Answer:
(a): [tex]\rm meter/ second^2.[/tex]
(b): [tex]\rm meter/ second^3.[/tex]
(c): [tex]\rm 2ct-3bt^2.[/tex]
(d): [tex]\rm 2c-6bt.[/tex]
(e): [tex]\rm t=\dfrac{2c}{3b}.[/tex]
Explanation:
Given, the position of the particle along the x axis is
[tex]\rm x=ct^2-bt^3.[/tex]
The units of terms [tex]\rm ct^2[/tex] and [tex]\rm bt^3[/tex] should also be same as that of x, i.e., meters.
The unit of t is seconds.
(a):
Unit of [tex]\rm ct^2=meter[/tex]
Therefore, unit of [tex]\rm c= meter/ second^2.[/tex]
(b):
Unit of [tex]\rm bt^3=meter[/tex]
Therefore, unit of [tex]\rm b= meter/ second^3.[/tex]
(c):
The velocity v and the position x of a particle are related as
[tex]\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.[/tex]
(d):
The acceleration a and the velocity v of the particle is related as
[tex]\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.[/tex]
(e):
The particle attains maximum x at, let's say, [tex]\rm t_o[/tex], when the following two conditions are fulfilled:
[tex]\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.[/tex][tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}<0.[/tex]Applying both these conditions,
[tex]\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.[/tex]
For [tex]\rm t_o = 0[/tex],
[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c[/tex]
Since, c is a positive constant therefore, for [tex]\rm t_o = 0[/tex],
[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0[/tex]
Thus, particle does not reach its maximum value at [tex]\rm t = 0\ s.[/tex]
For [tex]\rm t_o = \dfrac{2c}{3b}[/tex],
[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.[/tex]
Here,
[tex]\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}<0.[/tex]
Thus, the particle reach its maximum x value at time [tex]\rm t_o = \dfrac{2c}{3b}.[/tex]
The units of constant c are in meters per second squared per second squared (m/s^2). The units of constant b are in meters per second cubed (m/s^3). The velocity can be found by taking the derivative of the position function, and the acceleration can be found by taking the derivative of the velocity function. To find the time when the particle reaches its maximum x value, set the velocity equal to zero and solve for t.
Explanation:(a) The units of constant c can be determined by analyzing the equation x = ct2 - bt3. Since x is in meters and t is in seconds, the units of x/t2 will be in meters per second squared. Therefore, the units of constant c will be in meters per second squared per second squared (m/s2).
(b) Similar to part (a), the units of constant b can be determined by analyzing the equation x = ct2 - bt3. Since x is in meters and t is in seconds, the units of x/t3 will be in meters per second cubed. Therefore, the units of constant b will be in meters per second cubed (m/s3).
(c) The velocity v can be found by taking the derivative of the position function x with respect to time t. In this case, v = d/dt(ct2 - bt3) = 2ct - 3bt2.
(d) The acceleration a can be found by taking the derivative of the velocity function v with respect to time t. In this case, a = d/dt(2ct - 3bt2) = 2c - 6bt.
(e) To find the time when the particle reaches its maximum x value, we can set the velocity v equal to zero and solve for t. 2ct - 3bt2 = 0. Solving this equation will give us the time t when the particle reaches its maximum x value.
According to the Creative Curriculum model, the _____ sets the context for teaching and learning
A. structure of the classroom
B. children
C. Centers used in the classroom
D. teacher
Answer:
A. structure of the classroom
Explanation:
According to the Creative Curriculum model, the A. structure of the classroom sets the context for teaching and learning.
Creative curriculum model stands to enhance the skill of the students through a structured class room. Creative curriculum goes beyond the rote learning and focuses on big ideas and individual passion of each students.
Two spaceships A and B move in opposite directions from earth at the same speed as measured by an earthbound observer. If an observer on spaceship A measures the velocity of spaceship B to be 0.99c, what is the speed of each spaceship as seen from earth (in units of c)?
Answer:
the magnitude of each spaceships velocity is 0.495c
Explanation:
This is a relative velocity problem, that means that the velocity is relative to the observer, in this case, there is an observer on earth, from this reference point the velocity of each spaceship are the same, but for the observer on a spaceship would be different, we can see it as a velocity difference :
[tex]V=Va-Vb\\[/tex]
because the spaceship B is going on the opposite direction from A, the velocity is negative, the observer on earth says the velocity is the same for both o them, so:
[tex]0.99c=va-(-va)\\0.99c=2va\\va=0.495c[/tex]
vb=-va
During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot can fall when the acceleration is more than 8g, with g being the gravitational constant. Find the radius of the smallest circle that the pilot can take. (HINT: Think about the centripetal acceleration).
Answer:
1472.98 m
Explanation:
Data provided:
Speed of circular looping, v = 340 m/s
Acceleration, a = 8g
here,
g is the acceleration due to the gravity = 9.81 m/s²
Now,
the centripetal acceleration is given as,
[tex]a=\frac{v^2}{r}[/tex]
r is the radius of the loop
on substituting the respective values, we get
[tex]8\times9.81=\frac{340^2}{r}[/tex]
or
r = 1472.98 m
Solids in which electrons are able to move easily from one atom to another are: a) conductors b) electrically charged c) capacitors d) insulators
Answer:
Option a) conductors
Explanation:
Conductors are the solids which allows the movement of electrons from atom to atom.
