Answer:
C . Magnetic dipoles is the correct
Answer:
b). movement of charged particles.
Explanation:
These charges create the nagnetic dipoles.
A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that the spring is compressed by 0.090 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise
After being released, the restoring force exerted by the spring performs
1/2 (5200 N/m) (0.090 m)² = 12.06 J
of work on the block. At the same time, the block's weight performs
- (0.260 kg) g (0.090 m) ≈ -0.229 J
of work. Then the total work done on the block is about
W ≈ 11.83 J
The block accelerates to a speed v such that, by the work-energy theorem,
W = ∆K ==> 11.83 J = 1/2 (0.260 kg) v ² ==> v ≈ 9.54 m/s
Past the equilibrium point, the spring no longer exerts a force on the block, and the only force acting on it is due to its weight, hence it has a downward acceleration of magnitude g. At its highest point, the block has zero velocity, so that
0² - v ² = -2gy
where y is the maximum height. Solving for y gives
y = v ²/(2g) ≈ 4.64 m
An electron is moving at speed of 6.3 x 10^4 m/s in a circular path of radius of 1.7 cm inside a solenoid the magnetic field of the solenoid is perpendicular to the plane of the electron's path. Find its relevatn motion.
Answer:
Here, m=9×10
−31
kg,
q=1.6×10
−19
C,v=3×10
7
ms
−1
,
b=6×10
−4
T
r=
qB
mv
=
(1.6×10
−19
)(6×10
−4
)
(9×10
−31
)×(3×10
7
)
=0.28m
v=
2πr
v
=
2πm
Bq
=
2×(22/7)×9×10
−31
(6×10
−4
)×(1.6×10
−19
)
=1.7×10
7
Hz
Ek=
2
1
mv
2
=
2
1
×(9×10
−31
)×(3×10
7
)
2
J
=40.5×10
−17
J=
1.6×10
−16
40.5×10
−17
keV
=2.53keV
An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]
Apply the principle of conservation of energy;
ΔK.E = ΔU
[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]
Therefore, the speed of the electron at the given position is 106.2 m/s
on a horizontal axis whose unit is the meter, a linear load ranging from 0 to 1 ma a linear load distribution = 2 nC / m.
determine the modulus of the electric field created by the previous loaded bar at the point A of abscissa 2m (we have to find the relation between l, which is the distance between the elementary bar and the point A and x which sweeps the segment [0: 1]
Answer:
The correct answer is - 8.99N/C
Explanation:
[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,
[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]
I assume the path itself is a line segment, which can be parameterized by
[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]
with 0 ≤ t ≤ 1. Then the work performed by F along C is
[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]
answer bhejo please please please
Answer:
Various uses of water :
1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.
2. Water used as a universal solvent.
3. water maintains the temperature of our body.
4. Water helps in digestion in our body.
5 .water is used in factories and industries.
6. Water is used to grow plants , vegetables and crops.
What are the systems of units? Explain each of them.
THERE ARE COMMONLY THREE SYSTEMS OF UNIT. THEY ARE:-
• CGS System- (Centimeter-Gram-Second system) A metric system of measurement that uses the centimeter, gram and second for length, mass and time.
• FPS System- (Foot–Pound–Second system).
The system of units in which length is measured in foot , mass in pound and time in second is called FPS system. It is also known as British system of units.
• MKS System- (Meter-Kilogram-Second system) A metric system of measurement that uses the meter, kilogram, gram and second for length, mass and time. The units of force and energy are the "newton" and "joule."
A 700N marine in basic training climbs a 10m vertical rope at constant speed in 8sec. what is power put
Answer:
875 Watts
Explanation:
P = W/t = mgh/t = 700(10)/8 = 875 Watts
Define wave length as applied to wave motion
Answer: Wavelength can be defined as the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave.
Explanation:
Wavelength refers to the length or distance between two identical points of neighboring cycles of a wave signal traveling in space or in any physical medium. ... The wavelength of a signal is inversely proportional to its frequency, that is, the higher the frequency, the shorter the wavelength.
