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.
Explanation: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:
Determine the number of moles of oxygen: Use the molar mass of oxygen (32.0 g/mol) to convert grams to moles.Calculate the energy transferred as heat: Use the specific heat capacity of oxygen and the temperature change to find the amount of energy transferred to the oxygen (this could require accessing the specific heat capacity from a table or reference).Find the fraction of heat used to raise the internal energy: Since the oxygen molecules rotate but do not oscillate, we use the degrees of freedom for a diatomic molecule in the rotational energy mode to determine the internal energy change.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.
A 2 kg ball of putty moving to the right at 3 m/s has a perfectly inelastic, head-on collision with a 1 kg ball of putty moving at 2 m/s. What is the final magnitude and direction (left or right) of the velocity of the balls after the collision?
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]
Over a span of 8.0 seconds, an object moves 16 m to the left relative to where it began. If we treat left as the direction towards increasingly negative displacement, what is the average velocity of the object during this motion?
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
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. When each of the spheres has lost half its initial charge, the magnitude of the electrostatic force will be
a) 1/16 Fb) 1/8 Fc) 1/2 Fd) 1/4 F
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
Theophylline, an asthma medicine, is to be prepared from an elixir with a drug concentration of 5 mg/mL and a cherry-flavored syrup that is to be added to hide the taste of the drug. How much of each must be used to prepare 100 milliliters of solution with a drug concentration of 2 mg/mL?
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!
An infinite line charge of linear density λ = 0.30 µC/m lies along the z axis and a point charge q = 6.0 µC lies on the y axis at y = 2.0 m. The electric field at the point P on the x axis at x = 4.0 m is approximately.a. (4.2 kN/C) b. (4.2 kN/C) i + (0.64 kN/C) j c. (-0.96 kN/C) j d. (2.8 kN/C) i + (0.64 kN/C) j e. (5.2 kN/C) i = (2.3 kN/C) j
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]
You measure water in two containers: a 10-mL graduated cylinder with marks at every mL, and a 1-mL pipet marked at every 0.1 mL. If you have some water in each of the containers and add them together, to what decimal place could you report the total volume of water?
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
The amount of steering wheel movement needed to turn will ____________ the faster you go.
Answer:
The answer to your question is Decrease
Answer:
The amount of steering wheel movement needed to turn will decrease the faster you go.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.
High-speed stroboscopic photographs show that the head of a 160-g golf club is traveling at 43 m/s just before it strikes a 46-g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 31 m/s. Find the speed of the golf ball just after impact.
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]
As an electron moves through a region of space, its speed decreases from 6.03 × 106 m/s to 2.45 × 106 m/s. the electric force is the only force acting on the electron. (a) did the electron move to a higher potential or a lower potential? (b) across what potential difference did the electron travel?
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
A car traveling initially at +7.9 m/s accelerates uniformly at the rate of +0.72 m/s2 for a distance of 265 m. What is its velocity at the end of the acceleration? Answer in units of m/s. 10.0 points What is its velocity after it accelerates for 114 m? Answer in units of m/s
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]
Tarzan is testing the strength of a particular vine, which is 7 m long. As he is hanging on the vine, what is the magnitude of the tension force in the vine? (Assume that Tarzan's mass is 80 kg.)
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.
What is tension force?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.
To learn more about the tension force refer to the link;
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The sun is 1.5 × 108 km from Earth. The index of refraction for water is 1.349. How much longer would it take light from the sun to reach Earth if the space between them were filled with water rather than a vacuum?
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
A loaf of bread (volume 3100 cm3) with a density of 0.90 g/cm3 is crushed in the bottom of the grocery bag into a volume of 1240 cm3. What is the density of the mashed bread? . g/cm3
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.
On a velocity vs time graph, what does a line of constant acceleration look like?
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.
A water rocket can reach a speed of 95 m/s in 0.050 seconds from launch.
What is its average acceleration?
An electric field is constant at every point on a square surface that is 0.80 m on a side. This field has a magnitude of 3.5 N/C and is oriented at an angle of 35° with respect to the surface, as the drawing shows. Calculate the electric flux ΦE passing through the surface.
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.
The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate parameter of 1% per day. A sample of this radioactive substance has an initial mass of 2.5kg. Find the mass of the sample after five days.
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.
Explanation: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').
Learn more about Exponential Decay here:https://brainly.com/question/2193799
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A car travels in the +x-direction on a straight and level road. For the first 3.00 s of its motion, the average velocity of the car is υav−x = 6.31 m/s .
How far does the car travel in 4.00s?
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.
Explanation: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.
What is the boiling point of an benzene solution that freezes at 0.3 degrees C? Normal freezing point is 5.5 and the normal boiling point is 80.1 degrees C. Kfp = 5.12 K/m and Kbp = 2.53 K/m. Enter your answer to 2 decimal places.
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]
Jane is riding in a hot air balloon that is rising vertically at a constant speed of 3 m/s over a lake. She reaches out and drops a rock from the balloon when the distance from the rock to the water is 50 m. Use g = 10 m/s2, and let the up direction be positive. About how long after Jane drops the rock will it splash into the water?
Answer:
The rock will splash into the water [tex]3.47s[/tex] after Jane drops it.
Explanation:
Hi
Known data[tex]v_{i}=3m/s, y_{i}=50, y_{f}=0m[/tex] and [tex]g=10m/s^{2}[/tex].
We are going to use the formula below[tex]y_{f}=y_{i}+v_{i}t-\frac{1}{2} gt_^{2}[/tex]
Letting up direction be positive and computing with the known data, we have[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.
A powerboat, starting from rest, maintains a constant acceleration. After a certain time 't', its displacement and velocity are 'r' and 'v'. At time '2t', what will be its displacement and velocity, assuming the acceleration remains constant?
a) 2r and 2v
b) 2r and 4v
c) 4r and 2v
d) 4r and 4v
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.
A person cannot see clearly objects closer than 2.00 m from her/him. What is the refractive power (in diopters) of the correction lenses that will allow her/him to clearly see objects located 30.0 cm, but not closer, from her eyes? Ignore the distance between her/his eyes and the lenses.
Answer:24
Explanation:
The refractive power of the correction lenses is - 2.833 diopters.
What is refractive power of lens?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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A ball is thrown upward at initial velocity of 20 m/s at the edge of a cliff 150 m high. 1. Find time it takes to reach the top of the path.2. Find total time it takes to reach bottom.3. Find final velocity at bottom.
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.
Consider two reaction vessels, one containing A and the other containing B with equal concentrations at t= 0/ If both substances decompose by first order kinetics where ka =4.50e-4s^-1 kb=3.70e-3s^-1 how much time must pass to reach a condition such that [A] = 4.00[B]?
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: