Answer:
v_r = 1.268 × 10⁸ mi/hr
Explanation:
We are given;
wavelength of the red light; λr = 693 nm = 693 × 10^(-9) m
wavelength of the yellow light; λy = 582 nm = 582 × 10^(-9) m
Frequency is given by the formula;
f = v/λ
Where v is speed of light = 3 x 10^(8) m
Frequency of red light; f_o = [3 x 10^(8)]/(693 × 10^(-9)) = 4.33 x 10¹⁴ Hz
Similarly, Frequency of yellow light;
f = [3 x 10⁸]/(582 × 10^(-9)) = 5.15 x 10¹⁴ Hz
To find the speed of the car, we will use the formula;
f = f_o[(c + v_r)/c)]
Where c is speed of light and v_r is speed of car.
Making v_r the subject;
cf/f_o = c + v_r
v_r = c(f/f_o - 1)
So, plugging in the relevant values, we have;
v_r = 3 × 10⁸[((5.15 x 10¹⁴)/(4.33 x 10¹⁴)) - 1]
v_r = 3 × 10⁸(0.189)
v_r = 5.67 x 10⁷ m/s
Converting to mi/hr, 1 m/s = 2.23694 mile/hr
So, v_r = 5.67 × 10⁷ × 2.23694
v_r = 1.268 × 10⁸ mi/hr
A substance that produces H+ ions in solution is a.
A)soap
B)acid
C)salt
D)base
A boat crosses a 200m wide river at 3m/s, north relative to water. The river flows at 1m/s as shown. What is the velocity of the boat as observed by a stationary observer on the river back from the boat departed?
Answer:
The resultant velocity = 3.16 m/s
Explanation:
Since the boat is moving North of the direction of the riverflow, the river would either be flowing westward or Eastward. The two motions form a right angle triangle with the resultant velocity being the hypotenuse of the traingle.
The resultant velocity will be given as ;
R² = B² + r²
Where B is the velocity of the boat and r is the velocity of the river
R² = 3² + 1²
R² = 10
R = √10 = 3.16 m/s
Therefore, the resultant velocity = 3.16 m/s
You are working in a biology lab and learning to use a new ultracentrifuge for blood ts. The specifications for the centrifuge say that a red blood cell rotating in the ultracentrifuge moves at 470 m/s and has a radial acceleration of 150,000 g's (that is, 150,000 times 9.8 m/s2). The radius of the centrifuge is 0.15 m. You wonder if this claim is correct.
What is the radial acceleration of the ultra-centrifuge using calculations ?
Answer:
The radial acceleration is [tex]a_r = 1472667 \ m/s^2[/tex]
Explanation:
From the question we are told that
The velocity of the red blood cell is [tex]v_r = 470 \ m/s[/tex]
The radial acceleration of the red blood cell is [tex]a_r = 150000g = 150000*9.8 = 1470000 \ m/s^2[/tex]
The radius of the centrifuge is [tex]r = 0.15 \ m[/tex]
Generally radial acceleration is mathematically represented as
[tex]a_r = \frac{v^2}{r}[/tex]
substituting values
[tex]a_r = \frac{470^2}{0.15}[/tex]
[tex]a_r = 1472667 \ m/s^2[/tex]
The initial mass of a radioisotope is 10.0 g. If the radioisotope has a half life of 2.75 years, how much remains after four half lives?
After four half-lives, the amount that remains is (1/2)⁴ of the original sample.
That's (1/16) of 10 g, or 0.625 g.
It doesn't matter how long the half-life is.
A radioisotope has a starting mass of 10.0 g. Half lives of the isotopes are 0-10, 1-5, 2-2.25, 3-1.25, and 4-0.625 if the radioisotope has a half life of 2.75 years.
Who or what are radioisotopes?A chemical element in an unstable state that emits radiation as it decomposes and becomes more stable. Both the natural environment and a laboratory can produce radioisotopes. They are utilized in imaging studies and therapy in medicine.
Why do you use the term "half-life"?In radioactivity, the half-life is the amount of time needed for half of a radioactive sample's atomic nuclei to spontaneously decay (convert into different nuclear species by emitting particles and energy), or, more precisely, the amount of time needed for the number of disintegrations per second.
