Answer:
12.27 ft/s
Explanation:
At 72 ft above the ground, the balloons height increases at a rate of 4ft/s. For 66s, vertical distance moved, y = 4ft/s × 66 s = 264 ft. When the balloon is at 72 ft above the ground, just below it, the bicycle which moves at a rate of 12 ft/s in 66 s, covers a horizontal distance, x = 12ft/s 66 = 792 ft.
The distance between the bicycle and the balloon 66 s later is given by
s = √(x² + (y + 72)²) = √(792² + (264 + 72)²) = √(792² + 336²) = √740160 ft = 860.33 ft
From calculus
The rate of change of the distance between the balloon and bicycle s is obtained by differentiating s with respect to t. So,
ds/dt = (1/s)(xdx/dt + ydy/dt)
dx/dt = 12 ft/s, x = 792 ft, dy/dt = 4 ft/s, y = 264 ft, s = 860.33. These are the values of the variables at t = 66 s.
So, substituting these values into ds/dt, we have
ds/dt = (1/860.33)(792 ft × 12 ft/s + 264 ft × 4ft/s) = (1/860.33)(9504 + 1056) = 10560/860.33 = 12.27 ft/s
If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force F120 on the lid when the air inside the cooker had been heated to 120∘C? Assume that the temperature of the air outside the pressure cooker is 20∘C (room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be pa
Answer:
The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is [tex]\frac{135.9}{A}N[/tex]
Explanation:
Here we have
Initial temperature of air T₁ = 20 °C = 293.15 K
Final temperature of air T₁ = 120 °C = 393.15 K
Initial pressure P₁ = 1 atm = 101325 Pa
Final pressure P₂ = Required
Area = A
Therefore we have for the pressure cooker, the volume is constant that is does not change
By Chales law
P₁/T₁ = P₂/T₂
P₂ = T₂×P₁/T₁ = 393.15 K× (101325 Pa/293.15 K) = 135,889.22 Pa
∴ P₂ = 135.88922 KPa = 135.9 kPa
Where Force = [tex]\frac{Pressure}{Area}[/tex] we have
Force = [tex]F_{120}=\frac{135.9}{A}N[/tex].
You drop an irregular piece of metal into a container partially filled with water and measure that the water level rises 4.8 centimeters. The square base of the container has a side length of 8 centimeters. You measure the mass of the metal to be 450 grams. What is the density of the metal?
Answer:
Density = 1464.8kg/m3
the density of the metal is 1464.8kg/m^3
Explanation:
Given;
Mass m = 450g
Density = Mass/Volume = m/V
Volume V = change in height × base area = ∆h × A
∆h = 4.8cm
A = 8×8 = 64cm^2
V = 4.8×64 = 307.2cm^3
Density = 450g/307.2cm^3
Density = 1.4648g/cm^3
Density = 1.4648 × 1000kg/m^3
Density = 1464.8kg/m3
the density of the metal is 1464.8kg/m^3
Answer:
1464.84 kg/m³
Explanation:
Density = mass/volume.
D = m/v................. Equation 1
But from Archimedes principle,
every object immersed in water will displaced an amount of water equal to its own volume
Therefore,
v = v'................... Equation 2
Where v = volume of the irregular object, v' = volume of water displaced.
Since the base of the container is a square,
Then,
v' = L²(d)...................... Equation 3
Where L = length of the square base of the container, d = rise in water level.
Substitute equation 3 into equation 1
D = m/L²d......................... Equation 4
Given: m = 450 g = 0.45 kg, L = 8 cm = 0.08 m, d = 4.8 cm = 0.048 m
Substitute into equation 4
D = 0.45/(0.08²×0.048)
D = 0.45/0.0003072
D = 1464.84 kg/m³
What bodie is outside the solar system
Answer:
Oumuamua is outside the solar system which is in cigar shaped.
Hope it will help you :))
Every star, except the Sun, is outside the solar system.
A ball thrown upward near the surface of the Earth with a velocity of 50 m/s will come to rest about 5 seconds later. If the ball were thrown up with the same velocity on Planet X, after 5 seconds it would be still moving upwards at nearly 31 m/s. The magnitude of the gravitational field near the surface of Planet X is what fraction of the gravitational field near the surface of the Earth? Using the MathType functions, be sure to show your work
Answer:
19/49
Explanation:
Using v = u + at where v = velocity of ball after 5 s on planet X = 31 m/s, u = initial velocity of ball on planet X = 50 m/s , a = acceleration due to gravity on planet X and t = 5 s
So, 31 = 50 - a × 5 = 50 - 5a
31 - 50 = 5a
-19 = 5a
a = -19/5 = -3.8 m/s²
So, the magnitude of a = 3.8 m/s²
So a/g = 3.8/9.8 = 19/49
The fraction of the magnitude of the gravitational field near the surface of Planet X to the gravitational field near the surface of the Earth is 0.39.
Given the following data:
Time = 5 secondsInitial velocity = 50 m/sFinal velocity = 30 m/sWe know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex]
To determine what fraction is the magnitude of the gravitational field near the surface of Planet X to the gravitational field near the surface of the Earth:
First of all, we would calculate the acceleration due to gravity (g) on Planet X by using first equation of motion:
Mathematically, the first equation of motion is calculated by using the formula;
[tex]V = U-at[/tex]
Where:
V is the final velocity.U is the initial velocity.a is the acceleration.t is the time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]31=50-a(5)\\\\5a =50-31\\\\5a=19\\\\a=\frac{19}{5}[/tex]
Acceleration, a = 3.8 [tex]m/s^2[/tex]
For the ratio:
[tex]Fraction = \frac{a}{g} \\\\Fraction = \frac{3.8}{9.8}[/tex]
Fraction = 0.39
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The idea that little children are continually coming up with ideas and testing them is called the ______ theory.
atomic
theory
moral foundations
general
Answer:
The idea that little children are continually coming up with ideas and testing them is called the general theory.
A wheel, starting from rest, rotates with a constant angular acceleration of 2.80 rad/s2. During a certain 5.00 s interval, it turns through 65.0 rad. (a) How long had the wheel been turning before the start of the 5.00 s interval? (b) What was the angular velocity of the wheel at the start of the 5.00 s interval?
Answer:
a) time t1 = 2.14s
b) initial angular speed w1 = 6 rad/s
Explanation:
Given that;
Initial Angular velocity = w1
Angular distance = s = 65 rad
time = t = 5 s
Angular acceleration a = 2.80 rad/s^2
Using the equation of motion;
s = w1t + (at^2)/2
w1 = (s-0.5(at^2))/t
Substituting the values;
w1 = (65 - (0.5×2.8×5^2))/5
w1 = 6rad/s
Time to reach w1 from rest;
w1 = at1
t1 = w1/a = 6/2.8 = 2.14s
a) time t1 = 2.14s
b) initial angular speed w1 = 6 rad/s
Answer:
a. The wheel was turning 2.14 s before the start of the 5.00 s interval.
b. The angular velocity of the wheel at the start of the 5.00 s interval was 6.00 rad/s.
Explanation:
At the start of the 5.00s interval, the wheel might have an initial angular velocity [tex]\omega_0[/tex]. We can obtain it from the kinematic equation:
[tex]\theta=\omega_0t+\frac{1}{2}\alpha t^{2}\\\\\implies \omega_0=\frac{\theta}{t}-\frac{1}{2}\alpha t[/tex]
Plugging in the known values for the time interval, the angular displacement and the angular acceleration, we get:
[tex]\omega_0=\frac{65.0rad}{5.00s}-\frac{1}{2}(2.80rad/s^{2})(5.00s)\\\\\omega_0=6.00rad/s[/tex]
It means that the angular velocity of the wheel at the start of the 5.00 s interval was 6.00 rad/s (b).
The time it took for the wheel to reach that angular velocity can be obtained from another kinematic equation:
[tex]\omega = \omega_0+\alpha t\\\\\implies t=\frac{\omega-\omega_0}{\alpha}[/tex]
It is important to take in account that in this case, the initial angular velocity is zero as the wheel started from rest, and the final angular velocity is the one we got in the previous question:
[tex]t=\frac{6.00rad/s-0}{2.80rad/s^{2}}\\\\t=2.14s[/tex]
Finally, the wheel was turning 2.14 s before the start of the 5.00 s interval (a).
how does an antenna produce radio waves?
Answer:
As the electrons (tiny particles inside atoms) in the electric current wiggle back and forth along the antenna, they create invisibleelectromagnetic radiation in the form ofradio waves. ... 1) Electricity flowing into the transmitter antenna makes electrons vibrate up and down it, producing radio waves.
Answer:
As the electrons (tiny particles inside atoms) in the electric current wiggle back and forth along the antenna, they create invisible electromagnetic radiation in the form of radio waves.
Explanation:
Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Explanation:
Let us assume that the mass of a pitched ball is 0.145 kg.
Initial velocity of the pitched ball, u = 47.5 m/s
Final speed of the ball, v = -51.5 m/s (in opposite direction)
We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :
[tex]\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s[/tex]
So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.
Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :
[tex]J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s[/tex]
Hence, this is the required solution.
The magnitude of the change in momentum of the ball is 99 m/s. The magnitude of the impulse applied to the ball by the bat is also 99 m/s.
Explanation:To find the magnitude of the change in momentum of the ball, we can use the equation:
Magnitude of change in momentum = magnitude of final momentum - magnitude of initial momentum
Given that the initial velocity of the ball is 47.5 m/s and the final velocity of the batted ball is 51.5 m/s in the opposite direction, the magnitude of the change in momentum is:
Magnitude of change in momentum = 51.5 m/s - (-47.5 m/s) = 99 m/s
The magnitude of the impulse applied to the ball by the bat is equal to the magnitude of the change in momentum. Therefore, the magnitude of the impulse applied to the ball is 99 m/s.
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In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is
Fext=dp/dt=m dv/dt+v dm/dt
Suppose the mass of the raindrop depends on the distance x that it has fallen. Then m = kx, where k is a constant, and dm/dt=kv
dm/dt=kv This gives, since Fext=mg
Fext=mg,
mg=m dv/dt+v(kv)
Or, dividing by k,
xg=x dv/dt+v2
This is a differential equation that has a solution of the form v = at, where a is the acceleration and is constant. Take the initial velocity of the raindrop to be zero.
(a) Using the proposed solution for v find the acceleration a.
(b) Find the distance the raindrop has fallen in t = 3.00 s.
(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.
The acceleration of the raindrop is equal to the acceleration due to gravity. The distance the raindrop has fallen in 3.00 seconds is 44.1 meters. The mass of the raindrop at 3.00 seconds is 88.2 grams.
Explanation:To find the acceleration, we can use the given proposed solution for velocity, v = at. Taking the derivative of this equation with respect to time gives us dv/dt = a. Substituting this into the differential equation, we have xg = x(a) + v^2. Since the initial velocity is zero, v = 0 and the equation simplifies to xg = xa. Dividing by x, we get g = a. Therefore, the acceleration is equal to the acceleration due to gravity, g.
To find the distance the raindrop has fallen in t = 3.00 s, we can use the equation x = (1/2)at^2. Since we know the acceleration is g, we plug in the values into the equation: x = (1/2)(g)(3.00 s)^2 = (1/2)(9.8 m/s^2)(9.00 s^2) = 44.1 meters.
To find the mass of the raindrop at t = 3.00 s, we can use the given equation m = kx. Plugging in the values, m = (2.00 g/m)(44.1 m) = 88.2 grams.
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A(n)______is a device intended for the protection of personnel that functions to deenergize a circuit or portion thereof within an established period of time when a current to ground exceeds the values established for a Class A device.
Answer:
The answer is GFCI
(ground fault circuit interrupter)
Explanation:
A ground fault circuit interrupter (GFCI) can help prevent electrocution.
If a person's body starts to receive a shock, the GFCI senses this and cuts off the power before he/she can get injured. GFCIs are generally installed where electrical circuits may accidentally come into contact with water.
What is a circuit breaker?
It is an automatically operated electrical switch designed to protect an electrical circuit from damage caused by excess current from an overload or short circuit. Its basic function is to interrupt current flow after a fault is detected.
Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 1.4×109 electrons from one disk to the other causes the electric field strength between them to be 1.9×105 N/C
Answer:
r = 6.5*10^-3 m
Explanation:
I'm assuming you meant to ask the diameters of the disk, if so, here's it
Given
Quantity of charge on electron, Q = 1.4*10^9
Electric field strength, e = 1.9*10^5
q = Q * 1.6*10^-19
q = 2.24*10^-10
E = q/ε(0)A, making A the subject of formula, we have
A = q / [E * ε(0)], where
ε(0) = 8.85*10^-12
A = 2.24*10^-10 / (1.9*10^5 * 8.85*10^-12)
A = 2.24*10^-10 / 1.6815*10^-6
A = 1.33*10^-4 m²
Remember A = πr²
1.33*10^-4 = 3.142 * r²
r² = 1.33*10^-4 / 3.142
r² = 4.23*10^-5
r = 6.5*10^-3 m
An electric circuit consists of a variable resistor connected to a source of constant potential difference. If the resistance of the resistor is doubled, then
50 kg grandma is roller skating down the sidewalk at 2 m/s. Suddenly, someone throws her a 10 kg sack of oranges. How fast does she travel afterwards? (Inelastic)
Answer:
v2 = 1.67m/s
she will travel at 1.67m/s afterwards
Explanation:
Given;
Mass of grandma m1 = 50kg
Mass of sack m2 = 10kg
Initial speed of grandma v1 = 2m/s
Final velocity of both = v2
The momentum equation of the initial and final momentum can be written as;
m1v1 = m1v2 + m2v2 = (m1+m2)v2
(Since the collision is inelastic they both move at the same velocity v2 after the collision)
making v2 the subject of formula;
v2 = (m1v1)/(m1+m2)
Substituting the values;
v2 = (50×2)/(50+10)
v2 = 1.67m/s
she will travel at 1.67m/s afterwards
A parallel-plate capacitor has plates of area 0.30 m2 and a separation of 2.10 cm. A battery charges the plates to a potential difference of 190 V and is then disconnected. A dielectric slab of thickness 4 mm and dielectric constant 4.8 is then placed symmetrically between the plates.
(a) What is the capacitance before the slab is inserted?
(b) What is the capacitance with the slab in place?
(c) What is the free charge q before and after the slab is inserted?
(d) What is the magnitude of the electric field in the space between the plates and dielectric and in the dielectric itself?
(f) With the slab in place, what is the potential difference across the plates?
(g) How much external work is involved in inserting the slab?
Answer:
a) 1.26e^-10F
b) 1.47e^-10F
c) 2.39e^-8C 2.89e^-8C
d) E=4500.94N/C
e) E'=5254.23N/C
f) 100.68V
g) 1.65e^-10J
Explanation:
To compute the capacitance we can use the formula:
[tex]C=\frac{k\epsilon_o A}{d}[/tex]
where k is the dielectric constant of the material between the plates. d is the distance between plates and A is the area.
(a) Before the material with dielectric constant is inserted we have that k(air)=1. Hence, we have:
[tex]k=1\\A=0.30m^2\\d=0.021m\\e_o=8.85*10^{-12}C^2/(Nm^2)\\\\C=\frac{(1)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{0.021m}=1.26*10^{-10}F[/tex]
(b) With the slab we have that k=4.8 and the thickness is 4mm=4*10^{-3}m. In this case due to the thickness of the slab is not the same as d, we have to consider the equivalent capacitance of the series of capacitances:
[tex]C=(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_1})\\\\\\C_1=\frac{(1)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{8.5*10^{-3}m}=3.1*10^{-10}F\\\\C_2=\frac{(4.8)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{4*10^{-3}m}=3.186*10^{-9}F\\\\C=1.47*10^{-10}F[/tex]
(c)
The charge between the plates for both cases, with the slab is given by:
Q : without the slab
Q': with the slab
[tex]Q=CV=(1.26*10^{-10}F)(190V)=2.39*10^{-8}C\\\\Q'=C'V=(1.47*10^{-10F})(190V)=2.79*10^{-8}C\\[/tex]
(d) The electric field between the plate is given by:
[tex]E=\frac{Q}{2\epsilon_o A}[/tex]
E: without the slab
E': with the slab
[tex]E=\frac{2.39*10^{-8}C}{2(8.85*10^{-12}C^2/Nm^2)(0.30m^2)}=4500.94N/C\\\\E'=\frac{2.79*10^{-8}C}{2(8.85*10^{-12}C^2/Nm^2)(0.30m^2)}=5254.23N/C\\[/tex]
(f) We can assume the system as composed by V=V1+V'+V1 as in (c). By using the equation V=Ed we obtain:
[tex]V=2(4500.94)(8.85*10^{-3}m)+(5254.23)(4*10^{-3}m)=100.68V[/tex]
(g) External work is the difference between the energies of the capacitor before and after the slab is placed between the parallels:
[tex]\Delta E=\frac{1}{2}[(1.26*10^{-10}F)(120V)-(1.47*10^{-10})(100.6V)]=1.65*10^{-10}J[/tex]
A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k= 160 w/m·k). the fin diameter is d= 4.2 mm, and the fin is exposed to convective conditions characterized by h= 231 w/m2·k. it is reported that the fin efficiency is ηf= 0.65. determine (a) the fin length l and the fin effectiveness ɛf. account for tip convection.
Answer:
length of fin = 34.417
effectiveness = 33.77
Explanation:
the pictures attached below shows the whole explanation
how many days are in a year on mars
During which phase of the moon may a solar eclipse occur?
Answer:
New moon
Explanation:
A solar eclipse can only take place at the phase of new moon, when the moon passes directly between the sun and Earth and its shadows fall upon Earth's surface.
A solar eclipse happens when the moon is perfectly aligned between the Earth and the Sun, thus obscuring it.
If the moon is between the Earth and the Sun, the far side of the moon is lighted, and thus you have a new moon.
A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed of 10 m/s. If air resistance is negligible, the distance from the base of the cliff to where the rock hits the level ground below the cliff is most nearly
a.5m
b.10m
c.20m
d.40m
e.200m
Answer:
c.20
Explanation:
Which situations contain unbalanced forces? Check all that apply.
Answer:
Explanation:
Without seeing the options i can tell you this:
A balanced force is equal on both sides
The friction would equal the force
The amount on the left (friction side) would equal the amount on right (force side)
Or the net force would equal zero
You should post the answer options for more help
Answer: A book falls to the floor.
A car skids to a stop.
A foam ball launches
Explanation:
Provide 2 examples in sport or rehab where instantaneous velocity is more important than average velocity and 2 examples where average velocity is more important. Provide rationale.
Answer:
In high jump as well as long jump, instateneous velocity is more important than average velocity.
In relay races and 400m races, average is more important than instateneous velocity.
Explanation:
Instantaneous velocity is the velocity of anything in motion at a specific point in time. This is determined quite similarly to average velocity, however, we look at the period of time so that it approaches zero. If there is a standard velocity over a period of time, its average and instantaneous velocities may be the same. Instantaneous velocity is calculated as the limit as t approaches zero of the change in d over the change in t.
The range or length of long jump depends on the instantenous velocity of the jump and the height of high jump depends on the instantenous velocity of the height.
A person with greater average velocity wins a race. The average velocity of anything or object is referred to as its total displacement divided by the total time taken. That is to say, it is the rate at which an object changes its position from one place to another. Average velocity is also a Vector quantity. Meters per second is the SI unit. Although, any distance unit per any time unit can be used when necessary, such as miles per hour (mph) or kilometer per hour (kmph)
Instantaneous velocity is more important in specific instances like a tennis stroke or a football collision, while average velocity is more important for calculating average rates of change in situations like road trips or rehab.
Explanation:Instantaneous velocity is more important than average velocity in situations where we need to know the velocity at a specific instant in time. Examples of this include a tennis player's stroke in which they aim to hit the ball on the sweet spot of the racket for maximum velocity and minimal vibration, and in a collision in football where a player with the same velocity but greater mass has a greater impact due to their greater momentum.
On the other hand, average velocity is more important in situations where we need to calculate the average rate of change of position over a given time interval. Examples include calculating the average velocity of a car during a road trip to determine the time taken to reach a destination, and in rehab where the average velocity of a patient's movement is measured to track progress over time.
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What are some of the phenotypic plant traits that scientists are investigating? Why do you think these traits were chosen?
The term phenotype in a more specific context to describe the collective expression of the genotype in conjunction with the environment on a plant's observable characteristics.
Explanation:
Phenotype, any of the measurable features of an organism arising from its genotype association (total genetic inheritance) with the environment. Behaviour, biochemical properties, colour, form, and scale are examples of measurable characteristics.An allele is another type of one gene. A phenotype triggered by the expression of a gene or group of genes is the external manifestation of a human trait. A trait is an attribute unique to an organism. The genotype's external expression is called the phénotype.Some of the phenotypic traits includes Height, color, shape.
What is a phenotype?Phenotype is a physical properties or expression of a gene in an individual.
Phenotype are features that is observable or can be seen. This could include the physical appearance where we have height, color, shape.
Phenotypic traits are the observable features such as plant height and eye color, hair color.
The genetic make up of an individual and environment determines the phenotype.Scientific investigation is based on the interest and expected output or result. For a trait or many traits to be investigated it must be related to the research to get the expected outcome.
Therefore, some of the phenotypic traits that scientist can investigate are height, shape and color.
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In the space below, explain why you agree or disagree with the first statement: Each person in a family has the same traits. There are no differences in traits between parents and offspring or among siblings.
Answer:
I disagree.
Explanation:
Yes, traits may be similar, but it all depends on the dominant and recessive alleles that are passed on. No one person can look alike. Even with twins, a widow's peak or close lobes can be different.
I hope this was the brainliest answer! Thank you for letting me help you.
Which of the following treatments would enhance the level of the Pfr form of phytochrome?A) exposure to far-red lightB) exposure to red lightC) long dark periodD) inhibition of protein synthesisE) synthesis of phosphorylating enzymes
Answer:
B) exposure to red light
Explanation:
Plants use a phytochrome system to sense the level, intensity, duration, and color of environmental light do as to adjust their physiology.
The phytochromes are a family of chromoproteins with a linear tetrapyrrole chromophore, similar to the chlorophyll. Phytochromes have two photo-interconvertible forms: Pr and Pfr. Pr absorbs red light (~667 nm) and is immediately converted to Pfr. Pfr absorbs far-red light (~730 nm) and is quickly converted back to Pr. Absorption of red or far-red light causes a massive change to the shape of the chromophore, altering the conformation and activity of the phytochrome protein to which it is bound. Together, the two forms represent the phytochrome system.
A 2.05 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0390 m. The spring has force constant 830 N/m. The coefficient of kinetic friction between the floor and the block is 0.380. The block and spring are released from rest and the block slides along the floor.What is the speed of the block when it has moved a distance of 0.0200 m from its initial position?
Answer:
0.552 m/s
Explanation:
Given,
Mass of the block = 2.05 Kg
initial compression, x₁ = 0.03690 m
Spring constant, k = 830 N/m
coefficient of friction of the block, μ = 0.380
distance moved by the block,x₂ = 0.20 m
Speed, v = ?
Using conservation of energy
Initial spring energy + Work done by friction = Final spring energy + kinetic energy
[tex]\dfrac{1}{2}kx_1^2 - \mu mg x_2 = \dfrac{1}{2}kx_2^2 + \dfrac{1}{2}mv^2[/tex]
[tex]\dfrac{1}{2}\times 830\times 0.039^2 - 0.38\times 2.05\times 9.81\times 0.02 = \dfrac{1}{2}\times 830 \times 0.02^2 + \dfrac{1}{2}\times 2.05\times v^2[/tex]
v² = 0.3047
v = 0.552 m/s
Hence, the speed of the block is equal to 0.552 m/s
Two lab carts are pushed together with a spring mechanism compressed between them. Upon release, the 5.0 kg cart repels one way with a velocity of 0.12 m/s. The 2.0 kg cart goes the other way. At what velocity does it travel?
Answer:
V = -0.3 m/sec.
Explanation:
5.0 x 0.12 + 2.0 x v = 0. Which means that V = -0.3 m/sec.
The -ve sign shows it moves in the opposite direction.
Describe what is happening within the system when it is at equilibrium in terms of concentrations
Answer:
When a system is at equilibrium in terms of concentration what happens is that the rate of change of the concentration of the product and the reactants does not vary or change with time.
Explanation:
What is equilibrium?
A chemical reaction is in equilibrium when the concentrations of reactants and products are constant - their ratio does not vary.
Equilibrium does not necessarily mean that reactants and products are present in equal amounts. It means that the reaction has reached a point where the concentrations of the reactant and product are unchanging with time, because the forward and backward reactions have the same rate.
Final answer:
A chemical system at equilibrium reflects no net change in reactant and product concentrations. Upon a concentration change, Le Chatelier's principle dictates that the system will adjust to partially counteract the change and set up a new equilibrium.
Explanation:
When a system is at equilibrium, it means no net change in the concentrations of reactants and products takes place. This is because the rate of the forward reaction, in which reactants are converted into products, is equal to the rate of the reverse reaction, in which products transform back into reactants. However, when there is a change in concentration of either reactants or products during equilibrium, the system responds according to Le Chatelier's principle. This principle states that the system will adjust to partially counteract the change and reestablish a new equilibrium state.
For example, if you were to increase the concentration of a reactant, the system tends to counter this by producing more product, essentially shifting the equilibrium to the right. Conversely, if you decrease the concentration of a product, the system will proceed to produce more of that product from the reactants, again shifting the reaction to the right, until a new equilibrium is established.
A student throws a 130 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is 0.18 s?
Answer:
4.7 N
Explanation:
130 g = 0.13 kg
The momentum of the snowball when it's thrown at the wall is
[tex]p = mv = 0.13*6.5 = 0.845 kgm/s[/tex]
Which is also the impulse. From here we can calculate the magnitude of the average force F knowing the duration of the collision is 0.18 s
[tex]p = F\Delta t[/tex]
[tex]F*0.18 = 0.845[/tex]
[tex]F = 0.845 / 0.18 = 4.7 N[/tex]
Four long, parallel conductors carry equal currents of I = 5.00 A. The direction of the current is into the page at points A and B (indicated by the crosses) and out of the page at C and D (indicated by the dots). Calculate the magnitude and direction of the magnetic field at point P, located at the center of the square with an edge length 0.200 m.
Answer:
Explanation:
The magnetic field due to a straight wire carrying current is given by
B = μo• I / 2πr
Where,
μo = permeability of free space
μo = 4π×10^-7 Tm/A
I is the current in wire
I = 5A
r Is the distance of point from the wire.
The distance between the parallel conductors and the point is r/2
R = 0.283/2 = 0.1415m
Check attachment on how I used Pythagoras theorem to find the diagonal of the square..
Hence, the magnetic field at point P is
B = μo•I / 2πR
B = 4π × 10^-7 × 5 / 2π × 0.1415
B = 7.07 × 10^-6 T.
The time between a lightning flash and the following thunderclap may be used to estimate, in kilometers, how far away a storm is. How far away is a storm if 6 seconds elapse between the lightning and the thunderclap?
Given Information:
Elapsed time = t = 6 seconds
Required Information:
Distance = d = ?
Answer:
Distance = d = 2.058 km
Explanation:
We know that the speed of sound in the air is given by
v = 343 m/s
The relation between distance, speed and time is given by
distance = speed*time
substituting the given values yields,
distance = 343*6
distance = 2058 m
There are 1000 meters in 1 km so
d = 2058/1000
d = 2.058 km
Therefore, the storm is about 2.058 km away when elapse time between the lightning and the thunderclap is 6 seconds.
A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteries must be able to store and move in order to accelerate the 770 kg car from rest to 26 m/s, make it climb a 2.15 x 10^2 m high hill while maintaining that speed, and then cause it to travel at a constant 26 m/s by exerting a 5.3 x 10^2 N force for an hour
Answer:
4.29×10⁵ C
Explanation:
From the question,
The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.
E = 1/2mv²+mgh+Fd.................... Equation 1
Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.
But,
d = vt.................... Equation 2
Where v = velocity, t = time.
Substitute equation 2 into equation 1
E = 1/2mv²+mgh+F(vt)................... Equation 3
Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²
Substitute into equation 1
E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)
E = 260260+1622390+49608000
E = 51490650 J
Using,
E = qV................. Equation 4
Where q = charge of the battery, V = Voltage.
make q the subject of the equation
q = E/V............... Equation 5
Given: E = 51490650 J, V = 120 V
Substitute into equation 5
q = 51490650/120
q = 429088.75 C
q = 4.29×10⁵ C