The speed of the center of mass just before it hits the horizontal surface can be found using the principle of conservation of energy.
Explanation:To find the speed of the center of mass just before it hits the horizontal surface, we can use the principle of conservation of energy. When the rod falls freely, it gains gravitational potential energy which is converted into kinetic energy. At the lowest point, where the center of mass hits the horizontal surface, all the gravitational potential energy is converted into kinetic energy. Therefore, we can equate the gravitational potential energy at the top to the kinetic energy at the bottom:
mgh = (1/2)mv^2
Where m is the mass of the rod, g is the acceleration due to gravity, h is the height of the rod, and v is the speed of the center of mass. Solving for v:
v^2 = 2gh
v = sqrt(2gh)
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When a 0.350-kg package is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The package is now displaced from its equilibrium position and undergoes simple harmonic oscillations when released. What is the period of the oscillations?
To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.
We know that by Hooke's law
[tex]F=kx[/tex]
Where,
k = Spring constant
x = Displacement
Re-arrange to find k,
[tex]k= \frac{F}{x}[/tex]
[tex]k= \frac{mg}{x}[/tex]
[tex]k= \frac{(0.35)(9.8)}{12*10^{-2}}[/tex]
[tex]k = 28.58N/m[/tex]
Perioricity in an elastic body is defined by
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
Where,
m = Mass
k = Spring constant
[tex]T = 2\pi \sqrt{\frac{0.35}{28.58}}[/tex]
[tex]T = 0.685s[/tex]
Therefore the period of the oscillations is 0.685s
The period of the oscillations can be calculated using the formula T = 2π√(m/k), where m is the mass of the package and k is the spring constant. Plugging in the values mentioned in the question, the period can be determined.
Explanation:The period of the oscillations can be calculated using the formula T = 2π√(m/k), where T is the period, π is a constant (approximately 3.14), m is the mass of the package, and k is the spring constant.
In this case, the mass of the package is 0.350 kg and the spring stretches 12.0 cm (or 0.12 m). The spring constant can be determined using the equation k = mg/X, where g is the acceleration due to gravity (approximately 9.8 m/s2) and X is the displacement (0.12 m).
Plugging the values into the formula, we get T = 2π√(0.350/[(0.350 * 9.8)/0.12]). Solving this equation will give the period of the oscillations.
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An astronaut goes out for a space walk. Her mass (including space suit, oxygen tank, etc.) is 100 kg. Suddenly, disaster strikes and her tether line becomes disconnected, so she is stuck at rest a distance x away from the space craft!
Luckily, you have to know physics to be an astronaut. The astronaut takes off her 15 kg oxygen tank and throws it away from the spacecraft with a speed of 10 m/s. Due to conservation of momentum, she is propelled towards the spacecraft.
The astronaut has 1.5 minutes of oxygen remaining in her space suit to get her back to the craft. What is the maximum distance the astronaut can be away from the spacecraft to make it back before she runs out of oxygen?
Part A: What are the Known Variables and Unknown Variables? list them.
Part B: What are the equations needed to solve this problem?
Part C: Solve the problem.
Answer:
Part A:
Unknown variables:
velocity of the astronaut after throwing the tank.
maximum distance the astronaut can be away from the spacecraft to make it back before she runs out of oxygen.
Known variables:
velocity and mass of the tank.
mass of the astronaut after and before throwing the tank.
maximum time it can take the astronaut to return to the spacecraft.
Part B:
To obtain the velocity of the astronaut we use this equation:
-(momentum of the oxygen tank) = momentum of the astronaut
-mt · vt = ma · vt
Where:
mt = mass of the tank
vt = velocity of the tank
ma = mass of the astronaut
va = velocity of the astronaut
To obtain the maximum distance the astronaut can be away from the spacecraft we use this equation:
x = x0 + v · t
Where:
x = position of the astronaut at time t.
x0 = initial position.
v = velocity.
t = time.
Part C:
The maximum distance the astronaut can be away from the spacecraft is 162 m.
Explanation:
Hi there!
Due to conservation of momentum, the momentum of the oxygen tank when it is thrown away must be equal to the momentum of the astronaut but in opposite direction. In other words, the momentum of the system astronaut-oxygen tank is the same before and after throwing the tank.
The momentum of the system before throwing the tank is zero because the astronaut is at rest:
Initial momentum = m · v
Where m is the mass of the astronaut plus the equipment (100 kg) and v is its velocity (0 m/s).
Then:
initial momentum = 0
After throwing the tank, the momentum of the system is the sum of the momentums of the astronaut plus the momentum of the tank.
final momentum = mt · vt + ma · va
Where:
mt = mass of the tank
vt = velocity of the tank
ma = mass of the astronaut
va = velocity of the astronaut
Since the initial momentum is equal to final momentum:
initial momentum = final momentum
0 = mt · vt + ma · va
- mt · vt = ma · va
Now, we have proved that the momentum of the tank must be equal to the momentum of the astronaut but in opposite direction.
Solving that equation for the velocity of the astronaut (va):
- (mt · vt)/ma = va
mt = 15 kg
vt = 10 m/s
ma = 100 kg - 15 kg = 85 kg
-(15 kg · 10 m/s)/ 85 kg = -1.8 m/s
The velocity of the astronaut is 1.8 m/s in direction to the spacecraft.
Let´s place the origin of the frame of reference at the spacecraft. The equation of position for an object moving in a straight line at constant velocity is the following:
x = x0 + v · t
where:
x = position of the object at time t.
x0 = initial position.
v = velocity.
t = time.
Initially, the astronaut is at a distance x away from the spacecraft so that
the initial position of the astronaut, x0, is equal to x.
Since the origin of the frame of reference is located at the spacecraft, the position of the spacecraft will be 0 m.
The velocity of the astronaut is directed towards the spacecraft (the origin of the frame of reference), then, v = -1.8 m/s
The maximum time it can take the astronaut to reach the position of the spacecraft is 1.5 min = 90 s.
Then:
x = x0 + v · t
0 m = x - 1.8 m/s · 90 s
Solving for x:
1.8 m/s · 90 s = x
x = 162 m
The maximum distance the astronaut can be away from the spacecraft is 162 m.
The solar system is 25,000 light years from the center of our Milky Way galaxy. One light year is the distance light travels in one year at a speed of 3.0×108m/s. Astronomers have determined that the solar system is orbiting the center of the galaxy at a speed of 230 km/s.
