Answer:
F' = F
Hence, the magnitude of the attractive force remains same.
Explanation:
The force of attraction between two bodies is given by Newton's Gravitational Law:
F = Gm₁m₂/r² --------------- equation 1
where,
F = Force of attraction between balls
G = Universal Gravitational Constant
m₁ = mass of first ball
m₂ = mass of 2nd ball
r = distance between balls
Now, we double the masses of both balls and the separation between them. So, the force of attraction becomes:
F' = Gm₁'m₂'/r'²
here,
m₁' = 2 m₁
m₂' = 2 m₂
r' = 2 r
Therefore,
F' = G(2 m₁)(2 m₂)/(2 r)²
F' = Gm₁m₂/r²
using equation 1:
F' = F
Hence, the magnitude of the attractive force remains same.
The inner and outer surface temperature of a glass window 10 mm thick are 25 and 5 degree-C, respectively. What is the heat loss through a 1 m x 3 m window
Answer:
The heat loss is [tex]H = 8400\ W[/tex]
Explanation:
From the question we are told that
The thickness is [tex]t = 10 \ mm = 0.01 \ m[/tex]
The inner temperature is [tex]T_i = 25 ^oC[/tex]
The outer temperature is [tex]T_o = 5 ^oC[/tex]
The length of the window is L = 1 m
The width of the window is w = 3 m
Generally the heat loss is mathematically represented as
[tex]H = \frac{k * A * \Delta T}{t}[/tex]
Where k is the thermal conductivity of glass with value [tex]k = 1.4\ W/m \cdot K[/tex]
and A is the area of the window with value
[tex]A = 1 * 3[/tex]
[tex]A = 3 \ m^2[/tex]
substituting values
[tex]H = \frac{1.4 * 3 * (23-5)}{0.01}[/tex]
[tex]H = 8400\ W[/tex]
Suppose you have two point charges of opposite sign. As you move them farther and farther apart, the potential energy of this system relative to infinity:_____________.
(a) stays the same.
(b) Increases.
(c) Decreases.
(d) The answer would depend on the path of motion
Answer:
(b) Increases
Explanation:
The potential energy between two point charges is given as;
[tex]U = F*r = \frac{kq_1q_2}{r}[/tex]
Where;
k is the coulomb's constant
q₁ ans q₂ are the two point charges
r is the distance between the two point charges
Since the two charges have opposite sign;
let q₁ be negative and q₂ be positive
Substitute in these charges we will have
[tex]U = \frac{k(-q_1)(q_2)}{r} \\\\U = - \frac{kq_1q_2}{r}[/tex]
The negative sign in the above equation shows that as the distance between the two charges increases, the potential energy increases as well.
Therefore, as you move the point charges farther and farther apart, the potential energy of this system relative to infinity Increases.
A scooter is traveling at a constant speed v when it encounters a circular hill of radius r = 480 m. The driver and scooter together have mass m = 159 kg.
(a) What speed in m/s does the scooter have if the driver feels weightlessness (i.e., has an apparent weight of zero) at the top of the hill?
(b) If the driver is traveling at the speed above and encounters a hill with a radius 2r,
Answer:
68.585m/sec , 779.1 N
Explanation:
To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.
From centripetal motion.
F =( mv^2)/2
But since we are dealing with weightlessness
r = 480m
g = 9.8m/s^2
M also cancels, so forget M.
V^2 = Fr
V = √ Fr
V =√ (9.8 x 480) = 4704
= 68.585m/sec.
b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960
= 4.9m/sec^2.
Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)
159kg × ( 9.8-4.9)
159kg × 4.9
= 779.1N
A) The speed of the scooter at which the driver will feel weightlessness is;
v = 68.586 m/s
B) The apparent weight of both the driver and the scooter at the top of the hill is;
F_net = 779.1 N
We are given;
Mass; m = 159 kg
Radius; r = 480 m
A) Since it's motion about a circular hill, it means we are dealing with centripetal force.
Formula for centripetal force is given as;
F = mv²/r
Now, we want to find the speed of the scooter if the driver feels weightlessness.