In conductors, electrons are free to move in the conductor's vicinity thus these free electrons contributes to the electrical conductivity of the conductor. A body is said to be electrically charged when the electrons in the atoms of the body are not evenly shared.Capacitors are the elements for the storage of electric charge. Insulators are the materials with no free electrons and have a high energy gap which do not allow the conduction of electricity.Answer:
a) Conductors.
Explanation:
Conductors are solids that have free electrons. Some of the examples of solids that are conductors are Steel, Iron, Silver, Copper etc. How best a metallic solid can conduct depends on the number of conduction electrons present in it.These electrons are called conduction electrons.
Not all solids are conductors. For example, wood is a solid, but it is not a conductor. It is an insulator.
Insulators are those materials that do not have free electrons.
Semiconductors are the materials that behave as conductors under certain conditions
Preston pushes a wheelbarrow weighing 500 N to the top of a 50.0-m ramp, inclined at 20.0° with the horizontal, and leaves it. Tamara accidentally bumps the wheelbarrow. It slides back down the ramp, during which an 80.0-N frictional force acts on it over the 50.0 m. What is the wheelbarrow's kinetic energy at the bottom at of the ramp
The wheelbarrow's final kinetic energy at the bottom of the ramp is 4500 J, found by calculating the work done on the wheelbarrow by gravity and friction and applying the work-energy theorem.
Explanation:To determine the wheelbarrow's kinetic energy at the bottom of the ramp, we need to calculate the work done on the wheelbarrow by both gravity and friction as it slides down the ramp, and then use this to find the change in kinetic energy.
The work done by gravity is simply the component of the wheelbarrow's weight (500 N) that acts parallel to the ramp (i.e., 500 N * sin(20) * 50 m = 8500 J).
The work done by friction is the frictional force (80 N) times the distance (50 m), but since friction opposes the motion it’s negative, so -4000 J.
By the work-energy theorem, the total work done on the wheelbarrow equals its change in kinetic energy. The wheelbarrow starts from rest, so its initial kinetic energy is zero. Hence, its final kinetic energy (at the bottom of the ramp) is the total work done on it: 8500 J - 4000 J = 4500 J.
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Say you have a differential drive robot that has an axle length of 30cm and wheel diameter of 10cm. Find the angular velocity for the left and right wheel if the robot is going to a. Spin in place at a rate of 6 rpm (revolutions per min), b. Drive a circle of radius 1 meter (measured center of circle to middle of axle) at 3 rpm, c. Drive a straight line at 1 meter/min.
Answer:
a) ω1 = 18rpm ω2 = -18rpm
b) ω1 = 102rpm ω2 = 138rpm
c) ω1 = ω2 = 3.18rpm
Explanation:
For the first case, we know that each wheel will spin in a different direction but with the same magnitude, so:
ωr = 6rpm This is the angular velocity of the robot
[tex]\omega = \frac{\omega r * D/2}{r_{wheel}}[/tex] where D is 30cm and rwheel is 5cm
[tex]\omega = \frac{6 * 30/2}{5}=18rpm[/tex] One velocity will be positive and the other will be negative:
ω1 = 18rpm ω2 = -18rpm
For part b, the formula is the same but distances change. Rcircle=100cm:
[tex]\omega 1 = \frac{\omega r * (R_{circle} - D/2)}{r_{wheel}}[/tex]
[tex]\omega 2 = \frac{\omega r * (R_{circle} + D/2)}{r_{wheel}}[/tex]
Replacing values, we get:
[tex]\omega 1 = \frac{6 * (100 - 30/2)}{5}=102rpm[/tex]
[tex]\omega 2 = \frac{\omega r * (100 + 30/2)}{5}=138rpm[/tex]
For part c, both wheels must have the same velocity:
[tex]\omega = \frac{V_{robot}}{r_{wheel}}=20rad/min[/tex]
[tex]\omega = 20rad/min * \frac{1rev}{2*\pi rad}=3.18rpm[/tex]
Calculate the net charge on a substance consisting of a combination of 1.3 × 10^13 protons and 4.4x 10^13 electrons. The elemental charge is 1.6 × 10^-19 C. Answer in units of C
Answer:
The charge transferred equals [tex]-4.96\times 10^{-6}Columbs[/tex]
Explanation:
The charge quantizatrion principle states that the charge is always transferred in terms of product of fundamental charges.
Mathematically [tex]Q=ne[/tex]
where
'e' is the fundamental magnitude of charge
Thus charge transferred by [tex]4.4\times 10^{13}[/tex] electrons equals
[tex]Q_{1}=4.4\times 10^{13}\times -1.6\times 10^{-19}=-7.04\times 10^{-6}Columbs[/tex]
Similarly charge transferred by [tex]1.3\times 10^{13}[/tex] protons equals
[tex]Q_{2}=1.3\times 10^{13}\times 1.6\times 10^{-19}=2.08\times 10^{-6}Columbs[/tex]
Thus the net charge transferred equals
[tex]Q_{1}+Q_{2}=-7.04\times 10^{-6}+2.08\times 10^{-6}\\\\=-4.96\times 10^{-6}columbs[/tex]
Change a speed of 72.4 miles per hour to its equivalent in meters per second.
