To return to the tree after finding the treasure, you need to calculate the total displacement vector from the tree to the treasure by adding up the vectors of each leg of the journey. The direction and magnitude of the opposite of this vector will provide the direction and distance to get back to the tree. A graphical representation of the vectors will help confirm this.
Explanation:The problem is about determining the direction and distance to return to the original location after walking certain distances at specific angles. This involves the understanding of vectors and how they are applied in real life.
Given the three legs of the journey:
1st leg: 825 m directly south, which can be represented as a vector with a magnitude of 825 m and a direction of 180 degrees.2nd leg: 1.25 km at a 30-degree angle west of north, which can be represented as a vector with north and west components. Using trigonometry, we get a north-component of 1.25 km * cos(30), and a west-component of 1.25 km * sin(30).3rd leg: 1.00 km at a 32-degree angle north of east, which can be represented with east and north components. Calculating similarly, we get the east-component as 1.00 km * cos(32), and the north-component as 1.00 km * sin(32).
Summing up these components for all the legs, we get the total displacement vector from the tree to the treasure. To get back to the tree, you would need to walk in the opposite direction of this vector. The magnitude of this vector gives the total distance to be covered.
Comparing this calculated result with a graphical solution will help validate the projected direction and distance.
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You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height (h) from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 2.0 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 30.0 m/s when they reach the bottom of the ramp. You determine that for a 85.0 kg skier with good form, friction and air resistance will do total work of magnitude 4000 J on him during his run down the slope.What is the maximum height (h) for which the maximum safe speed will not be exceeded?
Answer:
h = 50.49 m
Explanation:
Data provided:
Speed of skier, u = 2.0 m/s
Maximum safe speed of the skier, v = 30.0 m/s
Mass of the skier, m = 85.0
Total work = 4000 J
Height from the starting gate = h
Now, from the law of conservation of energy
Total energy at the gate = total energy at the time maximum speed is reached
[tex]\frac{1}{2}mu^2+mgh=4000J+\frac{1}{2}mv^2[/tex]
where, g is the acceleration due to the gravity
on substituting the values, we get
[tex]\frac{1}{2}\times85\times2.0^2+85\times9.81\times h=4000J+\frac{1}{2}\times85\times30^2[/tex]
or
170 + 833.85 × h = 4000 + 38250
or
h = 50.49 m
Calculate the work done in compressing adiabatically 3kg of helium (He) to one fifth of its original volume if it is initially at 13°C. Find the change in internal energy of the gas resulting from the compression. (cp/cv for monatomic gases is 1.667; gas constant for helium is 2079 K^-1 kg^-1)
Answer:
Work done,[tex]w=5.12\times10^{6}\ \rm J[/tex]
change in internal Energy ,[tex]\Delta U=5.12\times10^6\ \rm J[/tex]
Explanation:
Given:
Mass of helium gas [tex]m=3\ \rm kg[/tex]initial temperature [tex]T_i=286\ \rm K[/tex]Since It is given that the process is adiabatic process it means that there is no exchange of heat between the system and surroundings
[tex]T_iV_i^{\gamma -1}=T_fV_f^{\gamma -1}\\\\286\times V_i^{\gamma -1}=T_f \left( \dfrac{V_i}{5} \right )^{\gamma -1}\\T_f=840.76\ \rm K[/tex]
Let n be the number of moles of Helium given by
[tex]n=\dfrac{m}{M}\\n=\dfrac{3\times10^3}{4}\\n=0.75\times10^3[/tex]
Work done in Adiabatic process
Let W be the work done
[tex]W=\dfrac{nR(T_1-T_2)}{\gamma-1}\\W=\dfrac{0.75\times10^3\times8.314(286-840.76)}{1.67-1}\\W=-5.12\times 10^6\ \rm J[/tex]
The Internal Energy change in any Process is given by
Let [tex]\Delta U[/tex] be the change in internal Energy
[tex]\Delta U=nC_p\Delta T\\\Delta U=0.75\times10^3\times1.5R\times(840.76-286)\\\Delta U=5.12\times10^6\ \rm J[/tex]
I have a device that can generate sounds with frequencies between 800 Hz and 1600 Hz. I also have an unlabeled tuning fork that I need to work out the fundamental frequency of. I find that it resonates at 920 Hz and 1380 Hz, but no other frequencies in that range. What's the lowest frequency that it will resonate at?
Answer:
460 Hz
Explanation:
the given resonating frequency of the device
f₁ = 920 Hz and f₂ = 1380 Hz
fundamental frequency of the device is
f₀ = n₂ - n₁
= 1380 - 920
= 460 Hz
expression of frequency of organic pipe open at both ends
[tex]f_0=n\dfrac{\nu}{2l}[/tex]
at n = 1
[tex]f_0=\dfrac{\nu}{2l} = 460 Hz[/tex]
the frequency ratios of the closed pipe
[tex]f_0:f_1:f_2: ...... =[1:2:3:.........]f_0[/tex]
=[1:2:3:.........]460 Hz
= 460 Hz : 920 Hz : 1380 Hz
so, the lowest frequency for the pipe open at both end is 460 Hz
The output of an ac generator connected to an RLC series combination has a frequency of 12 kHz and an amplitude of 28 V. If R = 4.0 Ohms, L = 30 μH, and C = 8 μF, find a. The impedance
b. The amplitude for current
c. The phase difference between the current and the emf of the generator
Please show all steps and units. Thank you.
