Answer:
C. Mass
Explanation:
Tay quay OB quay đều quanh trục cố định đi qua O với vận tốc góc không đổi ω. Con lăn A chuyển động trong rãnh thẳng đứng. Tại vị trí trên hình vẽ thì thanh OB thẳng đứng, OA có phương nằm ngang. Hãy xác định vận tốc góc thanh AB, vận tốc của con lăn A; gia tốc góc của thanh AB, gia tốc của con lăn A. Cho ω = 1,5 rad/s, r = 1 m.
A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s. (a) Determine the magnitude of the emf induced in the loop. (b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B
This question is incomplete, the complete question is;
A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s.
(a) Determine the magnitude of the emf induced in the loop.
(b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 1.5 T, if the induced emf is to be zero? (Give the magnitude of the rate of change of the area.) (m2/s)
Answer:
a) the magnitude of the emf induced in the loop is 0.003 V
b) dA/dt = 0.002 m²/s
Explanation:
Area of the loop wire A = 0.015 m²
magnitude of the field is increasing dB/dt = 0.20 T/s
a)
Determine the magnitude of the emf induced in the loop.
V = A( dB/dt )
we substitute
V = 0.015 m² × 0.20 T/s
V = 0.003 V
Therefore, the magnitude of the emf induced in the loop is 0.003 V
b) the induced emf is;
V = B( dA/dt ) + A( dB/dt )
given that; induced emf is 0, B = 1.5
so we substitute
0 = [ 1.5T × ( dA/dt ) ] + [ 0.015 m² × 0.20 T/s ]
-[ 1.5T × ( dA/dt )] = 0.003 m²T/s
dA/dt = -[ 0.003 m²T/s / 1.5T ]
dA/dt = -0.002 m²/s
the negative shows that the area is decreasing
hence, dA/dt = 0.002 m²/s
A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
PLEASE EXPLAIN HOW YOU GOT THE ANSWER THANK YOU SO MUCH
Answer:
0
Explanation:
The speed of the ball when it reaches the floor is 0 because when an object is at rest or in uniform motion, it has no speed/velocity
The final speed of the ball when it reaches the floor is 7.10 m/s.
What is the conservation of energy?The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but only converted from one form to another or transferred from one system to another. In other words, the total amount of energy in a closed system remains constant over time, even though it may be converted from one form to another.
This principle is based on the first law of thermodynamics, which states that the total energy of a closed system is always conserved, and can only be changed by the transfer of heat, work, or matter into or out of the system. The conservation of energy has important applications in various fields of physics, including mechanics, thermodynamics, and electromagnetism, and is a fundamental principle in the understanding of the natural world.
Here in the Question,
We can use the conservation of energy to solve this problem. Initially, the ball has kinetic energy due to its motion on the tabletop, but no potential energy since it is at a constant height. When the ball rolls off the edge of the table, it loses some kinetic energy due to friction but gains potential energy as it moves upward. When it reaches the floor, it has gained potential energy but lost kinetic energy due to friction. We can assume that the energy lost due to friction is converted to thermal energy, so the total energy of the system is conserved.
Let's start by calculating the potential energy gained by the ball as it moves from the edge of the table to the floor:
ΔPE = mgh
where ΔPE is the change in potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical distance traveled by the ball.
ΔPE = (0.50 kg)(9.81 m/s^2)(1.0 m) = 4.905 J
Now we can use the conservation of energy to find the final kinetic energy of the ball, which will allow us to calculate its final speed:
KEi + ΔPEi = KEf + ΔPEf
where KEi and ΔPEi are the initial kinetic and potential energies of the ball, respectively, and KEf and ΔPEf are the final kinetic and potential energies of the ball, respectively.
Since the ball is not bouncing, we can assume that its initial and final potential energies are zero. Therefore:
KEi = KEf + ΔKE
where ΔKE is the change in kinetic energy due to friction.
We can assume that the coefficient of kinetic friction between the ball and the incline is constant, and use the work-energy principle to find ΔKE:
Wfric = ΔKE
where Wfric is the work done by friction.
The work done by friction can be expressed as:
Wfric = ffricd
where ffric is the force of friction and d is the distance traveled by the ball on the incline.
The force of friction can be expressed as:
ffric = μmg
where μ is the coefficient of kinetic friction, and m and g have their usual meanings.
Putting it all together, we get:
KEi = KEf + ffricd
KEi = KEf + μmgd
(1/2)mv^2 = (1/2)mu^2 + μmgd
v^2 = u^2 + 2gd
where u is the initial speed of the ball on the tabletop, and v is the final speed of the ball on the floor.