Final answer:
To convert 72.4 miles per hour to meters per second, multiply by 1609.34 to convert miles to meters, then divide by 3600 to convert hours to seconds. The result is approximately 32.34 meters per second.
Explanation:
To convert a speed of 72.4 miles per hour to its equivalent in meters per second, you can use the conversion factors that 1 mile is equal to 1609.34 meters and 1 hour is equal to 3600 seconds. Since the conversion from kilometers per hour to meters per second involves dividing by 3.6, and knowing from the reference material that 60 miles per hour is approximately equal to 27.8 meters per second, we can calculate the speed in meters per second as follows:
72.4 mi/h * (1609.34 m / 1 mi) * (1 h / 3600 s) = 72.4 * 1609.34 / 3600 m/s.Doing the math:
72.4 * 1609.34 / 3600 = 32.34 m/s (rounded to two decimal places).Check if the answer is reasonable: Since 72.4 mi/h is a little more than 60 mi/h, and we know that 60 mi/h is approximately 27.8 m/s, it seems reasonable that 72.4 mi/h would be slightly over 27.8 m/s, so our answer of 32.34 m/s seems plausible within the context.
One cubic meter (1.00 m) of aluminum has a mass of 2.70 10 kg, and the same volume of iron has a mass of 7.8610'kg. Find the radus of a solid aluminum sphere that will balance a sod irn sphere o radu 14 squal-a balano rm
Answer:
Radius of aluminium sphere which has same mass as of sphere of iron with radius 14 m is 40.745 meters.
Explanation:
Let the radius of aluminium sphere be [tex]r_{a}[/tex]
From the relation between density, mass and volume we know that
[tex]Mass=Density\times volume...............(i)[/tex]
Applying equation 'i' separately to iron and aluminium sphere we get
[tex]M_{a}=\rho _{a}\times V_{a}[/tex]
[tex]M_{ir}=\rho _{ir}\times V_{ir}[/tex]
Equating the masses of iron and aluminium spheres we get
[tex]M_{a}=M_{ir}[/tex]
[tex]\rho _{a}\times \frac{4\pi r_a^3}{3}=\rho _{ir}\times \frac{4\pi r_{ir}^3}{3}[/tex]
[tex]\rho _{a}\times r_{a}^3=\rho _{ir}\times r_{ir}^{3}\\\\\therefore r_{a}=(\frac{\rho _{ir}}{\rho _{a}})^{1/3}\cdot r_{ir}\\\\r_{a}=\frac{7.8610}{2.701}\times 14\\\\\therefore r_a=40.745m[/tex]
Consider a gray squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel's velocity (in m/s) just before hitting the ground, assuming it fell from a height of 3.0 m
Answer: 7.66 8 m/s
Explanation:
Since the squirrel fell out and it is not thrown, we assume the initial velocity is zero ([tex]V_{o}=0[/tex]). On the other hand, the squirrel only experiences the acceleration due gravity, which is constant and in the downward direction [tex]g=-9.8 m/s^{2}[/tex]. So, the following equation will be useful to find the squirrel's final velocity [tex]V_{f}[/tex]:
[tex]{(V_{f})}^{2}={(V_{o})}^{2}-2gd[/tex]
Where [tex]d=3 m[/tex] is the height from which the squirrel fell
As [tex]V_{o}=0[/tex]:
[tex]{(V_{f})}^{2}=-2gd[/tex]
Then:
[tex]{(V_{f})}^{2}=-2(-9.8 m/s^{2})(3 m)[/tex]
[tex]V_{f}=\sqrt{58.8 m^{2}/s^{2}}[/tex]
Finally:
[tex]V_{f}=7.668 m/s[/tex]
A plane wall of thickness 0.1 mm and thermal conductivity 25 W/m K having uniform volumetric heat generation of 0.3 MW/m3 is insulated on one side, while the other side is exposed to a fluid at 92 oC. The convection heat transfer coefficient between the wall and the fluid is 500 W/m2 K. Determine the maximum temperature in the wall.
Answer:
The maximum temperature is 90.06° C
Explanation:
Given that
t= 0.1 mm
Heat generation
[tex]q_g=0.3\ MW/m^3[/tex]
Heat transfer coefficient
[tex]h=500\ W/m^2K[/tex]
Here one side(left side) of the wall is insulated so the all heat will goes in to right side .
The maximum temperature will at the left side.
Lets take maximum temperature is T
Total heat flux ,q
[tex]q=q_g\times t[/tex]
[tex]q=0.3\times 1000000\times 0.1 \times 10^{-3}\ W/m^2[/tex]
[tex]q=30\ W/m^2[/tex]
So the total thermal resistance per unit area
[tex]R=\dfrac{t}{K}+\dfrac{1}{h}[/tex]
[tex]R=\dfrac{0.1\times 10^{-3}}{25}+\dfrac{1}{500}[/tex]
R=0.002 K/W
We know that
q=ΔT/R
30=(T-90)/0.002
T=90.06° C
The maximum temperature is 90.06° C
Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way between the plates?
Answer: 8*10^-15 N
Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that, F=q*E
The electric field between the plates is given by:
E = ΔV/d = 500 V/0.01 m=5*10^3 N/C
the force applied to the electron is: F=e*E=8*10^-15 N
A parallel-plate capacitor is formed from two 2.0 cm x 2.0 cm electrodes spaced 2.2 mm apart. The electric field strength inside the capacitor is 1.0×10^6 N/C. Part A What is the charge (in nC) on the positive electrode? Express your answer in nanocoulombs.
