Answer:
a
Explanation:
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field towards the other end. If they move fast enough when they strike the positive electrode at the other end, they will give up their energy as X-Rays
(a) Through what potential difference should electrons be accelerated so that their speed is 1% of the speed of light?
(b) What potential difference would be needed to give the protons same kinetic energy as electrons?
(c) What speed would this potential difference give to the protons, both in m/s and as a % of the speed of light.
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV = [tex]\frac{1}{2}[/tex] mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV = [tex]\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }[/tex]
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
[tex]K_e = K_p[/tex]
K_p = ½ m v_e²
K_p = [tex]\frac{1}{2}[/tex] 9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² = [tex]\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }[/tex]
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c = [tex]\frac{7 \ 10^4 }{ 3 \ 10^8}[/tex]
v/c= 2.33 10⁻⁴
A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.76 rad/sec. The moment of inertia of the student plus the stool is 5 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.33 m from the rotation axis.
Required:
a. Find the new angular speed of the student.
b. Find the kinetic energy of the student before and after the objects are pulled in.
Answer:
a) the new angular speed of the student is 0.9642 rad/s
b)
the kinetic energy of the student before the objects are pulled in is 1.9119 J
the kinetic energy of the student after the objects are pulled in is 2.4252 J
Explanation:
Given that;
mass of each object m = 1 kg
distance of objects from axis of rotation r = 0.9 m
Moment of inertia of each object initially [tex]I_{oi}[/tex]
[tex]I_{oi}[/tex] = mr² = 1kg ×(0.9m)² = 1 kg × 0.81 m² = 0.81 kg.m²
moment of inertia of each object finally [tex]I_{of}[/tex]
[tex]I_{of}[/tex] = mr² = 1kg × (0.33 m)² = 0.1089 kg.m²
Now
moment of inertia of student plus stool [tex]I_{}[/tex] = 5 kg.m²
initial angular speed ω₀ = 0.76 rad/sec
final angular speed ω = ?
Now using conservation of angular momentum;
([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀ = ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω
so we substitute
(5 + 2 (0.81) )0.76 = (5 + 2 (0.1089) )ω
5.0312 = 5.2178 ω
ω = 5.0312 / 5.2178
ω = 0.9642 rad/s
Therefore, the new angular speed of the student is 0.9642 rad/s
b)
K.E of student before = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀²
= (0.5) (5 + 2 (0.81) )(0.76)²
= 0.5 × 6.62 × 0.5776
= 1.9119 J
Therefore, the kinetic energy of the student before the objects are pulled in is 1.9119 J
KE of student finally = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω²
= (0.5) (5 + 2 (0.1089) ) (0.9642)²
= 0.5 × 5.2178 × 0.9296
= 2.4252 J
Therefore, the kinetic energy of the student after the objects are pulled in is 2.4252 J
keli learned that an air mass is a very large body of air with similar temperature humidity and pressure and the air mass are constantly in motion she knows that you're messing depending on the temperature and moisture content tent of region where they form she looked up more information about what makes them move what are the major causes for moving & Masten North America choose two that apply.
Answer choices
A. changing humidity
B. low temperature
C. jet storm
D. prevailing westerlies
Air masses from the tropics and the equator are warm as they form over lower latitudes. The major causes for moving air masses North America exists jet storm.
What is meant by air mass?An air mass is a volume of air that in meteorology is identified by its temperature and humidity. Many hundreds or thousands of square miles are covered by air masses, which adjust to the properties of the land underneath them. Latitude and their continental or maritime source regions are used to categories them.
Warmer air masses are referred to as tropical, whilst colder air masses are referred to as polar or arctic. Superior and maritime air masses are moist, whereas continental and superior air masses are dry. Air masses with various densities are divided by weather fronts. Once an air mass has left its original location, nearby plants and bodies of water can quickly change the way it behaves. Classification systems address both the properties and modification of an air mass.