To know more about Radioisotope, visit-
https://brainly.com/question/19472025
#SPJ2
The density of gasoline is 730 kg/m3 at 00C. Its volume expansion coefficient is 0.000960C-1. If one gallon of gasoline occupies 0.0038 m3, how many extra kilograms of gasoline are obtained when 10 gallons of gasoline are bought at 00C rather than at 200C
Answer:
Explanation:
volume expansion coefficient = .000960 C⁻¹
d₀ = d ( 1 + .000960 x t ) , d is density at t and d₀ is density at 0 degree Celsius .
if t = 200
d₀= d ( 1 + .00096 x 200 )
= d ( 1 + .00096 x 200 )
= 1.192d
10 gallons of gasolene
= 10 x .0038 m³ of gasolene
= .038 m³
difference of mass = volume x difference of density
= .038 x ( d₀ - d )
.038 x ( d₀ - d₀ / 1.192 )
= .038 x ( 1 - 1 / 1.192 ) d₀
= .00612 x 730
= 4.468 kg .
Whale sharks swim forward while ascending or descending. They swim along a straight-line path at a shallow angle as they move from the surface to deep water or from the depths to the surface. In one recorded dive, a shark started 50 m below the surface and swam at 0.85 m/s along a path tipped at a 13° angle above the horizontal until reaching the surface.
Required:
a. What was the horizontal distance between the shark's starting and ending positions?
b. What was the total distance that the shark swam?
c. How much time did this motion take?
Answer:
a)217m
b)222m
c)261s
Explanation:
Considering the figure in the attachment
depth 'h'=50m
speed 'v'=0.85m/s
angle 'θ'=13°
a) we have
[tex]\frac{h}{x}[/tex]=tanθ
x = h / tanθ => 50/tan13°
x=216.57≈217m
b)we have
[tex]\frac{h}{d}[/tex]=sinθ
d=h / sinθ => 50 / sin13°
d=222.27≈222m
c)Time needed = d / v = 222/0.85
261.18≈261s
The frames for a pair of eyeglasses have a radius of 2.14 cm at 20.0°C. Lenses with radius of 2.16 cm have to be inserted into these frames. To what temperature must the technician heat the frames to accommodate the lenses? The frames are made of a material whose thermal expansion coefficient is 1.30 10-4/°C.
Answer:
The technician must heat the frame to a temperature of T₂ = 0.0072°C
Explanation:
Change in one-dimension of an object upon heating is given by the following equation:
ΔL = α L ΔT
For the radius this equation can be re-written as follows:
ΔR = α R ΔT
where,
ΔR = Change in Radius = 2.16 cm - 2.14 cm = 0.02 cm
α = Thermal Expansion Coefficient = 1.3 x 10⁻⁴ /°C
R = Original Radius = 2.14 cm
ΔT = Change in Temperature = T₂ - 20°C
Therefore,
0.02 cm = (1.3 x 10⁻⁴ /°C)(2.14 cm)(T₂ - 20°C)
T₂ - 20°C = (0.02 cm)/(1.3 x 10⁻⁴ /°C)(2.14 cm)
T₂ = 0.0072°C + 20°C
T₂ = 20.0072°C
The Michelson interferometer can be used to measure the index of refraction of a gas by placing an evacuated transparent tube in the light path along one arm of the device. Fringe shifts occur as the gas is slowly added to the tube. Assume 560-nm light is used, the tube is 5.10 cm long, and 168 bright fringes pass on the screen as the pressure of the gas in the tube increases to atmospheric pressure. What is the index of refraction of the gas
Answer:
The index of refraction of the gas is [tex]n_g = 1.000922[/tex]
Explanation:
From the question we are told that
The wavelength of light is [tex]\lambda = 560 \ nm = 560 *10^{-9} \ m[/tex]
The length of tube is [tex]L = 5.10 \ cm = \frac{5.10 }{100} =0.0510 \ m[/tex]
The number of fringes is [tex]N = 168[/tex]
Generally the index of refraction of the gas is mathematically repreented as
[tex]n_g = 1 + \frac{N \lambda }{2 L }[/tex]
substituting values
[tex]n_g = 1 + \frac{168 *[ 560* 10^{-9}] }{2 * 0.0510 }[/tex]
[tex]n_g = 1.000922[/tex]
Suppose the kicker launches the ball at 60∘ instead of 30∘. Assuming that the goal is 4.55 m high and 40 m away, what minimum initial speed v0 would the ball need to have in order to just clear the goal?