Answer:
3.54 nC.
Explanation:
Given:
Dimensions of the electrodes of parallel-plate capacitor = [tex]2.0\ cm \times 2.0\ cm.[/tex]Separation between the electrodes, [tex]d=2.2\ mm = 2.2\times 10^{-3}\ m.[/tex]Electric field strength inside the capacitor, [tex]E = 1.0\times 10^6\ N/C.[/tex]The surface area of each of the electrode is given by
[tex]A=2.0\ cm \times 2.0\ cm = 4.0\ cm^2 = 4.0\times 10^{-4}\ m^2.[/tex]
The strength of the electric field inside a capacitor is given by
[tex]E = \dfrac{\sigma}{\epsilon_o}[/tex]
where,
[tex]\epsilon_o[/tex] = electrical permittivity of free space = [tex]8.85\times 10^{-12}\ C^2N^{-1}m^{-2}[/tex][tex]\sigma[/tex] = surface charge density of the electrode = [tex]\dfrac qA[/tex][tex]q[/tex] = charge on the electrode.Therefore,
[tex]E = \dfrac{q}{A\epsilon_o}\\q=EA\epsilon_o\\=1.0\times 10^6\times 4.0\times 10^{-4}\times 8.85\times 10^{-12}\\=3.54\times 10^{-9}\ C\\=3.54\ nC.[/tex]
It is the charge on the positive electrode.
The value of an electric field at a particular point is known as the strength of the electric field. the value of the charge will be 3.54 NC.
What is the strength of the electric field?
The electric field is the area around an electric charge where its impact may be felt. The force encountered by a unit positive charge put at a spot is the electric field strength at that point.
Electrode dimension = 2.0 cm×2.0 cm
Area of electrode = 4.0 cm²
seperated distance beween electrode = 2.2 mm = 2.2 × 10⁻³ m
strength of electric field = E = 1.0 × 10⁶ N/C
Electric field strength is given by
[tex]\rm E = \frac{\sigma}{\varepsilon_0}[/tex]
where,
[tex]\varepsilon_0[/tex]= electrical permittivity of free space
[tex]\sigma[/tex] = surface charge density of the electrode
q=Electrode charge
[tex]\rm \sigma =\RM{ \frac{q}{A} }[/tex]
[tex]\rm E = \frac{q}{A\varepsilon_0}[/tex]
[tex]\rm q = \sigma{A\varepsilon_0}[/tex]
[tex]\rm q =1.0\times 10^-6\times 4.0\times10^{-4}\times 8.85 \times 10^-12[/tex]
[tex]\rm q = 3.54\times 10^-9 \; C[/tex]
[tex]\rm q = 3.54\; nC[/tex]
Hence the value of the charge will be 3.54 NC.
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Three moles of ideal gas is inside a sealed rectangular container of size 0.2 m x 0.2 m x 0.4 m. If the pressure inside the container is 498,840 N/m^2, what is the temperature of the gas inside the container?
Answer:
320 K
Explanation:
Given:
Number of moles of gas , n = 3
Dimensions of the rectangular container = 0.2 m × 0.2 m × 0.4 m
Thus,
Volume of the container = 0.2 m × 0.2 m × 0.4 m = 0.016 m³
Pressure inside the container, P = 498,840 N/m²
Now,
From the ideal gas law, we have
PV = nRT
here,
T is the temperature
R is the ideal gas constant = 8.314 J / (mol·K)
on substituting the values, we get
498,840 × 0.016 = 3 × 8.314 × T
or
T = 320 K
Hence, the pressure inside the container is 320 K
If an amplitude modulation (AM) has an amplitude modulating signal of 0.5 V and carrier amplitude of 1.3 V, what is the modulation index?
Answer:
The modulation index in the amplitude modulation will be 0.384
Explanation:
We have given amplitude of adulating signal [tex]A_m=0.5volt[/tex]
Amplitude of carrier signal [tex]A_C=1.3volt[/tex]
We have to find modulation index
Modulation index is the ratio of amplitude of modulating signal and amplitude of carrier signal
So modulation index [tex]m=\frac{A_m}{A_c}=\frac{0.5}{1.3}=0.384[/tex]
So the modulation index in the amplitude modulation will be 0.384
You comb your hair and the comb becomes negatively charged. Strictly speaking, how will the mass of your hair change? A. it wil increase B. it will decrease C. it wil not change
Answer:
B. it will decrease
Explanation:
If the comb is negatively charged, it's because electrons went from the hair to the comb. These electrons have a certain mass, that is why the mass of the hair will decrease.
The mass of your hair will not change significantly when combing and the comb becomes negatively charged because the mass of electrons is extremely small. Small pieces of paper will be attracted to a negatively charged comb due to electrostatic forces.
Explanation:When you comb your hair and the comb becomes negatively charged, it does so by gaining excess electrons from your hair. However, the mass of electrons is extremely small, so when considering the mass change of your hair due to the transfer of electrons, it's accurate to say that the mass will not change significantly enough to be noticeable or measurable with standard equipment. Therefore the correct answer is C. the mass will not change.
