Answer:
21.28 m
Explanation:
height, h = 71 m
velocity of raft, v = 5.6 m/s
let the time taken by the stone to reach to raft is t.
use second equation of motion for stone
[tex]h = ut + \frac{1}{2}at^{2}[/tex]
u = 0 m/s, h = 71 m, g = 9.8 m/s^2
71 = 0 + 0.5 x 9.8 x t^2
t = 3.8 s
Horizontal distance traveled by the raft in time t
d = v x t = 5.6 x 3.8 = 21.28 m
Determine the velocity and position as a function of time for the time force F(t)=F Cos^2(WT). Generate plots for the resulting equation.
Answer with Explanation:
We know from newton's second law that acceleration produced by a force 'F' in a body of mass 'm' is given by
[tex]a=\frac{Force}{Mass}=\farc{F}{m}[/tex]
In the given case the acceleration equals
[tex]a=\frac{F_{o}cos^{2}(\omega t)}{m}[/tex]
Now by definition of acceleration we have
[tex]a=\frac{dv}{dt}\\\\\int dv=\int adt\\\\v=\int adt\\\\v=\int \frac{F_{o}cos^{2}(\omega t)}{m}\cdot dt\\\\v=\frac{F_{o}}{m}\int cos^{2}(\omega t)dt\\\\v=\frac{F_{o}}{\omega m}\cdot (\frac{\omega t}{2}+\frac{sin(2\omega t)}{4}+c)[/tex]
Similarly by definition of position we have
[tex]v=\frac{dx}{dt}\\\\\int dx=\int vdt\\\\x=\int vdt[/tex]
Upon further solving we get
[tex]x=\int [\frac{F_{o}}{\omega m}\cdot (\frac{\omega t}{2}+\frac{sin(2\omega t)}{4}+c)]dt\\\\x=\frac{F_{o}}{\omega m}\cdot ((\int \frac{\omega t}{2}+\frac{sin(2\omega t)}{4}+c)dt)\\\\x(t)=\frac{F_{o}}{\omega m}\cdot (\frac{\omega t^{2}}{4}-\frac{cos(2\omega t)}{8\omega }+ct+d)[/tex]
The plots can be obtained depending upon the values of the constants.
An electric field of intensity 3.22 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m long if the following conditions are true. (a) The plane is parallel to the yz.plane. (b) The plane is parallel to the xy.plane. (C) The plane contains the yaxis, and its normal makes an angle of 34.2* with the x-axis. The electric field everywhere on the surface of a charged sphere of radius 0.205 m has a magnitude of 570 N/C and points radially outward from the center of the sphere. (a) What is the net charge on the sphere? nC (b) What can you conclude about the nature and distribution of charge inside the sphere?
Answer:
Part I:
(a): [tex]7.889\times 10^2\ Nm.[/tex]
(b): [tex]0\ Nm.[/tex]
(c): [tex]6.52\times 10^2\ Nm.[/tex]
Part II:
(a): [tex]2.664\times 10^{-9}\ C.[/tex]
(b): Charge on the sphere is positive and only distributed on the surface of the sphere and not inside the sphere.
Explanation:
Part I:
Assuming,
[tex]\hat i,\ \hat j,\ \hat k[/tex] are the unit vectors along positive x, y and z axes respectively.
Given, the electric field is of intensity 3.22 kN/C and is along x-axis.
Therefore,
[tex]\vec E = 3.22\ kN/C\ \hat i=3.22\times 10^3\ N/C\ \hat i.[/tex]
The magnetic flux through a surface is defined as
[tex]\phi = \vec E\cdot \vec A[/tex]
where,
[tex]\vec A[/tex] is the area vector of the surface which is directed along the normal to the plane of the surface and its magnitude is equal to the area of the surface.
[tex]A = \text{length of the plane }\times \text{width of the plane}\\=0.700\times 0.350\\=0.245\ m^2.[/tex]
(a):
When the plane is parallel to yz plane, then its normal is along x axis, therefore,
[tex]\vec A = A\ \hat i.[/tex]
Electric flux,
[tex]\phi = \vec E\cdot \vec A \\=(3.22\times 10^3)\hat i\cdot (0.245)\hat i = 3.22\times10^3\times 0.245\times (\hat i\cdot \hat i) = 7.889\times 10^2\ Nm.[/tex]
(b):
When the plane is parallel to xy plane, then its normal is along z axis, therefore,
[tex]\vec A = A\ \hat k.[/tex]
Electric flux,
[tex]\phi = \vec E\cdot \vec A \\=(3.22\times 10^3)\hat i\cdot (0.245)\hat k = 3.22\times10^3\times 0.245\times (\hat i\cdot \hat k) = 0\ Nm.[/tex]
(c):
When the normal to the plane makes an angle [tex]34.2^\circ[/tex] with the x-axis.
Electric flux through the plane is given by,
[tex]\phi = \vec E\cdot \vec A\\=EA\cos(34.2^\circ) \\=(3.22\times 10^3)\times (0.245)\times \cos(34.2^\circ)\\=6.52\times 10^2\ Nm.[/tex]
Part II:(a):
Given:
Electric field everywhere on the surface of the sphere, [tex]E = 570\ N/C.[/tex]Radius of the sphere, [tex]R = 0.205\ m.[/tex]The electric field is directed radially outward from the surface of the sphere at its every point and the normal to the surface of the sphere is also radially outwards at its every point therefore both the electric field and the area vector of the sphere is along the same direction.