Polarized sunglasses:
a. block most sunlight because sunlight is polarized
b. are better but work the same way as non-polarized sunglasses
c. are polarized to filter out certain wavelengths of light
d. block reflected light because reflected light is partially polarized.
Polarized sunglasses creates filter of vertical openings for light. The light rays will reach the eyes of human vertically only.
The sun rays will not reach human eye directly which will create a shield for sun light burden on human eye.
Polarized sunglasses are best used for blocking and eliminating certain wavelengths of light.
Therefore the correct answer is option C. Polarizes Sunglasses are polarized and it filter out certain wavelengths of light.
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Choose the appropriate explanation how such a low value is possible given Saturn's large mass - 100 times that of Earth.
a. This low value is possible because the magnetic field of Saturn is so strong.
b. This low value is possible because the magnetic field of Saturn is so weak.
c. This low value is possible because the density of Saturn is so high.
d. This low value is possible because the density of Saturn is so low.
Answer:
Explanation:
That is an amazing fact.
The minus sign is what you have to pay attention to. The earth has a mass of 100 times that of Saturn. As someone on here once noted, Saturn has such a low density that it would float in water.
The answer is D
A cylindrical swimming pool has a radius 2m and depth 1.3m .it is completely filled with salt water of specific gravity 1.03.The atmospheric preassure is 1.013 x 10^5 Pa.
a.calculate the density of salt water.
Answer:
the density of the salt water is 1030 kg/m³
Explanation:
Given;
radius of the cylindrical pool, r = 2 m
depth of the pool, h = 1.3 m
specific gravity of the salt water, γ = 1.03
The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa
Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³
The density of the salt water is calculated as;
[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]
Therefore, the density of the salt water is 1030 kg/m³
You are to connect resistors R1 andR2, with R1 >R2, to a battery, first individually, then inseries, and then in parallel. Rank those arrangements according tothe amount of current through the battery, greatest first. (Useonly the symbols > or =, for exampleseries>R1=R2>parallel.)
Answer:
The current is more in the parallel combination than in the series combination.
Explanation:
two resistances, R1 and R2 are connected to a battery of voltage V.
When they are in series,
R = R1 + R2
In series combination, the current is same in both the resistors, and it is given by Ohm's law.
V = I (R1 + R2)
[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)
When they are connected in parallel.
the voltage is same in each resistor.
The effective resistance is R.
[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]
So, the current is
[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)
So, the current is more is the parallel combination.
A uniform disk turns at 3.6 rev/s around a frictionless spindle. A non rotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk . They then both turn around the spindle with their centers superposed.
What is the angular frequency in rev/s of the combination?
please express answer in proper significant figures and rounding.
Answer:
ω₁ = 2.2 rev/s
Explanation:
Conservation of angular momentum
moment of inertia uniform disk is ½mR²
moment of inertia uniform rod about an end mL²/3
We can think of our rod as two rods of mass m/2 and length R
L = ½mR²ω₀
L = (½mR² + 2(m/2)R²/3)ω₁
ω₁ = ω₀(½mR² / (½mR² + mR²/3))
ω₁ = ω₀(½ / (½ + 1/3))
ω₁ = 0.6ω₀
ω₁ = 2.16
A 207-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.750 rev/s in 2.00 s
Answer:
366 N
Explanation:
τ = Iα
FR = ½mR²α
F = ½mR(Δω/t)
F = ½(207)(1.50)(0.75)(2π) /2.00
F = 365.79919...
If ∆H = + VE , THEN WHAT REACTION IT IS
1) exothermic
2) endothermic
Answer:
endothermic
Explanation:
An endothermic is any process with an increase in the enthalpy H (or internal energy U) of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transfer into the system.