In the contexts of the related information provided, when you bring a negatively charged comb close to small pieces of paper, they will be attracted to the comb because of electrostatic forces. This is because the paper, being neutral, becomes polarized in the presence of the charged comb, leading to an attraction. Thus, the observed effect would be b. Pieces of paper are attracted to the comb.
A car traveling 85 km/h slows down at a constant 0.47 m/s^2 just by "letting up on the gas." Calculate the distance it travels during the first second. Express your answer to two significant figures and include the appropriate units. Calculate the distance it travels during the fifth second.
Answer:
Explanation:
u = 85 km / h = 23.61 m /s
Acceleration a = - 0.47 m /s²
Distance travelled in n th sec
S_n = u + ( 2t -1 )a/2
S_n is distance travelled in n th seond
Distance traveled in 1 st second
= 23.61 - .5 x .47
= 23.37 m
=23 m
Distance travelled when t = 5 th
= 23.61 - (2x5 -1)/2 x .47
= 23.61 - 4.5 x .47
23.61 -2.115
=21.495
21 m is distance travelled in 5 th second .
Which of the following is not a fundamental quantity? A. time B. weight C. mass D. length E. electric charge
Answer:
Option B
Explanation:
The fundamental quantities are the basic quantities from which other quantities are derived.
There are a total of 7 fundamental quantities, namely Length, Mass, Time, Luminous Intensity, Amount of substance, Electric Current, Temperature.
From these 7 quantities, the other quantities are derived.
Weight is the product of its mass and the gravitational acceleration 'g' which acts on that mass.
Where mass is the fundamental quantity and 'g' is also derived and hence weight is a derived quantity.
what is the necessary condition on a force the result the conservation of angular momentum for a particle affected by that force?
Answer:
The necessary condition is that the force that is applied should not be able to produce any torque about the axis of rotation which is achieved only if the force passes through the axis of rotation.
Explanation:
Mathematically the angular momentum is related to force as
[tex]\overrightarrow{F}\times \overrightarrow{r}=\frac{d\overrightarrow{l}}{dt}[/tex]
where
'r' is arm of the force 'F'
For angular momentum to be conserved the right hand side of the above equation should be zero which can happen only if:
1) F = 0
2) r = 0
3) F and r are parallel.
Thus fulfilling any of the above condition shall conserve the angular momentum of a particle.
What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle? a. 18.9 m/s b. 26.7 m/s с. 37.8 m/s d. 51.9 m/s
Answer:
Option (a) is correct.
Explanation:
The angle of banking of curved path is given as
tan θ[tex]=\frac{v^{2} }{rg}[/tex]
Here, v is linear velocity, r is radius of curved path, θ is bank angle and g is acceleration due to gravity.
We have, θ= 20.0°, r = 100 m and take [tex]g=9.8 m/s^{2}[/tex]
Substituting these values in above formula, we get
[tex]tan 20 =\frac{v^{2} }{100 m*9.8m/s^{2} }[/tex]
[tex]v^{2} =356.69[/tex]
[tex]v=\sqrt{356.69} =18.9m/s[/tex]
Thus, the ideal speed is 18.9 m/s.
The ideal speed to take a 100 m radius curve banked at a 20.0° angle is 18.9 m/s. The correct Option is a.
What is speed?The speed of any moving object is the ratio of the distance covered and the time taken to cover that distance.
Speed s = distance d / time t
Given a 100 m radius curve banked at a 20.0° angle.
The angle of banking of curved path is given as
tan θ= V² /2g
Here, V is linear velocity, r is radius of curved path, θ is bank angle and g is acceleration due to gravity.
Substituting the values in above formula, we get
tan 20° = V² /2x9.81
V = 18.9 m/s
Thus, the correct option is a.
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Part A Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depression constant (Kf) for water is 1.86 °C/m. Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depression constant () for water is 1.86 °C/m. +0.23 °C -0.23 °C 1.23 °C -0.45 °C +0.45 °C
Answer: [tex]-0.45^0C[/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=(0-T_f)^0C[/tex] = Depression in freezing point
i= vant hoff factor = 2 (for electrolyte undergoing complete dissociation, i is equal to the number of ions produced)
[tex]KCl\rightarrow K^++Cl^-[/tex]
[tex]K_f[/tex] = freezing point constant = [tex]1.86^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (water)= 550 g = 0.55 kg (1kg=1000g)
Molar mass of solute (KCl) = 74.5 g/mol
Mass of solute (KCl) = 5.0 g
[tex](0-T_f)^0C=2\times 1.86\times \frac{5g}{74.5g/mol\times 0.55kg}[/tex]
[tex](0-T_f)^0C=0.45[/tex]
[tex]T_f=-0.45^0C[/tex]
Thus the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water is [tex]-0.45^0C[/tex]
The freezing point of a solution : -0.45 °C
Further explanationSolution properties are the properties of a solution that don't depend on the type of solute but only on the concentration of the solute.