Air masses from the tropics and the equator are warm as they form over lower latitudes. They move poleward along the southern edge of the subtropical ridge and are drier and hotter than those that originate over seas. Trade air masses are another name for tropical maritime air masses. The Caribbean Sea, southern Gulf of Mexico, and tropical Atlantic Oceans, east of Florida via the Bahamas, are the origins of maritime tropical air masses that have an impact on the United States.
Monsoon air masses are moist and unstable. Rarely do dry superior air masses touch the ground. A trade wind inversion, which is a warmer and drier layer over the more moderately moist air mass below, is typically created over maritime tropical air masses when they are located above them.
Therefore, the correct answer is option C. jet storm.
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Electric power is to be generated by installing a hydraulic turbine-generator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 2300 kg/s steadily. Determine the power generation potential.
Answer:
the power generation potential is 2.705 x 10⁶ J/s.
Explanation:
Given;
height below the free surface of a large water reservoir, h = 120 m
mass flow rate of the water, m' = 2300 kg/s
The power generation potential is calculated as;
[tex]Power = \frac{Energy}{time} = \frac{F\times h}{t} = \frac{(mg) \times h}{t} = \frac{m}{t}\times gh = m' \times gh\\\\Power = m' \times gh\\\\Power = 2300 \ kg/s \ \times \ 9.8 \ m/s^2 \ \times \ 120 \ m\\\\Power = 2.705 \times 10^6 \ J/s[/tex]
Therefore, the power generation potential is 2.705 x 10⁶ J/s.
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta leading away from the heart. Since blood within the heart is essentially stationary, this pressure difference can be inferred from a measurement of the speed of blood flow in the aorta. Take the speed of sound in stationary blood to be c.
a. Sound sent by a transmitter placed directly inline with the aorta will be reflected back to a receiver and show a frequency shift with each heartbeat. If the maximum speed of blood in the aorta is v, what frequency will the receiver detect? Note that you cannot simply use the textbook Doppler Shift formula because the detector is the same device as the source, receiving sound after reflection.
b. Show that in the limit of low blood velocity (v <
f= 2fo v/c
Answer:
a) f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex] , b) Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo[tex]\frac{c+v}{c}[/tex]
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’ [tex]\frac{c}{ c-v}[/tex]
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀ [tex]\frac{c}{c-v}[/tex]
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex]
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
[tex]( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}[/tex]
f ’’ = fo [tex]( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )[/tex]
f ’’ = fo [tex]( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )[/tex]
leave the linear term
f ’’ = f₀ + f₀ 2[tex]\frac{v}{c}[/tex]
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 520 lines/mm , and the light is observed on a screen 1.4 m behind the grating.
Required:
What is the distance between the first-order red and blue fringes?
Answer:
0.143 m
Explanation:
Since
d = 1/N = 1/520 = 1.92 * 10^-3 mm
For red light;
θ = sin^-1 (1 * λred/d) = sin^-1 (1 * 656 * 10^-9/1.92 * 10^-6) = 19.98°
L = 1.4 * (tan 19.98) = 0.509 m
For blue light;
θ = sin^-1 (1 * λblue/d) = sin^-1 (1 * 486 * 10^-9/1.92 * 10^-6) = 14.66°
L = 1.4 * (tan 14.66°) = 0.366 m
Distance between the first-order red and blue fringes= 0.509 m - 0.366 m = 0.143 m
A car is traveling along a straight road at a velocity of +30.0 m/s when its engine cuts out. For the next 1.79 seconds, the car slows down, and its average acceleration is . For the next 4.03 seconds, the car slows down further, and its average acceleration is . The velocity of the car at the end of the 5.82-second period is +18.4 m/s. The ratio of the average acceleration values is = 1.53. Find the velocity of the car at the end of the initial 1.79-second interval.
Answer:
first value+2nd +3rd
Explanation:
thug life and there
A circuit has 12 Amps and 220 Volts. What is the Resistance of the circuit?
Answer:
:To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)
To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)
To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)
To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)
You are working in a shoe test laboratory measuring the coefficients of friction for running shoes on a variety of surfaces. The shoes are pushed against the surface with a downward force of 400 N, and a sample of the surface material is then pulled out from under the shoe by a machine. The machine pulls with a force of 300 N before the material begins to slide. When the material is sliding, the machine only has to pull with a force of 200 N to keep the material moving. What is the coefficient of static friction between the shoe and the material
Answer:
0.75
Explanation:
Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.
Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).
So, μ = F/N
= 300 N/400 N
= 3/4
= 0.75
So, the coefficient of static friction μ = 0.75
how long it take a train to cover 630km having a speed of 30 km/hr
Answer:
21 hours
Explanation:
well 30 x 20 = 600 than 21 = 630
Which three statements are true of all matter?
A.
It is filled with air.
B.
It takes up space.
C.
It contains aluminum.
D.
It has mass.
E.
It is made up of atoms
Answer:
B, D and E, not all matter can be filled with air
What is the value of the angle of inclination of the slide?
Answer:
63°
that's my answer
but then I am sorry if I'm wrong
Explanation:
90-27 = 63°
The cart is given an initial push up the ramp. After this push, as the car moves up the ramp, the direction of the acceleration of the cart is ________ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _______.
Answer:
Explanation:
The only force acting on the cart is a component of its weight parallel to ramp downwards . No other force acts parallel to the ramp .
Even when the cart is moving up after the initial push , its weight is acting downwards so acceleration is acting downwards .
When the cart is stationary at the top position , its weight is acting downwards so acceleration is downwards at that moment also . When the cart is going downwards , still its weight is acting down so acceleration is acting downwards .
After this push, as the car moves up the ramp, the direction of the acceleration of the cart is _down _______ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is _down ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _downwards ______.
In a liquid with a density of 1500 kg/m3, longitudinal waves with a frequency of 410 Hz are found to have a wavelength of 7.80 m. Calculate the bulk modulus of the liquid.
Answer:
The bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²
Explanation:
Given;
density of the liquid, ρ = 1500 kg/m³
frequency of the wave, F = 410 Hz
wavelength of the sound, λ = 7.80 m
The speed of the wave is calculated as;
v = Fλ
v = 410 x 7.8
v = 3,198 m/s
The bulk modulus of the liquid is calculated as;
[tex]V = \sqrt{\frac{B}{\rho} } \\\\V^2 = \frac{B}{\rho}\\\\B = V^2 \rho\\\\B = (3,198 \ m/s)^2 \times 1500 \ kg/m^3\\\\B = 1.534 \ \times 10^{10} \ N/m^2[/tex]
Therefore, the bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²
Determine the gravitational potential energy, in kJ, of 1 m3 of liquid water at an elevation of 30 m above the surface of Earth. The acceleration of gravity is constant at 9.7 m/s2 and the density of the water is uniform at 1000 kg/m3. Determine the change in gravitational potential energy if the elevation decreases by 10 m.
Answer:
Explanation:
Gravitational potential energy = mgh where m is mass , g is acceleration due to gravity and h is height from the ground .
In the first case mass m = volume x density
= 1 x 1000 = 1000 kg
height h = 30 m
potential energy = 1000 x 30 x 9.8 = 294000 J = 294 kJ .
When height decreases by 10 m , potential decreases as follows .
Decrease in potential energy
= mass x gravitational energy x decrease in height
= 1000 x 9.8 x 10
= 98000 J
= 98 kJ .
A girl jogs around a horizontal circle with a constant speed. She travels one fourth of a revolution, a distance of 25 m along the circumference of the circle, in 5.0 s. The magnitude of her acceleration is
Answer:
The centripetal acceleration of the girl is 2.468 m/s²
Explanation:
Given;
number of turns, = ¹/₄ Revolution
distance traveled by the girl, d = 25 m
time of motion, t = 5.0 s
The linear speed of the of the girl is calculated as;
[tex]v = \omega \ r\\\\v =(\frac{1}{4}rev \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1}{5 \ s} ) (25 \ m)\\\\v = (0.3142 \ \frac{rad}{s} )(25 \ m)\\\\v = 7.855 \ m/s[/tex]
The centripetal acceleration of the girl is calculated as;
[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(7.855)^2}{25} \\\\a_c = 2.468 \ m/s^2[/tex]
Therefore, the centripetal acceleration of the girl is 2.468 m/s²
A characteristic of a nebula is that it-
Answer:
Center of solar system
Explanation:
Answer: b
Explanation:
A Car is moving at a speed of 20 m/s. How Much Distance it will cover in 1 min? Express the answer in km.
