves. So, if the model was right, how can it be that they don't repel each other? With the same logic, the protons don't repel each other either even though they are all together in the nucleus.

The second flaw, related to what we've just said, is that the electrons can move from one layer to another, but they will always stay at a minimum distance to the proton. How can it be so if it is known that they attract because of their charge signs?

Answer:

a.) [tex]f^{-1} (q) =\frac{C - 100}{2}[/tex]

b.) The inverse function allows me to understand the number of  newspapers articles I can produce using different amounts of money to spend on production costs.

Explanation:

Finding the inverse  function

The cost  of production function is the mathematical formula that  tells how much cost is required to produce certain amount of newspaper articles. The invere function then is the one that tell us the reversed concept:  how many newspapers article you can produce if you have certain amount of money for production. How do we get the inverse function? As the function is solved for C (cost of production), we need now to solve it for q (quantity of newspapers articles). This process is as follows:

That is okay, but what does it mean to us?

The inverse function would at first glance seem to mean the same than the original function but said in other way, but this function it is very useful by itself. Let's think about this case: If you are the head of the printing house and you have a limited budget to produce the articles. How do you know you have enough money to produce the number of articles for all your readers in the city? Just replace C  for your budget and see if the amount of articles will be enough. Just as simple as replacing values. Mathematics at service of printing houses.

Answer: Ok, so you know the acceleration, lets call it A.

now, the velocity will take the form of V= A*t + v0, where v0 is the inicial velocity, in this case the boat starts from the rest, so v0 = 0

integrating again you obtain R = (A*t*t)/2 + r0, and we will take r0 = 0.

so, at a time t₁ we have a velocity V = v = A*t₁

                                                           R = r = (A*t₁*t₁)/2

so a t₂=2*t₁

V= A*2*t₁= 2v

R= 0.5*A*t₁*t₁*4 = 4r

so the answer is c.

Answer:

[tex]E_{net} = (3.765\hat i - 1.207\hat j)kN/C[/tex]

Explanation:

Electric field due to long line charge on position of charge at x = 4 m is given as

[tex]E = \frac{2k\lambda}{r} \hat i[/tex]

so we have

[tex]\lambda = 0.30 \mu C/m[/tex]

now we have

[tex]E = \frac{2(9\times 10^9)(0.30 \mu C/m)}{4}[/tex]

[tex]E = 1350 N/C[/tex]

Now electric field due to the charge present at y = 2.0 m

[tex]E = \frac{kq}{r^2} \hat r[/tex]

[tex]E = \frac{(9\times 10^9)(6 \times 10^{-6})}{2^2 + 4^2}\times \frac{4\hat i - 2\hat j}{\sqrt{4^2 + 2^2}}[/tex]

[tex]E = 603.7 ( 4\hat i - 2\hat j)[/tex]

[tex]E = 2415\hat i - 1207.5 \hat j[/tex]

Now total electric field is given as

[tex]E_{net} = (1350\hat i) + (2415\hat i -1207.5\hat j)[/tex]

[tex]E_{net} = 3765\hat i - 1207.5 \hat j[/tex]

[tex]E_{net} = (3.765\hat i - 1.207\hat j)kN/C[/tex]

Answer:

a)V= 179.056 m/s

b)[tex]a_{r}=843.71 g \ m/s^2[/tex]

Explanation:

Given that

Length of blade = 3.8 m

Rotational speed N= 450 rev/min

We know that

[tex]\omega =\dfrac{2\pi N}{60}\ \frac{rad}{s}[/tex]

[tex]\omega =\dfrac{2\pi \times 450}{60}\ \frac{rad}{s}[/tex]

  ω=47.12 rad/s

Linear velocity

We know that linear velocity V =  ω x r

Here r = 3.8 m

So by putting the values

V =  ω x r

V = 47.12  x 3.8 m/s

V= 179.056 m/s

Radial acceleration  

[tex]a_{r}=\omega ^2r\ m/s^2[/tex]

[tex]a_{r}=47.12^2\times 3.8 \ m/s^2[/tex]

[tex]a_{r}=8437.12 \ m/s^2[/tex]

[tex]a_{r}=843.71 g \ m/s^2[/tex]

The linear speed of the blade tip is approximately 178.6 m/s. The radial acceleration of the blade tip is approximately 864 times the acceleration of gravity.

The two parts of this question relate to basic concepts in physics, specifically relating to circular motion and the relationship between linear speed, radial acceleration, and the acceleration of gravity. The model of a helicopter rotor in question has four blades, each of length 3.80 m, and rotates at a speed of 450 rev/min.

a. What is the linear speed of the blade tip? The linear velocity (v) of the tip of the blade can be calculated using the formula v = rω, where r is the radial distance (half the length of the blade) and ω is the angular velocity. The angular velocity can be converted to rad/s by multiplying the rotational speed (450 rev/min) by 2π rad/rev and then by 1/60 min/s, giving approximately 47 rad/s. Then, v = (3.8 m) * (47 rad/s) gives a linear velocity of approximately 178.6 m/s.

b. What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity? The radial (or centripetal) acceleration (a_r) can be calculated using the formula a_r = ω²r. Substituting the known values gives a_r = (47 rad/s)² * 3.8 m = approximately 8460 m/s². As the acceleration of gravity is approximately 9.8 m/s², the radial acceleration is therefore about 864 times the acceleration of gravity.

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Answer:

2,38kg

Explanation:

Mass in function of time can be found by the formula: [tex]m_{(t)} =m_{0} e^{-kt}[/tex], where [tex]m_{0}[/tex] is the initial mass, t is the time and k is a constant.

Given that a sample decay 1% per day, that means that after first day you have 99% of mass.

[tex]m_{(1)} =m_{0} e^{-k(1)}[/tex], but [tex]m_{(1)}=\frac{99m_{0} }{100}[/tex], so we have [tex]\frac{99m_{0} }{100}=m_{0}e^{-k}[/tex], then [tex]k=-ln(\frac{99}{100})=0.01[/tex]

Now using k found we must to find [tex]m_{(5)}[/tex].

[tex]m_{(5)}=m_{0}e^{-(0.01)5}=2.5kge^{-0.05} =2.5x0.951=2.38kg[/tex]

The formula for exponential decay can be used to calculate the mass of a radioactive sample after a certain time. The equation N = N0 * e^{(-decay rate*t)} can be used, with the decay rate expressed as a negative number. Substituting the given values will give the remaining mass after 5 days.

The mass of a radioactive substance after a certain period can be calculated using the formula for exponential decay, which in this context states that the remaining mass of the substance is equal to its initial mass multiplied by e to the power of the decay rate multiplied by the time.

Given the initial mass (N0) is 2.5kg, the decay rate is 1% (expressed as -0.01 in the formula), and the time (t) is 5 days, the equation is:

N = N0 * e^{(-decay rate*t)}

This simplifies to:

N = 2.5 * e^{(-0.01*5)}

You can calculate this on a scientific calculator or use a programming language with a command for the base of natural logarithms (often designated by 'e').

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Answer:

25.24 m

Explanation:

We are given that a car travels in the positive x direction on  a straight and level road.

We have to find the distance travel by car in 4 s.

Average velocity of car=6.31 m/s

Time =3 s

Distance traveled by the car in 3 sec=[tex]velocity\times time[/tex]

Distance traveled by the  car  in 3 sec=[tex]6.31\times 3=18.93 m[/tex]

Distance traveled by the car in 1 sec=6.93 m

Distance traveled by the car in 4 sec=[tex]6.93\times 4=25.24 m[/tex]

Hence,distance traveled by the car in 4 sec=25.24 m

Using the constant velocity of 6.31 m/s, the car travels a total of 25.24 meters in 4.00 seconds.

To determine how far the car travels in 4.00 seconds, we can use the equation for displacement given that the car is moving with a constant velocity. Since the average velocity over the first 3.00 seconds is given as νav-x = 6.31 m/s, and there's no indication that the velocity changes, we can assume that the car continues to move at this velocity for the remaining 1.00 second.