The flux through the surface is given by
[tex]\phi = \vec E \cdot \vec A \\= EA\cos0^\circ\\=EA\\=E\ 4\pi R^2\\=570\times 4\pi \times 0.205^2\\=3.01\times 10^2\ Nm.[/tex]
According to Gauss law of electrostatics,
[tex]\phi = \dfrac q{\epsilon_o}[/tex]
where,
[tex]q[/tex] = charge enclosed by the sphere.[tex]\epsilon_o[/tex] = electric permittivity of free space =[tex]8.85\times10^{-12}\ C^2N^{-1}m^{-2}.[/tex]Therefore, the net charge on the sphere is given by
[tex]q=\epsilon_o\times \phi\\=8.85\times 10^{-12}\times 3.01\times 10^2\\=2.664\times 10^{-9}\ C.[/tex]
(b):
Since, the sphere is charged all of the charge resides on its surface. The electric field through the sphere is radially outwards which means the charge on the sphere should be positive.
There is no charge distributed inside the sphere.
Write down an equation describing a sinusoidal traveling wave (in 1-D). Tell us (words and/or equations) what in your equation tells us the speed and direction of the wave? [Hint: you can google this if you do not know the answer. Be sure you understand it though!]
Answer:
Explanation:
The standard equation of the sinusoidal wave in one dimension is given by
[tex]y = A Sin\left ( \frac{2\pi }{\lambda }\left ( vt-x \right )+\phi \right )[/tex]
Here, A be the amplitude of the wave
λ be the wavelength of the wave
v be the velocity of the wave
Φ be the phase angle
x be the position of the wave
t be the time
this wave is travelling along positive direction of X axis
The frequency of wave is f which relates with velocity and wavelength as given below
v = f x λ
The relation between the time period and the frequency is
f = 1 / T.
When the electric field strength is large, electric field lines are: a) close together.
b) perpendicular to each other.
c) electric field lines are not related to the field strength.
d) far apart.
e) None of the above.
Answer: a) close together
Explanation: The electric field lines also represent the intensity of the field, in this sense for strong electric fields it is usually draw the lines close to each other. In constrast when they are far apart the electric field is weak.
Final answer:
The electric field lines are close together when the electric field strength is large, indicating a stronger field due to the direct proportionality between field strength and the density of the field lines. The correct answer is a) close together.
Explanation:
When the electric field strength is large, electric field lines are close together. This is because the strength of the electric field is directly proportional to the number of lines per unit area perpendicular to the field lines. The more electric field lines there are in a given area, the stronger the electric field at that point.
Hence, near a positive charge or around a negative charge, where the field lines begin or terminate, the field lines are densely packed when the charge is large, indicating a strong electric field. Field lines provide a visual representation of the field's strength and direction, and they will never intersect because the electric field at a point must have a unique direction.
An equilateral triangle has a height of 3.32 cm. Draw the picture and use this information to determine the length of the sides of the triangle.
Answer:
The answer is a=b=c=3.833 cm
Explanation:
Lets call the variables a=side a b=side b c=side c
We have that the formula of the equilateral triangle for its height is:
1)h=(1/2)*root(3)*a
2) If we resolve the equation we have
2.1)2h=root(3)*a
2.2)(2h/root(3))=a
3) After the replacement of each value we have that
a=2*3.32/1.73205
a=3.833 cm
And we know that the equilateral triangle has the same value for each side so a=b=c=3.833 cm
A 4000-kg truck traveling with a velocity of 20 m/s due south collides head-on with a 1350-kg car traveling with a velocity of 10 m/s due north. The two vehicles stick together after the collision. What is the momentum of each vehicle prior to the collision?
Answer:Momentum of Truck[tex]=4000\times 20=80,000 kg-m/s[/tex]
Momentum of car[tex]=1350\times 10=13,500 kg-m/s[/tex]
Explanation:
Given
mass of truck(m)=4000 kg
Velocity of Truck is ([tex]V_T[/tex])=[tex]-20\hat{j}[/tex]
mass of car ([tex]m_c[/tex])=1350 kg
Velocity of car[tex](V_c)=10 \hat{j}[/tex]
Conserving momentum
[tex]4000\times \left ( -20\right )+1350\left ( 10\right )=5350v[/tex]
[tex]v=\frac{66,500}{5350}=12.42 m/s[/tex]
Momentum of Truck[tex]=4000\times 20=80,000 kg-m/s[/tex]
Momentum of car[tex]=1350\times 10=13,500 kg-m/s[/tex]
Answer:
1) Momentum of truck before collision is [tex]\overrightarrow{p_{1}}=-80000kgm/s[/tex]
2) Momentum of car before collision is [tex]\overrightarrow{p_{2}}=13500kgm/s[/tex]
Explanation:
The momentum of an object with mass 'm' travelling with speed 'v' is mathematically given by
[tex]\overrightarrow{p}=mass\times \overrightarrow{v}[/tex]
In this problem we shall assume that direction's coincide with the Cartesian axis for simplicity
Thus for Truck we have
Mass = 4000 kg
Velocity = [tex]-20\widehat{j}[/tex]m/s
Thus momentum of truck becomes
[tex]\overrightarrow{p_{1}}=4000\times -20\overrightarrow{j}=-80000kgm/s[/tex]
Similarly for car we have
Mass = 1350 kg
Velocity = [tex]10\widehat{j}[/tex]m/s
Thus momentum of car becomes
[tex]\overrightarrow{p_{2}}=1350\times 10\overrightarrow{j}=13500kgm/s[/tex]
onsider a 20-cm-thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held constant at 50°C, whereas the right face is exposed to a flow of 22°C air with a convection heat transfer coefficient of 15 W/m2·K. Neglecting heat transfer by radiation, find the right wall surface temperature and the heat flux through the wall.
