Answer:
A torque of 102.5375 Nm must be exerted by the fireman
Explanation:
Given:
The rate of water flow = 6.31 kg/s
The speed of nozzle = 12.5 m/s
Now, from the Newton's second law we have
The reaction force to water being redirected horizontally (F) = rate of change of water's momentum in the horizontal direction
thus we have,
F = 6.31 kg/s x 12.5m/s
or
F = 78.875 N
Now,
The torque (T) exerted by water force about the fireman's will be
T = (F x d)
or
T = 78.875 N x 1.30 m
T = 102.5375 Nm
hence,
A torque of 102.5375 Nm must be exerted by the fireman
The torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.
What is Newtons second law of motion?Newtons second law of motion shows the relation between the force mass and acceleration of a body. It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.
It can be given as,
[tex]F=ma[/tex]
Here, (m) is the mass of the body and (a) is the acceleration.
For the flow of water, the second law of motion can be given as,
[tex]F=Q\times v[/tex]
As he rate of water flow is 6.31 kg/s, and the nozzle speed is 12.5 m/s. Thus the force of this can be given as,
[tex]F=6.31\times12.5\\F=78.875\rm N[/tex]
The torque for the fireman exert on the hose is equal to the product of force applied and the distance traveled. Therefore the value of torque is,
[tex]\tau=78.875\times1.30\\\tau=102.5375\rm Nm[/tex]
Thus, the torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.
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Which of the following objects has the greatest kinetic energy?A. a car traveling at 35 miles per hour B. a horse sprinting at 35 miles per hour C. a baseball flying at 35 miles per hour D. an insect flying at 35 miles per hour E. All four objects have the same kinetic energy.
Answer: The car if it weighs more than the horse
Explanation: Kinetic energy is proportional to the object's mass and velocity, so higher the mass and velocity the more kinetic energy it will have. In this case as all the objects have the same velocity we have to consider their masses and thus, the car assumingly has the greatest kinetic energy
Answer:
A car traveling at 35 miles per hour
Explanation:
Kinetic energy is the energy possessed by an object having motion. The kinetic energy depends on the mass and speed of the object:
K.E. = 0.5 mv²
In the given options, all the things have same speed. So, the kinetic energy can be compared on the basis of the mass. The car is the heaviest among the given four options. Thus, the car traveling at the 35 miles per hour has the maximum kinetic energy.
A series RC circuit, which is made from a battery, a switch, a resistor, and a 3.0 µF capacitor, has a time constant of 10.0 ms. If an additional 8.0 µF is added in series to the 3.0 µF capacitor, what is the resulting time constant?
Answer:
New time constant is 7.27 ms
Explanation:
Time constant of RC circuit is given as
[tex]\tau = RC[/tex]
now we know that time constant is given as 10 ms for capacitor of 3 micro farad
then it is given as
[tex]10\times 10^{-3} s = (3\mu F)(R)[/tex]
[tex]R = 3333.33 ohm[/tex]
now the capacitor is connected with 8 micro farad capacitor in series then we have
[tex]C = \frac{(3\mu F)(8\mu F)}{3\mu F + 8\mu F}[/tex]
[tex]C = 2.18 \mu F[/tex]
now the new time constant is given as
[tex]\tau' = (2.18\mu F)(3333.3)[/tex]
[tex]\tau' = 7.27 ms[/tex]
The resulting time constant after adding an additional 8.0 µF capacitor in series to the original 3.0 µF capacitor is approximately 7.26 ms.
The time constant of an RC circuit is determined by the product of resistance (R) and capacitance (C), denoted by the equation T = RC. When capacitors are connected in series, the total capacitance (Ctotal) is decreased compared to individual capacitors. The formula for capacitors in series is 1/Ctotal = 1/C1 + 1/C2 + ... + 1/Cn. For the given initial time constant of 10.0 ms and an original capacitor of 3.0 µF, we can determine the original resistance (R). When an additional 8.0 µF capacitor is added in series, the new time constant can be calculated by finding the new total capacitance and then applying the RC formula.
Calculating the initial resistance:
R = T / Coriginal = 10.0 ms / 3.0 µF = 3.33 kΩ (approx.)
Calculating the total capacitance with the additional capacitor:
1/Ctotal = 1/3.0 µF + 1/8.0 µF = 0.333 µF-1 + 0.125 µF-1 = 0.458 µF-1
Ctotal = 1 / 0.458 µF-1 ≈ 2.18 µF
Now, calculate the new time constant:
Tnew = R * Ctotal = 3.33 kΩ * 2.18 µF ≈ 7.26 ms (approx.)
if we want to describe work, we must have
Energy and time since,
[tex]W=\dfrac{E}{t}[/tex]
Hope this helps.
r3t40
At what distance r from a point charge is the electric potential due to that point charge 0 v?