Answer:
The time period of the solar system's orbit is [tex]2.05\times10^{8}\ year[/tex]
Explanation:
Given that,
Distance = 25000 light year
Speed of light [tex]c=3\times10^{8}\ m/s[/tex]
Speed of astronomers = 230 m/s
Suppose the orbit is circular, we need to find the time period of the solar system's orbit.
We need to calculate the time period
Using formula of time period
[tex]t = \dfrac{d}{v}[/tex]
[tex]t = \dfrac{2\pi r}{v}[/tex]
r = distance
t = time
v = speed
Put the value into the formula
[tex]t=\dfrac{2\pi\times25000\times9.46\times10^{15}}{230\times10^{3}}[/tex]
[tex]t=6.460\times10^{15}\ sec[/tex]
Time in years
[tex]t=\dfrac{6.460\times10^{15}}{3.15\times10^{7}}[/tex]
[tex]t=2.05\times10^{8}\ year[/tex]
Hence, The time period of the solar system's orbit is [tex]2.05\times10^{8}\ year[/tex]
The solar system is about 25,000 light years away from the centre of the Milky Way galaxy. It orbits the galaxy at a speed of 230 km/s, which translates to a galactic year of approximately 225 million Earth years. The galaxy's mass, including dark matter, influences the orbital velocities of all celestial bodies.
Explanation:
The solar system, orbiting at a speed of 230 km/s, is located approximately 25,000 light years from the Milky Way galaxy's center. A light year is defined as the distance light travels in one year with a speed of 3.0×108m/s. Understanding this concept helps us comprehend the vast size of our galaxy. Additionally, the solar system's movement around the Milky Way is referred to as a galactic year, which lasts approximately 225 million years in Earth's time.
The sun, and all stars in the galaxy, orbit the galactic center in a nearly circular path lying in the galaxy's disk. Moreover, the total mass of our Galaxy can be determined by measuring the orbital velocities of stars and interstellar matter, including the Sun. This vast mass, around 2 × 1012 Msun, includes a significant part composed of dark matter that emits no electromagnetic radiation.
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Two speakers separated by a distance of D = 3.90 m emit sound with opposite phase. A person listens from a location d1 = 2.75 m in front of one of the speakers. What is the lowest (nonzero) frequency that gives destructive interference in this case?
Answer:
Hence lowest (nonzero) frequency that gives destructive interference in this case = 3400 Hz
Explanation:
Since, the two are in out of phase,
their path difference is
d= nλ
[tex]d_2-d_1= n\lambda[/tex]
Given d1= 2.75 m
D= 3.90 m
[tex]d_2= \sqrt{D^2- d_1^2}[/tex]
[tex]d_2= \sqrt{3.90^2- 2.75^2}[/tex]
d_2= 2.76 m
2.76-2.75= 1×λ
λ= 0.01 m
0.01= 1*λ
λ =0.01
frequency ν = v/λ = 340/0.01
f= 3400 Hz
Hence lowest (nonzero) frequency that gives destructive interference in this case = 3400 Hz
An air bubble released by a remotely operated underwater vehicle, 120 m below the surface of a lake, has a volume of 1.40 cm3. The surface of the lake is at sea level, and the density of the lake water can be approximated as that of pure water. As the bubble rises to the surface, the temperature of the water and the number of air molecules in the bubble can each be approximated as constant. Find the volume (in cm3) of the bubble just before it pops at the surface of the lake.
Answer:
17.7 cm^3
Explanation:
depth, h = 120 m
density of water, d = 1000 kg/m^3
V1 = 1.4 cm^3
P1 = P0 + h x d x g
P2 = P0
where, P0 be the atmospheric pressure
Let V2 be the volume of the bubble at the surface of water.
P0 = 1.01 x 10^5 Pa
P1 = 1.01 x 10^5 + 120 x 1000 x 9.8 = 12.77 x 10^5 Pa
Use
P1 x V1 = P2 x V2
12.77 x 10^5 x 1.4 = 1.01 x 10^5 x V2
V2 = 17.7 cm^3
Thus, the volume of bubble at the surface of water is 17.7 cm^3.
The volume (in cm3) of the bubble just before it pops at the surface of the lake is mathematically given as
V2 = 17.7 cm^3
What is the volume (in cm3) of the bubble just before it pops at the surface of the lake.?Question Parameter(s):
An air bubble released by a remotely operated underwater vehicle, 120 m
below the surface of a lake, has a volume of 1.40 cm3.
Generally, the equation for the initial Pressure is mathematically given as
P1 = P0 + h x d x g
Where,atmospheric pressure
P0 = 1.01 x 10^5 Pa
Therefore
P1 = 1.01 * 10^5 + 120 * 1000 &* 9.8
P1= 12.77 * 10^5 Pa
In conclusion
P1 x V1 = P2 x V2
12.77 x 10^5 x 1.4 = 1.01 x 10^5 x V2
V2 = 17.7 cm^3
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A sample of a low-density gas is initially at room temperature and has pressure p0 . The gas is warmed at constant volume until the pressure is 2p0 .
Compared to the initial Celsius temperature of the gas, the final Celsius temperature is
A. greater by a factor of more than 2.
B. greater by a factor of 2.
C. greater by a factor between 1 and 2.
D. the same.
E. less
The final Celsius temperature is C. greater by a factor between 1 and 2.
Explanation:
We can solve this problem by applying the pressure law, which states that for an ideal gas kept at constant volume, the ratio between the pressure and the absolute temperature of the gas is constant:
[tex]\frac{p}{T}=constant[/tex]
or
[tex]\frac{p_1}{T_1}=\frac{p_2}{T_2}[/tex]
where in this problem:
[tex]p_1 = p_0[/tex] is the initial pressure of the gas
[tex]T_1[/tex] is the initial absolute temperature (in Kelvin)
[tex]p_2 = 2 p_0[/tex] is the final pressure
[tex]T_2[/tex] is the final temperature in Kelvin
Re-arranging the equation,
[tex]\frac{p_0}{T_1}=\frac{2p_0}{T_2}\\T_2 = 2T_1[/tex]
So, the temperature in Kelvin has doubled. We can now rewrite this relationship by rewriting the Kelvin temperature in Celsius degrees:
[tex]T_{C2}-273 = 2(T_{C1}-273)\\T_{C2}=2T_{C1}-273[/tex]
where [tex]T_{C1}, T_{C2}[/tex] are the initial and final temperatures in Celsius degrees.
