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.
Explanation: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.
Learn more about Projectile Motion here:https://brainly.com/question/29545516
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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.
A water rocket can reach a speed of 95 m/s in 0.050 seconds from launch.
What is its average acceleration?
The cost of producing q newspaper articles is given by the function C = f(q) = 100 + 2q. a.) Find a formula for the inverse function. b.) Explain in practical terms what the inverse function tells you.
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:
[tex]C = 100 + 2q[/tex] (That is the cost function)Substract 100 from the function: [tex]C - 100 = 2q[/tex]Divide by 2: [tex]\frac{c - 100}{2} = q[/tex]Reorganizing: [tex]q = \frac{C - 100}{2}[/tex]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.
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.
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
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
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 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]
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]
Imagine that you are driving in your car with a heavy bag of groceries on the front seat and you slam on your brakes to stop. The car stops but the heavy bag continues to move forward. The concept that best explains why your bag of groceries moves forward is _
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!
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.
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]
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.
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 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.
Which of the following is the best example of a proper hypothesis? Orange popsicles taste sweeter than grape popsicles. If two objects are dropped at the same height, the heavier object will hit the ground first. Children who wear backpacks do better academically. I think there is no effect on hearing if you listen to loud music only three hours a day.
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?
Orange popsicles taste sweeter than grape popsicles.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.
Children who wear backpacks do better academically.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.
I think there is no effect on hearing if you listen to loud music only three hours a day.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:
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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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:
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').
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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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Suppose 16.0 g of oxygen (O2) is heated at constant atmospheric pressure from 26.4°C to 111°C. (a) How many moles of oxygen are present? (Take the molar mass of oxygen to be 32.0 g/mol) (b) How much energy is transferred to the oxygen as heat? (The molecules rotate but do not oscillate.) (c) What fraction of the heat is used to raise the internal energy of the oxygen?
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 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.
An aluminum clock pendulum having a period of 1.0000 s keeps perfect time at 20.0°C. When placed in a room at a temperature of -10.2°C, will the clock gain time (run fast) or lose time (run slow)? use 24 x 10^-6 for the expansion coefficient of aluminum.
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.
Explanation: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.
Explanation: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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A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting?
(c) What total distance did it travel?AAI8yoZA helicopter is ascending vertically with a speed of 5.40 m/s. At a height of 105 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground?
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.
Explanation: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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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
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!
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]
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]
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
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