Answer:
For 0.24 sec the person was in the air.
Explanation:
Given that,
Height = 1 m
Initial velocity = 3 m/s
We need to calculate the time
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Where, u = initial velocity
s = height
Put the value into the formula
[tex]1 =3\times t+\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]4.9t^2+3t-1=0[/tex]
[tex]t = 0.24\ sec[/tex]
On neglecting the negative value of time
Hence, For 0.24 sec the person was in the air.
Two point charges are located on the x-axis. One has a charge of 1.77 μC and is located at x = 0.0 m, and the other has a charge of -4.09 μC and is located at x = 15.1 m. At what location on the x-axis (other than at infinity) would the electric force on a third point charge of 3.32 μC be zero?
Answer:
x = -29.032m
Explanation:
Since the third charge is positive, it cannot be between the other two charges, because it would be repelled by the positive one and attracted by the negative one, so the electric force would never be zero.
This leaves only two options: To the left of the positive one or to the right of the negative one.
If it was located on the right of the negative charge, the force of the positive charge would be weaker because of both the distance is larger and its charge is smaller than the negative charge. So, there is no point the would make the result force equal zero.
This means that the third charge has to be at the left of the positive charge. With this in mind, we make the calculations:
[tex]F_{13}=K*\frac{Q_{1}*Q_{3}}{d^{2}} =F_{23}=K*\frac{Q_{2}*Q_{3}}{(x_{2}+d)^{2}}[/tex]
Replacing the values of Q1=1.77, Q2=4.09, X2=15.1, we solve for d and get two possible results:
d1 = 29.032m and d2 = -5.99m
Since we assumed in our formula that the third charge was on the left of the positive charge, the distance d has to be positive so that our final result can be a negative position. This is X = -d
This way, we get:
X = -29.032m
A banked circular highway is designed for traffic moving
at60km/h. The radius of the curve is 200m. Traffic ismoving along
the highway at 40km/h on a rainy day. What isthe minimum
coefficient of friction between tires and road thatwill allow cars
to negotiate the turn without sliding off theroad?
Answer:
0.063
Explanation:
velocity of the car, v = 40 km/h = 11.11 m/s
radius, r = 200 m
Let the coefficient of friction is μ.
The coefficient of friction relates to the velocity on banked road is given by
[tex]\mu =\frac{v^{2}}{rg}[/tex]
where, v is the velocity, r be the radius of the curve road and μ is coefficient of friction.
By substituting the values, we get
[tex]\mu =\frac{11.11^{2}}{200\times 9.8}[/tex]
μ = 0.063
A 4.0 µC point charge and a 3.0 µC point charge are a distance L apart. Where should a third point charge be placed so that the electric force on that third charge is zero? (Use the following as necessary: L.)
Answer:
Q must be placed at 0.53 L
Explanation:
Given data:
q_1 = 4.0 μC , q_2 = 3.0μC
Distance between charge is L
third charge q be placed at distance x cm from q1
The force by charge q_1 due to q is
[tex]F1 = \frac{k q q_1}{x^2}[/tex]
[tex]F1 = \frac{k q ( 4.0 μC )}{ x^2}[/tex] ----1
The force by charge q_2 due to q is
[tex]F2 = \frac{k q q_2}{(L-x)^2}[/tex]
[tex]F2 = \frac{kq (3.0 μC)}{(L-x)^2}[/tex] --2
we know that net electric force is equal to zero
F_1 = F_2
[tex]\frac{k q ( 4.0 μC )}{x^2} =\frac{k q ( 3.0 μC )}{(l-x)^2}[/tex]
[tex]\frac{4}{3}*(L-x)^2 = x^2[/tex]
[tex]x = \sqrt{\frac{4}{3}*(L - x)[/tex]
[tex]L-x = \frac{x}{1.15}[/tex]
[tex]L = x + \frac{x}{1.15} = 1.86 x[/tex]
x = 0.53 L
Q must be placed at 0.53 L
The third charge should be placed either between the two charges or on the extended line of the two charges to make the electric force on that charge zero. The exact position depends on the charges involved and can be calculated using the principle of superposition and Coulomb's Law.
Explanation:The problem deals with the principle of superposition in electrostatics and the force on a charge due to other charges nearby. The force on any charge due to a number of other charges is simply the vector sum of the forces due to individual charges. Starting from this principle, we can try to figure out where the third charge must be placed so that the net force on it is zero.
The two possible positions along the line of the charges are on either side of the two given charges, let's call them 4.0 µC (charge_1) and 3.0 µC (charge_2). These positions can be calculated using the formula of force between two point charges (Coulomb's Law): F = k(q1 x q2)/r² where F is force, k is Coulomb's constant, q1 and q2 are charges and r is distance between charges.
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Which is true concerning the acceleration due to Earth's gravity (ge) ? It decreases with increasing altitude. B. It is different for different objects in free fall. C. It is a fundamental quantity. D. It increases with increasing altitude. E. all of these
Answer:
Option A decreases with increase in altitude
Explanation:
This can be explained as the value of gravitational acceleration, 'g' is not same everywhere.
It has its maximum value at poles of the Earth and minimum on its equator.
Thus a person will weigh more at poles than equator.
This variation is in accordance to:
[tex]g = \frac{GM_{E}}{radius^{2}}[/tex]
Thus the gravitational acceleration changes as inverse square of the Radius of the Earth.
Thus as we move away from the Earth's center, gravitational acceleration, g decreases.
Magnetic field is measured in SI units of a tesla (T), and a current through a wire generates a field around the wire. The largest fields that we can make with most normal conducting materials carrying current are limited by the resistance of the wire and are around 1 tesla. For comparison, Earth's magnetic field is roughly 50 micro tesla You may have heard of a field unit called a gauss. It takes 10,000 G to make 1 T. What would be the maximum force on a wire 0.1 m long carrying a current of 80 A in a uniform magnetic field of 1 T? 8 N when the wire is perpendicular to the field
80 N when the wire is perpendicular to the field
8 N when the wire is parallel to the field
80 N when the wire is parallel to the field
Answer:
The answer is 8 N
Explanation:
The Lorentz force for a current carrying wire is
f = I * L x B
So, for magnetic forces to manifest the current must not be parallel to the magnetic field. So the cases where the wire is parallel to the field would result in a force of zero applied on the wires by the magnetic field because the cross product becomes zero.
For the perpendicular cases:
f = I * L * B
f = 80 * 0.1 * 1 = 8 N
two punds of water vapor at 30 psia fill the 4ft3 left chmaber of a partitioned system. The right chmaber has twice the volume of the left and is initially evacuated. Detrmine the pressure of water after the paertiion has been removed nd enough heat has been transfered so that the temperature of the water is 40F.
Answer:
pressure of water will be 49.7 atm
Explanation:
given data
pressure = 30 psi = 2.04 atm
water = 2 pound = 907.18
mole of water vapor = 907.19 /2 = 50.4 mole
volume = 4 ft³ = 113.2 L
temperature = 40 F = 277.59 K
to find out
pressure of water
solution
we will apply here ideal gas condition
that is
PV = nRT .......................1
put here all value and here R = 0.0821 , T temperature and V volume and P pressure and n is no of mole
and we get here temperature
PV = nRT
2.04 × 113.2 = 50.4×0.0821×T
solve it and we get
T = 55.8 K
so we have given right chamber has twice the volume of the left chamber i.e
volume = twice of volume + volume
volume = 2(113.2) + 113.2
volume = 339.6 L
so from equation 1 pressure will be
PV = nRT
P(339.6) = 50.4 × ( 0.0821) × (277.59)
P = 3.3822 atm = 49.7 atm
so pressure of water will be 49.7 atm
Suppose you exert a 400-N force on a wall but the wall does not move. The work you are doing on the wall is : unknown, because the mass of the wall is not given
zero, because the wall is not moving
3920 J
400 J
Answer:
W = 0 :The work done on the wall is zero,because the wall is not moving
Explanation:
Work theory
Work is the product of a force applied to a body and the displacement of the body in the direction of this force.
W= F*d Formula (1)
W: Work (Joules) (J)
F: force applied (N)
d=displacement of the body (m)
The work is positive (W+) if the force goes in the same direction of movement.
The work is negative (W-)if the force goes in the opposite direction to the movement
Data
F= 400-N
d= 0
Problem development
We apply formula (1) to calculate the work done on the wall:
W= 400*0
W=0
A glass vessel that can be repeatedly filled with precisely the same volume of liquid is called a pycnometer. A certain pycnometer, when empty and dry, weighed 25.296 g. When filled with water at 25 oC the pycnometer and water weighed 34.914 g. When filled with a liquid of unknown composition the pycnometer and its contents weighed 33.485 g. At 25 oC the density of water is 0.9970 g/ml. What is the density of the unknown liquid?
Answer:
density of liquid 0.848 g/ml
Explanation:
from the information given in the question
mass of water = 34.914 - 25.296 = 9.618 g
volume of pycnometer = volume of water
which will be equal to [tex]= \frac{ mass}{density}[/tex]
[tex]= \frac{9.618}{0.9970} = 9.646 ml[/tex]
mass of liquid =33.485-25.296 = 8.189 ml
density of liquid[tex]= \frac{mass}{volum\ of\ liquid}[/tex]
= [tex]\frac{8.189}{9.646} =0.848 g/ml[/tex]
A glider moves along an air track with constant acceleration a. It is projected from the start of the track (x = 0 m) with an initial velocity of v0. At time t = 8s, it is at x = 100 cm and is moving along the track at velocity vt = − 0.15 m/s. Find the initial velocity v0 and the acceleration a.
Answer:
vo = 0.175m/s
a = -0.040625 m/s^2
Explanation:
To solve this problem, you will need to use the equations for constant acceleration motion:
[tex]x = \frac{1}{2}at^2 +v_ot+x_o \\v_f^2 - v_o^2 = 2(x-x_o)a[/tex]
In the first equation you relate final position with the time elapsed, in the second one, you relate final velocity at any given position. In both equations, you will have both the acceleration a and the initial velocity vo as variables. We can simplify with the information we have:
1. [tex]x = \frac{1}{2}at^2 +v_ot+x_o\\0.1m = \frac{1}{2}a(8s)^2 +v_o(8s)+0m \\0.1 = 32a + 8v_o[/tex]
2. [tex]v_f^2 - v_o^2 = 2(x-x_o)a\\(-0.15m/s)^2 - v_o^2 = 2(0.1m-0m)a\\0.0225 - v_o^2 = 0.2a\\a = \frac{0.0225 - v_o^2}{0.2} = 0.1125 - 5v_o^2[/tex]
Replacing in the first equation:
[tex]0.1 = 32(0.1125 - 5v_o^2) + 8v_o\\0.1 = 3.6 - 160v_o^2 + 8v_o\\160v_o^2 - 8v_o - 3.5 = 0[/tex]
[tex]v_0 = \frac{-(-8) +- \sqrt{(-8)^2 - 4(160)(-3.5)}}{2(160)} \\ v_o = 0.175 m/s | -0.125 m/s[/tex]
But as you are told that the ball was projected om the air track, it only makes sense for the velocity to be positive, otherwise it would have started moving outside the air track, so the real solution is 0.175m/s. Then, the acceleration would be:
[tex]a = 0.1125 - 5v_o^2\\a = -0.040625 m/s^2[/tex]
A cyclist reaches the bottom of a hill with a speed of 18 m/s. Neglecting air resistance and other friction, to what maximum height can they coast up the hill without pedalling? a) 19 m b) 20 m c) 17 m d) 18 m
Answer:
c) 17 m
Explanation:
From the conservation of energy.