Answer:
q=heat flux=216.83W/m^2
Ts=36.45C
Explanation:
What you should do is raise the heat transfer equation from the inside of the granite wall to the air.
You should keep in mind that in the numerator you place the temperature differences, while the denominator places the sum of the thermal resistances by conduction and convection.
When you have the heat value, all you have to do is use the convection equation q = h (ts-t) and solve for the surface temperature.
I attached procedure
One cubic meter (1.00 m3) of aluminum has a mass of 2.70 103 kg, and the same volume of iron has a mass of 7.86 103 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 1.82 cm on an equal-arm balance.
Answer:
[tex]r_{al}=2.598cm[/tex]
Explanation:
Density =mass/volume
[tex]D_{al}=2.70103kg/m^{3}[/tex]
[tex]D_{iron}=7.86103/m^{3}[/tex]
condition for balance:
[tex]M_{iron}=M_{al}[/tex]
M=D*Volum
then:
[tex]D_{iron}*4/3*pi*r_{ir}^{3}=D_{al}*4/3*pi*r_{al}^{3}[/tex]
[tex]r_{al}=r_{ir}*\sqrt[3]{\frac{D_{iron}}{D_{al}}}[/tex]
[tex]r_{al}=2.598cm[/tex]
Answer:
1.28 cm
Explanation:
Density (ρ) is an intensive property resulting from the quotient of mass (m) and volume (V).
The density of Al is ρAl = 2.70 × 10³ kg/m³
The density of Fe is ρFe = 7.86 × 10³ kg/m³
A Fe sphere of radius 1.82 cm has the following volume.
V = 4/3 × π × r³
V = 4/3 × π × (0.0182 cm)³
V = 2.53 × 10⁻⁵ m³
The mass corresponding to this sphere is:
2.53 × 10⁻⁵ m³ × 2.70 × 10³ kg/m³ = 0.0683 kg
The Al sphere has the same mass, so its volume is:
0.0683 kg × (1 m³/7.86 × 10³ kg) = 8.69 × 10⁻⁶ m³
The radius corresponding to an Al sphere of 8.69 × 10⁻⁶ m³ is:
V = 4/3 × π × r³
8.69 × 10⁻⁶ m³ = 4/3 × π × r³
r = 0.0128 m = 1.28 cm
A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in the car's momentum?
Answer:
Change in momentum is [tex]2.667\times 10^{4} kg.m/s[/tex]
Solution:
The momentum of any body is the product of its mass and the velocity associated with the body and is generally given by:
[tex]\vec{p} = m\vec{v}[/tex]
Now, as per the question:
Mass of the car, M = 1500 kg
The velocity in the east direction, [tex]v\hat{i} = 40\hat{i} km/h[/tex]
The velocity in the north direction, [tex]v\hat{j} = 50\hat{j} km/h[/tex]
Now, the momentum of the car in the east direction:
[tex]p\hat{i} = mv\hat{i} = 1500\times 40\hat{i} = 60000\hat{i} kg.km/h[/tex]
Now, the momentum of the car in the north direction:
[tex]p\hat{j} = mv\hat{j} = 1500\times 50\hat{j} = 75000\hat{j} kg.km/h[/tex]
Change in momentum is given by:
[tex]\Delta p = p\hat{i} - p\hat{j} = 60000\hat{i} - 75000\hat{j}[/tex]
Now,
[tex]|\Delta p| = |60000\hat{i} - 75000\hat{j}|[/tex]
[tex]|\Delta p| = \sqrt{60000^{2} + 75000^{2}}[/tex]
[tex]|\Delta p| = 96046 kg.km/hr = \frac{96046}{3.6} = 2.667\times 10^{4} kg.m/s[/tex]
(Since, [tex]1kg.km/h = \frac{1}{3.6} kg.m/s[/tex])
A quantity of 30.1 cm^3 of water at 22.4°C is placed in a freezer compartment and allowed to freeze to solid ice at -7.2°C. How many joules of energy must be withdrawn from the water by the refrigerator?
Answer:
13330.86 J
Explanation:
mass of volume of 30.1 cc of water = density x volume
= 1 x 30.1 = 30.1 gm
= 30.1 x 10⁻³ kg
Heat to be withdrawn to cool water from 22.4 degree to 0 degree
= mass x specific heat x fall of temperature
= 30.1 x 10⁻³ x 4186 x 22.4
= 2822.36 J
Heat to be withdrawn to cool water 0 degree ice
mass x latent heat of freezing
30.1 x 10⁻³ x 334000
= 10053.4 J
Heat to be withdrawn to cool ice from 0 degree to -7.2 degree
mass x specific heat of ice x fall of temperature
= 30.1 x 10⁻³ x 2100 x 7.2
= 455.1 J
Total heat to be withdrawn
= 2822.36 + 10053.4 + 455.1
= 13330.86 J
If the price of gasoline at a particular station in Europe is 5 euros per liter. An American student in Europe is allowed to use 40 euros to buy gasoline. Knowing that 4 quarts make a gallon, and that 1 liter is about 1 US liquid quart, about how many gallons can the student buy?