The total displacement x, can thus be calculated using the equation x = xo + νavt. Assuming the car starts from position xo = 0, the displacement after 4.00 seconds is x = (6.31 m/s) × (4.00 s) = 25.24 m. Therefore, the car travels 25.24 meters in 4.00 seconds.

Answer:

1. 2 second

2. 7.83 second

3. 58.31 m/s

Explanation:

initial velocity, u = 20 m/s

g = 10 m/s^2

1. Let it takes time t1 to reach to maximum height.

At maximum height the final velocity of the ball is zero.

use first equation of motion, we get

v = u + at

0 = 20 - 10 t1

t1 = 2 second

Thus, the time taken to reach to maximum height is 2 second.

2. The maximum height above the cliff is h

Use third equation of motion

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

0 = 20 x 20 - 2 x 10 x h

400 = 20 h

h = 20 m

The total height is 20 + 150 = 170 m

let the time taken by the ball to reach to bottom from maximum height is t2.

use third equation of motion

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

170 = 0 + 0.5 x 10 x t2^2

t2 = 5.83 second

thus, the total time to reach to bottom, t = t1 + t2 = 2 + 5.83 = 7.83 second.

3. Let v be the velocity with which the ball strikes the ground.

Use third equation of motion, we get

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

[tex]v^{2}=0^{2}+2\times 10 \times 170[/tex]

v = 58.31 m/s

Thus, the ball reaches the ground with velocity of 58.31 m/s.

The aluminum pendulum clock would gain time and run faster in a colder environment due to decreased thermal expansion. The period of the pendulum shortens due to the decrease in length of the pendulum. The exact time difference can be calculated using the expansion coefficient of aluminum and the given formulae.

The behavior of the pendulum clock in different temperatures is subject to thermal expansion. The formula P=2π√L/g helps us understand that the period of the pendulum (P) is directly related to the pendulum's length (L). When the temperature decreases, the aluminum pendulum's length will also decrease due to the lower thermal expansion. Consequently, the pendulum's period will shorten causing the clock to gain time, or run faster.

For a more precise calculation, we can use the expansion coefficient of aluminum provided in the question which is 24 x 10^-6. The change in length due to temperature change can be calculated by using the formula ΔL = L0αΔT, where L0 is the original length, α is the linear coefficient of expansion and ΔT is the change in temperature. Hence, the time the clock gains or loses can be calculated with the help of these relations.

Note: It's crucial to recall that while this principle holds, in reality, pendulum clocks are often adjusted to maintain accuracy in varying temperatures.

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The drop in temperature will cause the aluminum pendulum to contract, making it shorter. This will result in the clock running faster and thus gain time.

The question you're asking relates to the phenomenon of thermal expansion and its effect on a pendulum clock. In short, the temperature change will affect the accuracy of the clock, due to the expansion coefficient of aluminum.

When the temperature decreases from 20.0°C to -10.2°C, the aluminum pendulum will contract, due to the negative temperature change multiplied by the expansion coefficient of aluminum (24 x 10^-6). As a result, the length of the pendulum decreases which in turn, using the period equation P = 2π√L/g results in a shorter period.

As a result, the clock will run fast, or gain time. To understand why, consider that when the pendulum is shorter, it swings faster and hence takes less time to complete one swing. That means the clock ticks faster than normal, making it gain time.

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Answer:

The answer to your question is Decrease

Answer:

Explanation:

A steering wheel, a driving wheel or a hand wheel allow us to control the vehicle, it's part of the steering. So the faster the vehicle goes, less movement of the wheel is needed, because the movement of the vehicle makes easier to handle the wheel. Also, when the vehicle is almost not moving or moving slowly, we have to put more effort to move the wheel.

In addition, this want of the reasons why when we are driving too fast, we must put attention on our wheel, because a simple and short movement can get us out of the road.

Answer: Option a.

Explanation:

Kepler's 2nd law of planetary motion states:

A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

It tells us that it doesn't matter how far Earth is from the Sun, at equal times, the area swept out by Earth's orbit it's always the same independently from the position in the orbit.