Answer:22 m/s
Explanation:
Given
launch angle [tex]\theta =60^{\circ}[/tex]
height of goal [tex]h=4.55\ m[/tex]
and horizontal distance [tex]x=40\ m[/tex]
Suppose initial speed is [tex]u[/tex]
Trajectory of a Projectile is given by
[tex]y=x\tan \theta -\frac{1}{2}\frac{gx^2}{u^2\cos ^2\theta }[/tex]
substituting the values we get
[tex]4.55=40\tan (60)-0.5\times \frac{9.8\times (40)^2}{u^2\cdot \cos ^260 }[/tex]
[tex]4.55=69.28-0.5\times \frac{15,680}{u^2\cdot 0.25}[/tex]
[tex]\frac{31,360}{u^2}=69.28-4.55[/tex]
[tex]\frac{31,360}{64.73}=u^2[/tex]
[tex]u^2=484.47[/tex]
[tex]u=22.01\ m/s[/tex]
So, initial launch speed is [tex]22\ m/s[/tex]
Two fledglings leave a nest which is 2.50 m off the ground. One drops from rest and then 0.10 s later the second charges out of the nest with a velocity with horizontal and
downward components of 1.00 and 1.50 m/s, respectively.
1. Which fledgling hits the ground first (first or second)?
2. With what speed, in m/s, does the first fledgling hit the ground?
3. With what speed, in m/s, does the second fledgling hit the ground?
4. Which fledgling travels the greater displacement?
5. How far does a parent bird have to travel on the ground between the two fledglings, in m, to make sure they are alright?
t Answer:
1) the time of the pigeon 1 is less, so it comes first
2) v = - 6,997 m / s , 3) v = 10.15 m / s ,
4) the displacement of the second point in greater
5) x = 0.883 m
Explanation:
For this exercise we will use the kinematics equations
1) ask which chick reaches the ground first
we calculate for the first chick that has zero initial velocity
y = y₀ + v₀ t - ½ g t²
0 = yo - ½ g t²
t = √ 2 y₀ / g
let's calculate
t = √ (2 2.50 / 9.8)
t = 0.714 s
We calculate the time it takes for the second chick that has velocity v = (1 i ^ - 1.5 j⁾ m / s
y = y₀ + v₀t - ½ g t²
0 = 2.5 - 1.5 t - ½ 9.8 t²
4.9 t² + 1.5 t - 2.5 = 0
t² + 0.306 t - 0.510 = 0
we solve the quadratic equation
t = [0.306 ± √ (0.306² - 4 (-0.510))] / 2
t = [0.306 ± 1.46] / 2
The results are
t₁ = -0.577 s
t₂ = 0.883 m / s
we take positive time as correct
the time of the pigeon 1 is less, so it comes first
2) the speed of the first chick is
v = v₀ - g t
we can see that
v = -gt
v = - 9.8 0.714
v = - 6,997 m / s
the negative sign indicates that the speed is down
3) the speed of the other bird is
v = -1.5 - 9.8 0.883
v = 10.15 m / s
4) which chick has the greatest displacement. The first point falls vertically and its displacement is y₀
The second point describes a parabola and its displacement is
d = √ (x² + y₀²)
therefore we see that the displacement of the second point in greater
5) calculate the horizontal displacement of the second point
x = vx t
x = 1 0.883
x = 0.883 m
What are the 2 types of electricity
In order for an external force to do work on a system, Question 9 options: a) there must be a component of the force perpendicular to the motion. b) there must be a component of force parallel to the motion of the object. c) the force can be at any angle relative to the motion of the object.
Answer:
b) there must be a component of force parallel to the motion of the object.
Explanation:
We know that work done on a body by an external force is calculated by the formula given below:
W = F.d = Fd Cos θ
where,
W = Work Done by the force on the body
F = Magnitude of force
d = displacement of the body
θ = The angle between the direction of motion of the body and the force applied
It is clear from the formula of the work done, that "F Cosθ" represents the component of the force, that is acting in the direction of motion of the object or parallel to the direction of motion of the object. So, if there is no component of force parallel to motion of object, then this factor will become zero. As a result, the work done will also be zero.
Therefore, the correct option will be:
b) there must be a component of force parallel to the motion of object.
The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal chain on the scale, it read 02.1kg. The physicist, then decided to test Einstein’s equation, and compressed the spring and tied it with the chain and placed it on the scale. It read 02.7kg. Which of the following conclusions is the most likely the physicist will come to?
a. Einstein's equation has an error
b. The scale is broken
c. Compressing the spring didn't add energy
d. The scale's resolution is too low to read the change in mass
Answer:
d. The scale's resolution is too low to read the change in mass
Explanation:
If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:
F = kx
since,
w = Fd
dw = Fdx
integrating and using value of F, we get:
ΔE = (0.5)kx²
where,
ΔE = Energy added to spring
k = spring constant
x = displacement
The spring constant is typically in range of 4900 to 29400 N/m.
So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:
ΔE = (0.5)(29400 N/m)(1 m)²
ΔE = 1.47 x 10⁴ J
Now, if we convert this energy to mass from Einstein's equation, we get:
ΔE = Δmc²
Δm = ΔE/c²
Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²
Δm = 4.9 x 10⁻¹³ kg
As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.
Since, the scale only gives the mass value upto 1 decimal place.
Thus, it can not determine such a small change. So, the correct option is:
d. The scale's resolution is too low to read the change in mass
A television weighs 8.50 pounds. How many grams is this? (Hint: You need to
use two unit conversion fractions. 1 pound equals about 0.454 kg.)
Answer:
3859 grams
Explanation:
Given: Weight of a television = 8.50 pounds
To find: Weight of a television in grams
Solution:
1 pound = 0.454 kg and 1 kg = 1000 g
So,
1 pound = 0.454 × 1000 = 454 grams
8.50 pounds = 8.50 × 454 = 3859 grams
Therefore,
Weight of television in grams = 3859 grams
4. An electric iron has a
power rating of 750W
a. How many joules of
electric energy does it
change into heat energy
every second?
b. How many joules of
work can it do in 3
seconds
c. How long does it take
the iron to do 1500J of
work?
5. Use the kinetic particle
theory to explain why a
solid has a definite shape
and liquid has none.
Answer:
4.
a) W = 750 J
b) W = 2250 J
c) t = 2 sec
5. Answered in explanation
Explanation:
4.
The formula of power is given as:
P = W/t
where,
P = Power
W = Work Done
t = Time Taken
a)
Here,
P = 750 W
t = 1 sec
W = ?
Therefore,
750 W = W/1 sec
W = 750 J
b)
Here,
P = 750 W
t = 3 sec
W = ?
Therefore,
750 W = W/3 sec
W = (750 W)(3 sec)
W = 2250 J
c)
Here,
P = 750 W
t = ?
W = 1500 J
Therefore,
750 W = 1500 J/t
t = 1500 J/750 W
t = 2 sec
5.
According to Kinetic Particle Theory, the molecules are tightly packed with each other, by strong inter-molecular forces and they can only vibrate at their position. While, molecules or particles in liquids have lesser attractive forces among them. They can move in layers and can take the shape of any container. This is the reason why solid has a definite shape and liquid has none.
A particle moves with a velocity v in a circle of radius r, then its angular velocity is equal
to………….. and acts along the…………..
Answer:
Given that,
Speed = v
Radius = r
We have to ascertain the precise speed
Utilizing equation of speed
Where, v = velocity/speed
r = radius
= precise speed/ angular velocity
Angular Velocity :
The precise speed is characterized by the speed of turn.
The precise speed is straightforwardly corresponding to the direct speed and contrarily relative to the range of the molecule.
Subsequently, The precise speed is v/r
A girl walks South at 2.7 m/s. What is the y component of her velocity
Answer:
-2.7m/s
Explanation:
...............