Answer:
Number of gallons =2 gallon
Explanation:
given data:
rate of gasoline ineurope = 5 euro per liter
total money to buy gasoline = 40 euro
total gasoline an american can buy in europe [tex]= \frac{40}{5}[/tex]
= 8 litres of gasoline
As given in the question 1 ltr is 1 quarts therefore
Total no. of quarts is 8 quarts
As from question 4 quarts is equal to one gallon, hence
Number of gallons[tex]= \frac{8}{4} = 2 gallon[/tex]
A nervous squirrel gets startled and runs 5.0\,\text m5.0m5, point, 0, space, m leftward to a nearby tree. The squirrel runs for 0.50\,\text s0.50s0, point, 50, space, s with constant acceleration, and its final speed is 15.0\,\dfrac{\text m}{\text s}15.0 s m 15, point, 0, space, start fraction, m, divided by, s, end fraction. What was the squirrel's initial velocity before getting startled
Answer:
−5.0
Explanation:
Answer:
The squirrel's initial velocity is 5 m/s.
Explanation:
It is given that,
Distance covered by the squirrel, d = 5 m
Time taken, t = 0.5 s
Final speed of the squirrel, v = 15 m/s
To find,
The squirrel's initial velocity.
Solution,
Let a is the acceleration of the squirrel. Using the first equation of motion to find it :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{15-u}{0.5}[/tex]..............(1)
Let u is the initial speed of the squirrel. Using again third equation of kinematics to find it :
[tex]v^2-u^2=2ad[/tex]
Use equation (1) to put value of a
[tex](15)^2-u^2=2\times \dfrac{15-u}{0.5}\times 5[/tex]
[tex]225-u^2=20(15-u)[/tex]
On solving the above quadratic equation, we get the value of u as :
u = 5 m/s
Therefore, the squirrel's initial velocity before getting started is 5 m/s.
A soccer ball starts from rest and accelerates with an acceleration of 0.395 m/s^2 while moving down a 8.50 m long inclined plane. When it reaches the bottom, the ball rolls up another plane, where, after moving 14.75 m, it comes to rest. (a) What is the speed of the ball at the bottom of the first plane (in m/s)? (Round your answer to at least two decimal places.) (b) How long does it take to roll down the first plane (in s)?
Answer:
a) The speed of the ball at the bottom of the first plane is 2.59 m/s
b) It takes the ball 6.56 s to roll down the first plane.
Explanation:
The equations for the position and velocity of the ball are as follows:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position at time t
x0 = initial position
v0 = initial speed
t = time
a = acceleration
v = velocity at time t
b) First, let´s calculate the time it takes the ball to reach the bottom of the plane using the equation for the position:
x = x0 + v0 · t + 1/2 · a · t²
Placing the center of the frame of reference at the point where the ball starts rolling, x = 0. Since the ball starts from rest, v0 = 0. Then:
x = 1/2 · a ·t²
Let´s find the time when the ball reaches a position of 8.50 m
8.50 m = 1/2 · 0.395 m/s² · t²
t² = 2 · 8.50 m / 0.395 m/s²
t = 6.56 s
a) Now, using the equation of the velocity, we can calculate the velocity of the ball at the bottom of the plane (t = 6.56 s):
v = a · t
v = 0.395 m/s² · 6.56 s = 2.59 m/s
what are the components of friction?
Friction:
When an object slips on a surface, an opposing force acts between the tangent planes which acts in the opposite direction of motion. This opposing force is called Friction. Or in other words, Friction is the opposing force that opposes the motion between two surfaces.
The main component of friction are:
Normal Reaction (R):
Suppose a block is placed on a table in the above picture, which is in resting state, then two forces are acting on it at that time.
The first is due to its weight mg which is working from its center of gravity towards the vertical bottom.
The second one is superimposed vertically upwards by the table on the block, called the reaction force (P). This force passes through the center of gravity of the block.
Due to P = mg, the box is in equilibrium position on the table.
Coefficient of friction ( μ ):
The ratio of the force of friction and the reaction force is called the coefficient of friction.
Coefficient of friction, µ = force of friction / reaction force
μ = F / R
The coefficient of friction is volume less and dimensionless.
Its value is between 0 to 1.
Advantage and disadvantage from friction force:
The advantage of the force of friction is that due to friction, we can walk on the earth without slipping. Brakes in all vehicles are due to the force of friction. We can write on the board only because of the force of friction. The disadvantage of this force is that due to friction, some parts of energy are lost in the machines and there is wear and tear on the machines.How to reduce friction:
Using lubricants (oil or grease) in machines. Friction can be reduced by using ball bearings etc. Using a soap solution and powder.Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 of 1110°C fresh lava into 36.2°C surroundings, assuming lava's emissivity is 1.
The rate of heat loss by radiation is equal to -207.5kW
Why?
To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:
[tex]HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )[/tex]
Where,
E, is the emissivity of the body.
A, is the area of the body.
T, are the temperatures.
S, is the Stefan-Boltzmann constant, which is equal to:
[tex]5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }[/tex]
Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.
We know that:
[tex]K=Celsius+273.15[/tex]
So, converting we have:
[tex]T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K[/tex]
Therefore, substituting the given information and calculating, we have:
[tex]HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )[/tex]
[tex]HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW[/tex]
Hence, we have that the rate of heat loss is equal to -207.5kW.