This means that the temperature in Celsius increases by a factor between 1 and 2, so the correct answer is C.
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The final Celsius temperature of the gas after being warmed at constant volume to reach a pressure of 2p0 is greater by a factor between 1 and 2, according to Gay-Lussac's law. So the correct option is C.
Explanation:The question deals with the behavior of gas under the influence of temperature changes at constant volume, which is described by Gay-Lussac's law. This law states that for a given mass of gas at constant volume, the pressure of the gas is directly proportional to its Kelvin (absolute) temperature. Since the initial pressure of the gas is p0 and the final pressure is 2p0, and we know that the volume remains constant, the final Kelvin temperature must also double. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. If the initial temperature in Celsius T1 corresponds to the Kelvin temperature K1, and after the change, the final Kelvin temperature is 2K1, then when converted back to Celsius, the final temperature T2 will not be twice T1—because the relationship between Kelvin and Celsius is linear and not proportional. Therefore, the final Celsius temperature will be higher by a certain amount but not double.
Given this, the correct answer is that the final Celsius temperature of the gas is greater by a factor between 1 and 2.
You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner diameter. Compared the rate at which water flows through the two types of hoses (i.e. the ratio of rates).
To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,
Q = V*A
Where,
A= Cross-sectional Area
V = Velocity
The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,
[tex]Q_1 = Q_2[/tex]
[tex]V_1A_1=V_2A_2[/tex]
Our values are given as,
[tex]A_1=\frac{1}{2}^2*\pi=0.785 in^2[/tex]
[tex]A_2=\frac{5}{8}^2*\pi=1.227 in^2[/tex]
Re-arrange the equation to find the first ratio of rates we have:
[tex]\frac{V_1}{V_2}=\frac{A_2}{A_1}[/tex]
[tex]\frac{V_1}{V_2}=\frac{1.227}{0.785}[/tex]
[tex]\frac{V_1}{V_2}=1.56[/tex]
The second ratio of rates is
[tex]\frac{V2}{V1}=\frac{A_1}{A2}[/tex]
[tex]\frac{V2}{V1}=\frac{0.785}{1.227}[/tex]
[tex]\frac{V2}{V1}=0.640[/tex]
The ratio of the flow rate of a ½ inch hose to that of a 5/8 inch hose is 16:25. This means that under equal pressures, for every 16 gallons of water that pass through the ½ inch hose, 25 gallons could flow through the 5/8 inch hose.
Explanation:To compare the flow rates within the two hoses, we'll utilize a principle of fluid dynamics which states that the flow rate of an incompressible fluid (like water) is proportional to the cross-sectional area of the pipe. The area can be calculated using the formula for the area of a circle: A=πr², where r is the radius of the pipe.
To transform the diameters into radii, we divide them by 2: the radii are therefore 1/4 inch and 5/16 inch. As we're primarily interested in the ratio of areas (and consequently flow rates), we can ignore the π in the equation, leaving us with r² as directly representing 'area' for this comparison.
(1/4)² = 1/16 and (5/16)² = 25/256. Hence, the flow rate of the ½-inch hose to the 5/8 inch hose is 1/16: 25/256.
To simplify, we can multiply both portions of the ratio by 256 to achieve 16:25. This means for every 16 gallons per minute through the ½ inch hose, 25 gallons per minute could flow through the 5/8-inch hose, assuming the pressure in both systems is equal.
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Part A
Calculate the magnitude of the maximum orbital angular momentum Lmax for an electron in a hydrogen atom for states with a principal quantum number of 7.
Express your answer in units of ℏ to three significant figures.
Part B
Calculate the magnitude of the maximum orbital angular momentum Lmax for an electron in a hydrogen atom for states with a principal quantum number of 26.
Express your answer in units of ℏ to three significant figures.
Part C
Calculate the magnitude of the maximum orbital angular momentum Lmax for an electron in a hydrogen atom for states with a principal quantum number of 191.
Express your answer in units of ℏ to three significant figures.
Answer:
Explanation:
A )
[tex]L_{max} = \sqrt{l(l+1)}[/tex]ℏ
where l is orbital quantum number
l = n-1 where n is principal quantum no
Given n = 7
l = 7 - 1 = 6
[tex]L_{max} = \sqrt{6(6+1)}[/tex]ℏ
= 6.48ℏ
B)
Here
n = 26
l = 26 - 1
= 25
[tex]L_{max} = \sqrt{25(25+1)}[/tex]ℏ
= 25.49ℏ
= 25.5ℏ
C )
n = 191
l = 191 - 1
190
[tex]L_{max} = \sqrt{190(190+1)}[/tex]
= 190.499ℏ
= 191ℏ
The magnitude of the maximum orbital angular momentum Lmax for an electron in a hydrogen atom can be calculated using the formula Lmax = √(l*(l + 1)), where l is the orbital quantum number, which for maximum Lmax is n - 1 for principal quantum number n. The values for n = 7, 26, and 191 are approximately 16.9706 ℏ, 81.6144 ℏ, and 675.3918 ℏ, respectively.
Explanation:The magnitude of the maximum orbital angular momentum for an electron in a hydrogen atom can be expressed in terms of the principal quantum number n. The maximum value of the orbital quantum number, l, is n - 1.
The magnitude of the maximum orbital angular momentum Lmax is then given by the square root of l*(l + 1). Therefore:
For a principal quantum number n = 7, l = 7 - 1 = 6. Consequently, Lmax = √(l*(l + 1)) = √(6*7) = 16.9706 ℏ to three significant figures. For a principal quantum number n = 26, l = 26 - 1 = 25. Consequently, Lmax = √(l*(l + 1)) = √(25*26) = 81.6144 ℏ to three significant figures.For a principal quantum number n = 191, l = 191 - 1 = 190. Consequently, Lmax = √(l*(l + 1)) = √(190*191) = 675.3918 ℏ to three significant figures. Learn more about Orbital Angular Momentum here:https://brainly.com/question/35912005
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Question 18 (1 points) A step-up transformer connected to a 110-V line is used to supply a hydrogen-gas discharge tube with 5.0 kV (rms). The tube dissipates 75 W of power. What is the ratio of the number of turns in the secondary winding to the number of turns in the primary winding?
Answer:
N₂/N₁ = 45
Explanation:
The relation of the number of turns with the voltage is given by the EMF equation of transformer:
[tex] \frac{V_{2}}{V_{1}} = \frac{N_{2}}{N_{1}} [/tex] (1)
where V₁: voltage of the primary winding, V₂: voltage of the secondary winding, N₁: number of turns in the primary winding, and N₂: number of turns in the secondary winding
Assuming that the primary winding is connected to the input voltage supply, and using equation (1), we can calculate the ratio of N₂ to N₁:
[tex] \frac{N_{2}}{N_{1}} = \frac{V_{2}}{V_{1}} = \frac{5.0 \cdot 10^{3}V}{110V} = 45 [/tex]
So, the ratio N₂/N₁ is 45.
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A double-slit interference pattern is observed on a screen 1.0 m behind two slits spaced 0.30 mm apart. From the center of one particular fringe to the center of the ninth bright fringe from this one is 1.6 cm. What is the wavelength of the light? [530nm]
Answer:
The wavelength of the light is 530 nm.
Explanation:
Given that,
Distance D= 1.0 m
Distance between slits d= 0.30 mm
Number of fringe = 9
Width = 1.6 cm
We need to calculate the angle
Using formula of angle
[tex]\tan\theta=\dfrac{y}{D}[/tex]
[tex]tan\theta=\dfrac{1.6\times10^{-2}}{1.0}[/tex]
[tex]\theta=\tan^{-1}(\dfrac{1.6\times10^{-2}}{1.0})[/tex]
[tex]\theta=0.91^{\circ}[/tex]
We need to calculate the wavelength of the light
Using formula of wavelength
[tex]d\sin\theta=m\lambda[/tex]
[tex]\lambda=\dfrac{d\sin\theta}{m}[/tex]
Put the value into the formula
[tex]\lambda= \dfrac{0.30\times10^{-3}\times\sin0.91}{9}[/tex]
[tex]\lambda=5.29\times10^{-7}\ m[/tex]
[tex]\lambda=530\ nm[/tex]
Hence, The wavelength of the light is 530 nm.
Answer:
530 nm
Explanation:
Screen distance, D = 1 m
slit distance, d = 0.3 mm
n = 9 th bright
y = 1.6 cm
Let λ be the wavelength of light used.
y = n x D x λ / d
1.6 x 10^-2 = 9 x 1 x λ / (0.3 x 10^-3)
λ = 5.3333 x 10^-7 m
λ = 533.33 nm
λ = 530 nm ( by rounding off)
A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in our atmosphere is ozone, O3. In particular, ozone absorbs radiation with frequencies around 9.38×1014 Hz . What is the wavelength λ of the radiation absorbed by ozone?
Answer:
[tex]\lambda=3.20*10^{-7}m[/tex]
Explanation:
The wavelength is inversely proportional to the frequency. The wavelength is equal to the speed of the wave, divided by the frequency. In the case of electromagnetic waves like ultraviolet radiation, the speed of propagation is the speed of light.
[tex]\lambda=\frac{c}{f}\\\lambda=\frac{3*10^8\frac{m}{s}}{9.38*10^{14}Hz}\\\lambda=3.20*10^{-7}m[/tex]
Answer : The wavelength of the radiation absorbed by ozone is, [tex]3.20\times 10^{-7}m[/tex]
Explanation : Given,
Frequency = [tex]9.38\times 10^{14}Hz=9.38\times 10^{14}s^{-1}[/tex]
Formula used :
[tex]\nu=\frac{c}{\lambda}[/tex]
where,
[tex]\nu[/tex] = frequency
[tex]\lambda[/tex] = wavelength
c = speed of light = [tex]3\times 10^8m/s[/tex]
Now put all the given values in the above formula, we get:
[tex]9.38\times 10^{14}s^{-1}=\frac{3\times 10^8m/s}{\lambda}[/tex]
[tex]\lambda=3.20\times 10^{-7}m[/tex]
Therefore, the wavelength of the radiation absorbed by ozone is, [tex]3.20\times 10^{-7}m[/tex]
g A four bladed propeller on a cargo aircraft has a moment of inertia of 40 kg m2. If the prop goes from rest to 400 rpm in 14 sec, findA) the torque required andB) the number of revolutions turned as the prop achieves operating speed.
Answer:
a. T = 119.68 N.m
b. r = 140 rev
Explanation:
first we know that:
∑T = Iα
where ∑T is the sumatory of the torques, I is the moment of inertia and α is the angular aceleration.
so, if the prop goes from rest to 400 rpm in 14 seconds we can find the α of the system:
1. change the 400 rpm to radians as:
W = 400*2π/60
W = 41.888 rad /s
2. Then, using the next equation, we find the α as:
w = αt
solving for α
α = [tex]\frac{w}{t}[/tex]
note: t is the time, so:
α = [tex]\frac{41.888}{14}[/tex]
α = 2.992 rad/s^2
Now using the first equation, we get:
T = Iα
T = 40(2.992)
T = 119.68 N.m
On the other hand, for know the the number of revolutions turned as the prop achieves operating speed, we use the following equation:
θ = wt +[tex]\frac{1}{2}[/tex]α[tex]t^2[/tex]
Where w = 41.888 rad /s, α = 2.992 rad/s^2, t is the time and θ give as the number of radians that the prop made in the fisrt 14 seconds, so:
θ = (41.888)(14)+[tex]\frac{1}{2}[/tex](2.992)(14[tex])^2[/tex]
θ = 879.648 rad
and that divided by 2π give us the number of revolutions r, so:
r = 140 rev
An arteriole has a radius of 25 μm and it is 1000 μm long. The viscosity of blood is 3 x 10-3 Pa s and its density is 1.055 g cm-3 . Assume the arteriole is a right circular cylinder. A. Assuming laminar flow, what is the resistance of this arteriole?