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity [tex]u_{y}=0[/tex]

The initial velocity is the horizontal component [tex]u_{x}[/tex]

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = [tex]u_{y}[/tex] t + [tex]\frac{1}{2}[/tex] gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , [tex]u_{y}=0[/tex]

Substitute these values in the rule

→ -60 = 0 + [tex]\frac{1}{2}[/tex] (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* The ball is in the air for 3.5 seconds

The initial velocity is the horizontal component [tex]u_{x}[/tex]

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = [tex]u_{x}[/tex] t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = [tex]u_{x}[/tex] (3.5)

Divide both sides by 3.5

→ [tex]u_{x}[/tex] = 28.57 m/s

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is [tex]v_{y}[/tex]

→ [tex]v_{y}[/tex] = [tex]u_{y}[/tex] + gt

→ [tex]u_{y}[/tex] = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ [tex]v_{y}[/tex] = 0 + (-9.8)(3.5)

→ [tex]v_{y}[/tex] = -34.3 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity v is the resultant vector of  [tex]v_{x}[/tex] and [tex]v_{y}[/tex]

→ Its magnetude is [tex]v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}[/tex]

→ Its direction [tex]tan^{-1}\frac{v_{y}}{v_{x}}[/tex]

→ [tex]v_{y}[/tex] = 28.6 , [tex]v_{y}[/tex] = -34.3

Substitute this values in the rules above

→ [tex]v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66[/tex]

→ Its direction [tex]tan^{-1}\frac{-34.3}{28.6}=-50.18[/tex]

The negative sign means the direction is below the horizontal

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

The ball remained in the air for 3.49 seconds. The initial horizontal velocity was approximately 28.65 m/s. Just before hitting the ground, the ball had a vertical velocity of approximately 34.21 m/s downwards and an overall velocity of approximately 45.1 m/s at an angle of roughly 49.8 degrees below the horizontal.

The time a projectile, such as the ball in this case, is in the air is completely determined by vertical motion. To calculate this, we can use the equation, y = y0 + v0yt - 0.5gt^2, where y is the final position, y0 is the initial position (the height of the building, 60 m in this case), v0y is the initial vertical velocity (which is 0 since the ball is thrown horizontally), g is acceleration due to gravity (approximated as 9.8 m/s^2), and t is the time we are looking for. By setting y to 0 (the base of the building where the ball lands), we can solve for t which gives us 3.49 seconds.

Next, the initial horizontal component of the velocity can be calculated using distance/time, which in this case is 100 m (distance traveled by the ball) divided by 3.49 seconds (the time the ball was in the air), giving us approximately 28.65 m/s.

The vertical component of the velocity just before the ball hit the ground can be calculated using v = v0y + gt, where v0y is still 0 and t is the time the ball is in the air. This calculation gives us approximately 34.21 m/s, in the downward direction, taking into account that acceleration due to gravity acts downward.

By combining the horizontal and vertical components, we can determine the velocity of the ball just before it hits the ground. The magnitude of this velocity can be found by using the Pythagorean theorem (v = sqrt((v0x)^2 + (v0y)^2)) which will give us approximately 45.1 m/s and the angle it makes with the horizontal can be calculated through the formula theta = tan^-1(v0y/v0x) giving us approximately 49.8 degrees below the horizontal.

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Answer:

The correct answer is inertia.

Explanation:

The heavy bag of groceries is initially within the inertia frame of the car. This indicates that the heavy bag acquires the same speed as the car.

When the car stops, the heavy bag continues to move forward with the speed it had due to the principle of inertia, which states the property that the bodies cannot modify by themselves the state of rest or movement in which they are.

Have a nice day!

Explanation:

Given that,

Force with which a child pulls a sled, F = 60 N

It is at an angle of  39° above the horizontal. We need to find the horizontal and vertical components of the force.

The horizontal component is given by :

[tex]F_x=30\ cos(39)=23.31\ N[/tex]

The vertical component is given by :

[tex]F_y=30\ sin(39)=18.87\ N[/tex]        

So, the horizontal and vertical components of the force are 23.31 N and 18.87 N. Hence, this is the required solution.

To find the horizontal and vertical components of the force, use the formulas Fx = F * cos(angle) and Fy = F * sin(angle), respectively.

