Answer:
Final magnification = -375
Explanation:
The total magnification of a compound microscope is expressed mathematically as the product of the magnifying power of each of the lenses that are combined in the compound microscope.
Final magnification = (magnifying power of the objective lens) × (magnifying power of the eyepiece lens) = m₁ × m₂
Magnifying power of a lens is defined as the ratio of the least distance of distinct vision to the focal length of the lens.
For the eyepiece, the least distance of distinct vision is the distance from the object to the near point = 25 + 1 + 0.15 = 26.15 cm
Focal length = 1 cm
Magnifying power of the eyepiece length = (26.15/1) = 26.15
For the objective lens,
The focal length = 0.14 cm
Least distance of distinct vision = -(1 + 1.00 + 0.15 - 0.14) = -2.01 cm (the negative sign means the image seen is upside down)
Magnifying power of the objective lens = (-2.01/0.14) = -14.357
Final magnification = -14.357 × 26.15 = -375
Hope this Helps!!!
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm .What is the difference in maximum stored energy between the sprinters and the nonathlethes?
Complete Question:
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, [tex]31 {\rm {N}/{mm}}[/tex]. What is the difference in maximum stored energy between the sprinters and the nonathlethes?
Answer:
[tex]\triangle E = 12.79 J[/tex]
Explanation:
Sprinters' tendons stretch, [tex]x_s = 43 mm = 0.043 m[/tex]
Non athletes' stretch, [tex]x_n = 32 mm = 0.032 m[/tex]
Spring constant for the two groups, k = 31 N/mm = 3100 N/m
Maximum Energy stored in the sprinter, [tex]E_s = 0.5kx_s^2[/tex]
Maximum energy stored in the non athletes, [tex]E_m = 0.5kx_n^2[/tex]
Difference in maximum stored energy between the sprinters and the non-athlethes:
[tex]\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J[/tex]
A mercury barometer reads 745.0 mm on the roof of a building and 760.0 mm on the ground. Assuming a constant value of 1.29 kg/m3 for the density of air, determine the height of the building
Answer:
The height of the building is 158.140 meters.
Explanation:
A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:
[tex]\Delta P \propto \rho \cdot \Delta h[/tex]
[tex]\Delta P = k \cdot \rho \cdot \Delta h[/tex]
Where:
[tex]\Delta P[/tex] - Manometric pressure difference, measured in kilopascals.
[tex]\rho[/tex] - Fluid density, measured in kilograms per cubic meter.
[tex]\Delta h[/tex] - Height difference, measured in meters.
Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:
[tex]\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}[/tex]
[tex]\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}[/tex]
Where:
[tex]\Delta h_{air}[/tex] - Height difference of the air column, measured in meters.
[tex]\Delta h_{Hg}[/tex] - Height difference of the mercury column, measured in meters.
[tex]\rho_{air}[/tex] - Density of air, measured in kilograms per cubic meter.
[tex]\rho_{Hg}[/tex] - Density of mercury, measured in kilograms per cubic meter.
If [tex]\Delta h_{Hg} = 0.015\,m[/tex], [tex]\rho_{air} = 1.29\,\frac{kg}{m^{3}}[/tex] and [tex]\rho_{Hg} = 13600\,\frac{kg}{m^{3}}[/tex], the height difference of the air column is:
[tex]\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)[/tex]
[tex]\Delta h_{air} = 158.140\,m[/tex]
The height of the building is 158.140 meters.
158.13m
Explanation:
Force exerted over a unit area is called Pressure. Also, in a given column of air, the pressure(P) is given as the product of the density(ρ) of the air, the height(h) of the column of air and the acceleration due to gravity(g). i.e
P = ρhg
Let;
Pressure measured at the roof top = ([tex]P_{R}[/tex])
Pressure measured at the ground level = ([tex]P_{G}[/tex])
Pressure at the ground level = Pressure at the roof + Pressure at the column height of air.
[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P ---------------(i)
(a) P = ρhg -----------(***)
But;
ρ = density of air = 1.29kg/m³
h = height of column of air = height of building
g = acceleration due to gravity = 10m/s²
Substitute these values into equation (***)
P = 1.29 x h x 10
P = 12.9h Pa
(b) [tex]P_{G}[/tex] = ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{ground} }[/tex] x g ------------(*)
But;
ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³
h[tex]_{(mercury)_{ground} }[/tex] = height of mercury on the ground = 760.0mm = 0.76m
g = acceleration due to gravity = 10m/s²
Substitute these values into equation (*)
[tex]P_{G}[/tex] = 13600 x 0.76 x 10
[tex]P_{G}[/tex] = 103360 Pa
(c) [tex]P_{R}[/tex] = ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{roof} }[/tex] x g --------------(**)
But;
ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³
h[tex]_{(mercury)_{roof} }[/tex] = height of mercury on the roof = 745.0mm = 0.745m
g = acceleration due to gravity = 10m/s²
Substitute these values into equation (**)
[tex]P_{R}[/tex] = 13600 x 0.745 x 10
[tex]P_{R}[/tex] = 101320 Pa
(d) Now that we know the values of P, [tex]P_{G}[/tex] and [tex]P_{R}[/tex] , let's substitute them into equation (i) as follows;
[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P
103360 = 101320 + 12.9h
Solve for h;
12.9h = 103360 - 101320
12.9h = 2040
h = [tex]\frac{2040}{12.9}[/tex]
h = 158.13m
Therefore, the height of the building is 158.13m
Space vehicles traveling through Earth's radiation belts can intercept a significant number of electrons. The resulting charge buildup can damage electronic components and disrupt operations. Suppose a spherical metallic satellite 1.7 m in diameter accumulates 3.1 µC of charge in one orbital revolution. (a) Find the resulting surface charge density. (b) Calculate the magnitude of the electric field just outside the surface of the satellite, due to the surface charge.
Answer:
(a) σ = 3.41*10⁻7C/m^2
(b) E = 38,530.1 N/C
Explanation:
(a) In order to calculate the resulting surface charge density, you use the following formula:
[tex]\sigma=\frac{Q}{S}[/tex] (1)
σ: surface charge density
Q: charge of the satellite = 3.1 µC = 3.1*10^-6C
S: surface area of the satellite
The satellite has a spherical form, then, the area of the surface is given by:
[tex]S=4\pi r^2[/tex] (2)
r: radius of the satellite = d/2 = 1.7m/2 = 0.85m
You replace the equation (2) into the equation (1) and solve for the surface charge density:
[tex]\sigma=\frac{3.1*10^{-6}C}{4\pi (0.85m)^2}=3.41*10^{-7}\frac{C}{m^2}[/tex]
The surface charge density acquired by the satellite on one orbit is 3.41*10⁻7C/m^2
(b) The electric field just outside the surface is calculate d by using the following formula:
[tex]E=k\frac{Q}{R^2}[/tex] (3)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
R: radius of the satellite = 0.85m
[tex]E=(8.98*10^9Nm^2/C^2)\frac{3.1*10^{-6}C}{(0.85m)^2}=38530.1\frac{N}{C}[/tex]
The magnitude of the electric field just outside the sphere is 38,530.1 N/C
If vector C is added to vector D, the result is a third vector that is perpendicular to D and has a magnitude equal to 3D. What is the ratio of the magnitude of C to that of D?
a) 1.3
b) 1.6
c) 1.8
d) 2.2
e) 3.2
Answer:
(e) 3.2
Explanation:
We are given that vector C and D.
Let R be the magnitude of C+D.
According to question
R=3D
We have to find the ratio of the magnitude of C to that of D.
By using right triangle property
[tex]C^2=R^2+D^2[/tex]
[tex]C^2=(3D)^2+D^2[/tex]
[tex]C^2=9D^2+D^2[/tex]
[tex]C^2=10D^2[/tex]
[tex]C=\sqrt{10D^2}=3.2D[/tex]
[tex]\frac{C}{D}=3.2[/tex]
Hence, the ratio of the magnitude of C to that of D=3.2
(e) 3.2
How does sodium (Na) becomes an ion?
A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket motor is ignited and the rocket burns fuel at a rate of 95 kg/s. The exit speed of the exhaust gas relative to the rocket is 2900 m/s. Neglecting drag and friction forces, determine the acceleration and the velocity of the car at time t = 10 s. Plot the acceleration and velocity from time t0 to t = 10 s.
Answer: Acceleration of the car at time = 10 sec is 108 [tex]m/s^{2}[/tex] and velocity of the car at time t = 10 sec is 918.34 m/s.
Explanation:
The expression used will be as follows.
[tex]M\frac{dv}{dt} = u\frac{dM}{dt}[/tex]
[tex]\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt[/tex]
= [tex]u\int_{M_{o}}^{M_{f}} \frac{dM}{M}[/tex]
[tex]v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}[/tex]
[tex]v_{o} = 0[/tex]
As, [tex]v_{f} = u ln (\frac{M_{f}}{M_{o}})[/tex]
u = -2900 m/s
[tex]M_{f} = M_{o} - m \times t_{f}[/tex]
= [tex]2500 kg + 1000 kg - 95 kg \times t_{f}s[/tex]
= [tex](3500 - 95t_{f})s[/tex]
[tex]v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s[/tex]
Also, we know that
a = [tex]\frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}[/tex]
= [tex]\frac{u}{3500 - 95 t} \times (-95) m/s^{2}[/tex]
= [tex]\frac{95 \times 2900}{3500 - 95t} m/s^{2}[/tex]
At t = 10 sec,
[tex]v_{f}[/tex] = 918.34 m/s
and, a = 108 [tex]m/s^{2}[/tex]
4. How much force is required to stop a 60 kg person traveling at 30 m/s during a time of a)
5.0 seconds
b) 0.50 seconds
c) 0.05 seconds
Explanation:
F = ma, and a = Δv / Δt.
F = m Δv / Δt
Given: m = 60 kg and Δv = -30 m/s.
a) Δt = 5.0 s
F = (60 kg) (-30 m/s) / (5.0 s)
F = -360 N
b) Δt = 0.50 s
F = (60 kg) (-30 m/s) / (0.50 s)
F = -3600 N
c) Δt = 0.05 s
F = (60 kg) (-30 m/s) / (0.05 s)
F = -36000 N
360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.
To find the force, we need to know about the mathematical formulation of force.
What is force?According to Newton's second law of motion, force is defined as mass times acceleration. Its SI unit is Newton (N).What is the mathematical formulation of force?Mathematically, it is written as
F= m×a= m×(∆V/∆t)
What is the force needed to stop 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds?Here, initially the velocity of the person is 30m/s. But after applying the force, he came to rest. So his final velocity is 0 m/s. ∆V= 30m/s
When ∆t=5 seconds, F= 60×(30/5)=360N
When ∆t=0.5 seconds, F= 60×(30/0.5)=3600N
When ∆t=0.05 seconds, F= 60×(30/0.05)=36000N
Thus, we can conclude that 360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.
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1. The smallest shift you can reliably measure on the screen is about 0.2 grid units. This shift corresponds to the precision of positions measured with the best Earth-based optical telescopes. If you cannot measure an angle smaller than this, what is the maximum distance at which a star can be located and still have a measurable parallax
Answer:
Explanation:
each grid corresponding 0.1s⁻¹.
0.2grid unit = 0.2×0.1 =0.02s⁻¹
distance of the star from telescope
d = 1/p
d= 1/0.02= 50 par sec
1par sec = 3.26 light year
1 light year = 9.5×10¹²km
3.26ly=3.084×10¹³km
d= 50×3.084×10¹³=1.55×10¹⁵km
according to newtons second law of motion, what is equal to the acceleration of an object
Answer: According to Newtons second Law of motion ;
F = ma (Force equals mass multiplied by acceleration.)
The acceleration is directly proportional to the net force; the net force equals mass times acceleration; the acceleration in the same direction as the net force; an acceleration is produced by a net force
Explanation:
Professional baseball player Nolan Ryan could pitch a baseball at approximately 160.0 km/h. At that average velocity, how long (in s) did it take a ball thrown by Ryan to reach home plate, which is 18.4 m from the pitcher's mound
Answer:
t = 0.414s
Explanation:
In order to calculate the time that the ball takes to reach home plate, you assume that the speed of the ball is constant, and you use the following formula:
[tex]t=\frac{d}{v}[/tex] (1)
d: distance to the plate = 18.4m
v: speed of the ball = 160.0km/h
You first convert the units of the sped of the ball to appropriate units (m/s)
[tex]160.0\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=44.44\frac{m}{s}[/tex]
Then, you replace the values of the speed v and distance s in the equation (1):
[tex]t=\frac{18.4m}{44.44m/s}=0.414s[/tex]
THe ball takes 0.414s to reach the home plate
A disk-shaped merry-go-round of radius 2.83 m and mass 185 kg rotates freely with an angular speed of 0.701 rev/s . A 63.4 kg person running tangential to the rim of the merry-go-round at 3.51 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. Part A What is the final angular speed of the merry-go-round
Answer:
The final angular speed of the merry-go-round is [tex]3.118\,\frac{rad}{s}[/tex] [tex]\left(0.496\,\frac{rev}{s} \right)[/tex].
Explanation:
Given the absence of external forces, the final angular speed of the merry-go-round can be determined with the resource of the Principle of Angular Momentum Conservation, which is described in this case as:
[tex]I_{g, m} \cdot \omega_{o,m} + I_{g, p}\cdot \omega_{o,p} = (I_{g,m} + I_{g, p})\cdot \omega_{f}[/tex]
Where:
[tex]I_{g,m}[/tex] - Moment of inertia of the merry-go-round with respect to its axis of rotation, measured in [tex]kg\cdot m^{2}[/tex].
[tex]I_{g,p}[/tex] - Moment of inertia of the person with respect to the axis of rotation of the merry-go-round, measured in [tex]kg\cdot m^{2}[/tex].
[tex]\omega_{o, m}[/tex] - Initial angular speed of the merry-go-round with respect to its axis of rotation, measured in radians per second.
[tex]\omega_{o,p}[/tex] - Initial angular speed of the merry-go-round with respect to the axis of rotation of the merry-go-round, measured in radians per second.
[tex]\omega_{f}[/tex] - Final angular speed of the merry-go-round-person system, measured in radians per second.
The final angular speed is cleared:
[tex]\omega_{f} = \frac{I_{g,m}\cdot \omega_{o,m}+I_{g,p}\cdot \omega_{o,p}}{I_{g,m}+I_{g,p}}[/tex]
Merry-go-round is modelled as uniform disk-like rigid body, whereas the person can be modelled as a particle. The expressions for their moments of inertia are, respectively:
Merry-go-round
[tex]I_{g,m} = \frac{1}{2}\cdot M \cdot R^{2}[/tex]
Where:
[tex]M[/tex] - The mass of the merry-go-round, measured in kilograms.
[tex]R[/tex] - Radius of the merry-go-round, measured in meters.
Person
[tex]I_{g,p} = m\cdot r^{2}[/tex]
Where:
[tex]m[/tex] - The mass of the person, measured in kilograms.
[tex]r[/tex] - Distance of the person with respect to the axis of rotation of the merry-go-round, measured in meters.