This means that the centripetal force would be equal to the gravitational force.
Thus;
mg = mv²/r
m will cancel out to give;
v²/r = g
v² = gr
v = √(gr)
v = √(9.8 × 480)
v = √4704
v = 68.586 m/s
B) Now, he is travelling with speed of;
v = 68.586 m/s
And the radius is 2r
Let's first find the centripetal acceleration from the formula; α = v²/r
Thus; α = 4704/(2 × 480)
α = 4.9 m/s²
Now, since he has encountered a hill with a radius of 2r up the slope, it means that the apparent weight will now be;
F_app = m(g - α)
F_net = 159(9.8 - 4.9)
F_net = 779.1 N
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The process of star and planet formation begins with a large cloud of gas and dust called a solar nebula. Rank the formation events that occur within a cloud from earliest to latest.
Rank from earliest to latest. To rank items as equivalent, overlap them.
A. The cloud is large, cool, and slowly rotating
B. The cloud collapses into a disk
C. Competing rotational and gravitational forces begin to flatten the cloud
D. The cloud becomes denser, heats up, and rotates faster
E. The cloud starts to contract under the influence of gravity
A, B, E , C, D
What is Nebula?A nebula is an enormous cloud of dust and gas occupying the space between stars and acting as a nursery for new stars.
Nebulae are made up of dust, basic elements such as hydrogen and other ionized gases.
Nebula Formation:
In essence, a nebula is formed when portions of the interstellar medium undergo gravitational collapse.
Mutual gravitational attraction causes matter to clump together, forming regions of greater and greater density.
The formation events that occur within a cloud from earliest to latest are:
A. The cloud is large, cool, and slowly rotating
B. The cloud collapses into a disk.
E. The cloud starts to contract under the influence of gravity
C. Competing rotational and gravitational forces begin to flatten the cloud.
D. The cloud becomes denser, heats up, and rotates faster
Therefore , The rank from earliest to latest is A, B, E , C, D
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The temperature gradient between the core of Mars and its surface is approximately 0.0003 K/m. Compare this temperature gradient to that of Earth. What can you determine about the rate at which heat moves out of Mars’s core compared to Earth?
Answer:
The temperature gradient between the core of Mars and its surface is lower than that on Earth. So, heat moves outward more slowly on Mars than on Earth.
Explanation:
Answer:
The temperature gradient between the core of Mars and its surface is lower than that on Earth. So, heat moves outward more slowly on Mars than on Earth.
Explanation:
Edmentum sample answer
A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.17 s. What are the (a) period and (b) frequency? (c) The wavelength is 1.5 m; what is the wave speed?
Answer:
31
Explanation:
A plane is flying horizontally with a constant speed of 55 .0 m/s when it drops a
rescue capsule. The capsule lands on the ground 12.0 s later.
c) How would your answer to part b) iii change if the constant speed of the plane is
increased? Explain.
Answer:
therefore horizontal displacement changes increasing with linear velocity
Explanation:
Since the plane flies horizontally, the only speed that exists is
v₀ₓ = 55.0 m / s
the time is the time it takes to reach the floor, which we can find because the speed on the vertical axis is zero
y =y₀ + v₀ t - ½ g t2
0 = I₀ + 0 - ½ g t2
t = √ 2y₀o / g
time is that we use to calculate the x-axis displacement
The distance it travels to reach the floor is
x = v t
x = 55 12
x = 660 m
When the speed horizontally the time remains the same and 120
x ’= v’ 12
therefore horizontal displacement changes increasing with linear velocity
An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 2.14 mT. If the speed of the electron is 1.48 107 m/s, determine the following.