Answer:
(a) 4.04 ohm
(b) 6.93 A
(c) 8.53°
Explanation:
f = 12 kHz = 12000 Hz
Vo = 28 V
R = 4 ohm
L = 30 micro Henry = 30 x 10^-6 H
C = 8 micro Farad = 8 x 10^-6 F
(a) Let Z be the impedance
[tex]X_{L} = 2\pi fL=2\times3.14\times12000\times30\times10^{-6}= 2.26 ohm[/tex]
[tex]X_{c} = \frac{1}{2\pi fC}=\frac{1}{2\times3.14\times12000\times8\times10^{-6}}= 1.66 ohm[/tex]
[tex]Z = \sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{4^{2}+\left ( 2.26-1.66 \right )^{2}}[/tex]
Z = 4.04 Ohm
(b) Let Io be the amplitude of current
[tex]I_{o}=\frac{V_{o}}{Z}[/tex]
[tex]I_{o}=\frac{28}{4.04}[/tex]
Io = 6.93 A
(c) Let the phase difference is Ф
[tex]tan\phi = \frac{X_{L}-X_{C}}{R}[/tex]
[tex]tan\phi = \frac{2.26-1.66}{4}[/tex]
tan Ф =0.15
Ф = 8.53°
Which of the following would describe a length that is 2.0×10^-3 of a meter? a: 2.0 kilometers
b: 2.0 megameters
c: 2.0 millimeters
d: 2.0 micrometers
Answer:
Option (c) [tex]2\times 10^{-3}\ m=2\ millimeters[/tex]
Explanation:
Here, it is required to describe the given length in a particular unit. Firstly, we need to see the following conversions as :
1 meter = 0.001 kilometers
1 meter = 10⁻⁶ megameters
1 meter = 1000 millimeters
1 meter = 1000000 micrometers
From the given option, the correct one is (c) because, 1 meter = 1000 millimeters
So, [tex]2\times 10^{-3}\ m=2\ millimeters[/tex]
Hence, the correct option is (c). Hence, this is the required solution.
2.0×10^-3 of a meter is equivalent to 2.0 millimeters, hence option c is the correct answer.
Explanation:The given length, 2.0×10^-3 of a meter, corresponds to a unit of length commonly used in the metric system. In this system, 1 meter is equal to 10^3 millimeters. Therefore, 2.0×10^-3 of a meter equates to 2.0 millimeters. This means the best answer from the provided options would be option c: 2.0 millimeters.
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Calculate the The wavelength of the first Balmer series of hydrogen is 6562 following: a) The ionization potential, and b) The first excitation potential of the hydrogen atom.
Answer:
(a) 13.6 eV
(b) 10.2 V
Explanation:
a) Ionization potential energy is defined as the minimum energy required to excite a neutral atom to its ionized state i.e basically the minimum energy required to excite an electron from n=1 to infinity.
Energy of a level, n, in Hydrogen atom is, [tex]E_{n}=-\frac{13.6}{n^{2} }[/tex]
Now ionization potential can be calculated as
[tex]E_{\infty}- E_{0}[/tex]
Substitute all the value of energy and n in above equation.
[tex]=-\frac{13.6}{\infty^{2}}-(-\frac{13.6}{1^{2}})\\=13.6eV[/tex]
Therefore, the ionization potential is 13.6 eV.
b) This is the energy required to excite a atom from ground state to its excited state. When electrons jumps from ground state level(n=1) to 1st excited state(n=2) the corresponding energy is called 1st excitation potential energy and corresponding potential is called 1st excitation potential.
So, 1st excitation energy = E(n 2)- E(n = 1)
[tex]=-\frac{13.6}{2^{2}}-(-\frac{13.6}{1^{2}})\\=-3.4eV - (-13.6eV) \\=10.2eV[/tex]
Now we can find that 1st excitation energy is 10.2 eV which gives,
[tex]eV'=10.2eV\\V'=10.2V[/tex]
Therefore, the 1st excitation potential is 10.2V.
Starting from rest, a person runs with a constant acceleration, traveling 40 meters in 10 seconds. What is their final velocity?
To calculate the final velocity of someone running with constant acceleration, we first find the acceleration using the distance and time, and then apply it to determine the final velocity. The final velocity after 10 seconds is 8 m/s.
To find the final velocity of the person running with constant acceleration, we use the kinematic equation v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time. The person starts from rest, so u = 0. We need to find a first, and we can do so using the equation s = ut + (1/2)at², where s is the distance traveled.
