A pinball bounces around its machine before resting between two bumpers. Before the ball came to rest, its displacement (all angle measuresup from x axis) was recorded by a series of vectors.83 cm at 90 degrees59cm at 147 degrees69cm at 221 degrees45cm at 283 degrees69cm at 27 degreesWhat is the magnitude and the direction?

Answer:

175s

Explanation:

time it takes sunlight to reach the earth in  vacuum

C=light speed=299792458m/s

X=1.5x10^8km=1.5x10^11m

c=X/t

T1=X/c

T1=1.5X10^11/299792458=500.34s

time it takes sunlight to reach the earth in  water:

First we calculate the speed of light in water taking into account the refractive index

Cw=299792458m/s/1.349=222233104.5m/s

T2=1.5x10^11/222233104.5m/s=675s

additional time it would take for the light to reach the earth

ΔT=T2-T1=675-500=175s

Answer:

[tex]\phi = 1.28 Nm^2/C[/tex]

Explanation:

As we know that electric flux is defined as

[tex]\phi = \vec E . \vec A[/tex]

now we have

[tex]A = L^2[/tex]

[tex]A = 0.80^2 = 0.64 m^2[/tex]

[tex]E = 3.5 N/C[/tex]

also we know that electric field makes and angle of 35 degree with surface

so angle made by electric field with Area vector is given as

[tex]\theta = 90 - 35 = 55[/tex]

[tex]\phi = EAcos\theta[/tex]

[tex]\phi = (3.5)(0.64)cos55[/tex]

[tex]\phi = 1.28 Nm^2/C[/tex]

Final answer:

The electric flux through the surface is calculated using the formula ΦE = E · A · cos(θ), resulting in an electric flux of 1.84 N·m²/C for a uniform electric field of 3.5 N/C at an angle of 35° to a 0.80 m square surface.

Explanation:

The area of the surface (A) can be found by squaring the length of one side of the square. For a square that is 0.80 m on a side, A = (0.80 m)² = 0.64 m².

The electric flux is given by the equation ΦE = E · A · cos(θ), where E is the magnitude of the electric field, A is the area of the surface, and θ is the angle between the field and the normal to the surface.

Given that E = 3.5 N/C and θ = 35°, the flux through the surface is ΦE = 3.5 N/C · 0.64 m² · cos(35°).  

The cosine of 35° is approximately 0.819. So, the electric flux is ΦE = 3.5 · 0.64 · 0.819 = 1.84 N·m²/C.

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.

Answer:

2,25 g/cm3

Explanation:

Hi, you have to know one thing for this.. Density = mass/Volume,

When you have the loaf of bread with 3100 cm3 and a density of 0.90 g/cm3, the mass of that bread is 2790 g because of if you isolate the variable mass from the equation you get..  mass= density x volume

Later, have on account the mass never changes, so you crush the bread and the mass is the same.. so when you have the mashed bread.. you know that the mass is 2790 g and the volume of the bag is 1240 cm3, so you apply the main equation.... density=2790 g / 1240 cm3 , so density =  2,25 g/cm3

Final answer:

The density of the mashed bread is 2.25 g/cm³, calculated by dividing the mass of the original loaf (found by multiplying its original density and volume) by the new volume after being crushed.

Explanation:

The density of the original loaf of bread is given as 0.90 g/cm³, and its initial volume is 3100 cm³. When the bread is crushed, its volume is reduced to 1240 cm³. Density is the ratio of the mass of a substance to its volume. Because the mass of the bread remains the same even when it is crushed, we can find the new density by using the mass from the original loaf. The mass can be calculated by multiplying the original density by the original volume. The calculated mass is then divided by the new volume.

Mass = Original density × Original volume = 0.90 g/cm³ × 3100 cm³ = 2790 grams

Now, we divide the mass by the new volume to get the density of the mashed bread:

Density of mashed bread = Mass / New volume = 2790 g / 1240 cm³ = 2.25 g/cm³.

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:

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:

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:

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!

Answer:

The rock will splash into the water [tex]3.47s[/tex] after Jane drops it.

Explanation:

Hi

[tex]v_{i}=3m/s, y_{i}=50, y_{f}=0m[/tex] and [tex]g=10m/s^{2}[/tex].

[tex]y_{f}=y_{i}+v_{i}t-\frac{1}{2} gt_^{2}[/tex]

[tex]0=50+3t-\frac{1}{2} 10t_^{2}[/tex]

[tex]0=50+3t-5t_^{2}[/tex] a second-grade polynomial, we obtain two roots, so [tex]t_{1}=3.47s[/tex] and [tex]t_{2}=-2.87s[/tex], due negative root has no sense, we take the positive one, so the rock will splash into the water [tex]3.47s[/tex] after Jane drops it.