If [tex]M = 185\,kg[/tex], [tex]m = 63.4\,kg[/tex], [tex]R = r = 2.83\,m[/tex], the moments of inertia are, respectively:
[tex]I_{g,m} = \frac{1}{2}\cdot (185\,kg)\cdot (2.83\,m)^{2}[/tex]
[tex]I_{g,m} = 740.823\,kg\cdot m^{2}[/tex]
[tex]I_{g,p} = (63.4\,kg)\cdot (2.83\,m)^{2}[/tex]
[tex]I_{g,p} = 507.764\,kg\cdot m^{2}[/tex]
The angular speed experimented by the person with respect to the axis of rotation of the merry-go-round is:
[tex]\omega_{o,p} = \frac{v_{p}}{r}[/tex]
[tex]\omega_{o,p} = \frac{3.51\,\frac{m}{s} }{2.83\,m}[/tex]
[tex]\omega_{o,p} = 1.240\,\frac{rad}{s}[/tex]
Given that [tex]I_{g,m} = 740.823\,kg\cdot m^{2}[/tex], [tex]I_{g,p} = 507.764\,kg\cdot m^{2}[/tex], [tex]\omega_{o,m} = 4.405\,\frac{rad}{s}[/tex] and [tex]\omega_{o,p} = 1.240\,\frac{rad}{s}[/tex], the final angular speed of the merry-go-round is:
[tex]\omega_{f} = \frac{(740.823\,kg\cdot m^{2})\cdot \left(4.405\,\frac{rad}{s} \right)+(507.764\,kg\cdot m^{2})\cdot \left(1.240\,\frac{rad}{s} \right)}{740.823\,kg\cdot m^{2}+507.764\,kg\cdot m^{2}}[/tex]
[tex]\omega_{f} = 3.118\,\frac{rad}{s}[/tex]
[tex]\omega_{f} = 0.496\,\frac{rad}{s}[/tex]
The final angular speed of the merry-go-round is [tex]3.118\,\frac{rad}{s}[/tex] [tex]\left(0.496\,\frac{rev}{s} \right)[/tex].
A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mass. It has a rotation rate of 8.0 rev/s. Two antennas deploy in the plane of rotation extending from the center of mass of the satellite. Each antenna can be approximated as a rod of mass 150.0 kg and length 6.6 m. What is the new rotation rate of the satellite (in rev/s)
Answer:
Therefore, the new rotation rate of the satellite is 6.3 rev/s.
Explanation:
The expression for conservation of the angular momentum (L) is
[tex]L_{i} = L_{f} I_{i}\times\omega_{i} = I_{f}\times\omega_{f}[/tex]
Where
[tex]I_{i}\ and \ \omega_{i}[/tex] initial moment of inertia and angular velocity
[tex]I_{f}\ and \ \omega_{f}[/tex] is the final moment of inertia and angular velocity
The expression of moment of inertia of the satellite (a solid sphere) is
[tex]I_{i} = \frac{2}{5}m_{s}r^{2}[/tex]
Where [tex]m_{s}[/tex] is the satellite mass
r is the radus of the sphere
Substititute 1900kg for m and 4.6m for r
[tex]I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}[/tex]
The final moment of inertia of the satellite about the centre of mass
[tex]I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}[/tex]
Where [tex]m_{x}[/tex] is the antenna's mass and
I is the length of the antenna
[tex]I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}[/tex]
So, the Final rotation rate of the satellite is:
[tex]I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s[/tex]
Therefore, the new rotation rate of the satellite is 6.3 rev/s.
An astronaut is in an all-metal chamber outside the space station when a solar storm results in the deposit of a large positive charge on the station. Which statement is correct?
a. The astronaut must abandon the chamber immediately to avoid being electrocuted.
b. The astronaut will be safe only if she is wearing a spacesuit made of non-conducting materials.
c. The astronaut does not need to worry: the charge will remain on the outside surface.
d. The astronaut must abandon the chamber if the electric field on the outside surface becomes greater than the breakdown field of air.
d. The astronaut must abandon the chamber immediately because the electric field inside the chamber is non-uniform.
Answer:
c. The astronaut does not need to worry: the charge will remain on the outside surface.
Explanation:
The astronaut need not worry because according to Gauss's law of electrostatic, a hollow charged surface will have a net zero charge on the inside. This is the case of a Gauss surface, and all the charges stay on the surface of the metal chamber. This same principle explains why passengers are safe from electrostatic charges, in an enclosed aircraft, high up in the atmosphere; all the charges stay on the surface of the aircraft.
A skydiver of mass 87 kg falls straight down. If air resistance exerts 355 N of force on him as he falls, what is the net force on the skydiver?
Answer:
497.6 N
Explanation:
From the question,
The net force on the skydiver = weight of the skydiver- the resistive force of air
F' = W-F...................... Equation 1
Where W = weight of the skydiver, F = resistive force of air.
But,
W = mg................ Equation 2
Where m = mass of the skydiver, g = acceleration due to gravity.
Substitute equation 2 into equation 1
F' = mg-F............ Equation 3
Given: m = 87 kg, F = 355 N, g = 9.8 m/s²
Substitute these values into equation 3
F' = 87(9.8)-355
F' = 852.6-355
F' = 487.6 N
Answer:
498 N down
Explanation:
A 2.0-kg object moving at 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost by the system as a result of this collision.
Answer:
20 J
Explanation:
From the question, since there is a lost in kinetic energy, Then the collision is an inelastic collision.
m'u'+mu = V(m+m')........... Equation 1
Where m' = mass of the moving object, m = mass of the stick, u' = initial velocity of the moving object, initial velocity of the stick, V = common velocity after collision.
make V the subject of the equation above
V = (m'u'+mu)/(m+m')............. Equation 2
Given: m' = 2 kg, m = 8 kg, u' = 5 m/s, u = 0 m/s (at rest).
Substitute into equation 2
V = [(2×5)+(8×0)]/(2+8)
V = 10/10
V = 1 m/s.
Lost in kinetic energy = Total kinetic energy before collision- total kinetic energy after collision
Total kinetic energy before collision = 1/2(2)(5²) = 25 J
Total kinetic energy after collision = 1/2(2)(1²) +1/2(8)(1²) = 1+4 = 5 J
Lost in kinetic energy = 25-5 = 20 J
The collision is inelastic collision. As a result of collision the kinetic energy lost by the given system is 20 J.
Since there is a lost in kinetic energy, the collision is inelastic collision.
m'u'+mu = V(m+m')
[tex]\bold {V =\dfrac { (m'u'+mu)}{(m+m')} }[/tex]
Where
m' = mass of the moving object = 2 kg
m = mass of the stick = 8 kg,
u' = initial velocity of the moving object = 5 m/s
V = common velocity after collision= ?
u = 0 m/s (at rest).
put the values in the formula,
[tex]\bold {V = \dfrac {(2\times 5)+(8\times 0)}{(2+8)}}\\\\\bold {V = \dfrac {10}{10}}\\\\\bold {V = 1\\ m/s.}[/tex]
kinetic energy before collision
[tex]\bold { = \dfrac 1{2} (2)(5^2) = 25 J}[/tex]
kinetic energy after collision
[tex]\bold { = \dfrac 12(2)(1^2) + \dfrac 12(8)(1^2) = 5\ J}[/tex]
Lost in kinetic energy = 25-5 = 20 J
Therefore, As a result of collision the kinetic energy lost by the given system is 20 J.
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An enclosed amount of nitrogen gas undergoes thermodynamic processes as follows: from an initial state A to a state B to C to D and back to A, as shown in the P-V diagram. Assume that the gas behaves ideally. What is the change in internal energy of the gas for the entire process, A-B-C-D-A? (pressure at B is 10kPa)
Answer:
The total internal energy change for the entire process is -0.94 kJ
Explanation:
Process A to B is an isothermal process, therefore, [tex]u_A[/tex] - [tex]u_B[/tex] = 0
Process B to C
P = -mV + C
When P = 12, V = 0.12
When P = 4, V = 0.135
Therefore, we have;
12 = -m·0.12 + C
4 = -m·0.135 + C
Solving gives
m = 533.33
C = 76
[tex]T = \dfrac{1}{nR} \times (-533.33 \times V^2 + 76 \times V)[/tex]
p₂ = p₁V₁/V₂ = 12*0.1/0.12 = 10 kPa
The work done = 0.5*(0.135 - 0.12)*(4 - 10.0) = -0.045 kJ = -45 J
For heat supplied
Assuming an approximate polytropic process, we have;
Work done = (p₃×v₃ - p₂×v₂)/(n - 1)
Which gives;
-45 = (4*0.135 - 10*0.12)/(n -1)
∴ n -1 = (4*0.135 - 10*0.12)/-45 = 14.67
n = 15.67
Q = W×(n - γ)/(γ - 1)
Q = -45*(15.67 - 1.4)/(1.4 - 1) = -1,605.375 J
u₃ - u₂ = Q + W = -1,605.375 J - 45 J = -1650 J = -1.65 kJ
For the constant pressure process D to C, we have;
[tex]Q = c_p \times \dfrac{p}{R} \times (V_4 -V_3) = \dfrac{5}{2} \times p \times (V_4 -V_3)[/tex]
Q₄₋₃ = (0.1 - 0.135) * 4*5/2 = -0.35 kJ
W₄₋₃ = 4*(0.1 - 0.135) = -0.14 kJ
u₄ - u₃ = Q₄₋₃ + W₄₋₃ = -0.14 kJ + -0.35 kJ = -0.49 kJ
For the process D to A, we have a constant volume process
[tex]Q_{1-4} = \dfrac{c_v}{R} \times V \times (p_1 - p_4) = \dfrac{3}{2} \times 0.1 \times (12 - 4) = 1.2 \ kJ[/tex]
W₁₋₄ = 0 for constant volume process, therefore, u₁ - u₄ = 1.2 kJ
The total internal energy change Δ[tex]u_{process}[/tex] for the entire process is therefore;
Δ[tex]u_{process}[/tex] = u₂ - u₁ + u₃ - u₂ + u₄ - u₃ + u₁ - u₄ = 0 - 1.65 - 0.49 + 1.2 = -0.94 kJ.
A 50-loop circular coil has a radius of 3 cm. It is oriented so that the field lines of a magnetic field are perpendicular to the coil. Suppose that the magnetic field is varied so that B increases from 0.10 T to 0.35 T in 2 ms. Find the induced emf in the coil.