Change in kinetic energy= change in potential energy
Now,
[tex]\Delta KE=\Delta PE[/tex]
Now according to the situation.
[tex]\frac{1}{2}mv^{2}=mgh\\h=\frac{v^{2} }{2g}[/tex]
Here, v is the velocity, m is the mass of an object, h is the height, g is the acceleration due to gravity.
Given that the speed of the bicycle is
[tex]v=18m/s[/tex]
And the acceleration due to gravity is
[tex]g=9.8 m/s^{2}[/tex]
Substitute these values.
[tex]h=\frac{18^{2} }{2\times 9.8}\\h=16.53m[/tex]
Which means it is equivalent to 17 m
Therefore, the maximum height is 17 m
Explain why two equipotential lines cannot cross each other.
Answer:
Explained
Explanation:
Equipotential lines cannot cross each other because. The equipotential at a given point in space is has single value of potential throughout. If two equipotential lines intersect with each other, that would mean two values of potential, that would not mean equipotential line. Hence two equipotential lines cannot cross each other.
Equipotential lines are imaginary lines that connect points with the same electric potential. Two equipotential lines cannot cross each other because that would mean two different points on the same line have the same electric potential, which is not possible.
Explanation:Equipotential lines are imaginary lines that connect points with the same electric potential. Two equipotential lines cannot cross each other because that would mean two different points on the same line have the same electric potential, which is not possible.
For example, imagine a hill with contour lines representing lines of equal height. If two contour lines were to cross each other, it would mean that two different points on the hill have the same height, which is not possible.
In electricity, the electric potential difference between two points on an equipotential line is zero. If two equipotential lines were to cross, it would create a contradiction, as the potential difference between the two points of intersection would be both zero and non-zero.
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A charge 4q is at the origin, and a charge of -3q is on the positive x-axis at x = a. Where would you place a third charge so it would experience zero net electric force? (Note: Either enter your answer as an unevaluated expression, e.g., (3/2 + sqrt(3))a, or if evaluated, use 2 decimal places.) Your expression should be in terms of the given variable.
Answer:
6.45 a
Explanation:
Charge on O, q1 = 4q
Charge on A, q2 = - 3 q
OA = a
Let the net force is zero at point P, where AP = x , let a charge Q is placed at P.
The force on point P due to the charge q1 = The force on point P due to the
charge q2
By using Coulomb's law
[tex]\frac{Kq_{1}Q}{OP^{2}}=\frac{Kq_{2}Q}{AP^{2}}[/tex]
[tex]\frac {K4qQ}{(a+x)^{2}}= \frac {K3qQ}{x^{2}}[/tex]
[tex]\frac{a+x}{x}=\sqrt{\frac{4}{3}}[/tex]
[tex]\frac{a+x}{x}=1.155[/tex]
a + x = 1.155 x
0.155 x = a
x = 6.45 a
Thus, the force is zero at x = 6.45 a.
A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary, and the electron has a speed of 7.5 105 m/s. Find the radius between the stationary proton and the electron orbit within the hydrogen atom.
Answer:
450 pm
Explanation:
The electron is held in orbit by an electric force, this works as the centripetal force. The equation for the centripetal acceleration is:
a = v^2 / r
The equation for the electric force is:
F = q1 * q2 / (4 * π * e0 * r^2)
Where
q1, q2: the electric charges, the charge of the electron is -1.6*10^-19 C
e0: electric constant (8.85*10^-12 F/m)
If we divide this force by the mass of the electron we get the acceleration
me = 9.1*10^-31 kg
a = q1 * q2 / (4 * π * e0 * me * r^2)
v^2 / r = q1 * q2 / (4 * π * e0 * me * r^2)
We can simplify r
v^2 = q1 * q2 / (4 * π * e0 * me * r)
Rearranging:
r = q1 * q2 / (4 * π * e0 * me * v^2)
r = 1.6*10^-19 * 1.6*10^-19 / (4 * π * 8.85*10^-12 * 9.1*10^-31 * (7.5*10^5)^2) = 4.5*10^-10 m = 450 pm
A container is filled to the brim with 1.5 L of mercury at 20°C. As the temperature of the container and mercury is increased to 55°C, a total of 7.8 mL of mercury spill over the brim of the container. Determine the linear expansion coefficient of the material that makes up the container.
Answer:
The linear expansion coefficient of the material is [tex]4.952\times10^{-5}\ K^{-1}[/tex]
Explanation:
Given that,
Volume = 1.5 l
Temperature = 20°C
Increased temperature = 55°C
Volume = 7.8 mL
We need to calculate the linear expansion coefficient of the material
Using formula of linear expansion
[tex]\Delta V=V_{0}\beta \Delta T[/tex]
[tex]\Delta V=V_{0}(3\alpha \Delta T)[/tex]
[tex]\alpha =\dfrac{\Delta V}{3V_{0}\Delta T}[/tex]
Put the value into the formula
[tex]\alpha=\dfrac{7.8\times10^{-3}}{3\times1.5\times(55-20)}[/tex]
[tex]\alpha=0.00004952[/tex]
[tex]\alpha=4.952\times10^{-5}\ K^{-1}[/tex]
Hence, The linear expansion coefficient of the material is [tex]4.952\times10^{-5}\ K^{-1}[/tex]
Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up from left to right and a 36-N force is applied to the left side of the left cube causing all three cubes to accelerate to the right. If the cubes are each subject to a frictional force of 6.0 N, what is the magnitude of the force exerted on the right cube by the middle cube in this case
Answer:
Force exerted on the right cube by the middle cube:
F= 12.02N : in the positive direction of the x axis( +x)
Explanation:
We apply Newton's second law for forces in the direction of the x-axis.