At the intersection of Texas Avenue and University Drive,
ablue, subcompact car with mass 950 kg traveling East on
universitycollides with a amrron pickup truck with mass 1900 kg
that istraveling north on Texas and ran a red light. The
twovehicles stick together as a result of the collision and, after
thecollision, the wreckage is sliding at 16.0 m/s in the
direction24.0 degrees east of north. Calculate the speed of
eachvehicle before the collision. the collision occurs during
aheavy rainstorm; you can ignore friction forces between
thevehicles and the wet road.
Answer:
The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/sExplanation:
We can use conservation of momentum to find the initial velocities.
Taking the unit vector [tex]\hat{i}[/tex] pointing north and [tex]\hat{j}[/tex] pointing east, the final velocity will be
[tex] \vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )[/tex]
[tex] \vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]
The final linear momentum will be:
[tex]\vec{P}_f = (m_{car}+ m_{truck}) * V_f[/tex]
[tex]\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]
[tex]\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]
[tex]\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )[/tex]
As there are not external forces, the total linear momentum must be constant.
So:
[tex]\vec{P}_0= \vec{P}_f [/tex]
As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be
[tex]\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )[/tex]
so:
[tex]\vec{P}_0= \vec{P}_f[/tex]
[tex]( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )[/tex]
so
[tex]\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.[/tex]
So, for the truck
[tex]m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} [/tex]
[tex]1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} [/tex]
[tex] v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg} [/tex]
[tex]v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg} [/tex]
[tex]v_{truck} = 21.93 \frac{m}{s}[/tex]
And, for the car
[tex]950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}[/tex]
[tex]v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}[/tex]
[tex]v_{car}=19.524 \frac{m}{s}[/tex]
600-nm light is incident on a diffraction grating with a ruling separation of 1.7 10 m. The second order line occurs at a diffraction angle of: 42 45° 21° 10°
Answer:
45°
Explanation:
λ = 600 nm = 600 x 10^-9 m
d = 1.7 x 10^-6 m
For second order, N = 2
Use the equation
d Sinθ = N λ
1.7 x 10^-6 x Sinθ = 2 x 600 x 10^-9
Sinθ = 0.706
θ = 45°
Thus, the angle of diffraction is 45°.
The diffraction angle of the second-order line is approximately 44.22°.
Explanation:The diffraction angle of the second order line can be found using the formula:
dsinθ = mλ
Where d is the ruling separation of the diffraction grating, θ is the diffraction angle, m is the order of the line, and λ is the wavelength of light. Plugging in the values given in the question:
1.7 × 10-6 m × sinθ = 2 × 600 × 10-9 m
Simplifying the equation:
sinθ = 2 × 600 × 10-9 m / (1.7 × 10-6 m)
Calculating the value of sinθ, we find:
sinθ ≈ 0.7059
Finally, taking the inverse sine of 0.7059, we find:
θ ≈ 44.22°
The second-order line occurs at a diffraction angle of approximately 44.22°.
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In his Galileo inclined plane experiment he proved that: a. The distance is proportional to the square of time
b. The distance is proportional to time
c. The distance is proportional to a third of the time
d. Distance and time are the same
Answer:
Option a
Explanation:
Galileo's experiment of inclined plane provides the measurement of acceleration with accuracy with the help of simple equipment.
The ultimate objective of the experiment was to prove that when other external forces are absent like the one due to air resistance, all the objects fall at a constant accelerated rate towards the Earth.
Thus he measured the influence of gravity on the free falling object accurately and observed that the distance covered by these objects when released from rest and moving at a steady rate, then the distance traveled by the object is proportional to the square of the time taken.
From his findings:
[tex]D = \frac{gH}{2L}t^{2}[/tex]
where
D = distance
L = Inclined plane length
H = Height of the inclined plane
A sphere is charged with electrons to -6 x 10^-6C. How many electrons make up this charge? The elemental charge is 1.6 x 10^-19 C.
Answer:
[tex]3.75*10^{-13}[/tex] electrons
Explanation:
The total charge Q is the sum of the charge of the N electrons contained in the sphere:
[tex]Q=N*q_{e}[/tex]
[tex]q_{e}=-1.6*10^{-19}C[/tex] charge of a electron
We solve to find N:
[tex]N=\frac{Q}{q_{e}}=\frac{-6 x 10^{-6}}{-1.6 x 10^{-19}}=3.75*10^{-13}[/tex]
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 −14 µC of charge. The tissue has 5 g of mass. What is the minimum charge required to pick up the tissue paper?
We have that for the Question "What is the minimum charge required to pick up the tissue paper?" it can be said that minimum charge is
q=0.0014\muCFrom the question we are told
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 −14 µC of charge. The tissue has 5 g of mass. What is the minimum charge required to pick up the tissue paper?minimum chargeGenerally the equation for the Electro static force is mathematically given as
[tex]\frac{kqq'}{r^2}=mg\\\\Therefore\\\\q=\frac{5*10^{-3}*9.8*(0.06^)^2}{9*10^{9}*14*10^{-6}}\\\\q=1.4*10^{-9}C\\\\q=1.4mc\\\\[/tex]
q=0.0014\muC
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A car travels in a straight line from a position 50 m to your right to a position 210 m to your right in 5 sec. a. What is the average velocity of the car?b. Construct a position vs. time graph and calculate the slope of the line of best fit.
Answer:
The average velocity of the car is 32 m/s.