Answer:
Infinite Distance
Explanation:
The electric potential due to a point charge can be expressed by the following equation:
[tex]V=\frac{kQ}{r}[/tex]
Here,
V is the electric potential due to the point charge
k is the proportionality constant
Q is the magnitude of the point charge
r is the distance from the charge
As the value of r increases, the value of V decreases since there is an inverse relation between the two. The value of V can be absolutely 0 when the distance from the charge is infinite i.e. r is infinite. Mathematically, dividing a number by infinity results in zero. Also theoretically speaking, at infinite distance the electric field lines won't approach and hence the electric potential would be zero.
The electric potential due to a point charge is 0 V at a distance r from the charge when the charge is infinite.
Explanation:The electric potential due to a point charge is 0 V at a distance r from the charge when the charge is infinite. In other words, if the point charge is far away from the observation point, the electric potential becomes zero. This is because the electric potential decreases with distance.
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Which ion channels mediate the falling phase of an action potontial?
Answer:
Voltage-gated K+ channels
The correct answer is that potassium ion channels mediate the falling phase of an action potential.
During the falling phase of an action potential, the membrane potential of the neuron must return to its resting state after being depolarized. This is primarily achieved through the efflux of potassium ions (K^+) out of the cell. The opening of voltage-gated potassium channels allows for this efflux, which repolarizes the membrane potential back towards the resting membrane potential.
Here is the sequence of events during an action potential:
1. Resting potential: The neuron is at its resting membrane potential ,typically around -70 mV, due to the concentration gradients of ions across the membrane and the selective permeability of the membrane to potassium ions via leak channels.
2. Rising phase (Depolarization): When a stimulus reaches the threshold level, voltage-gated sodium channels open, allowing sodium ions (Na^+) to rush into the cell. This influx of positive charges depolarizes the membrane potential towards +30 mV.
3. Peak of the action potential: The membrane potential reaches its peak when the sodium channels become inactivated, stopping the influx of sodium ions.
4. Falling phase (Repolarization): Voltage-gated potassium channels open, and potassium ions move out of the cell, restoring the membrane potential towards the resting potential. The movement of potassium ions out of the cell is slower than the initial sodium influx, which is why the falling phase is slower than the rising phase.
5. Overshoot: Sometimes, the membrane potential temporarily overshoots the resting potential, becoming more negative than at rest, due to the continued efflux of potassium ion.
6. Return to resting potential: The potassium channels close, and the sodium-potassium pump restores the original ion distribution across the membrane, bringing the membrane potential back to the resting potential.
In summary, the falling phase of an action potential is mediated by the opening of voltage-gated potassium channels, which allows for the efflux of potassium ions, thereby repolarizing the neuron's membrane potential."
A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizontal. What is the horizontal force exerted by the ground on the ladder when an 81 kg object is 4.0 m from the bottom?
Answer:
266.5 newton
Explanation:
Weight of first object = 51 kg * 9.81
= 500 N
Weight of second object = 81 kg *9.81
= 794.61 N
As length of ladder is 15 m, the center of mass will be at 7.5 m.
Level arm is always perpendicular to the force. So the level of arm is at
7.5 sin θ.
τ ( wall) = τ (first object ) + τ ( second object)
N cos 30 ×15 = 794.61 sin 30 × 4 + ( 500 sin 30 ×7.5 )
13 N = 1589.22 + 1875
13 N = 3464.22
N= 266.5 newton
Answer:
Explanation:
Given that, .
Mass of ladder is 51kg
Then, it weight is
WL = mg = 51 × 9.81 = 500.31N
This weight will act at the midpoint of the ladder
Length of ladder is 15m
The ladder makes an angle 55°C with the horizontal
An object whose mass is 81kg is at 4m from the bottom of the ladder
Then, weight of object
Wo = mg = 81 × 9.81 = 794.61 N
Using newton second law
Check attachment
Ng is normal force on the ground
Ff is the horizontal frictional force
Nw Is the normal force on the wall
ΣFy = 0
Ng = Wo + WL
Ng = 794.61 + 500.31
Ng = 1294.92 N
Also
ΣFx = 0
Ff — Nw = 0
Then,
Ff = Nw
Now taking moment about point A.
Check attachment
using the principle of equilibrium
Sum of clockwise moment equals to sum of anti-clockwise moment
Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object
Clockwise = Anticlockwise
Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15
794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15
1589.22 + 1876.163 = 12.99•Nw
3465.383 = 12.99•Nw
Nw = 3465.383 / 12.99
Nw = 266.77 N
Since, Nw = Ff
Then, Ff = 266.77N
the horizontal force exerted by the ground on the ladder is 266.77 N
The pressure at the bottom of a glass filled with water (r 5 1 000 kg/m3 ) is P. The water is poured out and the glass is filled with ethyl alcohol (r 5 806 kg/m3 ). The pressure at the bottom of the glass is now (a) smaller than P (b) equal to P (c) larger than P (d) indeterminate.
Answer:
Option (a) is the correct answer
Explanation:
Option (a) smaller than P is the correct answer for the given question. This is because the pressure (P) at any depth 'd' from the top surface of the water with density (ρ) is given as:
P = ρ × g × d
where,
g is the acceleration due to the gravity
thus from the above equation it can be concluded that the pressure at any depth 'd' is directly proportional to the density of the water.
thus,
in the given case the density of the water is lower than that in the first case.
Hence, the option (a) is the correct answer
Pressure at the bottom of a container filled with liquid depends on the density of the liquid, among other factors. When water is replaced with ethyl alcohol, which has a lower density, the resulting pressure at the bottom decreases. Therefore, the correct answer is (a) smaller than P.
Explanation:The pressure at the bottom of a container filled with liquid is determined by the formula P = hρg, where P is the pressure, h is the height of the liquid, ρ is the density of the liquid, and g is the acceleration due to gravity. When the water (density 1000 kg/m³) is replaced with ethyl alcohol (density 806 kg/m³), the height of the liquid and gravity remain constant while the density decreases.
Since ρ in the term hρg has decreased, the overall product, i.e, the pressure at the bottom (P), also decreases. Therefore, when the glass is filled with ethyl alcohol, the pressure at the bottom of the glass is smaller than before. So, the correct answer is (a) smaller than P.
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Looking down on a Northern Hemisphere extratropical cyclone, surface winds blow ________ about the center. a.counterclockwise and inward b.clockwise and outward
In a Northern Hemisphere extratropical cyclone, the surface winds blow counterclockwise and inward due to the Coriolis effect, a result of the Earth's rotation.
Explanation:Looking down on a Northern Hemisphere extratropical cyclone, surface winds blow counterclockwise and inward about the center. This is due to the Coriolis effect, a phenomenon caused by the Earth's rotation which influences the direction that wind travels across the globe. It results in winds in the Northern Hemisphere curving to the right, thus causing cyclone winds to move counterclockwise.
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In a Northern Hemisphere extratropical cyclone, the surface winds blow in a counterclockwise and inward direction. This pattern is influenced by the Coriolis force, which causes winds to be deflected to the right, leading to this counterclockwise rotation. The phenomenon also aligns with the fact that air is attracted towards low-pressure centers like cyclones.
Explanation:In the Northern Hemisphere, when viewed from above, surface winds around an extratropical cyclone flow in a counterclockwise and inward direction. This phenomenon is influenced by the Coriolis force, which causes winds to be deflected to the right in the Northern Hemisphere. This deflection results in a counterclockwise rotation. The Coriolis force similarly influences tropical cyclones, causing them to display the same pattern of rotation.
A key point to note is that this rotation is associated with areas of low pressure, such as the center of these cyclone systems. This low-pressure center attracts air, causing winds to flow inward. As the air rises within these low-pressure zones, it cools and forms clouds, making such cyclonic weather patterns visible from space.
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A barr body is typically found in the nucleus of
Answer:
Neutrophils
Explanation:
When a balloon is deflating, why does air leave the balloon?
Answer:
When a balloon deflates air moves out of the balloon because the pressure inside the balloon is higher than the pressure outside the balloon.
Explanation:
An inflated balloon has a high pressure region on its inside. Gases always move from a region of high pressure to a region of low pressure. When a balloon is inflated its membrane stretches making it even more porous.
The gas molecules inside the balloon easily diffuse out through this membrane. The diffusion rate may differ depending on the type of gas filled inside the balloon and the material of the balloon. For example helium balloon deflates faster than common air balloon.
This is because helium is a light element and can escape easier than gases like nitrogen and oxygen through the porous membrane of the balloon.
Final answer:
Air leaves a deflating balloon as the cooler air inside has lower pressure than the outside, causing the higher pressure external air to push its way in, equalizing the pressure until it's the same inside and outside the balloon.
Explanation:
When a balloon is deflating, air leaves the balloon due to the difference in pressure between the air inside the balloon and the air outside. The air inside the balloon is initially at a higher pressure than the surrounding air. According to the principles of thermal expansion, the air in the balloon cools and its pressure decreases. As the pressure inside becomes lower than the pressure outside, air naturally flows out to equalize the pressure.
The buoyancy that lifts a hot-air balloon is based on the air's density inside being lower than outside. When a hot air balloon is inflating, the air inside becomes less dense due to heating, and it floats. However, as the air cools or if there is a leak in the balloon's material, the air inside loses its buoyancy and starts to deflate, releasing air until pressure equilibrium is reached or until the material cannot hold the air any longer.
The pressure inside a latex balloon is nearly the same as the pressure outside. If you let a helium balloon go, and if, as it rises, it stays at a constant temperature, the volume of the balloon will:A. stay the sameB. IncreaseC. Decrease
Answer:
Option (B)
Explanation:
As the balloon is filled with helium, it rises up because the density of helium is less than the density of air. As it rises up the outside pressure means atmospheric pressure goes on decreasing and thus the inside pressure increases. It results teh volme of the balloon increases.
A pair of narrow slits that are 1.8 mm apart is illuminated by a monochromatic coherent light source. A fringe pattern is observed on a screen 4.8 m from the slits. If there are 5.0 bright fringes/cm on the screen, what is the wavelength of the monochromatic light?
Answer:
750 mm
Explanation:
Given:
d = 1.8 mm
R = 4.8 m
m = 5
y = 1
Using the equation
y = (mLR)/d ,
where,
m gives a distance 'y' to that particular slit image.
R = distance from the double slits to the screen
d = double slit separation distance.
L = wavelength of the light.
substituting the values in the given equation
we get
L = [tex]\frac{1\times 1.8\times 10^{-3}}{5\times 4.8}[/tex]
or
L = 750 mm
Answer:
The wavelength of the monochromatic light is [tex]7.5\times10^{-7}\ m[/tex]
Explanation:
Given that,
Distance between the slits d = 1.8 mm
Distance of fringe from the slits D =4.8 m
Number of fringe m =1
Distance between the fringes = 1 cm
We need to calculate the wavelength of monochromatic light
Using formula of young's double slits
[tex]\lambda=\dfrac{Yd}{mD}[/tex]
Where, d = Distance between the slits
D = Distance of fringe from the slits
m = Number of fringe
y = Distance between the fringes
Put the value in to the formula
[tex]\lambda=\dfrac{1\times10^{-2}\times1.8\times10^{-3}}{5\times4.8}[/tex]
[tex]\lambda =7.5\times10^{-7}\ m[/tex]
Hence, The wavelength of the monochromatic light is [tex]7.5\times10^{-7}\ m[/tex]
According to the ___________________ hypothesis, each emotion comes with its own specific profile of autonomic activity (heart rate, skin conductance, and finger temperature).
Answer:
autonomic specificity
Explanation:
The hypothesis which states that the different emotions in our body involve different and unique physiological profiles is called autonomic specificity hypothesis.
Viewed from this hypothesis perspective, emotions in the human body can be seen as time-tested solutions to timeless problems and challenges. With emotions, evolution in the human body has provided the human with at least one generalized response to tackle these problems.
Ultrasound is the name given to frequencies above the human range of hearing True or false
Yes. That's a true statement. Ultrasound is indeed the name given to sounds with frequencies above the human range of hearing.
Ultrasound is not different from "normal" (audible) sound in its physical properties, except that humans can't hear it.
Suppose the gravitational acceleration on a certain planet is only 4.0 m/s2. A space explorer standing on this planet throws a ball straight upward with an initial velocity of 24 m/s. Assume that upward direction is positive. What is the velocity of the ball 3 seconds after it is thrown?
Using the formula for velocity (v = u + at) with the given values, we find that the velocity of the ball 3 seconds after it's thrown upwards on a planet with a gravitational acceleration of 4.0 m/s2 is 12 m/s in the upward direction.
Explanation:The question asks about the velocity of a ball 3 seconds after it's thrown upwards on a planet with a gravitational acceleration of 4.0 m/s2. We can use the formula for velocity which is the initial velocity plus acceleration times time (v = u + at). In this case, the initial velocity (u) is 24 m/s, the acceleration (a) is -4.0 m/s2 (negative because it's working against the upwards motion of the ball), and the time (t) is 3 seconds.
Substituting these values into the formula gives us v = 24 m/s + (-4.0 m/s2 * 3 s) = 24 m/s - 12 m/s = 12 m/s. So the velocity of the ball 3 seconds after it's thrown is 12 m/s in the upwards direction.
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A freight car moves along a frictionless level railroad track at constant speed. The car is open on top. A large load of coal is suddenly dumped into the car. What happens to the velocity of the car?
Answer:
Velocity of the car decreases.
Explanation:
We can understand the situation if we apply the conservation of energy principle to the situation
Let the initial mass of the freight be [tex]m_{f}[/tex]
Initial velocity of the freight be [tex]v_{fi}[/tex]
Thus the initial Kinetic energy of the freight will be [tex]K.E=\frac{1}{2}m_{f}v_{if}^{2}[/tex]
When a Coal Block of mass M falls into the freight it's energy will become
[tex]K.E=\frac{1}{2}(m_{f}+M)v_{ff}^{2}[/tex]
Equating both the energies we get final velocity as[tex]v_{ff}[/tex]
[tex]\frac{1}{2}m_{f}v_{if}^{2}=\frac{1}{2}(M+m_{f})v_{ff}^{2}\\\\v_{ff}=\sqrt{\frac{m_f}{(M+m_{ff})}}\cdot v_{if}[/tex]
As we see that [tex]\sqrt{\frac{m_f}{(M+m_{ff})}}[/tex] is less than 1 we can infer that velocity decreases.
Final answer:
When coal is dumped into a moving freight car on a frictionless track, its velocity decreases due to the conservation of linear momentum, and the kinetic energy of the car decreases as a result.
Explanation:
If a large load of coal is suddenly dumped into an open freight car moving at a constant speed, the car's velocity will decrease due to the conservation of momentum. Since we are ignoring air resistance and assuming the track is frictionless, this scenario can be analyzed with the principle of conservation of linear momentum, which asserts that the total momentum of a closed system is constant if external forces are absent. In this case, the load of coal suddenly added to the freight car significantly increases the mass of the system without adding any horizontal momentum, as the coal is initially at rest relative to the horizontal motion of the car.
Considering the professional example provided, the initial momentum of the 30,000-kg freight car at a velocity of 0.850 m/s is m₁v₁. The final momentum, after 110,000 kg of scrap metal is added, must be the same as the initial momentum due to conservation of momentum. Thus, the final velocity v₂ can be found using the equation m₁v₁ = (m₁ + m₂)v₂, where m₂ is the mass of the scrap metal.
For the kinetic energy, initially, the car has a certain amount of kinetic energy due to its motion. After the additional scrap metal is dumped, the total mass of the car increases, and its velocity decreases, which corresponds to a decrease in kinetic energy. This lost kinetic energy can be computed by subtracting the final kinetic energy from the initial kinetic energy of the car before the coal was added.
A 26.0 kg wheel, essentially a thin hoop with radius 1.30 m, is rotating at 297 rev/min. It must be brought to a stop in 23.0 s. (a) How much work must be done to stop it? (b) What is the required average power? Give absolute values for both parts.
Answer:
Work done on the loop to stop it = 21,269.1496 J
Average power P = 924.7456 watt
Explanation:
Mass of the wheel M = 26.0 Kg
Radius r = 1.30 m
The wheel is rotating at a speed of 297 rev/min
1 minute = 60 seconds,
So,
The wheel is rotating at a speed of 297/60 rev/sec
Initial angular speed (ω₁) = 2π(297/60) = 31.1143 rad/s
Final angular speed (ω₂) = 0 rad/s
Time taken to stop , (t) = 23 s econds
Moment of inertia I of a circular hoop around its central axis = Mr²
Where m is the mass of the wheel and r is the radius of the wheel
Thus, I = 26.0×(1.30)² Kgm² = 43.94 Kgm²
(a)
Work done to stop it is the difference in the kinetic energy of the initial and the final system. So,
Work done W = (1/2)I(ω₂² - ω₁²) = 0.5*43.94(0 - (31.1143)²) = -21,269.1496 J
Thus, work done on the loop to stop it = - 21,269.1496 J
The answer has to be answered in absolute value so, Work = |-21,269.1496 J| = 21,269.1496 J
(b)
Average power P = |W|/t = 21,269.1496 J/23.0 s = 924.7456 watt
Bob can row 14 mph in still water. The total time to travel downstream and return upstream to the starting point is 4 hours. If the total distance downstream and back is 42 miles, determine the speed of the river (current speed).
Answer:
7 mph
Explanation:
Let the speed of river is c mph.
The effective downstream speed = 14 + c
The effective upstream speed = 14 - c
Total time = 4 hrs
Total distance = 42 miles
Time for upstream + time for downstream = 4 hrs
21 / (14 + c) + 21 / (14 - c) = 4
21 (14 - c + 14 + c) = 4 (196 - c^2)
21 x 28 = 4 (196 - c^2)
c^2 = 196 -147 = 49
c = 7 mph
The speed of the river is 7 miles/h.
Speed of Bob in still water is 14 miles/h.
Speed of Bob downstream = (14 + x) miles/h
Speed of Bob upstream = (14 - x) miles/h
here x = speed of the river.
The total distance downstream and back is 42 miles. Hence,
Distance upstream = Distance downstream = 21 miles
Given that total time of journey upstream and downstream is 4 hours. Then, using the formula:
[tex]t = \frac{distance}{speed}[/tex], we get:
[tex]\frac{21 \hspace{0.5mm} miles} {(14 + x) \hspace{0.5mm} miles/h} + \frac{21 \hspace{0.5mm} miles}{(14 - x)\hspace{0.5mm}miles/h} =4 \hspace{0.5mm} hours[/tex]
or, [tex]\frac{2 \times 21 \times 14 \hspace{0.5 mm} h}{14^{2} -x^{2} } = 4 \hspace{0.5mm} hours[/tex]
or, 21 × 7 h = 14² - x²
or, x² = 49
or, x = 7 miles/h
A neutron star and a white dwarf have been found orbiting each other with a period of 28 minutes. If their masses are typical, what is their average separation? Compare the separation with the radius of the sun, or about 0.005 AU. (Hints: Refer to Kepler's third law with regard to mass. Assume the mass of the neutron star is 2.5 solar masses and the mass of the white dwarf is 0.3 solar mass.)
Answer:
The average separation is 0.002041 AU
Explanation:
Given data:
Mass of the neutron star, M₁ = 2.5[tex]M_{solar}[/tex]
Mass of the White dwarf, M₂ = 0.3[tex]M_{solar}[/tex]
Orbiting period (P)= 28 minutes
1 year = 365 × 24 × 60 minutes = 525600 minutes
or
1 minute = 1/525600 years
thus, 28 minutes = 28/525600 years = 5.327 × 10⁻⁵ years
now from the Kepler's third law we have,
MP² = a³
where, P is the period
M is the mass = M₁ + M₂
a is the size of the orbit
thus, by substituting the values in the equation we get
(2.5[tex]M_{solar}[/tex]+0.3[tex]M_{solar}[/tex])(5.327 × 10⁻⁵ years)² = a³
Also,
[tex]M_{solar}=1[/tex] when planets orbiting sun
thus,
2.8 ×(5.327 × 10⁻⁵ years)² = a³
or
a³ = 7.94 × 10⁻⁹
or
a = 1.99 × 10⁻³ AU
thus, the average separation is 0.001995 AU
Now
1 AU = 1.5 × 10⁸ km
thus,
0.001995 AU = 299281.61 km = 2.99 × 10⁵ km
in terms of sun's radius = (2.99 × 10⁵ km)/(7 ×10⁵) = 0.427
Thus, the this orbit system will fit inside the sun
Help me with this problem
Vector E is 0.111 m long in a 90.0 direction.Vector F is 0.234 m long in a 300 direction. What is the magnitude and direction of their vector sum?
Answer:
0.149 m, [tex]321.8^{\circ}[/tex]
Explanation:
Let's start by resolving each vector into its components along the x- and y- direction:
[tex]E_x = E cos \theta = (0.111) cos 90^{\circ}=0\\E_y = E sin \theta = (0.111) sin 90^{\circ}=0.111 m[/tex]
And
[tex]F_x = F cos \theta = (0.234) cos 300^{\circ} = 0.117 m\\F_y = F sin \theta = (0.234) sin 300^{\circ} = -0.203 m[/tex]
So the components of the vector sum are
[tex]R_x = E_x + F_x = 0+0.117 = 0.117 m\\R_y = E_y + F_y = 0.111 -0.203 = -0.092 m[/tex]
The magnitude of the vector sum is
[tex]R=\sqrt{R_x^2 +R_y^2 }=\sqrt{(0.117)^2+(-0.092)^2}=0.149 m[/tex]
And the direction is
[tex]\theta=tan^{-1} (\frac{|R_y|}{R_x})=-tan^{-1} (\frac{0.092}{0.117})=-38.2^{\circ}=321.8^{\circ}[/tex]
In acceleration, if you were taking distance messurements of a free falling object at 1 second intervals, you see the object moving _________ at each measurement interval
Answer:
Farther
Explanation:
In acceleration, if you were taking distance measurements of a free falling object at 1 second intervals, you see the object moving farther at each measurement interval.
Consider a mass-spring system. The spring has a spring constant 2.17e 3 N/m. On the end is a mass of 4.71 kg. You start at equilibrium with an initial velocity of 1.78 m/s. What is the maximum displacement
Answer:
7.4 cm
Explanation:
K = 2.17 x 10^3 N/m
m = 4.71 kg
v = 1.78 m/s (It is maximum velocity)
The angular velocity
[tex]\omega =\sqrt{\frac{k}{m}}[/tex]
[tex]\omega =\sqrt{\frac{2.71\times 10^{3}}{4.71}}[/tex]
ω = 24 rad/s
Maximum velocity, v = ω x A
Where, A be the maximum displacement
1.78 = 24 x A
A = 0.074 m = 7.4 cm
Final answer:
The maximum displacement of the mass-spring system is 0.035 m.
Explanation:
If we have a mass-spring system with a spring constant of 2.17e3 N/m and a mass of 4.71 kg attached to it, we can calculate the maximum displacement using the formula for potential energy in a spring: U = (1/2)kx^2, where U is the potential energy, k is the spring constant, and x is the displacement from equilibrium. At maximum displacement, the potential energy is converted to kinetic energy, which is given by K = (1/2)mv^2, where m is the mass and v is the velocity. Equating the two equations and solving for x, we can find the maximum displacement.
First, let's find the equilibrium position by taking the weight of the mass as force due to gravity: F = mg, where m is the mass and g is the acceleration due to gravity. Then, we can calculate the maximum displacement using the formula x = (U/mg)^0.5.
Plugging in the given values, we have: x = ((1/2)(2.17e3 N/m)(x)^2) / (4.71 kg)(9.8 m/s^2), which gives us x = 0.035 m.
A 25 kg circular disk has a diameter of 2.5 feet and a thickness of 2.5 cm. Find the density of the disk in kg/m3. Next, find the weight of the object. Then calculate the buoyant force on the disk if it is submerged under water. Finally, will the object sink or float?
Answer:
Assume that [tex]\rm g= 9.81\; N\cdot kg^{-1}[/tex]; [tex]\rho(\text{Water}) = \rm 1000\;kg\cdot m^{-3}[/tex].
Density of the disk: approximately [tex]\rm 2.19\times 10^{3}\; kg\cdot m^{-3}[/tex].
Weight of the disk: approximately [tex]\rm 245\;N[/tex].
Buoyant force on the disk if it is submerged under water: approximately [tex]\rm 112\; N[/tex].
The disk will sink when placed in water.
Explanation:
Convert the dimensions of this disk to SI units:
Diameter: [tex]d = \rm 25\; inches = (25\times 0.3048)\; m = 0.762\;m[/tex].Thickness [tex]h = \rm 2.5\; cm = (2.5\times 0.01)\; m = 0.025\;m[/tex].The radius of a circle is 1/2 its diameter:
[tex]\displaystyle r = \rm \frac{1}{2}\times 0.762\;m = 0.381\; m[/tex].
Volume of this disk:
[tex]V(\text{disk}) = \pi\cdot r^{2}\cdot h = \pi\times 0.381^{2}\times 0.025 \approx 0.0114009\; m^{3}[/tex].
Density of this disk:
[tex]\displaystyle \rho(\text{disk}) = \frac{m}{V} = \rm \frac{25\; kg}{0.0114009\; m^{3}} = 2.19\times 10^{3}\;kg\cdot m^{-3}[/tex].
[tex]\rho(\text{disk}) >\rho(\text{water})[/tex] indicates that the disk will sink when placed in water.
Weight of the object:
[tex]W(\text{disk}) = m\cdot g = \rm 25\times 9.81 = 245.25\; N[/tex].
The buoyant force on an object in water is equal to the weight of water that this object displaces. When this disk is submerged under water, it will displace approximately [tex]\rm 0.0114009\; m^{3}[/tex] of water. The buoyant force on the disk will be:
[tex]\begin{aligned}F(\text{buoyant force}) &= W(\text{Water Displaced}) \\& = \rho\cdot V(\text{Water Displaced})\cdot g\\ & = \rm 1\times 10^{3}\; kg\cdot m^{-3}\times 0.0114009\; m^{3}\times 9.81\; N\cdot kg^{-1}\\ &\approx \rm 112\; N\end{aligned}[/tex].
The size of this disk's weight is greater than the size of the buoyant force on it when submerged under water. As a result, the disk will sink when placed in water.
What is the fastest time trial for the first quarter checkpoint? _____ seconds
What is the slowest time trial for the first quarter checkpoint? _____ seconds
What is the range of times measured for this checkpoint? _____ seconds
Answer:
What is the fastest time trial for the first quarter checkpoint? 2.02 seconds
What is the slowest time trial for the first quarter checkpoint? 2.15 seconds
What is the range of times measured for this checkpoint? 0.13 seconds
Answer:
2.02
2.15
0.13
Explanation:
i did it on edge
Which statement is true about the four atoms shown in figures A, B, C, and D? Four atoms. Figure A has 3 electrons, 3 protons, 3 neutrons. Figure B has 1 electron, 3 protons, 4 neutrons. Figure C has 4 electrons, 4 protons, 3 neutrons. Figure B has 4 electrons, 4 protons, 4 neutrons. B and D are different elements, while A is an isotope of D. A and C are different elements, while D is an isotope of C. C and D are different elements, while B is an isotope of C. A and B are different elements, while C is an isotope of B.
Answer:
A and C are different elements, while D is an isotope of C
Explanation:
To identify an atom, we simply use the atomic number. The atomic number is the number of protons in an atom.
For a neutral atom, the number of protons is the same as the number of electrons in such an atom.
Isotopes of an element have the same atomic number (protons) but different mass number(proton + neutron) as a result of their different number of neutrons.
From the given information about the atoms, only the second option is correct:
A and C are different elements because their atomic number differs:
A : 3 electrons, 3 protons and 3 neutrons
C : 4 electrons, 4 protons and 3 neutrons
D is an isotope of C:
C : 4 electrons, 4 protons and 3 neutrons
D : 4 electrons, 4 protons and 4 neutrons
They have the same atomic number but the mass number differs due to their different number of neutrons.
Car A
Mass: 1,500 kg
Velocity: 10 m/s
Car B
Mass: 1,500 kg
Velocity: 25 m/s
Car C
Mass: 1,000 kg
Velocity: 10 m/s
which order shows decreasing momentum?
~A, B, C
~B, A, C
~C, B, A
Answer:
B,A,C
Explanation:
Hope this helps!!!
Answer:
B, A, C
Explanation:
Momentum:
[tex]P = m*v[/tex]
[tex]P = momentum[/tex]
[tex]m = mass[/tex]
[tex]v = velocity[/tex]
Momentum of car A
[tex]P_{A} = 1500kg*10m/s = 15,000kg*m/s[/tex]
Momentum of car B
[tex]P_{B} = 1500kg*25m/s = 37,500kg*m/s[/tex]
Momentum of car C
[tex]P_{C} = 1000kg*10m/s = 10,000kg*m/s[/tex]
In decreasing order:
B, A, C
A 73 kg man in a 7.4 kg chair tilts back so that all his weight is balanced on two legs of the chair. Assume that each leg makes contact with the floor over a circular area of radius 1.5 cm. Find the pressure exerted on the floor by each leg. The acceleration of gravity is 9.8 m/s 2 . Answer in units of Pa.A 73 kg man in a 7.4 kg chair tilts back so that all his weight is balanced on two legs of the chair. Assume that each leg makes contact with the floor over a circular area of radius 1.5 cm. Find the pressure exerted on the floor by each leg. The acceleration of gravity is 9.8 m/s 2 . Answer in units of Pa.