Answer: [tex]1.955(10)^{13} \frac{Pa.s}{m^{3}}[/tex]
Explanation:
This can be solved by the Poiseuille’s law for a laminar flow:
[tex]R=\frac{8 \eta L}{\pi r^{4}}[/tex]
Where:
[tex]R[/tex] is the resistance of the arteriole
[tex]\eta=3(10)^{-3} Pa.s[/tex] is the viscosity of blood
[tex]L=1000 \mu m=1000(10)^{-6}m[/tex] is the length of the arteriole
[tex]r=25 \mu m=25(10)^{-6}m[/tex] is the radius of the arteriole
[tex]R=\frac{8 (3(10)^{-3} Pa.s)(1000(10)^{-6}m)}{\pi (25(10)^{-6}m)^{4}}[/tex]
[tex]R=1.955(10)^{13} \frac{Pa.s}{m^{3}}[/tex]
The resistance of the arteriole is calculated using the Hagen-Poiseuille equation, resulting in an approximate value of 1.96 × 10¹⁵ Pa·s·m⁻⁴. The formula determines resistance by varying viscosity, length, and radius values, which is essential for understanding blood flow dynamics in physiological settings.
We can solve this problem using the formula for fluid resistance in a cylindrical tube (Hagen-Poiseuille equation):
[tex]R = (8 \times \eta \times L) / (\pi \times r_4)[/tex]
Where:
R is the resistance.η is the viscosity of the fluid (3 × 10⁻³ Pa·s).L is the length of the tube (1000 × 10⁻⁶ m).r is the radius of the tube (25 × 10⁻⁶ m).Substituting the values:
[tex]R = (8 \times 3 \times 10^{-3} Pa\cdot s \times 1000 \times 10^{-6} m) / (\pi \times (25 \times 10^{-6} m))[/tex]
Calculating the values inside the formula:
[tex]R = (24 \times 10^{-3} Pa\cdot s\cdot m) / (\pi \times 3.90625 \times 10^{-18} m^{4})[/tex][tex]R = (24 \times 10^{-3} Pa\cdot s\cdot m) / (1.227 \times 10^{-17} m^{4})[/tex][tex]R \approx 1.96 \times 1015 Pa\cdot s/m^{-4}[/tex]So, the resistance of the arteriole is approximately 1.96 × 10¹⁵ Pa·s·m⁻⁴.
As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150-T magnetic field near the center of the solenoid. You have enough wire for 3100 circular turns. This solenoid must be 57.0 cm long and 7.00 cm in diameter.What current will you need to produce the necessary field?
Answer:
I = 21.94 A
Explanation:
Given that
B= 0.15 T
Number of turns N= 3100
Length L = 57 cm
Diameter ,d= 7 cm
We know that magnetic field in the solenoid given as
[tex]B=\dfrac{\mu_0IN}{L}[/tex]
[tex]I=\dfrac{BL}{\mu_0N}[/tex]
Now by putting the values
[tex]I=\dfrac{0.15\times 0.57}{4\pi \times 10^{-7}\times 3100}[/tex]
I = 21.94 A
Therefore current need to produce 0.15 T magnetic filed is 21.94 A.
About once every 30 minutes, a geyser known as Old Faceful projects water 11.0 m straight up into the air. Use g = 9.80 m/s^2, and take atmospheric pressure to be 101.3 kPa. The density of water is 1000 kg/m^3. What is the speed of the water when it emerges from the ground?
Answer:
The speed of the water is 14.68 m/s.
Explanation:
Given that,
Time = 30 minutes
Distance = 11.0 m
Pressure = 101.3 kPa
Density of water = 1000 kg/m³
We need to calculate the speed of the water
Using equation of motion
[tex]v^2=u^2+2gs[/tex]
Where, u = speed of water
g = acceleration due to gravity
h = height
Put the value into the formula
[tex]0=u^2-2\times9.8\times11.0[/tex]
[tex]u=\sqrt{2\times9.8\times11.0}[/tex]
[tex]u=14.68\ m/s[/tex]
Hence, The speed of the water is 14.68 m/s.
Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e., protons with one orbiting electron) are fed into a chamber where there is a strong magnetic field. Electrons in this chamber are trapped in tight orbits, which greatly increases the chance that they will collide with a hydrogen atom and ionizes it. One such source uses a magnetic field of 90 mT, and the electrons' kinetic energy is 2.0 eV.If the electrons travel in a plane perpendicular to the field, what is the radius of their orbits?
Answer:
Explanation:
Electron's kinetic energy = 2 eV
= 2 x 1.6 x 10⁻¹⁹ J
1/2 m v² = 3.2 x 10⁻¹⁹
1/2 x 9.1 x 10⁻³¹ x v² = 3.2 x 10⁻¹⁹
v² = .703 x 10¹²
v = .8385 x 10⁶ m/s
Electrons revolve in a circular orbit when forced to travel in a magnetic field whose radius can be expressed as follows
r = mv / Bq
where m , v and q are mass , velocity and charge of electron .
here given magnetic field B = 90 mT
= 90 x 10⁻³ T
Putting these values in the expression above
r = mv / Bq
= [tex]\frac{9\times10^{-31}\times.8385\times10^6}{90\times10^{-3}\times1.6\times10^{-19}}[/tex]
= .052 mm.
A circular copper disk of diameter D=10 cm rotates at frequency =1800 rev/min about an axis through its center and at right angles to the disk. A uniform magnetic field B of 10,000 gauss is perpendicular to the disk. What potential difference e develops between the axis of the disk and its rim?
Answer:
e=2356.125 V
Explanation:
Given that
D= 10 cm = 0.1 m
R= 0.05
B= 10,000
Angular speed ,ω = 1800 rev/min
Speed in rad/s given as
[tex]\omega =\dfrac{2\pi N}{60}[/tex]
[tex]\omega =\dfrac{2\pi \times 1800}{60}[/tex]
ω= 188.49 rad/s
The potential difference given as
e= B R v
v=average speed
[tex]v=\dfrac{\omega}{2}R[/tex]
[tex]e=\dfrac{B}{2}R^2 {\omega}[/tex]
By putting the values
[tex]e=\dfrac{10000}{2}0.05^2 \times {188.49}[/tex]
e=2356.125 V
A small car and a heavy pickup truck are both out of gas. The truck has twice the mass of the car. After you push both the car and the truck for the same amount of time with the same force, what can you say about the momentum and kinetic energy (KE) of the car and the truck? Ignore friction.
a. They have the same momentum, but the car has more kinetic energy than the truck.
b. They have the same momentum and the same KE.
c. They have the same kinetic energy, but the truck has more momentum than the car.
d. The car has more momentum and more KE than the truck.
e. The truck has more momentum and more KE than the car.