Explanation:
Given that,
Initial position = 50 m
Final position = 210 m
Time = 5 sec
(a). We need to calculate the average velocity of the car
Using formula of average velocity
[tex]v_{avg}=\dfrac{v_{f}-v_{i}}{t}[/tex]
Put the value into the formula
[tex]v_{avg}=\dfrac{210-50}{5}[/tex]
[tex]v_{avg}=32\ m/s[/tex]
(b). We need to draw the position vs. time graph
We need to calculate the slope of the line of best fit.
Using formula of slop
[tex]slope =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
Put the value into the formula
[tex]slope =\dfrac{210-50}{5-0}[/tex]
[tex]slope =32[/tex]
Hence, The average velocity of the car is 32 m/s.
An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at that time, y0 is the initial height (at t=0), v0 is the initial velocity, and g=9.8m/s2 (the acceleration due to gravity). If you stand at the edge of a cliff that is 75 m high and throw a rock directly up into the air with a velocity of 20 m/s, at what time will the rock hit the ground? (Note: The Quadratic Formula will give two answers, but only one of them is reasonable.)
Answer:
[tex]t=6.4534 s[/tex]
Explanation:
This is an exercise where you need to use the concepts of free fall objects
Our knowable variables are initial high, initial velocity and the acceleration due to gravity:
[tex]y_{0}=75m[/tex]
[tex]v_{oy} =20m/s[/tex]
[tex]g=9.8 m/s^{2}[/tex]
At the end of the motion, the rock hits the ground making the final high y=0m
[tex]y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}[/tex]
If we evaluate the equation:
[tex]0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex]
This is a classic form of Quadratic Formula, we can solve it using:
[tex]t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]a=-4.9\\b=20\\c=75[/tex]
[tex]t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s[/tex]
[tex]t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s[/tex]
Since the time can not be negative, the reasonable answer is
[tex]t=6.4534s[/tex]
The rock will hit the ground approximately 2.37 seconds after it is thrown.
Given:
- Initial height, [tex]\( y_0 = 75 \)[/tex] m
- Initial velocity, [tex]\( v_0 = 20 \)[/tex] m/s (since the rock is thrown upwards, this velocity will be negative in the equation as it is in the opposite direction to the positive y-axis)
- Acceleration due to gravity, [tex]\( g = 9.8 \) m/s\( ^2 \)[/tex]
The motion equation becomes:
[tex]\[ 0 = 75 - 20t - \frac{1}{2}(9.8)t^2 \][/tex]
Multiplying through by 2 to clear the fraction, we get:
[tex]\[ 0 = 150 - 40t - 9.8t^2 \][/tex]
Rearranging the terms to align with the standard quadratic form [tex]\( at^2 + bt + c = 0 \)[/tex], we have:
[tex]\[ 9.8t^2 + 40t - 150 = 0 \][/tex]
Now we can apply the Quadratic Formula to solve for t:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where ( a = 9.8 ), ( b = 40 ), and ( c = -150 ). Plugging in these values, we get:
[tex]\[ t = \frac{-40 \pm \sqrt{40^2 - 4(9.8)(-150)}}{2(9.8)} \][/tex]
Solving this, we get two values for t, but only the positive value is physically meaningful since time cannot be negative.
Calculating the discriminant:
[tex]\[ \sqrt{40^2 - 4(9.8)(-150)} = \sqrt{1600 + 5880} = \sqrt{7480} \][/tex]
Now, we find the two possible values for t:
[tex]\[ t = \frac{-40 \pm \sqrt{7480}}{2(9.8)} \] \[ t = \frac{-40 \pm 86.52}{19.6} \][/tex]
The two solutions for t are:
[tex]\[ t_1 = \frac{-40 + 86.52}{19.6} \approx 2.37 \text{ s} \] \[ t_2 = \frac{-40 - 86.52}{19.6} \approx -6.43 \text{ s} \][/tex]
Since time cannot be negative, we discard [tex]\( t_2 \)[/tex] and take [tex]\( t_1 \)[/tex] as the time when the rock hits the ground.
The time taken is 2.37 s.
Two point charges are fixed on the y axis: a negative point charge q1 = -24 µC at y1 = +0.19 m and a positive point charge q2 at y2 = +0.33 m. A third point charge q = +8.0 µC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 18 N and points in the +y direction. Determine the magnitude of q2.
Answer:
4.51 * 10^{-5} C
Explanation:
The force between two charge q and q1 is given as
[tex]F = \frac{k*q*q1}{r^2}[/tex]
[tex]= \frac{(9.0 * 10^9)(24 * 10^{-6})(8 * 10^{-6} C)}{(0.19m)^2} [/tex]
= 47.86 N in the +y direction
We need the force between q and q2 to be (47.86 - 18) = 29.86 N in the other direction to get the desired result.
solving for q2,
[tex]q2 = \frac{Fr^2}{(kq)} [/tex]
[tex]= \frac{(29.86)(0.33 m)^2}{(9.0 * 10^9*8*10^{-6} C)}[/tex]
[tex]= 4.51 * 10^{-5} C[/tex]
Final answer:
To determine the magnitude of q2, we can use Coulomb's Law to calculate the net force exerted on charge q. By setting up an equation using the given information, we can find that the magnitude of q2 is approximately 4.84 μC.
Explanation:
To determine the magnitude of q2, we can use Coulomb's Law, which states that the electrostatic force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, we have two charges, q1 and q2, with a distance of 0.33 - 0.19 = 0.14m between them.
The net electrostatic force on charge q can be calculated using the equation F = k * (|q1 * q| / r1^2) + k * (|q2 * q| / r2^2), where k is the electrostatic constant (8.99 x 10^9 N*m^2/C^2).