Solution properties of electrolyte solutions differ from non-electrolyte solutions because electrolyte solutions contain a greater number of particles and break down into ions. So the Solution properties of electrolytes is greater than non-electrolytes
The term is used in the Solution properties
• 1. molal
that is, the number of moles of solute in 1 kg of solvent
[tex]\large {\boxed {\bold {m = mole. \frac {1000} {mass \: of \: solvent (in \: grams)}}}[/tex]
• 2. mole fraction
the ratio of the number of moles of solute to the mole of solution
[tex]\large {\boxed {\bold {Xa = \frac {na} {na + nb}}}[/tex]
a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depression constant () for water is 1.86
• Step 1: determine molal
molar mass KCl = 39 + 35.5 = 74.5
mole KCl = mass: molar mass
mole KCl = 5 gr: 74.5
mole KCl = 0.067
molal = m = 0.067 x (1000: 550 gr water)
molal = 0.122
• Step 2: determine the freezing point of the solution
i = 1 + (n-1) a
i = 1 + (2-1) 1
i = 2
[tex]\displaystyle \Delta T_f=K_f.m.i[/tex]
---> KCl is an electrolyte solution
[tex]\Delta T_f=1.86.0.122.2[/tex]
[tex]\displaystyle \Delta T_f=0.454[/tex]
freezing point of water = 0 °C
[tex]0.454=0-T~solution[/tex]
[tex]T_f~solution=- 0.454 ^oC[/tex]
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Keywords: Freezing Point Depression, Boiling Point Elevation, Solution Properties
You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler has - 12 μC of charge. The tissue has 3 g of mass. What is the minimum charge required to pick up the tissue paper?
Answer:
9.81 × 10⁻¹⁰ C
Explanation:
Given:
Distance between the tissue and the tip of the scale, r = 6 cm = 0.06 m
Charge on the ruler, Q = - 12 μC = - 12 × 10⁻⁶ C
Mass of the tissue = 3 g = 0.003 Kg
Now,
The force required to pick the tissue, F = mg
where, g is the acceleration due to gravity
also,
The force between (F) the charges is given as:
[tex]F=\frac{kQq}{r^2}[/tex]
where,
q is the charge on the tissue
k is the Coulomb's constant = 9 × 10⁹ Nm²/C²
thus,
[tex]mg=\frac{kQq}{r^2}[/tex]
on substituting the respective values, we get
[tex]0.003\times9.81=\frac{9\times10^9\times(-12\times10^{-6})\times q}{0.06^2}[/tex]
or
q = 9.81 × 10⁻¹⁰ C
Minimum charge required to pick the tissue paper is 9.81 × 10⁻¹⁰ C
A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m. If her piston is 19.0 cm in diameter, what is the diameter of the football players' piston?
Answer:
The diameter of the piston of the players equals 55.136 cm.
Explanation:
from the principle of transmission of pressure in a hydraulic lift we have
[tex]\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}[/tex]
Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get
[tex]\frac{57.0\times 9.81}{\frac{\pi \times (19.0)^{2}}{4}}=\frac{4\times 120\times 9.81}{\frac{\pi \times D_{2}^{2}}{4}}[/tex]
Solving for [tex]D_{2}[/tex] we get
[tex]D_{2}^{2}=\frac{4\times 120}{57}\times 19^{2}\\\\\therefore D_{2}=\sqrt{\frac{4\times 120}{57}}\times 19\\\\D_{2}=55.136cm[/tex]
The diameter of the football players' piston is [tex]\({55.4 \text{ cm}} \).[/tex]
To solve for the diameter of the football players' piston in the hydraulic lift, we start by using the principle of hydraulic systems, which states that pressure is constant throughout the fluid.
Given data:
- Mass of the cheerleader, [tex]\( m_1 = 57.0 \)[/tex] kg
- Mass of four football players, [tex]\( m_2 = 4 \times 120 \)[/tex] kg
- Height difference, [tex]\( h = 1.10 \)[/tex] m
- Diameter of the cheerleader's piston, [tex]\( D_1 = 19.0 \)[/tex] cm
First, convert the diameter of the cheerleader's piston into meters:
[tex]\[ D_1 = 19.0 \text{ cm} = 0.19 \text{ m} \][/tex]
The force exerted by the cheerleader's piston is:
[tex]\[ F_1 = m_1 g \][/tex]
The force exerted by the football players' piston is:
[tex]\[ F_2 = m_2 g \][/tex]
According to Pascal's principle:
[tex]\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \][/tex]
where [tex]\( A_1 \)[/tex] and [tex]\( A_2 \)[/tex] are the cross-sectional areas of the cheerleader's and football players' pistons, respectively.
The area [tex]\( A \)[/tex] of a piston is given by:
[tex]\[ A = \frac{\pi D^2}{4} \][/tex]
Now, calculate [tex]\( A_1 \)[/tex]:
[tex]\[ A_1 = \frac{\pi (0.19)^2}{4} \][/tex]
[tex]\[ A_1 = \frac{\pi \times 0.0361}{4} \][/tex]
[tex]\[ A_1 \approx 0.0284 \text{ m}^2 \][/tex]
Next, find [tex]\( F_1 \)[/tex]:
[tex]\[ F_1 = m_1 g = 57.0 \times 9.8 \][/tex]
[tex]\[ F_1 \approx 558.6 \text{ N} \][/tex]
Now, find [tex]\( A_2 \)[/tex]:
[tex]\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \][/tex]
[tex]\[ A_2 = \frac{F_2 \times A_1}{F_1} \][/tex]
Calculate [tex]\( F_2 \)[/tex]:
[tex]\[ F_2 = m_2 g = 480 \times 9.8 \][/tex]
[tex]\[ F_2 = 4704 \text{ N} \][/tex]
Now, find [tex]\( A_2 \)[/tex]:
[tex]\[ A_2 = \frac{4704 \times 0.0284}{558.6} \][/tex]
[tex]\[ A_2 \approx 0.2397 \text{ m}^2 \][/tex]
Finally, solve for [tex]\( D_2 \)[/tex]:
[tex]\[ A_2 = \frac{\pi (D_2)^2}{4} \][/tex]
[tex]\[ (D_2)^2 = \frac{4 \times 0.2397}{\pi} \][/tex]
[tex]\[ D_2 \approx \sqrt{\frac{0.9588}{\pi}} \][/tex]
[tex]\[ D_2 \approx 0.554 \text{ m} \][/tex]
Convert [tex]\( D_2 \)[/tex] from meters to centimeters:
[tex]\[ D_2 \approx 55.4 \text{ cm} \][/tex]
A 1.0-μm-diameter oil droplet (density 900 kg/m^3) is negatively charged with the addition of 27 extra electrons. It is released from rest 2.0 mm from a very wide plane of positive charge, after which it accelerates toward the plane and collides with a speed of 3.5 m/s . What is the surface charge density of the plane?