(a) the radius of the circular path ............ cm
(b) the time interval required to complete one revolution ............ s
Answer:
(a) 3.9cm
(b) 1.66 x 10⁻⁸s
Explanation:
Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton's second law of motion is given by,
F = m x a --------------(i)
Where;
m = mass of the particle
a = acceleration of the mass
The centripetal acceleration is given by;
a = v² / r [v = linear velocity of particle, r = radius of circular path]
Therefore, equation (i) becomes;
F = m v²/ r --------------------(ii)
The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz's law is given by;
F = qvBsinθ -------------(iii)
Where;
q = charge of the particle
v = velocity of the particle
B = magnetic field
θ = angle between the velocity and the magnetic field
Combine equations (ii) and (iii) as follows;
m (v² / r) = qvBsinθ [divide both side by v]
m v / r = qBsinθ [make r subject of the formula]
r = (m v) / (qBsinθ) ---------(iv)
(a) From the question;
v = 1.48 x 10⁷m/s
B = 2.14mT = 2.14 x 10⁻³T
θ = 90° [since the direction of velocity is perpendicular to magnetic field]
m = mass of electron = 9.11 x 10⁻³¹kg
q = charge of electron = 1.6 x 10⁻¹⁹C
Substitute these values into equation (iv) as follows;
r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)
r = 3.9 x 10⁻²m
r = 3.9cm
Therefore, the radius of the circular path is 3.9cm
(b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by
T = d / v --------------(*)
Where;
d = distance traveled in the circular path in one complete turn = 2πr
v = velocity of the motion = 1.48 x 10⁷m/s
d = 2 π (3.9 x 10⁻²) [Take π = 22/7 = 3.142]
d = 2(3.142)(3.9 x 10⁻²) = 0.245m
Substitute the values of d and v into equation (*) as follows;
T = 0.245 / 1.48 x 10⁷
T = 0.166 x 10⁻⁷s
T = 1.66 x 10⁻⁸s
Therefore, the time interval is 1.66 x 10⁻⁸s
One hundred turns of insulated copper wire are wrapped around an iron core of cross-sectional area 0.100m2. As the magnetic field along the coil axis changes from 0.5 T to 1.00T in 4s, the voltage induced is:
Answer:
The voltage induced in the coil is 1.25 V.
Explanation:
Given;
number of turns, N = 100 turns
cross sectional area of the copper coil, A = 0.1 m²
initial magnetic field, B₁ = 0.5 T
final magnetic field, B₂ = 1.00 T
duration of change in magnetic field, dt = 4 s
The induced emf in the coil is calculated as;
[tex]emf = -N\frac{\delta \phi}{\delta t} \\\\emf = - N (\frac{\delta B}{\delta t}) A\\\\emf = -N (\frac{B_1 -B_2}{\delta t} )A\\\\emf = N(\frac{B_2-B_1}{\delta t} )A\\\\emf = 100(\frac{1-0.5}{4} )0.1\\\\emf = 1.25 \ Volts[/tex]
Therefore, the voltage induced in the coil is 1.25 V.
Damon purchased a pair of sunglasses that were advertised as being polarized. Describe how Damon could test the sunglasses to verify they are polarized.
Answer:
To verify that they're polarized, he could hold the two lenses perpendicular (90 degrees) to each other, one lens in front of the other, and point it at a light source. If no light passes through then the lenses are polarized
The test of Polarization of pair of sunglasses is , hold the two lenses perpendicular to each other, one lens in front of the other, and point it towards a light source. If no light passes through then the lenses are polarized.
When a beam of light is reflected from a smooth surface, such as water or ice, it becomes polarized.Polarized light irritates the eyes and makes it hard to see clearly.For example, when fishing on a sunny day, you wouldn't see through the water. You would only see a reflection of the sun hitting the water.
Polarized lenses will neutralize the reflection of the water, and you will be able to into the water.To verify that pair of sunglasses are polarized, he could hold the two lenses perpendicular to each other, one lens in front of the other, and point it towards a light source. If no light passes through then the lenses are polarized.
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The flywheel of an engine has I of 1.60kg.m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rpm in 8.00s, starting from rest?
Answer:
Torque = 8.38Nm
Explanation:
Time= 8.00s
angular speed (w) =400 rpm
Moment of inertia (I)= 1.60kg.m2 about its rotation axis
We need to convert the angular speed from rpm to rad/ sec for consistency
2PI/60*n = 0.1047*409 = 41.8876 rad/sec
What constant torque is required to bring it up to an angular speed of 40rev/min in a time of 8s , starting from rest?