From the given information, we know that F = 18 N and points in the +y direction, so we can set up the following equation: 18 = k * (|(-24 μC) * (8 μC)| / (0.19m)^2) + k * (|q2 * (8 μC)| / (0.33m)^2).
Solving this equation, we can find the magnitude of q2, which is approximately 4.84 μC.
g n some distant dystopian future on the surface of a war ravaged Mars, a rocket is launched from top of a weapons platform located 8 m above the ground at an angle of 30◦ above the horizontal and with an initial speed of 12 m/s. The rocket is equipped with a special propulsion system which provides a constant horizontal acceleration of 2 m/s2, t seconds after the rocket has been launched. Assuming a constant downward acceleration of 4 m/s2 due to gravity on Mars, give a vector-valued function r(t) (with a horizontal component and a vertical component) which represents the position of the rocket relative to the point on the ground directly below the launch point of the rocket t seconds after launch.
Answer:
the vector position is r = (10.4 t + t²) i + (8+ 6t - 2 t²) j
Explanation:
In this problem we have acceleration in the x and y axes, therefore we will use the accelerated motion equations
X = Xo + Vox t + ½ aₓ t²
Y = I + I go t + ½ [tex]a_{y}[/tex] t²
We calculate the components of velocity using trigonometry
Vox = Vo cos θ = 12 cos 30
Voy = Voy sin θ = 12 sin 30
Vox = 10.4 m / s
Voy = 6 m / s
The value accelerations are data
aₓ = 2 m / s²
[tex]a_{y}[/tex] = - 4 m / s²
The initial position is
Xo = 0
Yo = 8 m
Since the launching point is 8 m above the ground. The acceleration is negative because it has a downward direction; with this data we write the equations of the position
X = 10.4 t + ½ 2 t²
Y = 8 + 6 t - ½ 4 t²
X = 10.4 t + t²
Y = 8 + 6 t - 2 t²
We write the position vector
r = x i + y j
r = (10.4 t + t²) i + (8+ 6t - 2 t²) j
this the vector position
A bicycle tire is inflated to a gauge pressure of 3.71 atm when the temperature is 16°C. While a man rides the bicycle, the temperature of the tire rises to 45°C. Assuming the volume of the tire does not change, find the gauge pressure in the tire at the higher temperature.
Answer:
The pressure of the tire will be 4.19 atm
Explanation:
The ideal gas equation is:
p0 * V0 / T0 = p1 * V1 / T1
Since V0 = V1 because the volume of the tire doesn't change
p0 / T0 = p1 / T1
p1 = p0 * T1 / T0
We need absolute temperatures for this equation
16 C = 289 K
45 C = 318 K
Now we replace the values:
p1 = 3.71 * 318 / 289 = 4.19 atm
Final answer:
To calculate the new gauge pressure of a heated bicycle tire with constant volume, we use Charles's law. With initial conditions of 3.71 atm and 16°C, and final temperature 45°C, the final gauge pressure is approximately 4.08 atm.
Explanation:
The student is asking to calculate the new gauge pressure of a bicycle tire after its temperature increases, assuming the volume and amount of gas remain constant.
This scenario is governed by the Gas Law, specifically Charles's law, which states that the pressure of a gas is directly proportional to its temperature when the volume is held constant.
Therefore, one can apply the formula P2 = P1 * (T2 / T1), where P1 and T1 represent the original pressure and temperature and P2 and T2 represent the final pressure and temperature, respectively, to calculate the new pressure at the higher temperature.
Considering the initial gauge pressure (P1) is 3.71 atm and the initial temperature (T1) is 16°C or 289 K (since temperatures must be in Kelvin for these calculations), and the final temperature (T2) is 45°C or 318 K, we can solve for P2.
First, add 273.15 to the temperature to convert it from Celsius to Kelvin.
T1 = 16 + 273.15 = 289.15 K
T2 = 45 + 273.15 = 318.15 K
Next, apply Charles's law to find P2:
P2 = P1 * (T2 / T1)
P2 = 3.71 atm * (318.15 K / 289.15 K)
P2 = 3.71 atm * 1.1004
P2 ≈ 4.08 atm
The final gauge pressure in the tire at the higher temperature of 45°C is approximately 4.08 atm.
A single point charge is placed at the center of an imaginary cube that has 19 cm long edges. The electric flux out of one of the cube's sides is -2.5 kN·m2/C. How much charge is at the center? .................. nC
Answer:
The amount of charge at the center is -132.7 nC
Solution:
As per the question:
Electric flux from one face of cube, [tex]\phi_{E} = - 2.5 kN.m^{2}/C[/tex]
Edge, a = 19cm
Since, a cube has six faces, thus total flux from all the 6 faces = [tex]6\times (-2.5) = - 15 kN.m^{2}/C[/tex]
Also, from Gauss' law:
[tex]\phi_{E,net} = \frac{1}{\epsilon_{o}}Q_{enc}[/tex]
[tex]Q_{enc} = 8.85\times 10^{- 12}\times - 15\times 10^{3}[/tex]
[tex]Q_{enc} = - 132.7 nC[/tex]
Which of the following statements is/are CORRECT? 1. Atoms are the smallest particles of an element that retain the element's chemical properties. 2. Substances composed of only one type of atom are classified as elements. 3. Of the 118 known elements, only 48 occur naturally.