Having successfully ransomed a nearby city with her rail gun, Prof. Marcia Grail is looking for new and exciting ways to wreck science based havok for profit. She constructs a device capable of creating a constant intensity of electromagnetic radiation of any frequency over a circular space of radius 50cm. In order to do the most damage to biological systems (e.g. make those filthy normal humans pay), what type of electromagnetic wave does she need to set the device for
Answer:
Gamma radiation
Explanation:
Electromagnetic radiations are waves that do not require material medium for their propagation, and they travel at the same velocity. When these waves are arranged with respect to increasing frequency or decreasing wavelength, it forms a spectrum.
Since Prof. Marcia Grail's device can create an electromagnetic radiation of any frequency, the appropriate radiation to achieve her goal would be gamma radiation.
Gamma radiation is a high penetrating radiation with lethal effects on exposed biological molecules, causing induction of cancer and genetic mutations even at low level.
If the molar mass of helium is 4.0 g/mol and the molar mass of neon is 20.2 g/mol, then
a. All the atoms have exactly the same velocity.
b. All the atoms have the same average speed.
c.The average speed of the helium atoms is greater than the average speed of the neon atoms.
d. The average speed of the neon atoms is greater than the average speed of the helium atoms.
e. The atoms diffuse from high temperature to low temperature.
How much would it cost to cover the entire land area of United States with dollar bills? To answer this question, you may find the following information useful: The contiguous United States (all of the states minus Alaska and Hawaii) can be approximated as a rectangle that measures 1000 miles north to south and 3000 miles east to west, while Alaska has about one-fifth the area of the contiguous states and Hawaii is small enough to be ignored for this rough calculation. Also, note that a dollar bill measures around 6.5cm×15.5cm6.5cm×15.5cm.
Answer:
It would cost approximately $925,455,484 million to cover continental US and Alaska with $1 bills.
Explanation:
the area of a one dollar bill = 6.5 cm x 15.5 cm = 100.75 sq cm
the approximate area of continental US + Alaska = (1,000 miles x 3,000 miles) x 1.2 = 3,600,000 sq miles
each sq mile is roughly 2.58999 sq km, so the total area in sq km = 9,323,964 sq km
1 sq km = 1,000,000 sq meters
each sq meter = 10,000 sq cm
1 sq m = 10,000 / 100.75 = 99.25558 bills
1 sq km = 99,255,583.13 bills
9,323,964 sq km = 925,455,483,900,000 bills
If the US includes the area of the Hawaii and the Alaska. It would be costing approx. $925,455,484 mn to cover continental with the $1 bills.
As the area of 1 dollar bill = 6.5 cm x 15.5 cm = 100.75 cm2 The area of US + Alaska = (1,000 miles x 3,000 miles) x 1.2 = 3,600,000 sq Thus the total area in km2 = 9,323,964 sq km 1 km2 equals to 1,000,000 sq meters each sq meter = 10,000 sq cm 1 sq m = 10,000 / 100.75 = 99.25558 bills 1 km2 equals to 99,255,583.13 bills 9,323,964 sq km = 925,455,483,900,000 bills .Learn more about the would it cost to cover the entire land area.
brainly.com/question/17333335.
Based on the information in the table, which two elements are most likely in the same group, and why?
A table with 5 columns and 5 rows labeled facts about 4 elements. The first column labeled element has entries bismuth (B i), nitrogen (N), oxygen (O), sodium (N a), thallium (T l). The second column labeled atomic mass (a m u) has entries 209, 14, 16, 23, 204. The third column labeled total electrons has entries 83, 7, 8, 11, 81. The fourth column labeled valence electrons has entries 5, 5, 6, 1, 3. The fifth column labeled year isolated has entries 1753, 1772, 1772, 1807, 1861.
bismuth and thallium, because their atomic masses are very similar
nitrogen and oxygen, because they were both first isolated in the same year
sodium and thallium, because their names both end in the same suffix: -ium
bismuth and nitrogen, because they have the same number of valence electrons
Answer: bismuth and nitrogen, because they have the same number of valence electrons
Explanation:
Elements are distributed in groups and periods in a periodic table.
Elements that belong to same groups will show similar chemical properties because they have same number of valence electrons.
The number of valence electrons in Bismuth and nitrogen are 5 and thus thus they will show similar chemical properties and thus belong to the same group.
The atomic masses of elements in a group will differ drastically.
The group number has got nothing to be the isolation year.
Thus bismuth and nitrogen belong to same group because they have the same number of valence electrons
Answer:
D
Explanation:
i got it right in my test
50 POINTS
What happens to kinetic energy of a moving object of the velocity decreases? (Look at photo multiple choice)
Answer: A Answers. Assuming that the terminal velocity doesn't change during the fall, then the kinetic energy would remain constant. However the terminal velocity decreases during the fall since the air becomes denser at lower altitudes.
Explanation:
What happens to the KE of an object when it slows down and heats up? - Quora. The kinetic energy goes down and the loss of the kinetic energy is through the production of heat energy. In real world this is due to friction, or an opposing force that decelerates the object, or a combination of both.
Answer:
Kinetic energy stays the same
Learning Goal: To understand the behavior ofthe electric field at the surface of a conductor, and itsrelationship to surface charge on the conductor.
A conductor is placed in an external electrostatic field. Theexternal field is uniform before the conductor is placed within it.The conductor is completely isolated from any source of current orcharge.
PART A)
Which of the following describes the electricfield inside this conductor?
It is in thesame direction as the original external field.
It is in theopposite direction from that of the original externalfield.
It has adirection determined entirely by the charge on itssurface.
It is alwayszero.
PART B)
The charge density inside theconductor is:
0
non-zero;but uniform
non-zero;non-uniform
infinite
PART C)
Assume that at some point just outside thesurface of the conductor, the electric field has magnitudeE and is directed toward thesurface of the conductor. What is the charge density eta on the surface of the conductor at thatpoint?
Express your answer in terms ofE and epsilon_0.
eta =
Complete Question
The complete question is shown on the first uploaded image
Answer:
a it is always zero
b 0
c [tex]\eta = -\epsilon _o E[/tex]
Explanation:ss
Here the net charge is on the outer surface of the conductor thus this means that the net charge inside the conductor is zero
Generally the charge density of a conductor is dependent on the charge per unit area which implies that the charge density is dependent on the net charge so this means that the charge density inside the conductor is zero
Generally the direction of electric field this from the positive charge to the negative charge so from the question we can deduce that the negative charge is located on the surface of the conductor
So We can mathematically define the charge density on the surface of the electric field as
∮[tex]E \cdot dA = \frac{-Q}{\epsilon _o}[/tex]
Where E is the electric field
[tex]dA[/tex] change in unit area
[tex]-Q[/tex] is the negative charge
[tex]\epsilon _o[/tex] is the permittivity of free space
So
[tex]EA = \frac{-Q}{\epsilon _o }[/tex]
[tex]\frac{Q}{A} = -\epsilon _o E[/tex]
[tex]\eta = -\epsilon _o E[/tex]
Where [tex]\eta[/tex] is the charge density
You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 1.80×107 m/s. Assume that the initial speed of the target nucleus is negligible and the collision is elastic.
A) Find the mass of one nucleus of the unknown element.
B) What is the speed of the unknown nucleus immediately after such a collision?
Answer:
a
The mass is [tex]m_2 =21.75*10^{-27} \ kg[/tex]
b
The velocity is [tex]v_2 = 3.0*10^{6} m/s[/tex]
Explanation:
From the question we are told that
The speed of the protons is [tex]u_1 = 2.10*10^{7} m/s[/tex]
The mass of the protons is [tex]m[/tex]
The speed of the rebounding protons are [tex]v_1 = -1.80 * 10^{7} \ m/s[/tex]
The negative sign shows that it is moving in the opposite direction
Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as
[tex]m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1[/tex]
Where [tex]m_1[/tex] is the mass of a single proton
So substituting values
[tex]m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1[/tex]
[tex]m_2 =13 m_1[/tex]
The mass of on proton is [tex]m_1 = 1.673 * 10^{-27} \ kg[/tex]
So [tex]m_2 =13 ( 1.673 * 10^{-27} )[/tex]
[tex]m_2 =21.75*10^{-27} \ kg[/tex]
Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as
[tex]m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2[/tex]
Now [tex]u_2[/tex] because before collision the the nucleus was at rest
So
[tex]m_1 u_1 = m_1 v_1 + m_2v_2[/tex]
=> [tex]v_2 = \frac{m_1(u_1 -v_1)}{m_2}[/tex]
Recall that [tex]m_2 =13 m_1[/tex]
So
[tex]v_2 = \frac{m_1(u_1 -v_1)}{13m_1}[/tex]
=> [tex]v_2 = \frac{(u_1 -v_1)}{13}[/tex]
substituting values
[tex]v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}[/tex]
[tex]v_2 = 3.0*10^{6} m/s[/tex]
Write the formula for the Newton’s Law of Gravitation.
please answer this question!!!