Then we need to use the formula below for our torque calculation
from basic equation T = J*dω/dt ...we get
Where : t= time in seconds
W= angular velocity
T = J*Δω/Δt = 1.60*41.8876/8.0 = 8.38 Nm
Therefore, constant torque that is required is 8.38 Nm
Torque can be defined as the twisting or turning force that tends to cause rotation around an axis. The required constant torque is 8.38 N-m.
Given-
Inertia of the flywheel is 1.60 kg m squared.
Angular speed of the flywheel [tex]n[/tex] is 400 rpm. Convert it into the rad/sec, we get,
[tex]\omega =\dfrac{2\pi }{60} \times n[/tex]
[tex]\omega =\dfrac{2\pi }{60} \times 400[/tex]
[tex]\omega = 41.89[/tex]
Thus, the angular speed of the flywheel [tex]\omega[/tex] is 41.89 rad/sec.
When a torque [tex]\tau[/tex] is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia [tex]I[/tex]. Mathematically,
[tex]\tau=\dfrac{\Delta \omega }{\Delta t} \times I[/tex]
[tex]\tau=\dfrac{ 41.89 }{8} \times 1.6[/tex]
[tex]\tau=8.38[/tex]
Hence, the required constant torque is 8.38 N-m.
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g The current in a series circuit is 15.0 A. When an additional 8.00-% resistor is inserted in series, the current drops to 12.0 A. What is the resistance in the original circuit
Answer:
Explanation:
Let the original resistance be R and voltage be V
Applying ohm's law
V / R = 15
V = 15 R
In second case
V / (R+8 ) = 12
V = 12 R + 96
15 R = 12 R + 96
3R = 96
R = 32 ohm .
5. A nail contains trillions of electrons. Given that electrons repel from each other, why do they not then fly out of the nail?
Answer:
Nails are made of iron. Iron consists of 26 protons and 26 electrons. protons are positively charged and electrons are negatively charged, so this force of attraction keeps the electrons together.
If electrons repel from each other, the positively charge protons and nucleus allow them to move in a definite orbit and prevent them flying out of the nail.
A mass M = 4 kg attached to a string of length L = 2.0 m swings in a horizontal circle with a speed V. The string maintains a constant angle \theta\:=\:θ = 35.4 degrees with the vertical line through the pivot point as it swings. What is the speed V required to make this motion possible?
Answer:
The velocity is [tex]v = 2.84 1 \ m/s[/tex]
Explanation:
The diagram showing this set up is shown on the first uploaded image (reference Physics website )
From the question we are told that
The mass is m = 4 kg
The length of the string is [tex]L = 2.0 \ m[/tex]
The constant angle is [tex]\theta = 35.4 ^o[/tex]
Generally the vertical forces acting on the mass to keep it at equilibrium vertically is mathematically represented as
[tex]Tcos (\theta ) - mg = 0[/tex]
=> [tex]mg = Tcos (\theta )[/tex]
Now let the force acting on mass horizontally be k so from SOHCAHTOA rule
[tex]sin (\theta ) = \frac{k }{T}[/tex]
=> [tex]k = T sin \theta[/tex]
Now this k is also equivalent to the centripetal force acting on the mass which is mathematically represented as
[tex]F_v = \frac{m v^2}{r}[/tex]
So
[tex]k = F_v[/tex]
Which
=> [tex]T sin \theta= \frac{ m v^2}{ r }[/tex]
So
[tex]\frac{Tsin (\theta )}{Tcos (\theta )} = \frac{mg}{ \frac{mv^2}{r} }[/tex]
=> [tex]Tan (\theta ) = \frac{v^2}{ r * g }[/tex]
=> [tex]v = \sqrt{r * g * tan (\theta )}[/tex]
Now the radius is evaluated using SOHCAHTOA rule as
[tex]sin (\theta) = \frac{ r}{L}[/tex]
=> [tex]r = L sin (\theta)[/tex]
substituting values
[tex]r = 2 sin ( 35.4 )[/tex]
[tex]r = 1.1586 \ m[/tex]
So
[tex]v = \sqrt{1.1586* 9.8 * tan (35.4 )}[/tex]
[tex]v = 2.84 1 \ m/s[/tex]
An object has an acceleration of 6.0 m/s/s. If the net force was tripled and the mass were doubled, then the new acceleration would be _____ m/s/s.