Answer:
They have the same momentum, but the car has more kinetic energy than the truck.
Explanation:
Let [tex]m_c[/tex] is the mass of the car and [tex]m_t[/tex] is the mass of truck such that,
[tex]m_t=2m_c[/tex]
You push both the car and the truck for the same amount of time with the same force. Momentum of an object is given by :
[tex]p=mv=F\times t[/tex]
Since, force and time are same, so they have same momentum. The kinetic energy of an object is given by :
[tex]K=\dfrac{1}{2}mv^2[/tex]
Since, the mass of truck is more, it will have maximum kinetic energy. So, the correct option is (a) "They have the same momentum, but the car has more kinetic energy than the truck".
The car has more momentum and more KE than the truck because car mass is lower so it moves fast as compared to truck.
Relationship between mass, momentum and kinetic energyMomentum is the product of its mass and velocity. When we compare car with truck , car has a lower mass so when the mass is lower the velocity will be higher so car has more momentum.
We also know that kinetic energy also depends on mass so less mass have more kinetic energy.
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What is the best choice for the shape of a Gaussian surface?
A) spherical
B) cylindrical
C) cubic
D) It should be one that encloses the smallest volume.
E) It should be one that matches the symmetry of the charge distribution.
Answer: It should be one that matches the symmetry of the charge distribution
Explanation: Gaussian surfaces are usually carefully chosen to exploit symmetries of a situation to simplify the calculation of the surface integral. If the Gaussian surface is chosen such that for every point on the surface the component of the electric field along the normal vector is constant, then the calculation will not require difficult integration as the constants which arise can be taken out of the integral.It is defined as the closed surface in three dimensional space by which the flux of vector field be calculated.
Final answer:
The best choice for a Gaussian surface is one that matches the symmetry of the charge distribution, for example, a spherical Gaussian surface for a point charge and a cylindrical Gaussian surface for a cylindrical charge distribution, as it simplifies the integral for the flux.
Explanation:
The best choice for the shape of a Gaussian surface is E) It should be one that matches the symmetry of the charge distribution. When dealing with point charges that are spherically symmetric, selecting a spherical Gaussian surface makes sense since the electric field is radial. Likewise, for a cylindrical charge distribution, a cylindrical Gaussian surface is most appropriate. The goal is to make the flux integral easy to evaluate and to find a surface over which the electric field is constant in magnitude and makes the same angle with every element of the surface.
Using a Gaussian surface that matches the symmetry of the charge distribution, such as a spherical shell for a point charge or spherical charge distribution, ensures the electric field (E) is constant in magnitude across the surface, simplifying the calculations. Conversely, using a less ideal surface, like a cylindrical surface to envelop a spherical charge, leads to a more complex integral due to the variations in electric field strength (E) across the surface.
A Martian leaves Mars in a spaceship that is heading to Venus. On the way, the spaceship passes earth with a speed v = 0.80c relative to it. Assume that the three planets do not move relative to each other during the trip. The distance between Mars and Venus is 1.20 × 1011 m, as measured by a person on earth. What does the Martian measure for the distance between Mars and Venus?
To find the relative distance from one point to another it is necessary to apply the Relativity equations.
Under the concept of relativity the distance measured from a spatial object is given by the equation
[tex]l = l_0 \sqrt{1-\frac{v^2}{c^2}}[/tex]
Where
[tex]l_0[/tex]= Relative length
v = Velocity of the spaceship
c = Speed of light
Replacing with our values we have that
[tex]l = l_0 \sqrt{1-\frac{v^2}{c^2}}[/tex]
[tex]l = 1.2*10^{11} \sqrt{1-\frac{0.8c^2}{c^2}}[/tex]
[tex]l = 1.2*10^{11} \sqrt{1-0.8^2}[/tex]
[tex]l = 7.2*10^{10}m[/tex]
Therefore the distance between Mars and Venus measured by the Martin is [tex]7.2*10^{10}m[/tex]
A guitar string is plucked and set into vibration. The vibrating string disturbs the surrounding air, resulting in a sound wave. Which of the following is correct? Group of answer choices If the temperature of air changes, the speed of the Wave in the string Doesn't change, but the speed of the Sound wave in air changes . Both the Wave in the string and the Sound wave in air are transverse waves . Both the Wave in the string and the Sound wave in air are longitudinal waves . The Wave in the string is longitudinal , the Sound wave in air is transverse .
Answer:
a) True. The speed of the wave in the string does not depend on the temperature, but the speed of the wave the air depends on the Temperatue.
Explanation:
Let's analyze the vibration of the string, when the string is touched a transverse wave is produced whose speed is determined by the tension and density of the string
V = √T/μ
This wave advances and bounces at one of the ends forming a standing wave that induces a vibration in the surrounding air that therefore also produces a longitudinal wave whose velocity is a function of time
v = 331 √( 1+ T/273)
With this information we can review the statements given
a) True. The speed of the wave in the string does not depend on the temperature, but the speed of the wave the air depends on the Temperatue.
b) False. The wave in the string is transverse, but the wave in the air is longitudinal
c) False. The wave in the string is transverse
d) False. It's the other way around
Answer:
A
Explanation:
Just got it right on edge
A circular coil 17.7 cm in diameter and containing 18 loops lies flat on the gound. The Earth's magnetic field at this location has magnitude 5.5E-5 T and points into the Earth at an angle of 66.0 degrees below the horizontal line and aimed due north. Determine the torque if a 7.4 A clockwise current passes through the coil.
Answer:
Explanation:
Diameter of the circular coil d = 17.7 cm = 0.177m
Current in the coil, I = 7.4 A
Number of loops in the coil, N = 18
Magnetic field, B = 5.5 x 10-5 T
Angle below line, (θ) = 66°
Write the expression for the torque in the coil
Torque = N x I x B x A x sin (θ)
Where A is the area of the coil and θ° is the angle between the magnetic field and the coil face.