Answer:
surface charge density = 5.91 µC/m²
Explanation:
given data
diameter = 1 μm
radius = 0.5 × [tex]10^{-6}[/tex] m
speed = 3.5 m/s
density 900 kg/m³
distance = 2 mm = 2 × [tex]10^{-3}[/tex] m
to find out
surface charge density of the plane
solution
volume is express as
volume = [tex]\frac{4}{3} \pi r^3[/tex]
volume = [tex]\frac{4}{3} \pi (0.5*10^{-6})^3[/tex]
volume = 5.23 × [tex]10^{-19}[/tex] m³
and
mass = density × volume
mass = 900 ×5.23× [tex]10^{-19}[/tex]
mass = 4.712 × [tex]10^{-16}[/tex] kg
and
from motion of equation
v² -u² = 2×a×s
here v is speed 3.5 and u initial is 0 and a is acceleration and s is distance
3.5² = 2×a×2 × [tex]10^{-3}[/tex]
acceleration = 3062.5 m/s²
so from newton second law
Force = mass × acceleration
force = 4.712 × [tex]10^{-16}[/tex] ×3062.5
force = 1.44305 × [tex]10^{-12}[/tex] N
so
surface charge density = [tex]\frac{2\epsilon F}{q}[/tex]
here q is charge with addition 27 extra electron and f is force and ∈ = 8.85× [tex]10^{-12}[/tex]
surface charge density = [tex]\frac{2\epsilon F}{q}[/tex]
surface charge density = [tex]\frac{2*8.85*10^{-12}*1.443*10^{-12}}{27*1.60*10^{-19}}[/tex]
surface charge density = 5.91 × [tex]10^{-6}[/tex] C/m²
so surface charge density = 5.91 µC/m²
How does a capacitor store energy? Can a resistor store energy?
Answer:
A capacitor stores energy in the form of an electric field.A resistor can’t store energy.Explanation:
A capacitor stores energy in the form of an electric field formed between the plates of the capacitor. To establish this electric field ( or to move against it) the charges must do work, and this work its stored in the capacitor.
A resistor can’t store energy. There is a loss of energy in a current passing a resistor, but this is thanks to the heat dissipated by the resistor, and can’t be recovered.
Final answer:
A capacitor stores energy in its electric field between its plates, whereas a resistor dissipates energy as heat.
Explanation:
A capacitor stores energy in the electric field between its plates. When a voltage is applied across the capacitor, it charges up and stores electrical potential energy.
The energy stored in a capacitor can be calculated using the formula:
E = ½CV²
Where E is the energy stored in joules, C is the capacitance of the capacitor in farads, and V is the voltage across the capacitor in volts.
On the other hand, a resistor dissipates energy as heat when a current passes through it and cannot store energy like a capacitor.
Consider flipping N fair coins (assume N is even) (a) 1 point] What is the probability of getting half heads and half tails, in no particular order? 1, simplify your answer, and show the limiting behavior of the (b) [1.5 points] Assume that N probability as N -> co.
Answer with Explanation:
The problem can be simplified as follows
The number of possible outcomes after tossing a coin N times is [tex]2^N[/tex] since for each toss 2 outcomes are possible
Since we need equal heads and equal tails the no of cases amont the [tex]2^n[/tex] cases are
[tex]\binom{N}{N/2}=\frac{N!}{(N-N/2)!(N/2)!}=\frac{n!^2}{(n/2)!^2}[/tex]
Thus the required probability is
[tex]P(E)=\frac{\frac{n!}{(n/2)!^2}}{2^n}=\frac{n!}{(n/2)!^2\cdot 2^n}[/tex]
Part 2)
For the limit as N approaches infinity we have
[tex]P(E')=\lim_{n\rightarrow \infty }\frac{n!}{(n/2)!^2\cdot 2^n}[/tex]
Using Stirling's approximation and solving we get
[tex]\lim_{n\rightarrow \infty }\binom{n}{i}\approx \sqrt{\frac{n}{2\pi i(n-i)}}\times \frac{n^n}{i^i(n-i)^i}\\\\\lim_{n\rightarrow \infty }\binom{n}{n/2}\approx \sqrt{\frac{2n}{\pi n^2}}\times \frac{n^n}{(n/2)^{n/2}\cdot (n/2)^{n/2}}\\\\\lim_{n\rightarrow \infty }\binom{n}{i}\approx \sqrt{\frac{2n}{\pi n^2}}\times 2^n\\\\P(E')=\frac{\sqrt{\frac{2n}{\pi n^2}}\times 2^n}{2^n}=\sqrt{\frac{2}{\pi N}}[/tex]
The car traveling 56.0 km/h is 25.0 m towards a haystack when the driver brakes. The car hits the haystack 2.11 s later. How fast is the car traveling at impact? Answer in m/s
Answer:
36.13 m/s
Explanation:
initial speed, u = 56 km/h = 15.56 m/s
distance, s = 25 m
t = 2.11 s
Let the speed of the car at the impact is v and the acceleration is a.