Answer:
The new acceleration would be 9 m/s².
Explanation:
Acceleration of an object is 6 m/s²
Net force is equal to the product of mass and acceleration i.e.
F = ma
[tex]a=\dfrac{F}{m}\\\\\dfrac{F}{m}=6\ m/s^2[/tex]
If the net force was tripled and the mass were doubled, it means,
F' = 3F
m' = 2m
Let a' is new acceleration. So,
[tex]a'=\dfrac{F'}{m'}\\\\a'=\dfrac{(3F)}{(2m)}\\\\a'=\dfrac{3}{2}\times \dfrac{F}{m}\\\\a'=\dfrac{3}{2}\times 6\\\\a'=9\ m/s^2[/tex]
So, the new acceleration would be 9 m/s².
7.00 kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -20.0°C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 0.90 kg of ice and 1.10 kg of liquid water.
Required:
What was the initial temperature of the piece of copper?
Answer:
122°C
Explanation:
From the data Final temperature is 0 deg C since there is 0.9kg of ice and 1.10kg of liquid water.
That means that 1.10kg of the ice undergoes Heat of Fusion which is 3.34x10^5 J/kg...
Heat lost by copper = Heat gained by ice + Heat of fusion
-> (7.0kg)(390J/kg*C)(0-T) = (2.00kg)(2100J/kg*C)(0 - (-20) + (1.10kg)(3.34x10^5 J/kg)
-> T(2730) = 334001
-> T = 122°C
the efficiency of a carnot cycle is 1/6.If on reducing the temperature of the sink 75 degrees celcius ,the efficiency becomes 1/3,determine he initial and final temperatures between which the cycle is working.
Answer:
450°C
Explanation: Given that the efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature .
η = 1 - T2 / T1
η = 1/6 initially
when T2 is reduced by 65°C then η becomes 1/3
Solution
η = 1/6
1 - T2 / T1 = 1/6 [ using the Formula ]........................(1)
When η = 1/3 :
η = 1 - ( T2 - 75 ) / T1
1/3 = 1 - (T2 - 75)/T1.........................(2)
T2 - T1 = -75 [ because T2 is reduced by 75°C ]
T2 = T1 - 75...........................(3)
Put this in (2) :
> 1/3 = 1 - ( T1 - 75 - 75 ) / T1
> 1/3 = 1 - (T1 - 150 ) /T1
> (T1 - 150) / T1 = 1 - 1/3
> ( T1 -150 ) / T1 = 2/3
> 3 ( T1 - 150 ) = 2 T1
> 3 T1 - 450 = 2 T1
Collecting the like terms
3 T1- 2 T1 = 450
T1 = 450
The temperature initially was 450°C
Two small plastic spheres are given positive electrical charges. When they are 20.0 cm apart, the repulsive force between them has magnitude 0.200 N.
1. What is the charge on each sphere if the two charges are equal? (C)
2. What is the charge on each sphere if one sphere has four times the charge of the other? (C)
Answer:
A. 2.97x 10^-6C
B. 1.48x10^ -6 C
Explanation:
Pls see attached file
Answer:
1) +9.4 x 10^-7 C
2) +4.72 x 10^-7 C and +1.9 x 10^-6 C
Explanation:
The two positive charges will repel each other
Repulsive force on charges = 0.200 N
distance apart = 20.0 cm = 0.2 m
charge on each sphere = ?