∑Fx= m*a
∑Fx: algebraic sum of forces ( + to the right, - to the left)
m: mass
a: acceleration
Forces (x) in total mass : Newton's second lawWe apply Newton's first law for forces in the direction of the y-axis.
∑Fx= mt*a , mt: total mass = 6*3= 18 kg
36-6= 18*a
[tex]a=\frac{36-6}{18} = 1.67 \frac{m}{s}[/tex]
Forces (y) in total mass : Newton's first law∑Fy= 0
Nt-Wt=0 , Nt=Wt
Nt: total normal , Wt= total Weight: mt*g , g: acceleration due to gravity
Wt=18*9,8=176.4 N
Nt=176.4 N
Calculation of the coefficient of kinetic frictionμk=Ff/Nt Ff: friction force = 6 N
μk=6/176.4 = 0.034
Forces (x) on the first block (on the right): Newton's second law∑Fx= m₁*a
F-Ff₁= 6*1.67 , Equation (1)
F:Force exerted on the right cube by the middle cube
Ff₁= μk*N₁ , N₁=W₁ = 9.8*6= 58.8 N
Ff₁= 0.034*58.8 = 2 N
In the Equation (1):
F-2= 10.02
F= 2+10.02= 12.02N
F= 12.02N
A car enters a freeway with initial velocity of 15.0 m/s and with con stant rate of acceleration, reaches a velocity of 22.5 m/s in a time interval of 3.50 s. a) Determine the value of the car's acceleration. b) Determine the distance traveled by the car in this 3.50 s time interval. c) Determine the average velocity of the car over this 3.50 s time interval.
Answer:
a) The acceleration is 2.14 m/s^{2}
b) The distance traveled by the car is 65.61 m
c) The average velocity is 18.75 m/s
Explanation:
Using the equations that describe an uniformly accelerated motion:
a) [tex]a=\frac{v_f - v_o}{t} =\frac{22.5m/s - 15.0 m/s}{3.50s}[/tex]
b) [tex]d= d_0 + v_0 t + \frac{1}{2} a t^{2} = 0 +15.0 x 3.5 + \frac{2.14x3.50^{2} }{2} = 65.61 m[/tex]
c) [tex]v_m =\frac{d}{t}=\frac{65.61}{3.5} =18.75 m/s[/tex]
A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same directionWith the given uncertainties, what is the smallest distance the student could possibly be from the starting point?
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
[tex]A\pm \Delta A[/tex] and [tex]B\pm \Delta B[/tex]
The sum is represented as
[tex]Sum=(A+B)\pm (\Delta A+\Delta B)[/tex]
For the the values given to us the sum is calculated as
[tex]Sum=(2.9+3.9)\pm (0.1+0.2)[/tex]
[tex]Sum=6.8\pm 0.3[/tex]
Now the since the uncertainity inthe sum is [tex]\pm 0.3[/tex]
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals [tex]6.8-0.3=6.5[/tex]meters
The fastest measured pitched baseball left the pitcher's hand at a speed of 42.0 m/s. If the pitcher was in contact with the ball over a distance of 1.50 m and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?
Answer:
The acceleration and time are 588 m/s² and 0.071 sec respectively.
Explanation:
Given that,
Speed = 42.0 m/s
Distance = 1.50 m
(a). We need to calculate the acceleration
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
Put the value in the equation
[tex]a=\dfrac{42.0^2-0}{2\times1.50}[/tex]
[tex]a=588\ m/s^2[/tex]
(b). We need to calculate the time
Using equation of motion
[tex]v = u+at[/tex]
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{42.0-0}{588}[/tex]
[tex]t=0.071\ sec[/tex]
Hence, The acceleration and time are 588 m/s² and 0.071 sec respectively.
If a wave vibrates up and down twice each second and travels a distance of 20 m each second and travels a distance of 20 m each second, what is its frequency? Its wave speed?
Answer:
Frequency is 0.5 Hz and the wave speed is 10 m/s.
Explanation:
As we know that frequency is defined as the how many times the no of cycles repeat in one second so if the wave is vibrating up and down twice during 1 second then the frequency in 1 second is
[tex]f=\frac{1}{2} hz\\F=0.5hz[/tex]
Therefore frequency is 0.5 Hz.
Now the distance of wawe in each second is,
[tex]d=20m[/tex]
Now the wave velocity is,
[tex]v=fd[/tex]
Here, f is frequency, d is the distance, v is the wave velocity.
Substitute all the variables
[tex]v=0.5\times 20\\v=10m/s[/tex]
Therefore the wave speed is 10 m/s.
In the given question, the wave's frequency is 2 Hertz (Hz), which means it oscillates twice per second. The wave speed, or distance covered by the wave per second, is identified as 20 m/s. These are fundamental concepts in the Physics of wave mechanics.
Explanation:In this question, we're dealing with the topics of wave frequency and wave speed. The frequency of a wave relates to how many cycles of the wave occur per unit of time - in this case, the wave is oscillating twice per every second, meaning that its frequency is 2 Hertz (Hz).
The wave speed is the speed at which the wave is travelling. Given that the wave travels a distance of 20 m each second, the speed of the wave is 20 m/s (meters per second).
It's important to note that these two properties are interconnected. In general, the speed of a wave (v) is calculated by multiplying its wavelength (λ) by its frequency (f). So, using the relationship v = λf, the properties of a wave can be determined if the other two are known.
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A world-class sprinter accelerates to his maximum speed in 3.8 s. He then maintains this speed for the remainder of a 100-m race, finishing with a total time of 9.3 s . What is the runner's average acceleration during the first 3.8 s ?