The cross- sectional area of the coil = (pie)(d/2)²
A = (π)(0.177/2)² = 0.0246 m²
The Earth’s magnetic field points into the earth at an angle 66 below the line. The line points towards the north
Hence to find the angle between the magnetic field and the coil face we need to subtract the given angle by 90°
Theta = 90 – 66 = 24°
Substitute the value to find out torque
Torque = 18 x 7.4 x 5.5 x 10⁻⁵ x 0.0246 x sin(24) = 7.33 x 10⁻⁵ N-m
The torque on the coil is 7.33 x 10⁻⁵ N-m
we must apply a force of magnitude 83 N to hold the block stationary at x = −2.0 cm. From that position we then slowly move the block so that our force does +4.8 J of work on the spring-block system; the block is then again stationary. What is the block's position x? (There are two answers.)
The concept required to develop this problem is Hook's Law and potential elastic energy.
By definition the force by Hooke's law is defined as
[tex]F = kx[/tex]
Where,
k = Spring Constant
x = Displacement
On the other hand, the elastic potential energy is defined as
[tex]E = \frac{1}{2} k\Delta x^2[/tex]
With the given values we can find the value of the spring constant, that is,
[tex]F = kx[/tex]
[tex]k=\frac{F}{x}[/tex]
[tex]k= \frac{83}{0.02}[/tex]
[tex]k = 4120N/m[/tex]
Applying the concepts of energy conservation then we can find the position of the block, that is,
[tex]E = \frac{1}{2} k\Delta x^2[/tex]
[tex]E = \frac{1}{2} k\Delta x^2[/tex]
[tex]4.8 = \frac{1}{2} (4120)(x^2-(-0.02)^2)[/tex]
[tex]x = \sqrt{\frac{2*4.8}{4120}+(-0.02)^2}[/tex]
[tex]x = \pm 0.05225m[/tex]
Therefore the position of the block can be then,
[tex]x_1 = 5.225cm[/tex]
[tex]x_2 = -5.225cm[/tex]
Unpolarized light is passed through an optical filter that is oriented in the vertical direction. 1) If the incident intensity of the light is 90 W/m2, what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)
Answer:
45 W/m^2
Explanation:
Intensity of light, Io = 90 W/m^2
According to the law of Malus
[tex]I=I_{0}Cos^{2}\theta[/tex]
The average value of Cos^θ is half
So, I = Io/2
I = 90 /2
I = 45 W/m^2
Unpolarized light, when passed through a polarizer, reduces its intensity by half. So, the intensity if the light that emerges from a vertical filter will be 45 W/m².
Explanation:Given that the incident intensity of the unpolarized light is 90 W/m², when passed through a vertically oriented optical filter, the emerging light will be polarized and will have its intensity halved as it's the property of a polarizing filter to decrease the intensity of unpolarized light by a factor of 2. The formula used in this process is I = Io cos² θ. In the case of unpolarized light passing through a single polarizer, θ is 0. So, the formula simplifies to I = Io/2.
Therefore, the intensity of the light that emerges from the vertically oriented optical filter is: I = 90 W/m² / 2 = 45 W/m².
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A small bolt with a mass of 33.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical direction with a frequency of 3.05 Hz. What is the maximum amplitude with which the piston can oscillate without the bolt losing contact with the piston's surface? Use g = 9.81 m/s^2 for the acceleration due to gravity.
To determine the maximum amplitude of the piston's oscillation without the bolt losing contact, we need to set the gravitational force equal to the spring force, and solve for the amplitude. Using the given values, including the mass of the bolt, the acceleration due to gravity, and the frequency of oscillation, we can calculate the maximum amplitude.
Explanation:In order for the bolt to stay in contact with the piston's surface, the maximum amplitude of the piston's oscillation needs to be determined.
For simple harmonic motion, the restoring force is proportional to the displacement and acts opposite to the direction of motion. In this case, the restoring force is provided by the weight of the bolt (mg) and the force exerted by the piston (kx), where x is the displacement from equilibrium and k is the spring constant of the piston.
At maximum amplitude, the net force acting on the bolt is zero, so we can set the gravitational force equal to the spring force: mg = kA. Rearranging this equation gives us the maximum amplitude A = mg/k.
Now we can calculate the maximum amplitude using the given values: mass of bolt = 33.0 g = 0.033 kg, acceleration due to gravity g = 9.81 m/s^2, and frequency of oscillation f = 3.05 Hz. The period of oscillation is T = 1/f.
Using the equation T = 2*pi*sqrt(m/k) for the period of a mass-spring system, we can solve for the spring constant k: k = (2*pi/T)^2*m = (2*pi/1/f)^2*0.033 = (2*pi*f)^2*0.033. Plugging in the given value for f, we find k = (2*pi*3.05)^2*0.033.
Finally, we can calculate the maximum amplitude A = mg/k = 0.033*9.81/(2*pi*3.05)^2*0.033. Evaluating this expression gives us the maximum amplitude of the piston's oscillation.
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The earth is about 1.50 X 1011 m from the sun and has a period of about 365 days orbiting around the sun. Suppose that the orbit of the earth is circular, what is the magnitude of the acceleration of the earth in m/s2?
To solve this problem it is necessary to apply the concepts presented in Kepler's third law in which the period is described, as well as the theorems developed for acceleration based on gravity.
Acceleration in gravitational terms can be expressed as
[tex]a = \frac{GM}{r^2}[/tex]
Where,
G = Gravitational Universal Constant
M = Mass of Earth
r = Distance
At the same time the Period by Kepler's law the Period is described as
[tex]T^2 = \frac{4\pi^2r^3}{GM}[/tex]
That is equal to
[tex]T^2 = \frac{4\pi^2r}{\frac{GM}{r^2}}[/tex]
Using the equation of acceleration,
[tex]T^2 = \frac{4\pi^2r}{a}[/tex]
Re-arrange to find a,
[tex]a = \frac{4\pi^2r}{T^2}[/tex]
Our values are given as
[tex]r = 1.5*10^{11}m[/tex]
[tex]T = 365days(\frac{86400s}{1days}) = 31536000s[/tex]
Replacing we have,
[tex]a = \frac{4\pi^2r}{T^2}[/tex]
[tex]a = \frac{4\pi^2(1.5*10^{11})}{(31536000)^2}[/tex]
[tex]a = 0.005954m/s^2[/tex]
Therefore the magnitude of the acceleration of the earth is [tex]0.005954m/s^2[/tex]
The magnitude of the acceleration of the Earth in its orbit around the sun, assuming a circular orbit, is approximately 0.0059 m/s².