Answer:
d=20m/sx60s=1200m=1200/1000Km=1.2km
Explanation:
what is the difference between mass and weight
Answer:
The mass of an object is a measure of the object's inertial property, or the amount of matter it contains. The weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it. The pull of gravity on the earth gives an object a downward acceleration of about 9.8 m/s2.
Answer:
Explanation:
The mass is essentially "how much stuff" is in an object. ... Weight: There is a gravitational interaction between objects that have mass. If you consider an object interacting with the Earth, this force is called the weight. The unit for weight is the Newton (same as for any other force).
An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA. The current density is uniformly distributed through the beam. The kinetic energy of each proton is 20 MeV. The beam strikes a metal target and is absorbed by the target. (a) What is the number density of the protons in the beam
Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = [tex]\frac{1}{2}[/tex][tex]m_{p}[/tex]v²
v = √( 2K / [tex]m_{p}[/tex] )
lets relate the cross-sectional area A of the beam to its diameter D;
A = [tex]\frac{1}{4}[/tex]πD²
now, we substitute for v and A
n = I / [tex]\frac{1}{4}[/tex]πeD² ×√( 2K / [tex]m_{p}[/tex] )
n = 4I/π eD² × √([tex]m_{p}[/tex] / 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
A cylinder of gas is at room temperature (20°C). The air conditioner breaks down, and the temperature rises
to 46°C. What is the new pressure of the gas relative to its initial pressure (Pi)?
Answer: we divide this equation by pV and use {C}_{p}={C}_{V}+ . ... The temperature, pressure, and volume of the resulting gas-air mixture
Explanation:
The new pressure of the gas relative to its initial pressure P₁ is P₁ × 1.09
From the question given above, the following data were obtained:
Initial temperature (T₁) = 20 °C = 20 + 273 = 293 K
Initial pressure (P₁) = P₁
New temperature (T₂) = 46 °C = 46 + 273 = 319 K
New pressure (P₂) =?The new pressure of the gas can be obtained as follow:
P₁ / T₁ = P₂ / T₂
P₁ / 293 = P₂ / 319
Cross multiply
293 × P₂ = P₁ × 319
Divide both side by 293
P₂ = (P₁ × 319) / 293
P₂ = P₁ × 1.09Thus, the new pressure relative to it's initial pressure is P₁ × 1.09
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How would you compare the acceleration between the unbalanced net force of 100 N and of 50 N
Answer:
The acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.
Explanation:
Given;
first net force, F₁ = 100 N
second net force, F₂ = 50 N
If we consider equal mass for the two net forces, and apply Newton's second law of motion, the acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.
Let a₁ be the acceleration produced by the first net force
then, a₂ be the acceleration produced by the second net force
Thus, a₁ = 2a₂
Explain how you could use iron filings and a piece of paper to help reveal the effect of a magnetic field.
Answer:
you could put the iron filings on the peace of paper and hover a magnet over top of the paper and the iron filings would stand up, or even stick to the magnet
Explanation:
PHYSICS QUESTION PLS HELP
The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is
mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J
The total energy is the same, 970,200 J.
Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.
At point B, the coaster has dropped to a height of 10 m, so it has PE
mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J
which means it must have KE
970,200 J = KE + 294,000 J → KE = 676,200 J
which gives the coast a speed v at point B of
1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J → v ≈ 21.2 m/s
At point C, the coaster has a speed of 16.0 m/s, so it has KE
1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J
and hence PE
970,200 J = 384,000 J + PE → PE = 586,200 J
This lets us determine the height h at C:
mgh = (3000 kg) (9.80 m/s²) h = 586,200 J → h ≈ 19.939 m
which means the loop has diameter h - 10 m ≈ 9.94 m.