What is his average acceleration during the last 5.5 s ?
What is his average acceleration for the entire race?
Answer:
Explanation:
Given
sprinter achieve maximum speed in 3.8 sec
Let v be the maximum speed and a be the acceleration in first 3.8 s
[tex]a=\frac{v-0}{3.8}[/tex]
distance traveled in this time span
[tex]x=ut+\frac{1}{2}at^2[/tex]
here u=0
[tex]x=\frac{1}{2}\times \frac{v-0}{3.8}\times 3.8^2[/tex]
[tex]x=\frac{3.8}{2}v[/tex]
remaining distance traveled in 9.3-3.8 =5.5 s
[tex]100-x=v\times 5.5[/tex]
put value of x
[tex]100-\frac{3.8}{2}v=5.5v[/tex]
100=1.9v+5.5v
100=7.4v
[tex]v=\frac{100}{7.4}=13.51 m/s[/tex]
Thus average acceleration in first 3.8 sec
[tex]a_{avg}=\frac{0+a}{2}[/tex]
and [tex]a=\frac{13.51}{3.8}=3.55 m/s^2[/tex]
[tex]a_{avg}=\frac{3.55}{2}=1.77 m/s^2[/tex]
Average acceleration during last 5.5 sec will be zero as there is no change in velocity.
Average acceleration for the entire race[tex]=\frac{13.51}{9.3}=1.45 m/s^2[/tex]
The average acceleration in the first 3.8 seconds would be the final speed divided by 3.8 s. During the next 5.5 seconds, the average acceleration is zero because there is no change in velocity. The average acceleration for the entire race can be calculated as the final velocity divided by total time.
Explanation:The average acceleration is calculated by the change in velocity divided by the change in time. In this case, for the first 3.8 seconds, the sprinter was accelerating, so the average acceleration was the final speed (which we do not know yet) divided by 3.8 s. However, in the remaining 5.5 seconds, the sprinter did not accelerate or decelerate, so the average acceleration is zero.
For the first part, we first need to calculate the steady speed. This is given by the distance covered (100 m minus the distance covered in first 3.8 seconds) divided by the time for this (5.5 seconds). We will assume a uniform acceleration in the first 3.8 seconds. His average speed in this period will then be half his maximum speed. So, maximum speed = (2 * distance in first 3.8 secs) / 3.8. Finally, the average acceleration for entire race can be calculated by the total change in velocity (which is the final velocity) divided by the total time (which is 9.3 s).
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Two 1.0 g spheres are charged equally and placed 2.2 cm apart. When released, they begin to accelerate at 180 m/s^2 . What is the magnitude of the charge on each sphere?
Answer:
[tex]q=9.83\times 10^{-8}\ C[/tex]
Explanation:
Given that,
Mass of the two spheres, m₁ = m₂ = 1 g = 0.001 kg
Distance between spheres, d = 2.2 cm = 0.022 m
Acceleration of the spheres when they are released, [tex]a=180\ m/s^2[/tex]
Let q is the charge on each spheres. The force due to motion is balanced by the electrostatic force between the spheres as :
[tex]ma=k\dfrac{q^2}{d^2}[/tex]
[tex]q=\sqrt{\dfrac{mad^2}{k}}[/tex]
[tex]q=\sqrt{\dfrac{0.001\times 180\times (0.022)^2}{9\times 10^9}}[/tex]
[tex]q=9.83\times 10^{-8}\ C[/tex]
So, the magnitude of charge on each sphere is [tex]9.83\times 10^{-8}\ C[/tex]. Hence, this is the required solution.
To find the magnitude of the charge on the spheres, we utilize Coulomb's Law and Newton's Second Law, set up an equation, and solve for the charge q. Ensure consistency in the units while solving.
Explanation:To find the magnitude of the charge on the spheres, we start by using Coulomb's Law, which is expressed as F = k*q1*q2/r^2. Here, F is the force, k is Coulomb's constant (8.99 * 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the separation. Given that the spheres are charged equally (q1 = q2), we can refer to them just as q.
The spheres start accelerating once released, and the only force in operation is the electrostatic force. Thus, according to Newton's second law (F = ma), the force can also be expressed as F = 2*(mass*acceleration) because two spheres are involved.
By equating both expressions for F, we have 2*(mass*acceleration) = k*q^2/r^2. From this equation, we can solve for q = sqrt((2*mass*acceleration*r^2)/k). Substituting given values, we have q = sqrt((2 * 1.0 g * 180 m/s^2 * (2.2 cm)^2)/(8.99 * 10^9 N m^2/C^2)).
Remember to convert grams to kilograms and centimeters to meters to maintain consistency in the units while solving.
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A jetliner rolls down the runway with constant acceleration from rest, it reaches its take off speed of 250 km/h in 1 min. What is its acceleration? Express in km/h^2, does this result make sense?
Answer:
Acceleration, [tex]a=14970.05\ km/h^2[/tex]
Explanation:
Given that,
Initially, the jetliner is at rest, u = 0
Final speed of the jetliner, v = 250 km/h
Time taken, t = 1 min = 0.0167 h
We need to find the acceleration of the jetliner. The mathematical expression for the acceleration is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{250}{0.0167}[/tex]
[tex]a=14970.05\ km/h^2[/tex]
So, the acceleration of the jetliner is [tex]14970.05\ km/h^2[/tex]. Hence, this is the required solution.
The sun is 60° above the horizon. Rays from the sun strike the still surface of a pond and cast a shadow of a stick that is stuck in the sandy bottom of the pond. If the stick is 19 cm tall, how long is the shadow?
Answer:
shadow length 7.67 cm
Explanation:
given data:
refractive index of water is 1.33
by snell's law we have
[tex]n_{air} sin30 =n_{water} sin\theta[/tex]
[tex]1*0.5 = 1.33*sin\theta[/tex]
solving for[tex] \theta[/tex]
[tex]sin\theta = \frac{3}{8}[/tex]
[tex]\theta = sin^{-1}\frac{3}{8}[/tex]
[tex]\theta = 22 degree[/tex]
from shadow- stick traingle
[tex]tan(90-\theta) = cot\theta = \frac{h}{s}[/tex]
[tex]s = \frac{h}{cot\theta} = h tan\theta[/tex]
s = 19tan22 = 7.67 cm
s = shadow length
The calculated shadow length is approximately 10.97 cm.
To determine the length of the shadow cast by the stick, we can use basic trigonometry.
Specifically, we'll use the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right-angled triangle.
Given:
The height of the stick (opposite side) is 19 cm.The angle of elevation of the sun is 60°.The formula for tangent is:
tan(θ) = opposite / adjacent
Here, the opposite side is the height of the stick, and the adjacent side is the length of the shadow. Thus, we can rearrange this formula to solve for the length of the shadow (adjacent side):
adjacent = opposite / tan(θ)
Substituting the given values:
adjacent = 19 cm / tan(60°)
We know that tan(60°) is √3 or approximately 1.732.
So:
adjacent = 19 cm / 1.732 ≈ 10.97 cm
Therefore, the length of the shadow cast by the stick is approximately 10.97 cm.
a bicyclist is riding to the left with a velocity of 14m/s after a steady gust of wind that lasts 3.5m/s the bicyclist its moving to the left with a velocity of 21m/s
Final answer:
To find the speed and direction of the wind as observed by a stationary observer when a cyclist is traveling southeast along a road and there is a wind blowing from the southwest, we can use vector addition. By breaking down the velocities of the cyclist and the wind into their components and adding them together, we can determine that the speed of the wind is 15 km/h in the direction of 135°.
Explanation:
The question is asking for the speed and direction of the wind as observed by a stationary observer when a cyclist is traveling southeast along a road at 15 km/h and there is a wind blowing from the southwest at 25 km/h. To calculate the speed and direction of the wind, we can use vector addition. The velocity of the cyclist is given as 15 km/h in the southeast direction. The velocity of the wind is given as 25 km/h in the southwest direction. To find the speed and direction of the wind, we need to add the velocities of the cyclist and the wind.
We can represent the velocity vectors as follows:
Velocity of cyclist = 15 km/h (southeast)
Velocity of wind = 25 km/h (southwest)
To add these vectors, we can break them down into their components:
Velocity of cyclist = 15 km/h * sin(45°) (south) + 15 km/h * cos(45°) (east)
Velocity of wind = 25 km/h * sin(225°) (south) + 25 km/h * cos(225°) (west)
Adding the components of the velocities:
Velocity of cyclist + Velocity of wind = (15 km/h * sin(45°) + 25 km/h * sin(225°)) (south) + (15 km/h * cos(45°) + 25 km/h * cos(225°)) (east)
Calculating the components:
Velocity of cyclist + Velocity of wind = (-10.61 km/h) (south) + (-10.61 km/h) (east)
To find the speed and direction of the wind, we can use the Pythagorean theorem and trigonometry:
Speed of wind = sqrt((-10.61 km/h)^2 + (-10.61 km/h)^2) = 15 km/h
Direction of wind = atan2((-10.61 km/h), (-10.61 km/h)) = 135°
Therefore, the speed of the wind as observed by the stationary observer is 15 km/h in the direction of 135°.
At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building that is 300 m tall. How much earlier does the thrown rock strike the ground? Neglect air resistance. Please show all work and formulas used thanks
Answer:
The thrown rock strike 2.42 seconds earlier.
Explanation:
This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:
[tex]x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration[/tex]
So now we have an equation and unkown value.
for the thrown rock
[tex]\frac{1}{2}(9.8)*t^2+29*t-300=0[/tex]
for the dropped rock
[tex]\frac{1}{2}(9.8)*t^2+0*t-300=0[/tex]
solving both equation with the quadratic formula:
[tex]\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}[/tex]
we have:
the thrown rock arrives on t=5.4 sec
the dropped rock arrives on t=7.82 sec
so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)
The thrown rock hits the ground approximately 1.89 seconds earlier than the dropped rock when released simultaneously from a height of 300 meters.
Explanation:To determine how much earlier the thrown rock strikes the ground compared to the dropped rock, we need to use the kinematic equation Projectile motion that relates displacement (Δx), initial velocity (vi), acceleration (a), and time (t): Δx = vi * t + 0.5 * a * t₂.
Since we're dealing with gravity, acceleration due to gravity (a) is 9.8 m/s². For the rock that is dropped, the initial velocity (vi) is 0 m/s, while the rock thrown downward has an initial velocity of 29 m/s.
First, we'll find the time it takes for the dropped rock to hit the ground:
0 = vi * t + 0.5 * a * t₂ - Δx0 = 0 * t + 0.5 * 9.8 * t₂ - 3000 = 4.9 * t₂ - 300t₂ = 61.22t ≈ √61.22 ≈ 7.83 secondsNext, we use the same equation to find the time for the thrown rock:
0 = vi * t + 0.5 * a * t₂ - Δx0 = 29 * t + 0.5 * 9.8 * t₂ - 3000 = 29 * t + 4.9 * t₂ - 300This forms a quadratic equation in the form of at₂ + bt + c = 0. To solve for t, we use the quadratic formula. We only need the positive root since time cannot be negative:
t = (-b ± √(b2 - 4ac))/(2a)t ≈ (5.94 seconds)Finally, the difference in time between when the two rocks hit the ground is:
Time difference = Time for dropped rock - Time for thrown rockTime difference ≈ 7.83 s - 5.94 s ≈ 1.89 secondsTherefore, the thrown rock hits the ground approximately 1.89 seconds earlier than the dropped rock.