Answer:
-17.8 V
Explanation:
The induced emf in a coil is given as:
[tex]E = \frac{-NdB\pi r^2}{dt}[/tex]
where N = number of loops
dB = change in magnetic field
r = radius of coil
dt = elapsed time
From the question:
N = 50
dB = final magnetic field - initial magnetic field
dB = 0.35 - 0.10 = 0.25 T
r = 3 cm
dt = 2 ms = 0.002 secs
Therefore, the induced emf is:
[tex]E = \frac{-50 * 0.25 * \pi * 0.03^2}{0.002} \\E = -17.8 V[/tex]
Note: The negative sign implies that the EMf acts in an opposite direction to the change in magnetic flux.
A 26 kg child is coasting at 2.0 m/s over flat ground in a 5.0 kg wagon. The child drops a 1.5 kg ball from the side of the wagon. What is the final speed (in m/s) of the child and wagon
Answer:
Final speed = 2.067 m/s
Explanation:
We are told that the child weighs 26 kg.
Also, that the wagon weighs 5kg.
Thus,initial mass of the child and wagon with ball is;
m_i = 26 + 5 = 31 kg.
Also, we are told that the child now dropped 1.5 kg ball from the wagon. So,
Final mass is;
m_f = 26 + 5 - 1 = 30 kg
Now, from conservation of linear momentum, we know that;
Initial momentum = final momentum
Thus;
m_i * v_i = m_f * v_f
Where v_i is initial velocity and v_f is final velocity.
Making v_f the subject, we have;
v_f = (m_i * v_i)/m_f
We are given that initial velocity v_i = 2 m/s
Plugging in the relevant values, we have;
v_f = (31 * 2)/30
v_f = 2.067 m/s
Dr. Jones performed an experiment to monitor the effects of sunlight exposure on stem density in aquatic plants. In the study, Dr. Jones measured the mass and volume of stems grown in 5 levels of sun exposure. The data is represented below.
Sun exposure Stem mass (g) Stem volume (mL)
30 275 1100
45 415 1215
60 563 1425
75 815 1610
90 954 1742
a. Convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters (mº).
b. Calculate the density of the samples using the equation d = m/v. d = density m = mass (kg) v = volume (m)
c. Convert the density values to scientific notation.
Given that,
Sun exposure = 30%, 45%, 60%, 75%, 90%
Stem mass (g) = 275, 415, 563, 815, 954
Stem volume (ml) = 1100, 1215, 1425, 1610, 1742
(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters
Using conversion of mass
[tex]1\ g=0.001\ kg[/tex]
Conservation of volume
[tex]1\ Lt=0.001\ m^3[/tex]
[tex]1\ mL=1\times10^{-6}\ m^3[/tex]
So, mass in kg
Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954
Volume in m³,
Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742
(b). We need to calculate the density of the samples
Using formula of density
[tex]\rho=\dfrac{m}{V}[/tex]
Where, m = mass
V = volume
If the m = 0.275 kg and V = 0.0011 m³
Put the value into the formula
[tex]\rho=\dfrac{0.275}{0.0011}[/tex]
[tex]\rho=250\ kg/m^3[/tex]
If the m = 0.415 kg and V = 0.001215 m³
Put the value into the formula
[tex]\rho=\dfrac{0.415}{0.001215}[/tex]
[tex]\rho=341.56\ kg/m^3[/tex]
[tex]\rho=342\ kg/m^3[/tex]
If the m = 0.563 kg and V = 0.001425 m³
Put the value into the formula
[tex]\rho=\dfrac{0.563}{0.001425}[/tex]
[tex]\rho=395.08\ kg/m^3[/tex]
If the m = 0.815 kg and V = 0.001610 m³
Put the value into the formula
[tex]\rho=\dfrac{0.815}{0.001610}[/tex]
[tex]\rho=506.21\ kg/m^3[/tex]
If the m = 0.954 kg and V = 0.001742 m³
Put the value into the formula
[tex]\rho=\dfrac{0.954}{0.001742}[/tex]
[tex]\rho=547.6\ kg/m^3[/tex]
[tex]\rho=548\ kg/m^3[/tex]
(c). We need to convert the density values to scientific notation
In scientific notation
The densities are
[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]
Hence, This is required solution.
Imagine that you want to make sure the battery for your string of lights will last as long as possible. A battery will last longer if it powers a circuit with low current. How could you hook up a battery and 2 light bulbs so the least amount of current flows through the battery
Answer:
Hooking up the bulb to the battery in a series arrangement will draw the least amount of current.
Explanation:
In this case now, the bulb will serve as the load on the battery (resistance).
For the current to last longer, the least amount of energy must be drawn.
The least amount of energy will be drawn when the arrangement provides the maximum resistance possible.
Let us take the resistance of each bulb as 'R'
If we arrange the bulbs in series, then, the total resistance will be
Rt = R + R = 2R
at a EMF of V from the battery, current I through the battery will be
I = V/2R
If we arrange the bulbs in parallel, then , the total resistance will be
1/Rt = 1/R + 1/R
1/Rt = 2/R
therefore
Rt = R/2
at an EMF of V from the battery, the current I that will be drawn through the battery will be
I = 2V/R
we see that arranging the bulbs in parallel draws 4 times the current compared to arranging the bulb in series
From the above, we see that arranging the bulbs in series provides the maximum resistance, which means a lesser amount of current is drawn from the battery
A double slit illuminated with light of wavelength 588 nm forms a diffraction pattern on a screen 11.0 cm away. The slit separation is 2464 nm. What is the distance between the third and fourth bright fringes away from the central fringe
Answer:
[tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]
Explanation:
Given that
Wavelength [tex]\lambda=588 \mathrm{nm}[/tex]
slit separation [tex]\mathrm{d}=2464 \mathrm{nm}[/tex]
slit screen distance [tex]\mathrm{D}=11 \mathrm{cm}[/tex]
We know that for double slit the maxima condition is that
[tex]\operatorname{dsin} \theta=m \lambda[/tex]
[tex]\sin \theta=\frac{m \lambda}{d}[/tex]
[tex]\theta=\sin ^{-1}\left(\frac{\mathrm{m} \lambda}{\mathrm{d}}\right)[/tex]
For small angle approximation, [tex]\sin \theta \approx \tan \theta \approx \theta[/tex]
[tex]\tan \theta=\frac{y_{m}}{D}[/tex]
[tex]y_{m}=D \times \tan \left[\sin ^{-1}\left(\frac{m \lambda}{d}\right)\right][/tex]
Now [tex]y_{4}[/tex] [tex]y_{4}=D \times \tan \left[\sin ^{-1}\left(\frac{4 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{4 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=35.22 \mathrm{cm}[/tex]
Again [tex]y_{3}=D \times \tan \left[\sin ^{-1}\left(\frac{3 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{3 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=11.27 \mathrm{cm}[/tex]
Hence [tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]
Differences between regular and irregular
Answer:
Differences between regular and irregular objects are:
Regular object Those substances which have fixed geometrical shape are called regular objects.For example: Books,pencils etc.Irregular objectsThose substances which do not have fixed geometrical shape are called irregular object.For example: A piece of stone, pieces of broken glass etc.Hope this helps...
Good luck on your assignment..
The Thomson model of a hydrogen atom is a sphere of positive charge with an electron (a point charge) at its center. The total positive charge equals the electronic charge e. Prove that when the electron is at a distance r from the center of the sphere of positive charge, it is attracted with a force F=\frac{e^2r}{4\pi\varepsilon_oR^3} where R is the radius of the sphere.
Answer:
E = (1 / 4π ε₀ ) q r / R³
Explanation:
Thomson's stable model that the negative charge is mobile within the atom and the positive charge is uniformly distributed, to calculate the force we can use Coulomb's law
F = K q₁ q₂ / r²
we used law Gauss
Ф = ∫ E .dA = q_{int} /ε₀
E 4π r² = q_{int} /ε₀
E = q_{int} / 4π ε₀ r²
we replace the charge inside
E = (1 / 4π ε₀ r²) ρ 4/3 π r³
E = ρ r / 3 ε₀
the density for the entire atom is
ρ = Q / V
V = 4/3 π R³
we substitute
E = (r / 3ε₀ ) Q 3/4π R³
E = (1 / 4π ε₀ ) q r / R³
Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are charged with equal amount of opposite charges, ±17 µC. The charges on the plates face each other. Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates when the normal to the circle makes an angle of 4° with a line perpendicular to the plates. Note that this angle can also be given as 180° + 4°. N · m2/C
Answer:
Φ = 361872 N.m^2 / C
Explanation:
Given:-
- The area of the two plates, [tex]A_p = 180 cm^2[/tex]
- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]
- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]
- The radius for the flux region, [tex]r = 3.3 cm[/tex]
- The angle between normal to region and perpendicular to plates, θ = 4°
Find:-
Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.
Solution:-
- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:
[tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]
- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):
σ = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]
σ = 0.00094 C / m^2
- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.
[tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]
- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:
[tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]
- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:
Φ = E_net * Ar * cos ( θ )
Φ = (106214689.26553) * (0.00342) * cos ( 5 )
Φ = 361872 N.m^2 / C
A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 43.0 vibrations in 33.0 s. Also, a given maximum travels 424 cm along the rope in 15.0 s. What is the wavelength
Answer:
0.218
Explanation:
Given that
Total vibrations completed by the wave is 43 vibrations
Time taken to complete the vibrations is 33 seconds
Length of the wave is 424 cm = 4.24 m
to solve this problem, we first find the frequency.
Frequency, F = 43 / 33 hz
Frequency, F = 1.3 hz
Also, we find the wave velocity. Which is gotten using the relation,
Wave velocity = 4.24 / 15
Wave velocity = 0.283 m/s
Now, to get our answer, we use the formula.
Frequency * Wavelength = Wave Velocity
Wavelength = Wave Velocity / Frequency
Wavelength = 0.283 / 1.3
Wavelength = 0.218
A piston of small cross-sectional area a is used in a hydraulic press to exert a small force f on the enclosed liquid. A connecting pipe leads to a larger piston of cross sectional area A. a) What force F will the larger piston sustain
Answer:
force on larger piston = [tex]\frac{fA}{a}[/tex]
Explanation:
we label the pistons as piston A and piston B
small piston A:
area = a
force = f
large piston B:
area = A
force = ?