Plugging in the given values, we get:
v^2 = (5.0 m/s)^2 + 2(9.81 m/s^2)(1.0 m)
v^2 = 50.405
v = 7.10 m/s
Therefore, the final speed of the ball when it reaches the floor is 7.10 m/s.
To learn more about the Law of Conservation of Momentum click:
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1. To avoid getting hurt, do not play with sharp simple machines. The highlighted statements best reminds us about ____.
A. Safety measures in handling machines
B. Gentle reminders when playing
C. Benefits of using machines
D. Disadvantages of simple machines
2. Road signs are example of advertisements that play important role in the promotion of road safety. If you are going to make an advertisement, which do you think is appropriate to make part of a road is temporarily closed for repair or renovation?
A. End of 4 lanes
B. No Parking
C. One lane, follow the traffic signal
D. Stop, look and listen
3. Wherever you go, you see mountains, plains and volcanoes. Which among the given ideas is being supported by the highlighted statement?
A. Earth's landmasses before and at present are the same.
B. Earth's land features are not charging
C. Internal forces such as plate tectonic movements do not result to various surface features
D. Internal and external forces contribute to the formation of various surface features of earth
4. Landslides, avalanche, tsunami, seiche and fire are some effects of strong earthquakes. If the huge inland wave produced during an earthquake is known as SEICHE (SAYSH) then what refers to the huge underwater wave produced when an earthquake occurs under the sea?
A. Avalanche
B. Landslide
C. Seiche
D. Tsunami
Answer:
1. A. Safety measures in handling machines
2. C. One lane, follow the traffic signal
3. D. Internal and external forces contribute to the formation of various surface features of the earth.
4. D. Tsunami
Explanation:
1. Safety measures are precautions that must be taken to avoid injuries when handling some potentially dangerous equipment. They are meant to protect us from harm.
2. If a part of the road is to be sealed up temporarily, road users should be alerted that there is just one lane ahead. Therefore, they should follow the traffic signal.
3. The pressure exerted on the earth through phenomena like plate tectonics, earthquakes, etc., causes the eruption and formation of different features on the earth.
4. A tsunami is a giant ocean wave caused by earthquakes and volcanic eruptions that occur under the sea. They occur with a lot of speed and cause an overflow of water onto land.
The water evaporated. How did that help ?
Answer:
hellooooo. hi hellooooo and programming laptops
Answer:
The molecules move and vibrate so quickly that they escape into the atmosphere as molecules of water vapor. Evaporation is a very important part of the water cycle. The Heat from the sun, or solar energy, powers the evaporation process. ... Once the water evaporates, it also helps form clouds.
Explanation:
1. Estimate the buoyant force that air exerts on a man. (To do this, you can estimate his volume by knowing his weight and by assuming that his weight density is about equal to that of water. Assume his weight is 940 N.) answer in N
2.On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total weight of the balloon, including its load and the hot air in it, is 17000 N.
(a) What is the weight of the displaced air?
answer in N
(b) What is the volume of the displaced air?
answer in m^3
Solution :
1. We know that : Buoyant force = weight of the liquid displace
= volume displaced x density of the fluid
Now volume of the man = [tex]$\frac{\text{mass}}{\text{density}}$[/tex]
Mass = weight / g
[tex]$=\frac{940}{9.8}$[/tex]
= 95.92 kg
And density = 1000 [tex]kg/m^3[/tex]
Therefore,
[tex]$\text{volume} = \frac{\text{mass}}{\text{density}}$[/tex]
[tex]$=\frac{95.92}{1000}$[/tex]
= 0.0959 [tex]m^3[/tex]
We know density of air = 1.225 [tex]kg/m^3[/tex]
∴ Mass of air displaced = 0.0959 x 1.225
= 0.1175 kg
Weight of the air displaced = 1.1515 N
Therefore, the buoyant force = 1.1515 N
2). As the balloon is not accelerated, the net force acting on it is zero.
Thus the weight that acts downwards = buoyant force upwards
So, the weight of the air displaced = weight of the balloon
= 17000 N
Therefore, the mass of the air displaced = volume of the air displaced (volume of the balloon) x density of air
[tex]$\frac{17000}{9.8} = \text{volume of air} \times 1.225$[/tex]
[tex]$\text{Volume of air displaced} = \frac{1700}{9.8 \times 1.225}$[/tex]
= 1416.0766 [tex]m^3[/tex]
There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12 V until it is loaded completely. Then the capacitor is disconnected from the voltage source. After this the two plates of the capacitor are driven apart until their distance is 5 mm. Now a positive test charge of 1 nC is brought from the positively charged plate to the negatively charges plate. How large is the kinetic energy of the test charge? The test charge of 1 nC can be regarded to be so small that it does not influence the electric field between the two plates of the capacitor.
Answer:
K = 2 10⁻⁸ J
Explanation:
Let's solve this exercise in parts, we start by finding the charge on each plate of the capacitor
C = Q / ΔV
C = ε₀ A / d
ε₀ A / d = Q / ΔV
Q = ε₀ A ΔV / d (1)
indicate the potential difference ΔV₁ = 12 V, the distance between the plates d₁ = 3 mm = 0.003 m,
as the power supply is disconnected and the capacitor is ideal the charge remains constant
in the second part we separate the plates at d₂ = 5 mm = 0.005 m, using equation 1
ΔV₂ = [tex]\frac{Q d_2}{ \epsilon_o A}[/tex]
we substitute the equation for Q
ΔV₂ = [tex]\frac{d_2}{\epsilon_o A} \ \frac{\epsilon_o A \Delta V }{d_1}[/tex]
ΔV₂ = [tex]\frac{d_2}{d_1} \ \Delta V_1[/tex]
in the third part we use the concepts of energy
starting point. Test charge near positive plate
Em₀ = U = q ΔV₂
final point. Test charge near negative plate
Em_f = K
energy is conserved
Em₀ = Em_f
q ΔV₂ = K
K = q ΔV₁ [tex]\frac{d_2}{d_1}[/tex]
we calculate
K = 1 10⁻⁹ 12 0.005/0.003
K = 2 10⁻⁸ J
basic source of magnetism is a) charged particles alone b)Movement of charged particles c) Magnetic dipoles d)magnetic domains
Answer:
C . Magnetic dipoles is the correct
Answer:
b). movement of charged particles.
Explanation:
These charges create the nagnetic dipoles.
15 . A scientist who studies the whole environment as a working unit .
Botanist
Chemist
Ecologist
Entomologist
Answer:
Ecologist.
Your answer is Ecologist.
(Ecologist) is a scientist who studies the whole environment as a working unit.
A 5.85-mm-high firefly sits on the axis of, and 13.7 cm in front of, the thin lens A, whose focal length is 5.01 cm. Behind lens A there is another thin lens, lens B, with focal length 25.9 cm. The two lenses share a common axis and are 62.5 cm apart. 1. Is the image of the firefly that lens B forms real or virtual?
a. Real
b. Vrtual
2. How far from lens B is this image located (expressed as a positive number)?
3. What is the height of this image (as a positive number)?
4. Is this image upright or inverted with respect to the firefly?
a. Upright
b. Inverted
Answer:
1. The image is real
2. 5.85
3. h' = 3.05 mm
4. The image is upright
Explanation:
1. Start with the first lens and apply 1/f = 1/p + 1/q
1/5.01 = 1/13.7 + 1/q
q = 7.90 cm
Since that distance is behind the first lens, and the second lens is 62.5 cm behind the first lens, that distance is 62.5 - 7.90 = 54.6 cm in front of the second lens, and becomes the object for that lens, thus,
1/25.9 = 1/54.6 + 1/q
q = 49.3 cm behind the second lens
Using that information, since q is positive, the image is real
2. Also, using that information, you have the second answer, which is 49.3 cm
The height can be found from the two magnifications.
m = -q/p
m1 = -7.9/13.7 = -.577
m2 = -49.3/54.6 = -.903
Net m = (-.577)(-.903) = .521
Then, m = h'/h
.521 = h'/5.85
3. h' = 3.05 mm
4. For the fourth answer, since the overall magnification is positive, the final image is upright
Calculate the electric field at point A, located at coordinates (0 m, 12.0 m ). Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures.
Answer:
The correct answer is "[tex](0,300\times 10^{-3} \ N/C)[/tex]".
Explanation:
The given problem seems to be incomplete. Please find the attachment of the complete query.