Insert the given values: 40m = 0 + (1/2)a(10s)², resulting in a = 0.8 m/s². Now apply the acceleration to the initial equation: v = 0 + 0.8 m/s² * 10s, which gives us v = 8 m/s. Therefore, the person's final velocity after 10 seconds is 8 m/s.
A rocket sled accelerates at 21.5 m/s^2 for 8.75 s. (a) What's its velocity at the end of that time? (b) How far has it traveled?
Answer:
(a ) vf= 188.12m/s : Final speed at 8.75 s
(b) d= 823.04 m : Distance the rocket sled traveled
Explanation:
Rocket sled kinematics :The rocket sled moves with a uniformly accelerated movement, then we apply the following formulas:
d =vi*t+1/2a*t² Formula (1)
vf= vi+at Formula(2)
Where:
vi: initial speed =0
a: acceleration=21.5 m/s²
t: time=8.75 s
vf: final speed in m/s
d:displacement in meters(m)
Calculation of displacement (d) and final speed (vf)
We replace data in formulas (1) and (2):
d= 0+1/2*21.5*8.75²
d= 823.04 m
vf= 0+21.5*8.75
vf= 188.12m/s
A 0.5 μF and a 11 μF capacitors are connected in series. Then the pair are connected in parallel with a 1.5 μF capacitor. What is the equivalent capacitance? Give answer in terms of mF.
Answer:
[tex]C_{eq}=1.97\ \mu F[/tex]
Explanation:
Given that,
Capacitance 1, [tex]C_1=0.5\ \mu F[/tex]
Capacitance 2, [tex]C_2=11\ \mu F[/tex]
Capacitance 3, [tex]C_3=1.5\ \mu F[/tex]
C₁ and C₂ are connected in series. Their equivalent is given by :
[tex]\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}[/tex]
[tex]\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}[/tex]
[tex]C'=0.47\ \mu F[/tex]
Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :
[tex]C_{eq}=C'+C_3[/tex]
[tex]C_{eq}=0.47+1.5[/tex]
[tex]C_{eq}=1.97\ \mu F[/tex]
So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.
Jack and Jill ran up the hill at 2.8 m/s . The horizontal component of Jill's velocity vector was 2.1 m/s . What was the angle of the hill?What was the vertical component of Jill's velocity?
Answer:[tex]\theta =41.409 ^{\circ}[/tex]
Explanation:
Given
Jack and Jill ran up the hill at 2.8 m/s
Horizontal component of Jill's velocity vector was 2.1 m/s
Let [tex]\theta [/tex]is the angle made by Jill's velocity with it's horizontal component
Therefore
[tex]2.8cos\theta =2.1[/tex]
[tex]cos\theta =\frac{2.1}{2.8}[/tex]
[tex]cos\theta =0.75[/tex]
[tex]\theta =41.409 ^{\circ}[/tex]
Vertical velocity is given by
[tex]V_y=2.8sin41.11=1.85 m/s[/tex]
The question can be solved using the Pythagorean theorem to calculate the vertical velocity component and apply trigonometry to find the angle of the hill. Remember the conversion from radians to degrees.
Explanation:The question asks about the angle of the hill that Jack and Jill climbed, as well as the vertical component of Jill’s velocity. We know that Jill ran up the hill at a velocity of 2.8 m/s and the horizontal component of her velocity was 2.1 m/s. These two components form a right angle triangle where the hypotenuse is the total velocity (2.8 m/s), one side is the horizontal velocity (2.1 m/s), and the other side, which we're finding, is the vertical component of the velocity.
We can find the angle of the hill, θ, using the tangent function of trigonometry. Tan θ = opposite side / adjacent side. In this case, the 'opposite side' is the vertical velocity component we're after, and the 'adjacent side' is the horizontal component (2.1 m/s). To find the vertical velocity component, you can use the Pythagorean theorem, which states that: (Hypotenuse)² = (Adjacent Side)² + (Opposite Side)² or (2.8 m/s)² = (2.1 m/s)² + (vertical velocity)².
Once you solve for the vertical velocity through the Pythagorean theorem, you then insert it into the tangent equation to solve for the angle θ. Remember that when you use the arctangent function on a calculator to find the angle, the answer will likely be in radians, so you might need to convert them into degrees.
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A group of humans traveling in space discover a habitable planet. They settle down there and start populating it. Their population grows at a steady rate and doubles every 10 years. Thousand years after the first group of humans arrived, the planet is filled to half its carrying capacity. How many more years are there to reach the carrying capacity of the planet? Assume there are no resource crisis and they grow at the same rate. A) 500 years
B) 1000 years
C) 5 years
D) 1 year
E) 10 years
Answer:
10 years
Explanation:
As you can understand from the question it is given that the planet is already filled to half of its capacity. Also the population doubles in 10 years. To fill up the planet completely the population needs to double only once. To do that only 10 years are required.
As it is mentioned there are no other factors affecting the growth rate, in 10years the planet will be filled to its carrying capacity.
What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.
Answer:
51.82
Explanation:
First of all, let's convert both vectors to cartesian coordinates:
Va = 36 < 53° = (36*cos(53), 36*sin(53))
Va = (21.67, 28.75)
Vb = 47 < 157° = (47*cos(157), 47*sin(157))
Vb = (-43.26, 18.36)
The sum of both vectors will be:
Va+Vb = (-21.59, 47.11) Now we will calculate the module of this vector:
[tex]|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82[/tex]
Two moles of He gas are at 25 degrees C and a pressure of 210 kPa. If the gas is heated to 48 degrees C and its pressure reduced to 37 % of its initial value, what is the new volume?
Answer:[tex]0.0686 m^3[/tex]
Explanation:
Given
No of moles=2
[tex]T=25^{\circ}\approx 298 K[/tex]
[tex]P_1=210 KPa[/tex]
[tex]T_2=48 ^{\circ}\approx 321 k[/tex]
[tex]P_2=77.7 KPa[/tex]
PV=nRT
Substitute
[tex]77.7\times V_2=2\times 8.314\times 321[/tex]
[tex]V_2=0.0686 m^3[/tex]
Newton's second law: Rudolph the red nosed reindeer is pulling a 25 kg sled across the snow in a field. The coefficient of kinetic friction is .12 The rope that is pulling the sled is coming off at a 29 degree angle above the horizontal. Find the force in the rope when the acceleration is .12 m/s^2.
Answer:
[tex]F=39,68N[/tex]
Explanation:
Data:
Mass [tex]m=25 Kg[/tex]
Coefficient of kinetic friction [tex]\mu=0.12[/tex]
Angle = [tex]29^{0}[/tex]
Acceleration = [tex]0.12 \frac{m}{s^{2} }[/tex]
Solution:
By Newton's first law we know that for the x-axis:
[tex]F_{rope_x}-F_f=F_R[/tex] Where [tex]F_R[/tex] is the resulting force, and [tex]F_f[/tex] is the friction force.
And for the y-axis:
[tex]F_{rope_y}+N=W[/tex], where N is the normal force, and W is the weight of the sled.
We know that the resulting force's acceleration is [tex]0.12 \frac{m}{s^{2} }[/tex], and by using Newton's second law, we obtain:
[tex]F=m.a[/tex]
[tex]F_R=25Kg. 0.12\frac{m}{s^2} \\ F_R=3N[/tex] .
Now, the horizontal component of the force in the rope will be given by
[tex]F_{rope_x}=F_{rope}.cos(29^0)=F_R+F_f[/tex], since the resulting force is completely on the x-axis, and the friction opposes to the speed of the sled.
To obtain the friction force, we must know the normal force:
[tex]F_f=\mu. N[/tex]
Clearing N in the y-axis equation:
[tex]N=W-F_{rope_y}=W-F_{rope}.sin(29^0)[/tex]
So we can express the x-axis equation as follows:
[tex]F.cos(29^0)=F_R+\mu.(W-F_{rope}.sin(29^0))[/tex]
Finally, solving for F we get
[tex]F = (F_R + \mu. m.g) / (cos (29^0) + \mu.sin (29^0))[/tex]
[tex]F=39,68N[/tex]
A 50kg trunk is pushed 6.0 m at a constant speed up a 30degree
incline by a constant horizontal force. Thecoefficient of kinetic
energy n\between the trunk and the inclineis 0.20. Calculate the
work done by:
a.) the applied force and
b.) by the weight of the trunk.
c.) How much energy was dissipated by the frictional
forceacting on the trunk?
Answer:
a) 2034 J
b) -1471 J
c) -509 J
Explanation:
The trunk has a mass of 50 kg, so its weight is
f = m * a
f = 50 * 9.81 = 490 N
If the incline is of 30 degrees, the force tangential to the incline is:
ft = f * sin(a)
ft = 490 * sin(30) = 245 N
And the normal force is:
fn = f * cos(a)
fn = 490 * cos(30) = 424 N
The frictional force is:
ff = μ * fn
ff = 0.2 * 424 = 84.8 N
To push the trunk up one must apply a force slightly greater than the opposing forces, the opposing are the tangential component of the weight and the friction force
fp = fn + ff
fp = 245 + 84.8 = 339 N
The work of the applied force is:
L = f * d
Lp = fp * d
Lp = 339 * 6 = 2034 J
The work of the weight is done by the tangential component:
Lw = -245 * 6 = -1471 J (it is negative because the weight was opposed to the direction of movement)
The work of the friction force is
Lf = -84.8 * 6 = -509 J
Using local acceleration of gravity data from the Internet, de- termine the weight, in N, of a person whose mass is 80 kg living in: a. Mexico City, Mexico b. Cape Town, South Africa c. Tokyo, Japan d. Chicago, IL e. Copenhagen, Denmark
Answer:
Mexico Citi: 782.32 NCape Town: 783.68 NTokyo: 783.84 NChicago: 784.24 NCopenhagen: 785.2 NExplanation:
The weight w its:
[tex]w \ = \ m \ a[/tex]
where m is the mass and a is the acceleration.