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:

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:

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: 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:

[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:

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: The boiling point of an benzene solution is [tex]82.57^0C[/tex]

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=T_f^0-T_f=(5.5-0.3)^0C=5.2^0C=5.2K[/tex] = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

[tex]K_f[/tex] = freezing point constant = [tex]5.12K/m[/tex]

m= molality

[tex]5.2=1\times 5.12\times m[/tex]

[tex]m=1.015[/tex]

Elevation in boiling point is given by:

[tex]\Delta T_b=i\times K_b\times m[/tex]

[tex]\Delta T_b=Tb-Tb^0=(Tb-80.1)^0C[/tex] = elevation in boiling point

i= vant hoff factor = 1 (for non electrolyte)

[tex]K_b[/tex] = boiling point constant = [tex]2.43K/m[/tex]

m= molality = 1.015

[tex](T_b-80.1)^0C=1\times 2.43\times 1.015[/tex]

[tex]T_b=82.57^0C[/tex]

Thus the boiling point of an benzene solution is [tex]82.57^0C[/tex]

Answer:

the tension generated on the vine is equal to 784.8 N

Explanation:

As Tarzan is testing the strength of vine

to test the vine Tarzan hangs on it.

given,

mass of Tarzan  = 80 kg

acceleration due to gravity = 9.81 m/s²

tension on the wire will be            

     T = m × g                    

     T = 80 kg × 9.81 m/s²                

     T = 784.8 kg.m/s²                

     T = 784.8 N                    

hence, the tension generated on the vine is equal to 784.8 N

Tension force is the force produced when a load is applied in a direction away. The magnitude of the tension force in the vine will be  784.8 N.

Tension force is the force produced when a load is applied in a direction away from one or more ends of a material, usually to the cross-section of the material.

Tension is frequently described as a "pulling" force. To constitute a tension force, the load supplied to the material must be exerted axially.

The given data in the problem is;

m is the mass of Tarzan  = 80 kg

g is the acceleration due to gravity = 9.81 m/s²

T is the tension on the wire +?

The tension of the wire is given by;

[tex]\rm T = m\times g \\\\ \rm T = 80\times 9.81 \\\\ \rm T =784.8 N[/tex]

Hence the magnitude of the tension force in the vine will be  784.8 N.

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Answer: c) 1/4F

Explanation: In order to explain this result we must use the Coulomb force expression, that is:

F=(k*q1*q2)/d^2 where d is the distance between spheres, k a constant equal to 9*10^9N^2.m^2/C

after time each sphere lost haft of original charge and taking into account that F is directly proportional to charge of each sphere,

we have Fafter=(k*q1/2*q2/2)/d^2

then Fafter=(1/4)(k*q1*q2)/d^2=(1/4)Finitial

Answer:

-2 m/sec

Explanation:

We have given time t = 8 sec

As the object move in left relative to where it began and it is increasingly negative direction

So displacement = -16 m

We have to find the velocity

So velocity [tex]v=\frac{displacement}{time}=\frac{-16}{8}=-2m/sec[/tex]

As the velocity is a vector quantity so negative sign has significance its the direction of the velocity

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.

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]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:24

Explanation:

The refractive power of the correction lenses is - 2.833 diopters.

Refractive power in optics refers to how much light is converged or diverged by a lens, mirror, or other optical system. It is equivalent to the device's focal length's reciprocal.

Too much power in a myopic eye causes light to focus in front of the retina. A negative power is shown for this. In contrast, a hyperopic eye lacks insufficient strength, causing light to focus behind the retina when the eye is relaxed.

According to the question:

Object distance: u = 30.0m.

Image distance: v = 2.0 m = 200.0 cm.

Let the focal length of the lens be f, then:

1/f = 1/v - 1/u

= 1/200 - 1/30

f = - 35.29 cm

Hence, the refractive power of the correction lenses is = 100/f = 100/(-35.29) = - 2.833 diopters.

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Answer:

in first case velocity =21.07 m/sec in second case velocity =15.05 m/sec

Explanation:

We have given initial velocity u = 7.9 m/sec

Acceleration [tex]a=0.72m/sec^2[/tex]

Distance S = 265 m

Now according to third law of motion [tex]v^2=u^2+2as[/tex] here v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance

So [tex]v^2=7.9^2+2\times 0.72\times 265=444.01[/tex]

v = 21.07 m/sec

In second case s =114 m

So [tex]v^2=7.9^2+2\times 0.72\times 114=226.57[/tex]

v =15.05 m/sec

Answer:

1. 21.07 m/s

2. 15.05 m/s

Given:

initial speed of the car, u = 7.9 m/s

distance covered by the car, d = 265 m

acceleration, a = 0.72[tex]m/s^{2}[/tex]

Solution:

To calculate the velocity at the end of acceleration, we use the third eqn of motion:

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

[tex]v^{2} = 7.9^{2} + 2\times 0.72\times 265[/tex]

[tex]v = \sqrt{7.9^{2} + 2\times 0.72\times 265} = 21.07 m/s[/tex]

Now,

Velocity after it accelerates for a distance for 114 m:

Here d = 114 m

Again, from third eqn of motion:

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

[tex]v = \sqrt{7.9^{2} + 2\times 0.72\times 114} = 15.05 m/s[/tex]

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:

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) 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:

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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