Pascal's law of pressure state that the pressure delivered to a liquid is transmitted undiminished to every portion of the fluid.
we know that pressure = force ÷ area
pressure of piston A = [tex]\frac{f}{a}[/tex]
pressure of piston B = [tex]\frac{(force on piston B)}{A}[/tex]
obeying Pascal's law, the system pressures must be equal. Therefore
[tex]\frac{f}{a} = \frac{(force on piston B)}{A}[/tex]
force on large piston (B) = F = [tex]\frac{fA}{a}[/tex]
A person is standing on an elevator initially at rest at the first floor of a high building. The elevator then begins to ascend to the sixth floor, which is a known distance h above the starting point. The elevator undergoes an unknown constant acceleration of magnitude a for a given time interval T. Then the elevator moves at a constant velocity for a time interval 4T. Finally the elevator brakes with an acceleration of magnitude a, (the same magnitude as the initial acceleration), for a time interval T until stopping at the sixth floor.
Answer:
The found acceleration in terms of h and t is:
[tex]a=\frac{h}{5(t_1)^2}[/tex]
Explanation:
(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)
We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.
Stage 1Constant acceleration, starts from rest.
Distance = [tex]y = \frac{1}{2}a(t_1)^2[/tex]
Velocity = [tex]v_1=at_1[/tex]
Stage 2Constant velocity where
Velocity = [tex]v_o=v_1=at_1[/tex]
Distance =
[tex]y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\[/tex]Stage 3Constant deceleration where
Velocity = [tex]v_0=v_1=at_1[/tex]
Distance =
[tex]y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2[/tex]
Total HeightTotal height = y₁ + y₂ + y₃
Total height = [tex]\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2[/tex]
AccelerationFind acceleration by rearranging the found equation of total height.
Total Height = h
h = 5a(t₁)²
[tex]a=\frac{h}{5(t_1)^2}[/tex]
A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 30.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 230 complete revolutions.
a) At what rate is the flywheel spinning when the power comes back on?
b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?
Answer:
Explanation:
Hello,
Let's get the data for this question before proceeding to solve the problems.
Mass of flywheel = 40kg
Speed of flywheel = 590rpm
Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.
Time = 30s = 0.5 min
During the power off, the flywheel made 230 complete revolutions.
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + ω₂) / 2] × 0.5
But ∇θ = 230 revolutions
∇θ/t = (530 + ω₂) / 2
230 / 0.5 = (530 + ω₂) / 2
Solve for ω₂
460 = 295 + 0.5ω₂
ω₂ = 330rpm
a)
ω₂ = ω₁ + αt
but α = ?
α = (ω₂ - ω₁) / t
α = (330 - 590) / 0.5
α = -260 / 0.5
α = -520rev/min
b)
ω₂ = ω₁ + αt
0 = 590 +(-520)t
520t = 590
solve for t
t = 590 / 520
t = 1.13min
60 seconds = 1min
X seconds = 1.13min
x = (60 × 1.13) / 1
x = 68seconds
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + 0) / 2] × 1.13
∇θ = 333.35 rev/min
1- A 30 gram bullet travels at 300 m/s. How much kinetic energy does it have?
Answer:
1.35 kJ
Explanation:
KE = ½mv² = ½ × 0.030 kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ
Given:-
Mass (m) of the bullet = 30 gramsVelocity of the bullet (v) = 300 m/sTo Find: Kinetic energy of the bullet.
We know,
Eₖ = ½mv²
where,
Eₖ = Kinetic energy,m = Mass &v = Velocity.thus,
Eₖ = ½(30 g)(300 m/s)²
= (15 g)(90000 m²/s²)
= 1350000 g m²/s²
= 1350 kg m²/s²
= 1350 J
= 1.35 kJ (Ans.)
The probability of nuclear fusion is greatly enhanced when the appropriate nuclei are brought close together, but their mutual coulomb repulsion must be overcome. This can be done using the kinetic energy of high temperature gas ions or by accelerating the nuclei toward one another.
Required:
a. Calculate the potential energy of two singly charged nuclei separated by 1.00×10^−12m
b. At what temperature will atoms of a gas have an average kinetic energy equal to this needed electrical potential energy?
Answer:
a
[tex]PE = 2.3 *10^{-16} \ J[/tex]
b
[tex]T = 1.1 *10^{7} \ K[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 1.00 *10^{-12} \ m[/tex]
Generally the electric potential energy can be mathematically represented as
[tex]PE = \frac{k * q_1 q_2 }{d}[/tex]
Given that in a nuclei the only charged particle is the proton who charge is
[tex]p = 1.60 *10^{-19} \ C[/tex]
Hence
[tex]q_1 = q_2 = 1.60 *10 ^{-19} \ C[/tex]
And k is the coulomb constant with values [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.N/A2[/tex]
So we have that
[tex]PE = \frac{9*10^9 * (1.60 *10^{-19})^2}{ 1.00*10^{-12}}[/tex]
[tex]PE = 2.3 *10^{-16} \ J[/tex]
The relationship between the electrical potential energy and the temperature is mathematically represented as
[tex]PE = \frac{3}{2} kT[/tex]
Here k is the Boltzmann's constant with value [tex]k = 1.38*10^{-23} JK^{-1}[/tex]
making T the subject
[tex]T = \frac{2}{3} * \frac{PE}{k}[/tex]
substituting values
[tex]T = \frac{2}{3} * \frac{2.30 *10^{-16}}{ 1.38 *10^{-23}}[/tex]
[tex]T = 1.1 *10^{7} \ K[/tex]