Answer:
55738.539 Pa
Explanation:
Gvien:
The mass of the man = 73 kg
Mass of the chair = 7.4 kg
radius of the leg of the chair = 1.5 cm = 0.015 m
Now,
the force of gravity due to both the masses, F = (73+7.4) kg x 9.8 = 787.92 N
Also,
Pressure = force / area
Area of a circle of leg = πr² = π x 0.015² = 7.068 x10⁻⁴ m2
Now,
there are 2 legs so the force will be divided evenly in each leg
thus,
F' = [tex]\frac{787.92}{2}=393.96N[/tex]
Hence, the pressure on each leg will be
Pressure = [tex]\frac{393.96N}{7.068\times 10^{-4}}=55738.539 Pa[/tex]
If a particle with a charge of +3.3 × 10?18 C is attracted to another particle by a force of 2.5 × 10?8 N, what is the magnitude of the electric field at this location? 8.3 × 10^-26 NC 1.8 × 10^10 NC 1.3 × 10^-10 N/C 7.6 × 10^9 N/C
Answer:
[tex]7.6\cdot 10^9 N/C[/tex]
Explanation:
The relationship between force exerted on a charge and strength of the electric field is given by
[tex]F=qE[/tex]
where
F is the strength of the electric force
q is the charge of the particle
E is the magnitude of the electric field
For the particle in the problem, we have
[tex]q=3.3\cdot 10^{-18} C[/tex]
[tex]F=2.5\cdot 10^{-8} N[/tex]
So the magnitude of the electric field at the location of the particle is
[tex]E=\frac{F}{q}=\frac{2.5\cdot 10^{-8}}{3.3\cdot 10^{-18}}=7.6\cdot 10^9 N/C[/tex]
Final answer:
The magnitude of the electric field at the given location is [tex]7.6 × 10^9 N/C.[/tex]
Explanation:To calculate the magnitude of the electric field at a location, we can use the formula E = F/q, where E represents the electric field, F represents the force, and q represents the charge. In this case, we are given the charge of one particle as [tex]+3.3 × 10^-18 C[/tex] between the particles as [tex]2.5 × 10^-8 N[/tex]hese values into the formula, we get:
[tex]E = (2.5 × 10^-8 N) / (3.3 × 10^-18 C)[/tex]
Evaluating this expression, we find that the magnitude of the electric field at this location is approximately [tex]7.6 × 10^9 N/C.[/tex]
An electron with a speed of 1.9 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 24 mm horizontally.
Explanation:
It is given that,
Speed of the electron in horizontal region, [tex]v=1.9\times 10^7\ m/s[/tex]
Vertical force, [tex]F_y=4.9\times 10^{-16}\ N[/tex]
Vertical acceleration, [tex]a_y=\dfrac{F_y}{m}[/tex]
[tex]a_y=\dfrac{4.9\times 10^{-16}\ N}{9.11\times 10^{-31}\ kg}[/tex]
[tex]a_y=5.37\times 10^{14}\ m/s^2[/tex]..........(1)
Let t is the time taken by the electron, such that,
[tex]t=\dfrac{x}{v_x}[/tex]
[tex]t=\dfrac{0.024\ m}{1.9\times 10^7\ m/s}[/tex]
[tex]t=1.26\times 10^{-9}\ s[/tex]...........(2)
Let [tex]d_y[/tex] is the vertical distance deflected during this time. It can be calculated using second equation of motion:
[tex]d_y=ut+\dfrac{1}{2}a_yt^2[/tex]
u = 0
[tex]d_y=\dfrac{1}{2}\times 5.37\times 10^{14}\ m/s^2\times (1.26\times 10^{-9}\ s)^2[/tex]
[tex]d_y=0.000426\ m[/tex]
[tex]d_y=0.426\ mm[/tex]
So, the vertical distance the electron is deflected during the time is 0.426 mm. Hence, this is the required solution.
Answer:
[tex]y = 4.24 *10^{-4} m[/tex]
Explanation:
the vertical accerlaration acting on the electron
[tex]a = \frac{f_y}{m}[/tex]
[tex]a = \frac{4.9*10^{-16}}{9.11*10^{-31}} = 5.37 *10^{14} m/s^{2}[/tex]
the time taken by the electron to cover the horizontal distance is
[tex]t =\frac{24*10^{-3}}{1.9*10^{7}}[/tex]
[tex]t = 1.26 *10^{-9} s[/tex]
[tex]v_i = o[/tex]
the vertical distance trvalled in time t is
[tex]y = v_i*t +\frac{1}{2} a_y*t^{2}[/tex]
[tex]y = 0 +\frac{1}{2}*(5.37 *10^{14})(1.26 *10^{-9})^{2}[/tex]
[tex]y = 4.24 *10^{-4} m[/tex]
Two football players are pushing a 60kg blocking sled across the field at a constant speed of 2.0 m/s. The coefficient of kinetic friction between the grass and the sled is 0.30. Once they stop pushing, how far will the sled slide before coming to rest
Answer:
The sled slides 0.68m before rest.
Explanation:
Vi= 2 m/s
m= 60kg
μ= 0.3
N= m * g = 60kg * 9.8 m/s²= 588 N
Fr= μ * N
Fr= 176.4 N
a= Fr/m
a= -2.94m/s²
t= Vi/a
t= 0.68 seg
d= Vi*t - a*t²/2
d= 0.68m