At point D, the coaster is 15 m above the ground so its PE at D is
mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J
and so its KE is
970,200 J = KE + 441,000 J → KE = 529,200 J
and hence has speed v at D
1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J → v ≈ 18.9 m/s
Explain what is happening in this picture
Answer:
in this video waves are coming up for the BOTTOM to the top of the sandbar
To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant angular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.8 m . Part A If the thrower takes 1.2 s to complete one revolution, starting from rest, what will be the speed of the discus at release
Answer:
9.42 m/s
Explanation:
a) Using Newton's law of motion formula:
[tex]\theta=\frac{(\omega+\omega_o)}{2}t\\\\where \ \theta=angular\ displacement=1\ rev =2\pi, w_o=initial\ velocity\ of\ discus\\=0\ rad/s, \omega=angular\ speed\ of\ discus\ at\ release,t=time\ = 1.2\ s.\\\\Hence:\\\\2\pi=\frac{(0+\omega)}{2}(1.2)\\\\\omega=\frac{2*2\pi}{1.2} \\\\\omega=10.47\ rad/s\\[/tex]
The speed of the discus at release (v) is:
v = ωr; where r = radius of discus
diameter = 1.8 m, r = diameter / 2= 1.6 / 2 = 0.9 m
v = ωr = 10.47 * 0.9
v = 9.42 m/s
The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons).
(a) What is the force between the two alpha particles when they are 6.60 ✕ 10−15 m apart? N
(b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s2
Answer:
a) F = 21.16 N, b) a = 3.17 10²⁸ m / s
Explanation:
a) The outside between the alpha particles is the electric force, given by Coulomb's law
F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]
in that case the two charges are of equal magnitude
q₁ = q₂ = 2q
let's calculate
F = [tex]9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }[/tex]
F = 21.16 N
this force is repulsive because the charges are of the same sign
b) what is the initial acceleration
F = ma
a = F / m
a = [tex]\frac{21.16}{4.0026 \ 1.67 \ 10^{-27} }[/tex]21.16 / 4.0025 1.67 10-27
a = 3.17 10²⁸ m / s
this acceleration is in the direction of moving away the alpha particles
Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun's electric field were 2.95 × 104 N/C? (b) What speed would the proton attain if the field accelerated the proton through a distance of 1.26 cm?
Answer:
(A) the acceleration experienced by the proton 2.821 x 10¹² m/s²
(B) the speed of the proton is 2.67 x 10⁵ m/s
Explanation:
Given;
electric field experienced by the proton, E = 2.95 x 10⁴ N/C
charge of proton, Q = 1.6 x 10⁻¹⁹ C
mass of proton, m = 1.673 x 10⁻²⁷ kg
distance moved by the proton, d = 1.26 cm = 0.0126 m
(a)
The force experienced by the proton is calculated as;
F = ma = EQ
where;
a is the acceleration experienced by the proton
[tex]a = \frac{EQ}{m} \\\\a = \frac{2.95\times 10^4 \ \times \ 1.6\times 10^{-19}}{1.673 \times 10^{-27}} \\\\a = 2.821 \times 10^{12} \ m/s^2[/tex]
(b) the speed of the proton is calculated;
v² = u² + 2ad
v² = 0 + (2 x 2.821 x 10¹² x 0.0126)
v² = 7.109 x 10¹⁰
v = √7.109 x 10¹⁰
v = 2.67 x 10⁵ m/s
True or false it is impossible to determine weather you are moving unless you can touch another object
Answer: false
Explanation:
Answer:
false
Explanation:
According to Newton's First Law of motion, an object remains in the same state of motion unless a resultant force acts on it.
Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, as if they had been migrating. For years curiosity mounted about why the stones moved. One explanation was that strong winds during the occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about 0.80. What horizontal force is needed on a 25 kg stone (a typical mass) to maintain the stone's motion once a gust has started it moving
Answer:
F = 196 N
Explanation:
For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically
Y axis
N- W = 0
N = mg
X axis
F -fr = ma
In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.
F- fr = 0
F = fr
the friction force has the equation
fr = μ N
fr = μ mg
we substitute
F = μ mg
let's calculate
F = 0.80 9.8 25
F = 196 N