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A golfer takes three strokes to putt a golf ball into a hole. On the first stroke, the ball moves 4.8 m due east. On the second, it moves 2.7 m at an angle 20° north of east. On the third, it moves 0.50 m due north. If the golfer had instead hit the ball directly into the hole on the first stroke, what would have been the magnitude (in m) and direction of the ball's displacement? Give the direction as a positive angle measured counterclockwise from due east.
The magnitude of the ball's displacement would have been approximately 7.579 meters, and the direction would have been approximately 11.08 degrees counterclockwise from due east.
1. First stroke:
[tex]Displacement \( \vec{d}_1 = 4.8 \, \text{m} \) due east.[/tex]
2. Second stroke:
[tex]Eastward component: \( 2.7 \, \text{m} \times \cos(20^\circ) \approx 2.564 \, \text{m} \) due east.\\ Northward component: \( 2.7 \, \text{m} \times \sin(20^\circ) \approx 0.924 \, \text{m} \) due north.\\\\ Displacement \( \vec{d}_2 = 2.564 \, \text{m} \) due east and \( 0.924 \, \text{m} \) due north.[/tex]
3. Third stroke:
Displacement[tex]\( \vec{d}_3 = 0.50 \, \text{m} \)[/tex] due north.
Now let's calculate the total eastward and northward displacements:
Total eastward displacement:[tex]\( 4.8 \, \text{m} + 2.564 \, \text{m} = 7.364 \, \text{m} \)[/tex]
Total northward displacement: [tex]\( 0.924 \, \text{m} + 0.50 \, \text{m} = 1.424 \, \text{m} \)[/tex]
Using the Pythagorean theorem, the magnitude of the total displacement is:
[tex]Magnitude = \( \sqrt{(\text{Total eastward displacement})^2 + (\text{Total northward displacement})^2} \)Magnitude = \( \sqrt{7.364^2 + 1.424^2} \approx 7.579 \, \text{m} \)[/tex]
The direction of the total displacement is given by:
[tex]Direction = \( \tan^{-1}\left(\frac{\text{Total northward displacement}}{\text{Total eastward displacement}}\right) \)\\Direction = \( \tan^{-1}\left(\frac{1.424}{7.364}\right) \approx 11.08^\circ \) counterclockwise from due east.[/tex]
So, if the golfer had hit the ball directly into the hole on the first stroke, the magnitude of the ball's displacement would have been approximately 7.579 meters, and the direction would have been approximately 11.08 degrees counterclockwise from due east.
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Final answer:
The golfer's ball would have a direct displacement of approximately 7.434 meters in a direction of roughly 11° north of east if it were hit directly into the hole.
Explanation:
To determine the direct displacement and direction of the ball if the golfer had hit it directly into the hole, we need to sum up the three given vectors. The first is 4.8 m due east, the second is 2.7 m at a 20° angle north of east, and the third is 0.50 m due north.
First, decompose the second stroke into its north and east components. The east component is 2.7 cos(20°) m and the north component is 2.7 sin(20°) m.
East component of the second stroke: 2.7 cos(20°) ≈ 2.536 m
North component of the second stroke: 2.7 sin(20°) ≈ 0.923 m
Now, sum the components in each direction:
Total east displacement: 4.8 m + 2.536 m = 7.336 m
Total north displacement: 0.923 m + 0.50 m = 1.423 m
Finally, calculate the magnitude and direction of the resultant displacement vector:
Magnitude (√(x² + y²)): √(7.336² + 1.423²) ≈ 7.434 m
Direction (tan⁻¹(y/x)): tan⁻¹(1.423/7.336) ≈ 11° north of east
The Mariana Trench in the western Pacific Ocean includes the greatest known ocean depth at approximately 6.8 miles. The atmosphere exerts a pressure of 14.7 lbf/in.2 at the ocean surface. Modeling the ocean seawater as static and assuming constant local acceleration of gravity of 32.1 ft/s2 and constant seawater density of 64.2 lb/ft3 , determine the absolute pressure, in lbf/in.2 , at this depth
Final answer:
The absolute pressure at the bottom of the Mariana Trench is calculated by adding the atmospheric pressure to the hydrostatic pressure due to the water column, resulting in approximately 509,367.85 lbf/in.².
Explanation:
To calculate the absolute pressure at the bottom of the Mariana Trench, we start by understanding that pressure in a static fluid increases linearly with depth. The increase in pressure, ΔP, due to the water column can be calculated using the formula ΔP = ρgh, where ρ is the density of seawater, g is the acceleration due to gravity, and h is the depth. Given the constants, ρ = 64.2 lb/ft³, g = 32.1 ft/s², and h = 6.8 miles (35,856 ft), we first convert the depth into feet as pressure calculations require consistent units. The calculation is as follows: ΔP = 64.2 lb/ft³ * 32.1 ft/s² * 35,856 ft = 73,346,473.6 lb/ft². Converting this to lbf/in.², we divide by 144 (since 1 ft² = 144 in.²), resulting in ΔP approximately 509,353.15 lbf/in.². Adding the atmospheric pressure of 14.7 lbf/in.² at the surface, the absolute pressure at the bottom of the Mariana Trench in lbf/in.² is approximately 509,367.85 lbf/in.².
The absolute pressure at the depth of the Mariana Trench is approximately [tex]\(7.42 \times 10^9 \, \text{lbf/in}^2\),[/tex]calculated using the hydrostatic pressure formula.
To find the absolute pressure at the depth of the Mariana Trench, we can use the hydrostatic pressure formula:
[tex]\[ P = P_0 + \rho \cdot g \cdot h \][/tex]
Where:
- P is the absolute pressure at the depth,
- P0 is the atmospheric pressure at the surface (given as 14.7 lbf/in²),
- rho is the density of seawater (given as 64.2 lb/ft³),
- g is the acceleration due to gravity (given as 32.1 ft/s²), and
- h is the depth of the trench (given as 6.8 miles).
First, let's convert the depth from miles to feet:
[tex]\[ 6.8 \text{ miles} \times 5280 \text{ ft/mile} = 35904 \text{ ft} \][/tex]
Now, we can plug in the values into the formula:
[tex]\[ P = 14.7 \text{ lbf/in}^2 + (64.2 \text{ lb/ft}^3) \times (32.1 \text{ ft/s}^2) \times (35904 \text{ ft}) \][/tex]
Let's calculate this value.
To find the absolute pressure at the depth of the Mariana Trench, we'll first calculate the pressure due to the water column using the hydrostatic pressure formula:
[tex]\[ P = P_0 + \rho \cdot g \cdot h \][/tex]
Where:
- [tex]\( P_0 = 14.7 \, \text{lbf/in}^2 \)[/tex] is the atmospheric pressure at the surface,
-[tex]\( \rho = 64.2 \, \text{lb/ft}^3 \)[/tex] is the density of seawater,
- [tex]\( g = 32.1 \, \text{ft/s}^2 \)[/tex] is the acceleration due to gravity, and
- [tex]\( h = 6.8 \, \text{miles} \times 5280 \, \text{ft/mile} = 35904 \, \text{ft} \)[/tex]is the depth of the trench in feet.
Plugging in the values:
[tex]\[ P = 14.7 + (64.2 \times 32.1 \times 35904) \][/tex]
Let's calculate this.
[tex]\[ P = 14.7 + (64.2 \times 32.1 \times 35904) \][/tex]
[tex]\[ P = 14.7 + (64.2 \times 32.1 \times 35904) \][/tex]
[tex]\[ P = 14.7 + (206368.8 \times 35904) \][/tex]
[tex]\[ P = 14.7 + 7417798272 \][/tex]
[tex]\[ P ≈ 7417798286.7 \, \text{lbf/in}^2 \][/tex]
Therefore, the absolute pressure at the depth of the Mariana Trench is approximately [tex]\( 7.42 \times 10^9 \, \text{lbf/in}^2 \).[/tex]
A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force of 4.7 x 10^11 N. How much charge is on the sphere and which has the lower amount of charge?
Answer:0.114 C
Explanation:
Given
Total 4.7 C is distributed in two spheres
Let [tex]q_1[/tex] and [tex]q_2[/tex] be the charges such that
[tex]q_1+q_2=4.7[/tex]
and Force between charge particles is given by
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
[tex]4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}[/tex]
[tex]q_1\cdot q_2=0.522[/tex]
put the value of [tex]q_1[/tex]
[tex]q_2\left ( 4.7-q_2\right )=0.522[/tex]
[tex]q_2^2-4.7q_2+0.522=0[/tex]
[tex]q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}[/tex]
[tex]q_2=0.114 C[/tex]
thus [tex]q_1=4.586 C[/tex]
A traveler covers a distance of 217 miles in a time of 8 hours 32 minutes. (a) What is the average speed for this trip?
Answer:
Average speed, v = 11.37 m/s
Explanation:
Given that,
The distance covered by the traveler, d = 217 miles = 349228 meters
Time taken, [tex]t = 8\ hours \ 32\ minutes =8\ h+\dfrac{32}{60}\ h=8.53\ h[/tex]
or t = 30708 s
We need to find the average speed for this trip. The average speed is given by :
[tex]v=\dfrac{d}{t}[/tex]
[tex]v=\dfrac{349228\ m}{30708\ s}[/tex]
v = 11.37 m/s
So, the average speed for this trip is 11.37 m/s. Hence, this is the required solution.
A 200g of iron at 120 degrees and a 150 g piece of copper at -50 degrees are dropped into an insulated beaker containing 300 g of ethyl alcohol at 20 degrees. What is the final temperature?
The final temperature can be calculated using the principle of conservation of energy. Using the formulas Q = mcΔT and Q = mcΔT, we can find the heat lost by the iron and the heat gained by the ethyl alcohol. The final temperature of all 3 substances are, T_f = 12.08 °C.
Explanation:To calculate the final temperature of the substances, we can use the principle of conservation of energy. The heat gained by one substance is equal to the heat lost by another substance. First, we calculate the heat lost by the iron using the formula Q = mcΔT, where Q is the heat lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Then, we calculate the heat gained by the ethyl alcohol using the same formula. Finally, we use the principle of conservation of energy to find the final temperature.
We have:
Iron: m = 200 g, c = 452 J/g °C, Ti = 120 °CCopper: m = 150 g, c = 387 J/g °C, Ti = -50 °CEthyl alcohol: m = 300 g, c = 2.30 J/g °C, Ti = 20 °CUsing the formula Q = mcΔT, we can calculate:
Heat lost by iron: Qiron = (200 g)(452 J/g °C)(120 °C - T_f)Heat gained by ethyl alcohol: Qethyl alcohol = (300 g)(2.30 J/g °C)(T_f - 20 °C)Since the heat lost by the iron is equal to the heat gained by the ethyl alcohol, we can set up the equation:
(200 g)(452 J/g °C)(120 °C - T_f) = (300 g)(2.30 J/g °C)(T_f - 20 °C)
To solve for T_f, we can simplify and rearrange the equation:
(90400 - 452T_f) J = (T_f- 13800) J
Combining like terms:
13800 J = 1142T_f J
Dividing both sides by 1142:
T_f = 12.08 °C
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