According to the question,
At point A, we have
⇒ [tex]E_x = \frac{k q_1}{d_1^2} Cos \theta_1 - \frac{k q_2}{d_2^2} Cos \theta_2[/tex]
or,
⇒ [tex]E_x = 9\times 10^9\times [\frac{6\times 10^{-9}}{15^2}\times \frac{9}{15}-\frac{8\times 10^{-9}}{20^2}\times \frac{16}{20} ][/tex]
[tex]=0[/tex]
and,
⇒ [tex]E_y = \frac{kq_1}{d_1^2}Sin \theta_1 +\frac{kq_2}{d_2^2}Sin \theta_2[/tex]
or,
⇒ [tex]E_y = 9\times 10^9\times [\frac{6\times 10^{-9}\times 12}{15^2\times 15}+ \frac{8\times 10^{-9}\times 12}{20^2\times 20} ][/tex]
[tex]=0.3 \ N/C[/tex]
A green object will absorb ____________________ light and reflect ____________________ light. (ref: p.447-455)
Answer:
A green object will absorb all light except for green light and reflect blue and yellow light.
g A thin uniform film of oil that can be varied in thickness covers a sheet of glass of refractive index 1.52. The refractive index of the oil is 1.64. Light of wavelength 555 nm is shone from air at normal incidence on the film. Starting with no oil on the glass, you gradually increase the thickness of the oil film until the first interference maximum in the reflected light occurs. What is the thickness of the oil film at that instant
Answer:
The right solution is "84.09 nm".
Explanation:
The given values are:
Refractive index of glass,
= 1.52
Refractive index of oil (n),
= 1.64
Wavelength (λ),
= 555 nm
Now,
The thickness of the film (t) will be:
= [tex]\frac{\lambda}{4n}[/tex]
= [tex]\frac{555}{4\times 1.65}[/tex]
= [tex]\frac{555}{6.6}[/tex]
= [tex]84.09 \ nm[/tex]
What is the energy of a photon with a frequency of 3.6 × 1015 Hz? Planck’s constant is 6.63 × 10–34 J•s.
1.8 × 10–49 J
2.4 × 10–19 J
1.8 × 10–18 J
2.4 × 10–18 J
We know
[tex]\boxed{\sf E=hv}[/tex]
[tex]\\ \sf\longmapsto E=6.63\times 10^{-34}J\times 3.6\times 10^{15}s^{-1}[/tex]
[tex]\\ \sf\longmapsto E=23.86\times 10^{-19}J[/tex]
[tex]\\ \sf\longmapsto E=2.38\times 10^{-18}J[/tex]
[tex]\\ \sf\longmapsto E=2.4\times 10^{-18}J[/tex]
Answer:
D!!!!!
Explanation:
difference between wavefront and wavelets
Answer:
A wavefront is the locus of all the particles which are in phase. A wavelet is an oscilation that starts from zero, then the amplitude increases and later decreases to zero
you happen to visit the moon when some people on earth see a total solar eclipse. who has a better experience of this event, you or the friends you left behind back on earth
Your friend would have a better experience of this event, than you .
What is an eclipse?An eclipse is produced when a planetary body moves in front of another planetary body and is visible from a third planetary body. Considering the sun, moon, and earth's locations in relation to one another during the time of the eclipse,
there are various types of eclipses in our solar system. For instance, a lunar eclipse occurs when the earth passes between the moon and the sun.
For the solar eclipse to happen the light from the sun is obstructed by the moon observing from the earth.
The buddies left Earth because they could view the whole eclipse, but you were on the moon and only saw parts of the eclipse turn black.
To learn more about the eclipse from here, refer to the link;
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Particle A has less mass than particle B. Both are pushed forward across a frictionless surface by equal forces for 1 s. Both start from rest. Which is true? A. A has more momentum. B. B has more momentum. C. A and B have the same momentum D. Not enough information.
Answer:
Both will have the same momentum.
P = M v momentum
v = a t for uniform acceleration
P = M a t
But a = F / M
P = M (F / M) t = F t so both have the same momentum
Light takes 1.2 sec to get from the moon to the Earth. Assume you are looking at the moon with noticeable earth shine. If the Sun burned out, you would eventually see the crescent of the moon disappear. The earth shine part of the moon would disappear Answer 2.4 s after the crescent disappeared.
Answer:
1.2 seconds
Explanation:
Answer to the following question is 1.2 seconds
Because light from the moon takes 1.2 seconds to reach Earth, the light released from the crescent immediately before it vanishes will also take 1.2 seconds to reach Earth. As a result, the earth-shine portion of the moon will vanish 1.2 seconds after the crescent has vanished.
1. A sequence of potential differences v is applied accross a wire (diameter =0.32 mm length = 11 cm and the resulting current I are measured as follows: V 0.1 0.2 0.3 0.4 0.5 I (MA) 72 144 216 288 360 2) a) plot a graph of v against I.
b) determine the wire's resistence , R.
c) State ohm's law and try to relate it . your results.