Electrical force on charged spheres at a distance is given as
F = [tex]\frac{kQq}{r^{2} }[/tex]
where F is the force on the spheres
k is the Coulomb's constant = 8.98 x 10^9 kg⋅m³⋅s⁻²⋅C⁻²
Q is the charge on of the spheres
q is the charge on the other sphere
r is their distance apart
since the charges are equal, i.e Q = q, the equation becomes
F = [tex]\frac{kQ^{2} }{r^{2} }[/tex]
making Q the subject of the formula
==> Q = [tex]\sqrt{\frac{Fr^{2} }{k} }[/tex]
imputing values into the equation, we have
Q = [tex]\sqrt{\frac{0.2*0.2^{2} }{8.98*10^{9} } }[/tex] = +9.4 x 10^-7 C
If one charge has four times the charge on the other, then
charge on one sphere = q
charge on the other sphere = 4q
product of both charges = [tex]4q^{2}[/tex]
we then have
F = [tex]\frac{4kq^{2} }{r^{2} }[/tex]
making q the subject of the formula
==> q = [tex]\sqrt{\frac{Fr^{2} }{4k} }[/tex]
imputing values into the equation, we have
q = [tex]\sqrt{\frac{0.2*0.2^{2} }{4*8.98*10^{9} } }[/tex] = +4.72 x 10^-7 C
charge on other sphere = 4q = 4 x 4.72 x 10^-7 = +1.9 x 10^-6 C
A plane progressive
the expression
in time, ys
where you
progressivo ware is no presented by
(At + A
y- 5 sin
in metre, t es in time the doplicensel
Calculate
the amplitude of the wave.
Answer:
Amplitude, A = 5 m
Explanation:
Let a progressive wave is given by equation :
[tex]y=5\sin (100\pi t-0.4\pi x)[/tex] .....(1)
The general equation of a progressive wave is given by :
[tex]y=A\sin (\omega t-kx)[/tex] ....(2)
Here,
A is the amplitude of the wave
[tex]\omega[/tex] is the angular frequency
k is propagation constant
We need to find the amplitude of the wave.
If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.
if you place 0°c ice into 0°c water in an insulated container, what will happen? Will some ice melt, will more water freeze, or will neither take place?
Answer:
neither will happen
Explanation:
cause the water is already defreezed
You connect three resistors with resistances R, 2R, and 3R in parallel. The equivalent resistance of the three resistors will have a value that is
Answer:
The equivalent is 6R/11Explanation:
We know that the equivalent resistance of resistors connected in parallel is expressed as
[tex]\frac{1}{Re} =\frac{1}{R1} +\frac{1}{R2}+\frac{1}{R3}\\\\\frac{1}{Re} =\frac{1}{R} +\frac{1}{2R}+\frac{1}{3R}\\[/tex]
the L.C.M is 6R
[tex]\frac{1}{Re} =\frac{6+3+2}{6R} = \frac{11}{6R} \\\\Re= \frac{6R}{11}[/tex]
calculate the upthrust aciting on a body if its
true weight is 550 N and apparent weight
lis 490 N
Answer:
As a body moving upward
T=real weight + apparent weight
T=550+490
T=1040
hope u will get the answer:)
Explanation:
radiation transfers energy through___. a metal. b liquid. c touch. d waves.
Answer:
Radiation is transferred through electromagnetic waves so D.
Explanation:
Answer:
D. Waves
Explanation:
a and b don't make much sense, conduction is transfer of energy through touch
Suppose a tank filled with water has a liquid column with a height of 19 meter. If the area is 2 square meters 2m squared, what’s the force of gravity acting on the column of water?
Answer:
372,400 N
Explanation:
The volume of the column is ...
V = Bh = (2 m^2)(19 m) = 38 m^3
If we assume the density is 1000 kg/m^3, then the mass of the water is ...
M = ρV = (1000 kg/m^3)(38 m^3) = 38,000 kg
The force of gravity on that mass is ...
F = Mg = (38,000 kg)(9.8 m/s^2) = 372,400 N
A vertically polarized light wave of intensity 1000 mW/m2 is coming toward you, out of the screen. After passing through this polarizing filter, the wave's intensity is
Answer:
The intensity is [tex]I = 500 mW/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]
Generally the intensity of the light emerging from the polarizer is mathematically represented as
[tex]I = \frac{I_o}{2}[/tex]
substituting values
[tex]I = \frac{1000 *10^{-3}}{2}[/tex]
[tex]I = 500 *10^{-3} W/m^2[/tex]
[tex]I = 500 mW/m^2[/tex]
an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?