The armature of an AC generator has 200 turns, which are rectangular loops measuring 5 cm by 10 cm. The generator has a sinusoidal voltage output with an amplitude of 18 V. If the magnetic field of the generator is 300 mT, with what frequency does the armature turn
Answer:
[tex]f=9.55Hz[/tex]
Explanation:
From the question we are told that:
Number of Turns [tex]N=200[/tex]
Length [tex]l=5cm to 10cm[/tex]
Voltage [tex]V=18V[/tex]
Magnetic field [tex]B=300mT[/tex]
Generally, the equation for Frequncy of an amarture is mathematically given by
[tex]f =\frac{ V}{(N B A * 2 pi )}[/tex]
[tex]f =\frac{ 18}{(200 300*10^{-3} (10*10^-2)(5*10^{-2}) * 2 *3.142 )}[/tex]
[tex]f=9.55Hz[/tex]
When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pulls on the cell phone from you, the Earth, and the Sun. Rank the gravitational forces on the phone from largest to smallest. Assume the Sun is roughly 109 times further away from the phone than you are, and 1028 times more massive than you. Rank the following choices in order from largest gravitational pull on the phone to smallest. To rank items as equivalent, overlap them.
a. Pull phone from you
b. Pull on phone from earth
c. Pull on phone from sun
Answer:
The answer is "Option b, c, and a".
Explanation:
Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.
We now understand the effects of gravity:
[tex]F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}[/tex]
The force of the sun is, therefore, [tex]10^{10}[/tex] times greater and the proper sequence, therefore, option steps are:
b. Pull-on phone from earth
c. Pull-on phone from sun
a. Pull phone from you
Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the sound intensity increase and then decrease periodically due to wave interference. How long does it take between successive instances of the sound intensity increasing
Answer:
[tex]T=1.54s[/tex]
Explanation:
From the question we are told that:
Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]
Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]
Therefore
Frequency of Motor 1 [tex]f_1=13.75[/tex]
Frequency of Motor 2 [tex]f_2= 13.1[/tex]
Generally the equation for Time Elapsed is mathematically given by
[tex]T=\frac{1}{df}[/tex]
Where
[tex]df=f_1-f_2[/tex]
[tex]df=13.75-13.1[/tex]
[tex]df=0.65Hz[/tex]
Therefore
[tex]T=\frac{1}{65}[/tex]
[tex]T=1.54s[/tex]
Which was a major effect of Pope Leo III crowning Charlemagne emperor of the Romans ?
Answer:
The crowning of Charlemagne by Pope Leo III was significant in a number of ways. For Charlemagne, it was necessary because it encouraged to give him higher reliability. It gave him the rank of a dictator, giving him the only ruler in Europe west of the Byzantine emperor in Constantinople.
a point object is 10 cm away from a plane mirror while the eye of an observer(pupil diameter is 5.0 mm) is 28 cm a way assuming both eye and the point to be on the same line perpendicular to the surface find the area of the mirror used in observing the reflection of the point
Answer:
1.37 mm²
Explanation:
From the image attached below:
Let's take a look at the two rays r and r' hitting the same mirror from two different positions.
Let x be the distance between these rays.
[tex]d_o =[/tex] distance between object as well as the mirror
[tex]d_{eye}[/tex] = distance between mirror as well as the eye
Thus, the formula for determining the distance between these rays can be expressed as:
[tex]x = 2d_o tan \theta[/tex]
where; the distance between the eye of the observer and the image is:
[tex]s = d_o + d_{eye}[/tex]
Then, the tangent of the angle θ is:
[tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex]
replacing [tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex] into [tex]x = 2d_o tan \theta[/tex], we have:
[tex]x = 2d_o \Big( \dfrac{R}{d_o+d_{eye}}\Big)[/tex]
[tex]x = 2(10) \Big( \dfrac{0.25}{10+28}\Big)[/tex]
[tex]x = 20\Big( \dfrac{0.25}{38}\Big) cm[/tex]
x = (0.13157 × 10) mm
x = 1.32 mm
Finally, the area A = π r²
[tex]A = \pi(\frac{x}{2})^2[/tex]
[tex]A = \pi(\frac{1.32}{2})^2[/tex]
A = 1.37 mm²
Two train 75 km apart approach each other on parallel tracks, each moving at
15km/h. A bird flies back and forth between the trains at 20km/h until the trains pass
each other. How far does the bird fly?
Answer:
The correct solution is "37.5 km".
Explanation:
Given:
Distance between the trains,
d = 75 km
Speed of each train,
= 15 km/h
The relative speed will be:
= [tex]15 + (-15)[/tex]
= [tex]30 \ km/h[/tex]
The speed of the bird,
V = 15 km/h
Now,
The time taken to meet will be:
[tex]t=\frac{Distance}{Relative \ speed}[/tex]
[tex]=\frac{75}{30}[/tex]
[tex]=2.5 \ h[/tex]
hence,
The distance travelled by the bird in 2.5 h will be:
⇒ [tex]D = V t[/tex]
[tex]=15\times 2.5[/tex]
[tex]=37.5 \ km[/tex]
Develop a hypothesis regarding one factor you think might affect the period of a pendulum or an oscillating mass on a spring. Potential factors include the mass, the spring constant, and the length of the pendulum's string. Write down your hypothesis. 2. Design a controlled experiment to test your hypothesis. Take extreme care to keep all factors constant except the variable you are testing.
Answer:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
Explanation:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
To test this hypothesis we can carry out a measurement of a simple pendulum keeping the angle fixed, in general the angle used is about 5º since when placing this value in radiand and the sine of this angle they differ little <5%. therefore measured the time of some oscillations, for example about 10 oscillations, changing the length of the pendulum to test the hypothesis.
If the hypothesis and the model used is correct, the relationship to be tested is
T² =(4π² /g) L
by making a graph of the period squared against the length if obtaining, os a line, the hypothesis is tested.
19 point please please answer right need help
Block on an incline
A block of mass m1 = 3.9 kg on a smooth inclined plane of angle 38is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 2.6 kg hanging vertically. Take the positive direction up the incline and use 9.81
m/s2 for g.
What is the tension in the cord to the nearest whole number?
Explanation:
We can write Newton's 2nd law as applied to the sliding mass [tex]m_1[/tex] as
[tex]T - m_1g\sin38 = m_1a\:\:\:\:\:\:\:(1)[/tex]
For the hanging mass [tex]m_2,[/tex] we can write NSL as
[tex]T - m_2g = -m_2a\:\:\:\:\:\:\:(2)[/tex]
We need to solve for a first before we can solve the tension T. So combining Eqns(1) & (2), we get
[tex](m_1 + m_2)a = m_2g - m_1g\sin38[/tex]
or
[tex]a = \left(\dfrac{m_2 - m_1\sin38}{m_1 + m_2}\right)g[/tex]
[tex]\:\:\:\:= 0.30\:\text{m/s}^2[/tex]
Using this value for the acceleration on Eqn(2), we find that the tension T is
[tex]T = m_2(g - a) = (2.6\:\text{kg})(9.51\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:=24.7\:\text{N}[/tex]
What is the magnitude of the force between a 25μC charge exerts on a -10μC charge 8.5cm away?
Answer:
Force,
[tex]F = \frac{kQ_{1} Q_{2} }{ {r}^{2} } \\ F = \frac{(9 \times {10}^{9}) \times (25 \times {10}^{ - 6}) \times (10 \times {10}^{ - 6} ) }{ {(0.85)}^{2} } \\ \\ F = 3.114 \: newtons[/tex]
The magnitude of the force between a 25μC charge exerts on a -10μC charge 8.5cm away would be 311.4 N.
What is Coulomb's Law?Coulomb's law can be stated as the product of the charges and the square of the distance between them determine the force of attraction or repulsion acting in a straight line between two electric charges.
The math mathematical expression for the coulomb's law given as
F= k Q₁Q₂/r²
where F is the force between two charges
k is the electrostatic constant which is also known as the coulomb constant,it has a value of 9×10⁹
Q₁ and Q₂ are the electric charges
r is the distance between the charges
As given in the problem two charges a 25μC charge exerts on a -10μC charge 8.5cm away
By substituting the respective values in the above formula of Coulomb law
F =9×10⁹×(25×10⁻⁶)×(-10×10⁻⁶)/(8.5×10⁻²)²
F= -311.4 N
A negative sign represents that the force is attractive in nature
Thus, the magnitude of the force is 311.4 N.