Mexico City, MexicoLocal acceleration of gravity = [tex]9.779 \frac{m}{s^2}[/tex]
weight :
[tex]w = \ 80\ kg \ * \ 9.779 \frac{m}{s^2}[/tex]
[tex]w = \ 782.32 \ N [/tex]
Cape Town, South AfricaLocal acceleration of gravity = [tex]9.796 \frac{m}{s^2}[/tex]
weight :
[tex]w = \ 80\ kg \ * \ 9.796 \frac{m}{s^2}[/tex]
[tex]w = \ 783.68 \ N [/tex]
Tokyo, JapanLocal acceleration of gravity = [tex]9.798 \frac{m}{s^2}[/tex]
weight :
[tex]w = \ 80\ kg \ * \ 9.798 \frac{m}{s^2}[/tex]
[tex]w = \ 783.84 \ N [/tex]
Chicago, ILLocal acceleration of gravity = [tex]9.803 \frac{m}{s^2}[/tex]
weight :
[tex]w = \ 80\ kg \ * \ 9.803 \frac{m}{s^2}[/tex]
[tex]w = \ 784.24 \ N [/tex]
Copenhagen, DenmarkLocal acceleration of gravity = [tex]9.815 \frac{m}{s^2}[/tex]
weight :
[tex]w = \ 80\ kg \ * \ 9.815 \frac{m}{s^2}[/tex]
[tex]w = \ 785.2 \ N [/tex]
Answer
mass of a person is 80 kg
to calculate weight of the person = ?
a) Mexico City, Mexico
acceleration due to gravity in mexico = 9.779 m/s²
weight = 80× 9.779 = 782.32 N
b) Cape Town, South Africa
acceleration due to gravity in cape town = 9.796 m/s²
weight = 80× 9.796 = 783.68 N
c) Tokyo, Japan
acceleration due to gravity in Tokyo = 9.798 m/s²
weight = 80× 9.798 = 783.84 N
d) Chicago, IL
acceleration due to gravity in Chicago = 9.80 m/s²
weight = 80× 9.80 = 784 N
e) Copenhagen, Denmark
acceleration due to gravity in Copenhagen = 9.81 m/s²
weight = 80× 9.81 = 784.8 N
An astronaut weights 588 N in Earth surface. How much the astronaut would weight in the surface in the moon in pounds(lbs)?
Answer:
22.02 lb
Explanation:
The weight of astronaut Earth (w) = 588 N
We know that,
[tex]4.45N = 1lb[/tex]
Thus,
[tex]588N = \frac{1}{4.45}\times 588lb[/tex]
588N = 132.13 lb
Acceleration due to gravity on Earth (g) = 9.8 m/s²
Acceleration due to gravity on Moon = g'
[tex]g'=\frac{g}{6}[/tex]
We know that weight of an object on Earth is,
[tex]w = m\times g[/tex]
[tex]m = \frac{w}{g}[/tex]
Similarly, weight on Moon will be
[tex]w' = m\times g'[/tex]
[tex]w' = \frac{w}{g}\times\frac{g}{6}[/tex]
[tex]w' = \frac{132.13}{6}[/tex]
[tex]w' = 22.02[/tex]
Thus the astronaut will weigh 22.02 lb on Moon.
a. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kinetic energy of the helium atoms?
c. What is the rms speed of the helium atoms?
Answer:
a) 3.39 × 10²³ atoms
b) 6.04 × 10⁻²¹ J
c) 1349.35 m/s
Explanation:
Given:
Diameter of the balloon, d = 29.6 cm = 0.296 m
Temperature, T = 19.0° C = 19 + 273 = 292 K
Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa
Volume of the balloon = [tex]\frac{4}{3}\pi(\frac{d}{2})^3[/tex]
or
Volume of the balloon = [tex]\frac{4}{3}\pi(\frac{0.296}{2})^3[/tex]
or
Volume of the balloon, V = 0.0135 m³
Now,
From the relation,
PV = nRT
where,
n is the number of moles
R is the ideal gas constant = 8.314 kg⋅m²/s²⋅K⋅mol
on substituting the respective values, we get
1.013 × 10⁵ × 0.0135 = n × 8.314 × 292
or
n = 0.563
1 mol = 6.022 × 10²³ atoms
Thus,
0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms
b) Average kinetic energy = [tex]\frac{3}{2}\times K_BT[/tex]
where,
Boltzmann constant, [tex]K_B=1.3807\times10^{-23}J/K[/tex]
Average kinetic energy = [tex]\frac{3}{2}\times1.3807\times10^{-23}\times292[/tex]
or
Average kinetic energy = 6.04 × 10⁻²¹ J
c) rms speed = [tex]\frac{3RT}{m}[/tex]
where, m is the molar mass of the Helium = 0.004 Kg
or
rms speed = [tex]\frac{3\times8.314\times292}{0.004}[/tex]
or
rms speed = 1349.35 m/s
If a beam passes from a material with a refractive index of 1.3 into a material with a refractive index of 1.7 at an angle of 25 degrees (from normal), what is the angle of refraction of the beam? Is the beam bent towards normal or away from it? Sketch a diagram of this problem with rays, angles, and the interface labeled.