To find the horizontal and vertical components of the force, we can use trigonometry. The horizontal component of the force can be found using the formula: Fx = F * cos(angle), where F is the magnitude of the force and angle is the angle above the horizontal. Plugging in the values, we get Fx = 60 N * cos(39°) = 45.85 N.

The vertical component of the force can be found using the formula: Fy = F * sin(angle), where F is the magnitude of the force and angle is the angle above the horizontal. Plugging in the values, we get Fy = 60 N * sin(39°) = 36.34 N.

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Answer:

If two objects are dropped at the same height, the heavier object will hit the ground first.

Explanation:

A proper hypothesis has the form:

"If ... , then ... ".

Now, why the other sentence aren't hypothesis?

This sentence its an OPINION. It may be true that, for the speaker/writer, orange popsicles taste sweeter, but this is an subjective opinion, and difficult to test.

This sentence its an STATEMENT. If we wish to make an proper hypothesis, it should take the form:

If children wear backpacks, then they do better academically.

This sentence its another OPINION.  This opinion COULD be tested, but the proper way of write it as an hypothesis would be:

If someone listed to loud music only three hours a day, then there shoul be no effect on his/her hearing.

Answer:

Answer is:If two objects are dropped at the same height, the heavier object will hit the ground first.

Explanation:

Answer: It appears curved

Explanation:

Answer: like a linear equation where the slope is the constant value of the acceleration.

Explanation:

As you know, acceleration is the rate of change of the velocity over time, so if we integrate the acceleration over the time, we obtain the equation for the velocity (plus a constant, which is the initial velocity)

If we have a constant acceleration;

a(t) = c

we integrate it and get:

v(t) = c*t  + constant.

This is a linear equation, so the graph where the acceleration is constant, looks like a linear equation with slope equal to the constant.

Answer:

Q = 62383.44 Joules

Explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water, [tex]T_i=4^{\circ} C[/tex]

Final temperature of water, [tex]T_f=25^{\circ} C[/tex]

The specific heat capacity of liquid water is, [tex]c=4.184\ J/g\ ^oC[/tex]

Heat transferred is given by :

[tex]Q=mc(T_f-T_i)[/tex]

[tex]Q=710\times 4.184\times (25-4)[/tex]

Q = 62383.44 Joules

So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.

Answer:

To a tenth of a mililiter (0.1mL)

Explanation:

When considering the precision of a measurement system, a more precise intrument needs to record smaller intervals of data; a better resolution. In this case we have a cylinder with a resolution of 1 mL. When we pour the amount of water we can't precisely measure anything less than 1 mL UNLESS it is added to the pipet described in this problem which records measurements of up to 0.1 mL (a tenth of a mililiter). Thus the total measurement system can only report a resolution of up to 0.1 mL

Answer:

[tex]a_c = 13.26 m/s^2[/tex]

Friction Force

Explanation:

As we know that centripetal force is the product of mass and centripetal acceleration

so we know that

[tex]a_c = \frac{v^2}{R}[/tex]

so here we have

[tex]v = 27.5 m/s[/tex]

[tex]R = 57 m[/tex]

so we have

[tex]a_c = \frac{27.5^2}{57}[/tex]

[tex]a_c = 13.26 m/s^2[/tex]

This acceleration is given by the force which may be towards the center of the circular path

Here in the above case it is possible due to friction force.

Final answer:

The centripetal acceleration of the race car is 3.45 m/s². Centripetal force is calculated to be 1725 N, and the normal force is responsible for the centripetal acceleration.

Explanation:

The centripetal acceleration of the race car is 3.45 m/s². Centripetal force on the race car can be calculated using the formula F = m * a, where m is the mass of the car and a is the centripetal acceleration. This makes the centripetal force 1725 N. In this case, the force responsible for the centripetal acceleration is the normal force.

The force needed to move a body in a curved way is understood as centripetal force. The centripetal force experienced by a sample will be 7.14 N.    

The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.

The direction of centripetal force is always in the path of the center of the course.