Answer:
a. Find the graph in the attachment
b. 720 kΩ
c. The ratio V/I gives us our resistance which is 720 kΩ
Explanation:
a) plot a graph of V against I.
To plot the graph of V against I, we plot the corresponding points against each other. With the voltage V measured in volts and the current I measured in mA, the plotted graph is in the attachment.
b) Determine the wire's resistance , R.
The resistance of the wire is determined as the gradient of the graph.
R = ΔV/ΔI = (V₂ - V₁)/(I₂ - I₁)
Taking the first two corresponding measurements. V₁ = 72 V, I₁ = 0.1 mA, V₂ = 144 V and I₂ = 0.2 mA
R = (144 V - 72 V)/(0.2 - 0.1) mA
R = 72 V/0.1 mA
R = 72 V/(0.1 × 10⁻³ A)
R = 720 × 10³ V/A
R = 720 kΩ
c) State ohm's law and try to relate it your results.
Ohm's law states that the current flowing through a conductor is directly proportional to the voltage across it provided the temperature and all other physical conditions remain constant.
Mathematically, V ∝ I
V = kI
V/I = k = R
Since the ratio V/I = constant, from our results, the ratio of V/I for each reading gives us the resistance. Since we have a linear relationship between V and I, the gradient of the graph is constant and for each value of V and I, the ratio V/I is constant. So, the ratio V/I gives us our resistance which is 720 kΩ.
Since V/I is constant, we thus verify Ohm's law.
the lamp cord is 85cm long and comprises cupper wire. Calculate the wire‘s resistance?
radius of a wire is 1.8mmm,Use value of resistivity for Cu as 1.75 × 10-8Ωm.
Answer:
R = 0.0015Ω
Explanation:
The formula for calculating the resistivity of a material is expressed as;
ρ = RA/l
R is the resistance
ρ is the resistivity
A is the area of the wire
l is the length of the wire
Given
l = 85cm = 0.85m
A = πr²
A = 3.14*0.0018²
A = 0.0000101736m²
ρ = 1.75 × 10-8Ωm.
Substitute into the formula
1.75 × 10-8 = 0.0000101736R/0.85
1.4875× 10-8 = 0.0000101736R
R = 1.4875× 10-8/0.0000101736
R = 0.0015Ω
A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced when a 60-kg person sits on one end and a 78-kg person sits on the other end.
Required:
Find a displacement of the center of mass of the system relatively to the seesaw's midpoint.
Answer:
x = 0.9 m
Explanation:
For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive
∑ τ = 0
60 1.5 - 78 1.5 + 30 x = 0
where x is measured from the left side of the fulcrum
90 - 117 + 30 x = 0
x = 27/30
x = 0.9 m
In summary the center of mass is on the side of the lightest weight x = 0.9 m
HELP FAST PLS asappppp
Answer:
(A) Series and Parallel
Explanation:
Circuit Component: These are electrical devices that makes up the circuit. They include, resistors, capacitors, inductors, voltmeters, ammeters, cell/batteries, earth connection, bulb, switch, connecting wire etc.
These component can either be connected in series or in parallel.
(1) Series Connection: This can be refered as end to end connection of electric component.
(2) Paralel Connection: This can be refered as the side to side connection of electric component.
From the question above,
A electric component in a circuit can be combined in series and in parallel.
The right option is (A) Series and Parallel
The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.18 Hz , and the acceleration of the top of the building can reach 1.9 % of the free-fall acceleration, enough to cause discomfort for occupants.
Required:
What is the total distance, side to side, that the top of the building moves during such an oscillation?
Answer:
The total distance, side to side, that the top of the building moves during such an oscillation is approximately 0.291 meters.
Explanation:
Let suppose that the building is experimenting a Simple Harmonic Motion due to the action of wind. First, we determine the angular frequency of the system ([tex]\omega[/tex]), in radians per second:
[tex]\omega = 2\pi\cdot f[/tex] (1)
Where [tex]f[/tex] is the frequency, in hertz.
If we know that [tex]f = 0.18\,hz[/tex], then the angular frequency of the system is:
[tex]\omega = 2\pi\cdot (0.18\,hz)[/tex]
[tex]\omega \approx 1.131\,\frac{rad}{s}[/tex]
The maximum acceleration experimented by the system is represented by the following formula, of which we estimate amplitude of the oscillation:
[tex]r\cdot g = \omega^{2}\cdot A[/tex] (2)
Where:
[tex]r[/tex] - Ratio of real acceleration to free-fall acceleration, no unit.