Answer:
1. Atoms are the smallest particles of an element that retain the element's chemical properties.
2. Substances composed of only one type of atom are classified as elements.
Explanation:
Statement 3 is wrong because out of the 118 identified elements, the first 94 of them occur naturally on Earth and the remaining 24 are synthetic elements.
Final answer:
Atoms are the smallest units retaining an element's properties, and elements consist exclusively of one type of atom. The provided statement about natural occurrence of elements is incorrect, as 92 elements, not 48, occur naturally.
Explanation:
Regarding the student's question on the properties of atoms and elements:
Atoms are the smallest particles of an element that retain the element's chemical properties. For instance, a hydrogen atom possesses all properties of the element hydrogen.
Substances made up of only one type of atom are considered elements. Each element has atoms with a characteristic mass and identical chemical properties.
The statement about the occurrence of elements is partially incorrect. Although 118 elements have been identified, some sources suggest that 92 occur naturally, not just 48.
A spherical water droplet of radius 35 um carries an excess 236 electrons. What vertical electric field in N/C) is needed to balance the gravitational force on the droplet at the surface of the earth? (Assume the density of a water droplet is 1,000 kg/m". Enter the magnitude.)
Answer:
4.66 x 10^7 N/C
Explanation:
radius of drop, r = 35 micro meter = 35 x 10^-6 m
number of electrons, n = 26
density of water, d = 1000 kg / m^3
The gravitational force is equal to the product of mass of drop and the acceleration due to gravity.
Mass of drop, m = Volume of drop x density of water
[tex]m = \frac{4}{3}\times 3.14 \times r^{3}\times d[/tex]
[tex]m = \frac{4}{3}\times 3.14 \times \left ( 35\times10^{-6} \right )^{3}\times 1000[/tex]
m = 1.795 x 10^-10 kg
Gravitational force, Fg = m x g = 1.795 x 10^-10 x 9.8 = 1.759 x 10^-9 N
Charge,q = number of electrons x charge of one electron
q = 236 x 1.6 x 10^-19 = 3.776 x 10^-17 C
Electrostatic force, Fe = q E
Where, E is the strength of electric field.
here electrostatic force is balanced by the gravitational force
Fe = Fg
3.776 x 106-17 x E = 1.759 x 10^-9
E = 4.66 x 10^7 N/C
If our Sun and solar system had formed at the same time as the very first stars in the universe, describe how and why our solar system would be different than the one we have today and the impact this would likely have had on the formation of life. (You will need to consider the state of the universe at the time of the formation of the first stars.
Answer:
How: It will have a very different chemical composition.
Why: Because the interstellar medium at the beginning of universe were compounded only of Hydrogen and Helium.
Explanation:
Once the Big Bang passed and the universe began its expansion, the temperature started to decrease, what allowed the combination of hydrogen and helium. Therefore, the chemical composition of the molecular cloud in which the first stars were formed only held those elements.
If the Sun and the solar system were formed from a molecular cloud with that composition, it will both have an absence of metallic elements (elements heavier than helium), since all of them are formed through nuclear reactions in the core of the stars, and each time a star dies, the interstellar medium for the next generation is enriched.
Key elements for life as Carbon, Nitrogen, Oxygen, Phosphorus and Sulfur will be missing for that hypothetical scenario.
For example, the lack of O3 (ozone) in the atmosphere of earth will allow that ultraviolet light hits the earth surface.
If our solar system formed with the first stars, the planets would mostly comprise hydrogen and helium, as heavier elements had not yet formed. The conditions would be hotter and harsher, making life as we know it unlikely to evolve. The Sun, too, would've been exhausted already due to its short lifespan.
Explanation:If our Sun and the solar system had formed around the same time as the first stars in the universe, our solar system would look very different. The universe at the time of the formation of the first stars was composed of only hydrogen and helium, with trace amounts of lithium. Thus, the planets in our system, formed from the leftover dust and gasses surrounding our Sun, would mostly be made up of hydrogen and helium.
Believed to have occurred just about 200 million years after the big bang, the first stars were much larger and hotter than the current generation of stars, including our Sun. Hence, the planets in our system could have been significantly hotter and larger. It's also worth noting that these first-generation stars had relatively short lifetimes and ended their lives in explosive supernovae, catalyzing the formation of heavier elements. But this means if our Sun was a first-generation star, it would have exhausted its nuclear fuel billions of years ago.
Due to these circumstances, life as we know it might not have been possible in our solar system. Life on Earth relies on heavier elements like carbon, nitrogen, and oxygen, which were only formed in the universe after generations of stellar evolution and supernova explosions. If our solar system was created at the same time as the first stars, these heavier elements wouldn't have existed yet. Therefore, life, at least life as we understand it, wouldn't likely evolve in our solar system.
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Given the wave function: Y(x,t) = 5sin27(0.2x - 3t); (x = meters, t = sec.): What are the amplitude, frequency, wavelength, angular frequency and phase velocity of the wave? In which direction is it traveling? (Note: this wave is not a light wave traveling in a vacuum so va = 3 x 108 is not valid)
Answer:
Explanation:
Y = 5 Sin27( .2x-3t)
= 5 Sin(5.4x - 81 t )
Amplitude = 5 m
Angular frequency ω = 81
frequency = ω / 2π
= 81 / (2 x 3.14 )
=12.89
Wave length λ = 2π / k ,
k = 5.4
λ = 2π / 5.4
= 1.163 m
Phase velocity =ω / k
= 81 / 5.4
15 m / s.
The wave is travelling in + ve x - direction.