The given data in the problem;

r is the radius =13.1 cm=0.13 m

n is the rpm = 50000.0 rev/min  

To find velocity first we have to go to angular velocity

[tex]\omega = \frac{2\pi n}{60} \\\\ \omega = \frac{2\times3.14\times50000}{60} \\\\\omega = 132.63\; rad/sec.[/tex]

The centripetal force is given as

[tex]\rm F_C= m\omega^{2} r\\\\F_C= 0.00310(132.63)^2 \times 0.131\\\\\rm{F_C=7.14 N[/tex]

Hence The centripetal force experienced by a sample will be 7.14 N.    

To learn more about the centripetal force refer to the link;

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Answer:

Vb = 41.74m/s

Explanation:

Let Voc be the velocity of the club before the collision, Vfc the velocity of the club after the collision, Vfb the velocity of the ball after the collision.

The masses of the club and the ball expressed in kg are:

mc = 0.16kg       mb = 0.046kg

By conservation of the moment:

Po = Pf

mc*Voc + mb *0 = mc*Vfc + mb*Vfb   Solving for Vfb:

[tex]Vfb = \frac{mc}{mb}*(Voc - Vfc) = 41.74m/s[/tex]

Answer:

a)n=0.5 mol

b)n=0.5 mol

c)U/Q=0.714

Explanation:

Given that

Mass of oxygen m= 16 g

Heated from 26.4°C to 111°C

Molar weight M=32 g/mol

a)

Number of mole,n

n=m/M

n=16/32 mol

n=0.5 mol

b)

Heat,Q

Q= n Cp ΔT

As we know that it is diatomic gas so

Cp=(7/2)R

Q=7 R n ΔT/2

Now by putting the values

[tex]\Delta U=\dfrac{7nR\Delta T}{2}[/tex]----1

[tex]Q=\dfrac{7\times 8.314\times 0.5\times (111-26.4)}{2}[/tex]

Q=1230.88 J

We know that internal energy

[tex]\Delta U=\dfrac{5nR\Delta T}{2}[/tex]--------2

From equation 1 and 2

U/Q= 5/7

U/Q=0.714

In 16.0 g of oxygen, there are 0.500 moles of oxygen. To calculate the energy transferred to the oxygen as heat and the fraction of heat used to raise the internal energy, specific heat capacity and the degree of freedom for rotational motion are required, which are not provided.

A student asked how many moles of oxygen are present in 16.0 g of O₂ and how much energy is transferred when the oxygen is heated from 26.4°C to 111°C at constant pressure, given that the molecules rotate but do not oscillate. The following steps provide the answers:

To answer (a), we use the formula:

16.0 g O₂ x (1 mol O₂ / 32.00 g O₂) = 0.500 mol O₂

The student now knows that there are 0.500 moles of oxygen in 16.0 g of O₂.

Parts (b) and (c) would require additional information such as the specific heat capacity of oxygen, which was not provided in the question. Generally, the energy transferred depends on the specific heat capacity, the change in temperature, and the number of moles. The internal energy for gases increases with temperature, but for a more detailed answer, specific values are needed.

Final answer:

The maximum distance traveled to the right of the origin will change if the initial velocity is doubled.

Explanation:

To find the maximum distance traveled by the person to the right of the origin, we can use the equation:

d = v0t + 0.5at2

Given the initial velocity (vx,0) of 16 m/s, we can substitute it into the equation along with the constant acceleration (-4 m/s2) and the time (t) to calculate the maximum distance.

d = (16 m/s)(t) + 0.5(-4 m/s2)(t2)

The maximum distance will change as we vary the time, but it will be greater than the distance traveled with an initial velocity of 8 m/s.

When heated, dry ice undergoes sublimation, transitioning directly from a solid to a gas. It bypasses the liquid phase, turning directly into carbon dioxide gas.

When the temperature of dry ice, which is the solid form of carbon dioxide, rises, it undergoes a process called sublimation. This is a phase transition where a substance moves directly from the solid phase to the gas phase, without passing through the liquid phase. So instead of melting and becoming liquid when heated, dry ice turns directly into carbon dioxide gas.

This is quite a unique characteristic, as most forms of matter have to pass through a liquid state before becoming a gas. The phenomenon of sublimation can also be observed in other substances like iodine and naphthalene (the main ingredient in mothballs).

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