[tex]g[/tex] - Free-fall acceleration, in meters per square second.
[tex]A[/tex] - Amplitude, in meters.
If we know that [tex]\omega \approx 1.131\,\frac{rad}{s}[/tex], [tex]r = 0.019[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the amplitude of the oscillation is:
[tex]A = \frac{r\cdot g}{\omega^{2}}[/tex]
[tex]A = \frac{(0.019)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{\left(1.131\,\frac{rad}{s} \right)^{2}}[/tex]
[tex]A \approx 0.146\,m[/tex]
The total distance, side to side, is twice the amplitude, that is to say, a value of approximately 0.291 meters.
Find out other examples of bodies showing more than one type of motion Tabulate your findings.
Answer:
down below
Explanation:
Image 1- wheels of train showing both translatory motion as well as rotatory motion.
Image 2- rotation of ball shows both rotatory motion as well as translatory motion.
Image 3- the earth rotates about its axis, same time it revolves around the sun thus showing both rotatory motion and curvilinear motion in a fixed time. (perodic motion)
Image 4- while cutting wood, the
carpenter's saw has both
translatory motion and oscillatory
motion, as it moves down while
oscillating.
Which nucleus completes the following equation?
39 17 CI-> 0 -1 e+?
Answer:
[tex]_{18}^{39} } Ar[/tex]
Explanation:
The given equation shows the disintegration of an unstable isotope of chlorine to beta particle and Argon nucleus. The nucleus undergoes the emission of a beta particle to form a more stable nucleus of Argon.
[tex]_{17} ^{39} Cl[/tex] ⇒ [tex]_{-1}^{0} e[/tex] + [tex]_{18}^{39} } Ar[/tex]
Argon is a stable gas and is found in the group 8 on the periodic table of elements.
Answer:
Answer is below
Explanation:
39 18 Ar
A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring the car to a complete stop in 25 meters. If the weight of the car is 4,729 newtons, how fast was the car going initially
Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s
An object moving with initial velocity 10 m/s is subjected to a uniform acceleration of 8 m/s ^² . The displacement in the next 2 s is: (a) 0m (b) 36 m (c) 16 m (d) 4 m
A wire long and with mass is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude . What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards
The question is incomplete. The complete question is :
A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.
Solution :
Given :
Length of the wire, L = 0.6 m
Mass of the wire length, m = 11 g
= [tex]11 \times 10^{-3}[/tex] kg
Magnetic field , B = 0.4 T
Know we know that :
ILB = mg
or [tex]$I=\frac{mg}{BL}$[/tex]
[tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]
[tex]I=0.44963\ A[/tex]
[tex]I = 449.63 \ mA[/tex]
A mass weighing 4 lb stretches a spring 4in. Suppose the mass is given an additional in displacement downwards and then released. Assuming no friction and no external force, the natural frequency W (measured in radians per unit time) for the system is? (Recall that the acceleration due to gravity is 32ft/sec2).
a) None of the other alternatives is correct.
b) W = v2 3
c)w=212
d) w = 4/6
e) w=213
Answer:
4√6 rad/s
Explanation:
Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.
So, W = F
mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft
making k the spring constant subject of the formula, we have
k = mg/x
substituting the values of the variables into the equation, we have
k = mg/x
k = 4 lb × 32 ft/s² ÷ 1/3 ft
k = 32 × 4 × 3
k = 384 lbft²/s²
Now, assuming there is no friction and no external force, we have an undamped system.
So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb
So, substituting the values of the variables into the equation, we have
ω = √(k/m)
ω = √(384 lbft²/s² ÷ 4 lb)
ω = √96
ω = √(16 × 6)
ω = √16 × √6
ω = 4√6 rad/s
A bus starts from rest and accelerates at 1.5m/s squared until it reaches a velocity of 9m/s .the bus continues at this velocity and then deccelerate at -2m/s squared until it comes to stop 400m from it's starting point. how much time did the bus takes to cover the 400m?
Answer:
23s
Explanation:
s=ut+1/2at^2
the distance (s) is 400, initial velocity (u) is 0, acceleration (a) is 1.5 therefore
400=0t+1/2(1.5)t^2
400/0.75=0.75t^2/0.75
t^2=√533.33
t=23s
I hope this helps and sorry if it's wrong