Answer:
F = G(m1m2)/R2.
Explanation:
Newton's law of universal gravitation:
Gravitational Force, in Newtons, between two objects =
(a constant)·(one mass)·(the other mass)/(distance between them)²
acting on EACH object, in the direction of the other object.
If the masses are in kilograms and the distance is in meters, then the constant is 6.67 x 10⁻¹¹ m³/kg-sec² .
I may be wrong, but I don't think Newton had any number to use for the constant. It had not been measured yet, the kilogram and the meter had not been invented yet, and there certainly was no unit called "a Newton" during his lifetime.
He might have been able to calculate the value of the constant by applying his law of gravity to the motion of one or two planets. But he would have needed to know the mass of the sun and the planets he used, and I don't think those were known yet in Newton's time.
standing on a hill that is 3.5 m high above level ground. A wall that is on level ground is 15 m away from the student. The student decides to throw a ball horizontally at a combined height of 4.5 m, at the wall. The student measures the time it takes to make an impact and finds it to be 0.65 s. At what height, from the ground, did the student hit the side of the wal
Answer:
The height will be "2.42 m".
Explanation:
The given values are:
time, t = 0.65 seconds
∴ g = 9.8
As we know,
⇒ [tex]x=vt+\frac{1}{2}gt^2[/tex]
∴ v = 0
On putting the estimated values, we get
⇒ [tex]=0+\frac{1}{2}\times 9.8\times (0.65)^2[/tex]
⇒ [tex]=\frac{1}{2}\times 9.8\times 0.4225[/tex]
⇒ [tex]=\frac{1}{2}\times 4.1405[/tex]
⇒ [tex]=2.07025 \ m[/tex]
Now,
Height, [tex]h=4.5-2.07025[/tex]
⇒ [tex]=2.42 \ m[/tex]
Helppppp please thanks
Answer:
It represents battery
Answer: I think #1 represent lights and #2 represent the battery
a block (mass 0.6kg) is released from rest at point A at the top of a ramp inclined at 36.9 degress above the horizontal. The block moves a distance of 4m along the ramp to point b at the bottom of the ramp. How much work is done by gravity as the block goes from a to b
Your friend says your body is made up of more than 99.9999% empty space. What do you think
Answer:false
Explanation:
A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by 1.8 cm. Voltage applied at the two ends is 5 V. A pulse arrives at the collection point at 0.608 ms, and the separation of the pulse is 180 sec. Calculate mobility and diffusion coefficient for minority carriers. Verify it from the Einstein relation.
Answer:
Mobility of the minority carriers, [tex]\mu_{n} =1184.21 cm^{2} /V-sec[/tex]
Diffusion coefficient for minority carriers,[tex]D_{n} = 29.20 cm^2 /s[/tex]
Verified from Einstein relation as [tex]\frac{D_{n} }{\mu_{n} } = 25 mV[/tex]
Explanation:
Length of sample, [tex]l_{s} = 2 cm[/tex]
Separation between the two probes, L = 1.8 cm
Drift time, [tex]t_{d} = 0.608 ms[/tex]
Applied voltage, V = 5 V
Mobility of the minority carriers ( electrons), [tex]\mu_{n} = \frac{V_{d} }{E}[/tex]
Where the drift velocity, [tex]V_{d} = \frac{L}{t_{d} }[/tex]
[tex]V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s[/tex]
and the Electric field strength, [tex]E = \frac{V}{l_{s} }[/tex]
E = 5/2
E = 2.5 V/cm
Mobility of the minority carriers:
[tex]\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec[/tex]
The electron diffusion coefficient, [tex]D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }[/tex]
[tex]\triangle x = (\triangle t )V_{d}[/tex], where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)
[tex]\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm[/tex]
[tex]D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s[/tex]
For the Einstein equation to be satisfied, [tex]\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V[/tex]
[tex]\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV[/tex]
Verified.