Answer:
18.86° , it will bend towards normal.
Explanation:
For refraction,
Using Snell's law as:
[tex]n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}[/tex]
Where,
[tex]{\theta_i}[/tex] is the angle of incidence ( 25.0° )
[tex]{\theta_r}[/tex] is the angle of refraction ( ? )
[tex]{n_r}[/tex] is the refractive index of the refraction medium (n=1.7)
[tex]{n_i}[/tex] is the refractive index of the incidence medium ( n=1.3)
Hence,
[tex]1.3\times {sin\ 25.0^0}={1.7}\times{sin\theta_r}[/tex]
Angle of refraction = [tex]sin^{-1}0.3232[/tex] = 18.86°
Since, the light ray is travelling from a material with a refractive index of 1.3 into a material with a refractive index of 1.7 or lighter to denser medium, it will bend towards normal.
The diagram is shown below:
Dry air will break down if the electric field exceeds 3.0*10^6 V/m. What amount of charge can be placed on a parallel-plate capacitor if the area of each plate is 73 cm^2?
Answer:
The charge on each plate of the capacitor is [tex]19.38 \mu C[/tex]
with one plate positive and one negative, i.e., [tex]\pm 19.38 \mu C[/tex]
Solution:
According to the question:
Critical value of Electric field, [tex]E_{c} = 3.0\times 10^{6} V/m[/tex]
Area of each plate of capacitor, [tex]A_{p} =73 cm^{2} = 73\times 10^{- 4} m^{2}[/tex]
Now, the amount of charge on the capacitor's plates can be calculated as:
Capacitance, C = [tex]\frac{epsilon_{o}\times Area}{Distance, D}[/tex] (1)
Also, Capacitance, C = [tex]\frac{charge, q}{Voltage, V}[/tex]
And
Electric field, E = [tex]\frac{Voltage, V}{D}[/tex]
So, from the above relations, we can write the eqn for charge, q as:
q = [tex]\epsilon_{o}\times E_{c}\times A_{p}[/tex]
q = [tex]8.85\times 10^{- 12}\times 3.0\times 10^{6}\times 73\times 10^{- 4}[/tex]
[tex]q = 19.38 \mu C[/tex]
A taxi traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 27.0 m/s. Then the vehicle travels for 41.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s. (a) How long is the taxi in motion (in s)? s (b) What is the average velocity of the taxi for the motion described? (Enter the magnitude in m/s.) m/s
Answer:
Part a)
[tex]T = 59.5 s[/tex]
Part b)
v = 22.8 m/s
Explanation:
Part a)
Time taken by the taxi to reach the maximum speed is given as
[tex]v_f - v_i = at[/tex]
[tex]27 - 0 = 2(t)[/tex]
[tex]t = 13.5 s[/tex]
now it moves with constant speed for next 41 s and then finally comes to rest in next 5 s
so total time for which it will move is given as
[tex]T = 13.5 s + 41 s + 5 s[/tex]
[tex]T = 59.5 s[/tex]
Part b)
Distance covered by the taxi while it accelerate to its maximum speed is given as
[tex]d_1 = \frac{v_f + v_i}{2} t[/tex]
[tex]d_1 = \frac{27 m/s + 0}{2} (13.5)[/tex]
[tex]d_1 = 182.25[/tex]
Now it moves for constant speed for next 41 s so distance moved is given as
[tex]d_2 = (27)(41)[/tex]
[tex]d_2 = 1107 m[/tex]
Finally it comes to rest in next 5 s so distance moved in next 5 s
[tex]d_3 = \frac{v_f + v_i}{2} t[/tex]
[tex]d_3 = \frac{27 + 0}{2}(5)[/tex]
[tex]d_3 = 67.5 m[/tex]
so here we have total distance moved by it is given as
[tex]d = 182.25 + 1107 + 67.5 [/tex]
[tex]d = 1356.75 m[/tex]
average speed is given as
[tex]v = \frac{1356.75}{59.5}[/tex]
[tex]v = 22.8 m/s[/tex]
Final answer:
The taxi is in motion for 59.5 seconds and its average velocity is 19.38 m/s.