Explanation:The concept at play here is centripetal acceleration, which is the rate of change of tangential velocity and points toward the center of the circle around which Earth, or any object, orbits. For Earth orbiting the sun, we can use the formula ac = v²/r (where v is velocity, r is radius), but given that v isn't directly stated, one must compute it first by v = 2πr/T (where T is the period of the orbit in seconds).
So, the velocity of the Earth in its orbit is approximated as 2π(1.50x10¹¹ m)/(365.25x24x60x60 s) = 29,785.6 m/s. To get the acceleration, plug this computed v value into the ac equation: ac = (29,785.6 m/s)²/(1.50 x 10¹¹ m) = 0.0059 m/s².
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According to the relationship between torque and angular acceleration, what happens when you have more torque (given a constant rotational inertia)?
A. The angular acceleration increases.
B. The angular acceleration decreases.
C. The angular acceleration stays constant.
D. Not enough information to know.
Answer:
The angular acceleration increases.
Explanation:
The relationship between the torque and angular acceleration is :
[tex]\tau=I\times \alpha[/tex]
Where
I is the moment of inertia
[tex]\alpha[/tex] is the angular acceleration
We can see that the torque is directly proportional to the angular acceleration. So, when we have more torque it means angular acceleration increases. Hence, the correct option is (A).
Final answer:
When there is more torque applied to an object with a constant rotational inertia, the angular acceleration of the object increases proportionally. So the correct option is A.
Explanation:
According to the relationship between torque and angular acceleration, when you have more torque (given a constant rotational inertia), the angular acceleration increases. This is because torque (τ) is directly related to angular acceleration (α) through the equation τ = Iα, where I represent the moment of inertia. If the moment of inertia remains constant, an increase in torque results in a proportional increase in angular acceleration, similar to how pushing a merry-go-round harder makes it accelerate faster.
Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° above the horizontal, in 3.00 s. You can neglect the person's kinetic energy. Express your answer in horsepower
Answer:
Power, P = 924.15 watts
Explanation:
Given that,
Length of the ramp, l = 12 m
Mass of the person, m = 55.8 kg
Angle between the inclined plane and the horizontal, [tex]\theta=25^{\circ}[/tex]
Time, t = 3 s
Let h is the height of the hill from the horizontal,
[tex]h=l\ sin\theta[/tex]
[tex]h=12\times \ sin(25)[/tex]
h = 5.07 m
Let P is the power output necessary for a person to run up long hill side as :
[tex]P=\dfrac{E}{t}[/tex]
[tex]P=\dfrac{mgh}{t}[/tex]
[tex]P=\dfrac{55.8\times 9.8\times 5.07}{3}[/tex]
P = 924.15 watts
So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.
We find the height of the 25° incline and then calculate the potential energy per kg, which is then turned into power by dividing by time. The result is 16.6 W or 0.022 hp. Therefore, the minimum average power output necessary for a person to run up a 12.0 m, 25° long hillside in 3.00 s is approximately 2.2% of a horsepower.
Explanation:This problem is solved in several steps. First, we need to determine the potential energy gain of the person running up the hillside. The formula for potential energy (PE) is PE = m*g*h, where m is mass, g is gravity (approximately 9.8 m/s²), and h is the height of the hill. Since we're not given the person's mass, we'll imagine the person has a mass of 1 kg just to calculate the potential energy gain per kg, but it isn't necessary to know the exact weight for calculating the minimum average power.
The height of the incline is given by 12.0m*sin(25°) = 5.09 m. So, the potential energy gain is PE = 1 kg * 9.8 m/s² * 5.09 m = 49.88 J. Converting this to power (P) by dividing energy by time, P = 49.88 J / 3.0 s = 16.6 W.
Since one horsepower (hp) is approximately 746 watts (W), the power in horsepower is 16.6 W / 746 W/hp = 0.022 hp, or 2.2% of a horsepower. This is the minimum average power output required for a person to climb this hill in 3 seconds.
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A mouse is running along the floor in a straight line at 1.3 m/s. A cat runs after it and, perfectly judging the distance d to the mouse ahead, springs up at a speed of 2.5 m/s and an angle 38 degree, landing right on top of the mouse.
What is d, the distance between the cat and mouse at the instant the cat springs into the air?
a)0.931 m
b)0.210 m
c) 0.401 m
d)0.552 m
e)0.641m
Answer:
Option b
Solution:
As per the question:
Speed of the mouse, v = 1.3 m/s
Speed of the cat, v' = 2.5 m/s
Angle, [tex]\theta = 38^{\circ}[/tex]
Now,
To calculate the distance between the mouse and the cat:
The distance that the cat moved is given by:
[tex]x = v'cos\theta t[/tex]
[tex]x = 2.5cos38^{\circ}\times t = 1.97t[/tex]
The position of the cat and the mouse can be given by:
[tex]x = x' + vt[/tex]
[tex]1.97t = x' + 1.3t[/tex]
x' = 0.67 t (1)
The initial speed of the cat ahead of the mouse:
u = [tex]v'sin\theta = 2.5sin38^{\circ} = 1.539\ m/s[/tex]
When the time is 0.5t, the speed of the cat is 0, thus:
[tex]0 = u - 0.5tg[/tex]
[tex]t = \frac{1.539}{0.5\times 9.8} = 0.314\ s[/tex]
Substituting the value of t in eqn (1):
x' = 0.67(0.314) = 0.210 m
Thus the distance comes out to be 0.210 m
Two blocks (m1 = 40 kg, m2 = 30 kg) are connected by a compressed spring, and are initially at rest. The spring breaks and the two objects fly apart from one another. If the speed of the first block is 15 m/s, what is the speed of the second block?
Answer:
Speed of second block will be 20 m /sec in opposite direction
Explanation:
We have given the mass of two blocks [tex]m_1=40kg[/tex] and [tex]m_2=30kg[/tex]
As both the blocks are initially at rest so [tex]u_1=0m/sec\ and\ u_2=0m/sec[/tex]
Speed of first block [tex]v_1=15m/sec[/tex]
We have to find the speed of second block, that is [tex]v_2[/tex]
From momentum conservation we know that
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
[tex]40\times 0+30\times 0=40\times 15+30\times v_2[/tex]
[tex]v_2=-20m/sec[/tex]
So speed of second block will be 20 m /sec in opposite direction.