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A man standing in an elevator holds a spring scale with a load of 5 kg suspended from it. What would be the reading of the scale, if the elevator is accelerating downward with an acceleration 3.8 m/s?.
Answer:
3.1 kg
Explanation:
Applying,
R = m(g-a)..................... Equation 1
Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.
From the question,
Given: m = 5 kg, a = 3.8 m/s²
Constant: g = 9.8 m/s²
Substitute these values into equation 1
R = 5(9.8-3.8)
R = 5(6)
R = 30 N
Hence the spring scale is
m' = R/g
m' = 30/9.8
m' = 3.1 kg
A spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight up in the classroom. The spring is initially compressed 0.2 m, and the block is initially at rest when it is released. When the block is 1.3 m above its starting position, what is its speed
Answer:
the speed of the block at the given position is 21.33 m/s.
Explanation:
Given;
spring constant, k = 3500 N/m
mass of the block, m = 4 kg
extension of the spring, x = 0.2 m
initial velocity of the block, u = 0
displacement of the block, d =1.3 m
The force applied to the block by the spring is calculated as;
F = ma = kx
where;
a is the acceleration of the block
[tex]a = \frac{kx}{m} \\\\a = \frac{(3500) \times (0.2)}{4} \\\\a = 175 \ m/s^2[/tex]
The final velocity of the block at 1.3 m is calculated as;
v² = u² + 2ad
v² = 0 + 2ad
v² = 2ad
v = √2ad
v = √(2 x 175 x 1.3)
v = 21.33 m/s
Therefore, the speed of the block at the given position is 21.33 m/s.
The speed of the block at a height of 1.3 m above the starting position is 21.33 m/s
To solve this question, we'll begin by calculating the acceleration of the block.
How to determine the acceleration Spring constant (K) = 3500 N/m Mass (m) = 4 KgCompression (e) = 0.2 mAcceleration (a) =?F = Ke
Also,
F = ma
Thus,
ma = Ke
Divide both side by m
a = Ke / m
a = (3500 × 0.2) / 4
a = 175 m/s²
How to determine the speed Initial velocity (u) = 0 m/sAcceleration (a) = 175 m/s²Distance (s) = 1.3 mFinal velocity (v) =?v² = u² + 2as
v² = 0² + (2 × 175 × 1.3)
v² = 455
Take the square root of both side
v = √455
v = 21.33 m/s
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If a jet travels 350 m/s, how far will it travel each second?
Answer:
It will travel 350 meters each second.
Explanation:
The unit rate, 350 m/s, tells us that the jet will travel 350 meters per every second elapsed.
Answer:
5.83 seconds
Explanation:
60 seconds in 1 minute
350 meters per second
350/60
=5.83
A 0.160 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s. It has a head-on collision with a 0.296 kg glider that is moving to the left with a speed of 2.23 m/s. Suppose the collision is elastic.
Required:
a. Find the magnitude of the final velocity of the 0.157kg glider.
b. Find the magnitude of the final velocity of the 0.306kg glider.
The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.
Momentum is conserved, so the total momentum of the system is the same before and after the collision:
m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'
==>
(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'
==>
-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'
where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.
Kinetic energy is also conserved, so that
1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' )² + 1/2 m₂ (v₂' )²
or
m₁ v₁² + m₂ v₂² = m₁ (v₁' )² + m₂ (v₂' )²
==>
(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
==>
1.55 kg•m²/s² ≈ (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is
v₁' ≈ -3.11 m/s
v₂' ≈ -0.167 m/s
and take the absolute values to get the magnitudes.
If you want to instead use the masses from the "Required" section, you would end up with
v₁' ≈ -3.18 m/s
v₂' ≈ -0.236 m/s
What is the volume of a metal block 3cm long by 2cm wide by 4cm high? What would be the volume of a block twice as long, wide, and high?
Answer:
Volume of a metal block = 24 cm^3
Volume of a block twice as long, wide and high = 192 cm^3
Explanation:
Volume of a block is measured in l*w*h and in the first block, the sides are 3, 2 and 4 and 3*2*4 = 24
Second block, just double each of the lengths to get 6*4*8 = 192