Explanation:
(a) To find the total time the taxi is in motion, we need to calculate the time it takes to accelerate to 27.0 m/s, the time it travels at constant speed, and the time it takes to stop. The time to accelerate is given by the equation:
t = (vf - vi) / a = (27.0 m/s - 0 m/s) / 2.00 m/s² = 13.5 s
The time traveling at constant speed is given as 41.0 s, and the time to stop is given as 5.00 s. Therefore, the total time the taxi is in motion is:
Total time = Time to accelerate + Time at constant speed + Time to stop = 13.5 s + 41.0 s + 5.00 s = 59.5 s
(b) The average velocity is given by the total displacement divided by the total time. The displacement during acceleration is given by:
Displacement = vi × t + 0.5 × a × t² = 0 m/s × 13.5 s + 0.5 × 2.00 m/s² × (13.5 s)² = 182.25 m
The displacement during constant speed is equal to the constant speed multiplied by the time:
Displacement = constant speed × time = 27.0 m/s × 41.0 s = 1,107 m
The displacement during deceleration is given by:
Displacement = vf × t + 0.5 × (-a) × t² = 27.0 m/s × 5.00 s + 0.5 × (-2.00 m/s²) × (5.00 s)² = -137.5 m
Therefore, the total displacement is 182.25 m + 1,107 m - 137.5 m = 1,152.75 m.
The average velocity is then:
Average velocity = Total displacement / Total time = 1,152.75 m / 59.5 s = 19.38 m/s
Draw, as best you can, what a velocity graph would look like as a function of time if the acceleration of the object is negative.
Answer:
Depends on the specific relation of acceleration as function of time, but it would always look like decresing or increasing negatively.
Explanation:
Acceleration is the derivative of velocity with respect to time. That means that it is the change. The fact that is negative means that it is decreasing, or increasing in the negative direction (i.e. going backwards faster).
Attached is the graph with constant negative acceleration, using kinematics relations, it would be a linear equation with negative slope.
Answer:a straight line with negative slope
Explanation:
A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of O 100 m/s. 60 m/s. 50 m/s 80 m/s. 110 m/s.
Answer:
The initial velocity of the ball should be 50 m/s.
Explanation:
Since the trip of the ball shall consist of upward ascend and the downward descend and since the ascend and the descend of the ball is symmetrical we infer that the upward ascend of the ball shall last for a time of 5 seconds.
Now since the motion of ball is uniformly accelerated we can find the initial speed of the ball using first equation of kinematics as
[tex]v=u+gt[/tex]
where,
'v' is final velocity of the ball
'u' is initial velocity of the ball
'g' is acceleration due to gravity
't' is the time of motion
Now we know that the ball will continue to ascend until it's velocity becomes zero hence to using the above equation we can write
[tex]0=u-9.81\times 5\\\\\therefore u=9.81\times 5=49.05m/s\approx 50m/s[/tex]
A boat’s speed in still water is 1.60 m/s . The boat is to travel directly across a river whose current has speed 1.05 m/s. Determine the speed of the boat with respect to the shore.
Express your answer using three significant figures and include the appropriate units
The speed of the boat with respect to the shore is 1.91 m/s.
From the information given, we have that;
A boat's speed in still water is 1.60 m/s
The boat is to travel directly across a river whose current has speed 1.05 m/s
We can see that the movement is in both horizontal and vertical directions.
Using the Pythagorean theorem, let use determine the resultant speed of the boat with respect to the shore, we have that;
Resultant speed² = √((boat's speed)² + (current's speed)²)
Substitute the value as given in the information, we have;
= (1.60)² + (1.05 )²)
Find the value of the squares, we get;
= (2.56 + 1.1025 )
Find the square root of both sides, we have;
= √3.6625
Find the square root of the value, we have;
= 1.91 m/s.
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A red ball is thrown down with an initial speed of 1.1 m/s from a height of 28 meters above the ground. Then, 0.5 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24.3 m/s, from a height of 0.9 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s^2. What is the height of the blue ball 2 seconds after the red ball is thrown?
How long after the red ball is thrown are the two balls in the air at the same height?
Answer:0.835 s
Explanation:
Given
Red ball initial velocity([tex]u_r[/tex])=1.1 m/s
height of building(h)=28 m
after 0.5 sec blue ball is thrown with a velocity([tex]u_b[/tex])=24.3 m/s
Height of blue after 2 sec red ball is thrown
i.e. height of blue ball at t=1.5 sec after blue ball is thrown upward
[tex]h=24.3\times 1.5-\frac{9.81\times 1.5^2}{2}=25.414 m[/tex]
therefore blue ball is at height of 25.414+0.9=26.314 from ground
moment after the two ball is at same height
for red ball
[tex]14=1.1\times \left ( t+0.5\right )+\frac{9.81\times \left ( t+0.5\right )^2}{2}-----1[/tex]
for blue ball
[tex]13.1=24.3\times t-\frac{9.81\times t^2}{2}-----2[/tex]
add 1 & 2
we get
[tex]27.1=1.1t+0.55+24.3t+\frac{g\left ( t+0.25\right )}{2}[/tex]
27.1=25.4t+0.55+4.905t+1.226
t=0.835 s
A rocket fired two engines simultaneously. One produces a thrust 725 N directly forward while the other gives a 513N thrust at 32.4degrees above the forward direction. Find the magnitude and direction ( relative to the forward direction) of the resultant force that these engines exert on the rocket
Answer:
Magnitude of resultant force is 1190.314 N
Direction of force is 13.352°
Explanation:
given data:
thrust force = 725 N
Angle = 32.4 degree
Let x is consider as positive direction
Resultant force in x direction is
Rx = 725 + 513cos32.4 = 1158.14 N
and Resultant force perpendicular to x direction is:
Ry = 513sin32.4 = 274.88 N
Magnitude of resultant force is
[tex]R=\sqrt{R_x^2+R_y^2} = 1190.314N[/tex]
and resultant force direction is
[tex]\theta=tan^{-1}\frac{R_y}{R_x} = 13.352\degree[/tex]
the driver of a car slams on her brakes to avoid collidingwith
a deer crossing the highway. what happens to the car's
kineticenergy as it comes to rest?
Answer:
Its dissipated by the brake.
Explanation:
Traditional brakes use friction to stop the wheels (or axis). This friction its dissipated in the form of heat. There are others mechanism to brake that don't dissipated the energy, they stored it. In electric cars (or hybrids), there are regenerative brakes, that store the kinetic energy as electrical energy.
As the driver slams the brakes, the kinetic energy of the car gets converted into other forms of energy until the car comes to a stop, at which point it becomes zero.
Explanation:When the driver of a car slams on her brakes to avoid a collision with a deer crossing the highway, the kinetic energy of the car decreases until the car comes to rest. This is due to the principle of energy conservation. Initially, when the car is moving, it has kinetic energy. However, when the brakes are applied, the kinetic energy gets converted into other forms of energy such as heat energy (due to friction between the brakes and the wheels) and potential energy (if the car is moving uphill). The kinetic energy keeps decreasing until the car comes to a complete stop. At that point, the kinetic energy of the car is zero because kinetic energy is associated with motion and the car is no longer moving.
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A charge of 7.0 μC is to be split into two parts that are then separated by 9.0 mm. What is the maximum possible magnitude of the electrostatic force between those two parts?
Answer:
[tex]F = 1361.1 N[/tex]
Explanation:
As we know that a charge is split into two parts
so we have
[tex]q_1 + q_2 = 7\mu C[/tex]
now we have
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we know
[tex]F = \frac{kq_1(7\mu C - q_1)}{r^2}[/tex]
here we have
r = 9 mm
now to obtain the maximum value of the force between two charges
[tex]\frac{dF}{dq_1} = 0[/tex]
so we have
[tex]7 \mu C - q_1 - q_1 = 0[/tex]
so we have
[tex]q_1 = q_2 = 3.5 \mu C[/tex]
so the maximum force is given as
[tex]F = \frac{(9\times 10^9)(3.5 \mu C)(3.5 \mu C)}{0.009^2}[/tex]
[tex]F = 1361.1 N[/tex]
You are on a train traveling east at speed of 28 m/s with respect to the ground. 1) If you walk east toward the front of the train, with a speed of 1.5 m/s with respect to the train, what is your velocity with respect to the ground? (m/s east)
2) If you walk west toward the back of the train, with a speed of 2.1 m/s with respect to the train, what is your velocity with respect to the ground? (m/s, east)
3) Your friend is sitting on another train traveling west at 22 m/s. As you walk toward the back of your train at 2.1 m/s, what is your velocity with respect to your friend? (m/s, east)
Answer:
A) 29.5m/s
B) 25.9m/s
C) 47.9 m/s
Explanation:
This is a relative velocity problem, which means that the velocity perception will vary from each observer at a different reference point.
We can say that the velocity of the train respect the ground is 28m/s on east direction If I walk 1.5m/s respect the train, the velocity of the person respect to the ground is the sum of both velocities:
Vgp=Vgt+Vtp
where:
Vgp=velocity of the person respect the ground
Vgt=Velocity of the train respect the ground
Vtp=Velocity of the person respect the train
1) Vgp=28m/s+1.5m/s=29.5m/s
2) here the velocity of the person respect the train is negative because it is going backward
Vgp=28m/s-2.1m/s=25.9m/s
3) the velocity of the train respect with your friend will be the sum of both velocities(Vft).
Vfp=Vft+Vtp
Vfp=velocity of the person respect the friend
Vft=Velocity of the train respect the friend
Vtp=Velocity of the person respect the friend
Vfp=(22m/s+28m/s)+(-2.1m/s)=47.9 m/s
Which of the following statements about electric field lines are true? (choose all that are true) a) They are only defined for positive charges.
b) They are always tangent to electric field vectors.
c) They are always perpendicular to charged surfaces.
d) They are a simple way to visualize the electric field vectors. e) None of the above.
Answer: b) TRUE and d) TRUE
Explanation: a) FALSE the electric field lines are used to represent the charges postives and negatives.
b) TRUE it is the definition of electric field lines , they are tangent to the electric field vector.
c) FALSE the electric field lines only for conductor are perpendicular to the surface in other any situation there is a tangencial electric field components.
d) TRUE since it is a way to describe and imagine the effect of vectorial fields.
e) FALSE