Use second equation of motion
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
[tex]25=1.56 \times 2.11+\frac{1}{2}\times a \times 2.11^{2}[/tex]
a = 9.75 m/s^2
Use first equation of motion
v = u + at
v = 15.56 + 9.75 x 2.11
v = 36.13 m/s
A european sports car dealer claims that his car
willaccelerates ar a constant rate from rest to 100 km/hr in8.00 s.
If so, what is the acceleration?
Answer:
Acceleration of the car, [tex]a=3.47\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the car, u = 0 (at rest)
Final speed of the car, v = 100 km/h = 27.77 m/s
Time, t = 8 s
We need to find the acceleration of the car. Mathematically, it is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{27.77\ m/s}{8\ s}[/tex]
[tex]a=3.47\ m/s^2[/tex]
So, the acceleration of the car is [tex]3.47\ m/s^2[/tex]. Hence, this is the required solution.
Answer:
acceleration is 3.47 m/s²
Explanation:
given data
speed = 100 km/hr = 27.78 m/s
time = 8 s
to find out
acceleration?
solution
we will apply here first equation of motion
v = u +at ..............1
here v is final speed and u is initial speed that is 0 and t is time that is 8 sec
so put all value in equation 1
v = u +at
27.78 = 0 +a(8)
a = [tex]\frac{27.78}{8}[/tex]
a = 3.47 m/s²
so acceleration is 3.47 m/s²
You are standing 1.2 m from a mirror, and you want to use a classic camera to take a photo of yourself. This camera requires you to select the distance of whatever you focus on. What distance do you choose?
Answer:
2.4 m
Explanation:
Plane Mirror: It is a type of mirror whose reflecting surface is plane. it produces an image of an object behind it at a distance equal to the object lying in front of the mirror.
According to the question, the person has to take the photo of himself who is looking at the mirror.
Let us assume that the person is standing in front of a plane mirror at a distance of 1.2 m.
Due to the reflection of light, the image of the person will be formed at a distance 1.2 m behind the mirror. This means the image of the person is 2.4 m apart from him.
As the person could take the picture of his own image from the classic camera in his hand. This means the image of the person will act as an object for the classic camera to which he is going to focus on which is at a distance of 2.4 m.
Hence, the person will choose a distance of 2.4 m.
Focus length is the distance of optics from the point of convergence of light rays to create a sharp image of the object.
The distance of the focus must be 2.4 meters.
How to select the distance for ideal focus length?
Focus length is the distance of optics from the point of convergence of light rays to create a sharp image of the object. Focus length represents in the millimetres.
In the given problem, the person is standing in front of the plane mirror. A plane mirror is the mirror with a flat reflective surface.
The image formed by the plane mirror is virtual, erect and the equal to the size of the person or thing.
Given information-
The distance of person from the mirror is 1.2 m.
As the image in the plane mirror appears at the equivalent distance to the object, thus the distance of the image of the person is,
[tex]d_i=1.2 \rm m[/tex]
Thus the distance of the image of the person appears is 1.2 meters behind the mirror.
To take the photo of this image, the distance of the focus should be equal to the sum of the distance of the image inside the mirror to the distance of person front of the mirror. Thus,
[tex]d_f=d_p+d_i\\d_f=1.2+1.2\\d_f=2.4 \rm m[/tex]
Hence the distance of the focus must be 2.4 meters.
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A bird flies 2.0km south and then 1.5km 37° east of south. How far will it have to fly to get back to its original place if it flies in a straight line?
Answer:
3.324 km
Explanation:
d1 = 2 km south
d2 = 1.5 km at 37° east of south
Write the displacements in vector form
[tex]\overrightarrow{d_{1}}=-2\widehat{j}[/tex]
[tex]\overrightarrow{d_{2}}=1.5\left (Sin37\widehat{i}-Cos37\widehat{j} \right )=0.9\widehat{i}-1.2\widehat{j}[/tex]
The resultant displacement is given by
[tex]\overrightarrow{d} = \overrightarrow{d_{1}}+ \overrightarrow{d_{2}}[/tex]
[tex]\overrightarrow{d} = \left ( 0.9 \right )\widehat{i}+\left ( -2-1.2 \right )\widehat{j}[/tex]
[tex]\overrightarrow{d} = \left ( 0.9 \right )\widehat{i}+\left ( -3.2\right )\widehat{j}[/tex]
The magnitude of displacement is given by
[tex]d=\sqrt{0.9^{2}+\left ( -3.2 \right )^{2}}=3.324 km[/tex]
Thus, the bird has to travel 3.324 km in a straight line to return to its original place.