Answer:
4°C
Explanation:
Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.
When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you reduce the charge of A to one-half its original value, and the charge of B to one-tenth, and reduce the distance between the objects by half, what will be the new force that they exert on each other in terms of force F
Answer:
F = F₀ 0.2
Explanation:
For this exercise we apply Coulomb's law with the initial data
F₀ = k q_A q_B / d²
indicate several changes
q_A ’= ½ q_A
q_B ’= 1/10 q_B
d ’= ½ d
let's substitute these new values in the Coulomb equation
F = k q_A ’q_B’ / d’²
F = k ½ q_A 1/10 q_B / (1/2 d)²
F = (k q_A q_B / d2) ½ 1/10 2²
F = F₀ 0.2
A person bends over to grab a 20 kg object. The back muscle responsible for supporting his upper body weight and the object is located 2/3 of the way up his back (where it attaches to the spine) and makes an angle of 12 degrees with the spine. His upper body weighs 36 kg. What is the tension in the back muscle
Answer:
T = 2689.6N
Explanation:
Considering the situation, one can say that torque due to tension in the spine is counter balanced by the torque due to weight of upper part of the body and the weight of the object. Hence, the tension force is acting at an angle of 12 degree
while both weight are acting perpendicular to the length. Hence we have :
Torque ( clockwise) = Torque ( anticlockwise)
m1g (L/2)+ m2g(L) = Tsin 12(2L/3)........1
Where m1 = 36kg
m2 = 20kg
g = 9.81m/s^2
Theta = 12
Substituting into equation 1
36(9.81) * (L/2)+20(9.81)(L) = Tsin12(2L/3)
353.16L/2+196.2L = T ×0.2079(2L/3)
176.58L+196.2L = T × 0.1386L
372.78L = 0.1386LT
T = 372.78L/0.1386L
T = 2689.6N
Help me with these question and please explainnn
Explanation:
1. Impulse = change in momentum
J = Δp
J = mΔv
In the x direction:
Jₓ = mΔvₓ
Jₓ = (0.40 kg) (30 m/s cos 45° − (-20 m/s))
Jₓ = 16.5 kg m/s
In the y direction:
Jᵧ = mΔvᵧ
Jᵧ = (0.40 kg) (30 m/s sin 45° − 0 m/s)
Jᵧ = 8.49 kg m/s
The magnitude of the impulse is:
J = √(Jₓ² + Jᵧ²)
J = 18.5 kg m/s
The average force is:
FΔt = J
F = J/Δt
F = 1850 N
2. Momentum is conserved.
m₁u₁ + m₂u₂ = (m₁ + m₂) v
In the x direction:
(1000 kg) (0 m/s) + (1500 kg) (-12 m/s) = (1000 kg + 1500 kg) vₓ
vₓ = -7.2 m/s
In the y direction:
(1000 kg) (20 m/s) + (1500 kg) (0 m/s) = (1000 kg + 1500 kg) vᵧ
vᵧ = 8 m/s
The magnitude of the final speed is:
v = √(vₓ² + vᵧ²)
v = 10.8 m/s
3. Momentum is conserved.
m₁u₁ + m₂u₂ = (m₁ + m₂) v
In the x direction:
(0.8 kg) (18 m/s cos 45°) + (0.36 kg) (9.0 m/s) = (0.8 kg + 0.36 kg) vₓ
vₓ = 11.6 m/s
In the y direction:
(0.8 kg) (-18 m/s sin 45°) + (0.36 kg) (0 m/s) = (0.8 kg + 0.36 kg) vᵧ
vᵧ = -8.78 m/s
The magnitude of the final speed is:
v = √(vₓ² + vᵧ²)
v = 14.5 m/s
A planet in another solar system orbits a star with a mass of 5.0 x 1030 kg. At one point in its orbit, it is 150 x 106 km from the star and is moving at 55 km/s. What is the semimajor axis of the planet's orbit
Answer:
32
Explanation: