Answer:
The magnetic field's direction is towards the north
Explanation:
The force on a positive charge in a uniform magnetic field is represented by the right hand rule. To determine the direction of the force, place your right hand with your palm up, and your thumb at 90° to the other fingers. If the fingers represent the magnetic field, and the thumb the direction of the positive charge, then the palm will push up in the direction of the force. But a negative charge like an electron pushes in exactly the opposite direction. So if you follow this rule, you will find that the magnetic field points towards the north.
The direction of the magnetic field is towards the North. This can be
determined using the right hand rule by Fleming.
The right hand rule states that to determine the direction of the magnetic
force, the right thumb should be pointed in in the direction of the velocity,
index finger in the direction of the magnetic field and middle finger in the
direction of magnetic force.
When this is applied, we will discover that the index finger will point towards
the north region.
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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.
Uses of pressure and the uses of density
Answer:
Pressure is a scalar quantity defined as per unit area.
Density is the objects ,times its the acceleration due to gravity.
Explanation:
Pressure is the alternative object increases the area of contact decrease .
Pressure is the force component to the surface used to calculate pressure.
pressure is that collisions of the gas to container as the per unit time .
pressure is an physical important quantity to play the solid and fluid .
Pressure is the expressed in a number of units depend the context use, pressure exerted by the liquid alone.
Density is the objects, times, volume of the object that times acceleration objects.
Density is the used to the system complex objects and materials.
Density force is the weight of a region or objects static fluid.
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.
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 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
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]
A toboggan is sliding down an icy slope. As it goes down, _________ does work on the toboggan and ends up converting __________ energy to _________ energy.
Answer:
As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.
1. weight
2. gravitational potential energy to kinetic energy.
Explanation:
As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.
work done by toboggan = weight × distance
W = mg and the distance is down the icy slope
By using law of conservation of energy, energy can neither be created nor destroyed, but can be conserve from one form to another in a closed system.
Toboggan converts gravitational potential energy (mgh) to kinetic energy(¹/₂mv²)
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
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.
Someone help find centripetal acceleration plus centripetal force!
Answer:Centripetal force that acts an object keep it along a moving circular path.
Explanation:Centripetal force along a path circular of radius(r) with velocity(V) acceleration the center of the path.
a=v/r
object will along moving continue a straight path unless by the external force.External force is the centripetal force.
Centripetal force is to moving in horizontal circle,Centripetal force is not a fundamental force.Gravitational force satellite and orbit of centripetal force.
Centripetal acceleration and centripetal force are used to calculate the motion of objects in circular motion. The main answer to the question is given below:The centripetal force is given by:F = mv²/rwhere m is the mass of the object, v is the speed of the object and r is the radius of the circle. The unit of centripetal force is Newtons (N).The centripetal acceleration is given by:a = v²/rThe unit of centripetal acceleration is meters per second squared
(m/s²).Explanation:When an object moves in a circular motion, there is a force that acts upon it. This force is called the centripetal force. This force always points towards the center of the circle. It is responsible for keeping the object moving in a circular motion.The centripetal force is related to the centripetal acceleration.
The centripetal acceleration is the acceleration of an object moving in a circle. It is always directed towards the center of the circle.The magnitude of the centripetal force is given by:F = mv²/rwhere F is the force, m is the mass of the object, v is the speed of the object and r is the radius of the circle.The magnitude of the centripetal acceleration is given by:a = v²/rwhere a is the acceleration, v is the speed of the object and r is the radius of the circle.
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How much heat is needed to melt 2.5 KG of water at its melting point? Use Q= mass x latent heat of fusion.
Answer:
Q = 832 kJ
Explanation:
It is given that,
Mass of the water, m = 2.5 kg
The latent heat of fusion, L = 333 kJ/kg
We need to find the heat needed to melt water at its melting point. The formula of heat needed to melt is given by :
Q = mL
[tex]Q=2.5\ kg\times 333\ kJ/kg\\\\Q=832.5\ kJ[/tex]
or
Q = 832 kJ
So, the heat needed to melt the water is 832 kJ.
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
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
Suppose that a 0.275 m radius, 500 turn coil produces an average emf of 11800 V when rotated one-fourth of a revolution in 4.42 ms, starting from its plane being perpendicular to the magnetic field.
Required:
Find the magnetic field strength needed to induce an average emf of 10,000 V.
Answer:
The magnetic field strength : 0.372 T
Explanation:
The equation of the induced emf is given by the following equation,
( Equation 1 ) emf = - N ( ΔФ / Δt ) - where N = number of turns of the coil, ΔФ = change in the magnetic flux, and Δt = change in time
The equation for the magnetic flux is given by,
( Equation 2 ) Ф = BA( cos( θ ) ) - where B = magnetic field, A = area, and θ = the angle between the normal and the magnetic field
The area of the circular coil is a constant, as well as the magnetic field. Therefore the change in the magnetic flux is due to the angle between the normal and the magnetic field. Therefore you can expect the equation for the change in magnetic flux to be the same as the magnetic flux, but only that there must be a change in θ.
( Equation 3 ) ΔФ = BA( Δcos( θ ) )
Now as the coil rotates one-fourth of a revolution, θ changes from 0 degrees to 90 degrees. The " change in cos θ " should thus be the following,
Δcos( θ ) = cos( 90 ) - cos( 0 )
= 0 - 1 = - 1
Let's substitute that value in the third equation,
( Substitution of Δcos( θ ) previously, into Equation 3 )
ΔФ = BA( - 1 ) = - BA
Remember the first equation? Well if the change in the magnetic flux = - BA, then through further substitution, the emf should = - N( - BA ) / Δt. In other words,
emf = - N( - BA ) / Δt,
emf = NBA / Δt,
B = ( emf )Δt / NA
Now that we have B, the magnetic field strength, isolated, let's solve for the area of the circular coil and substitute all known values into this equation.
Area ( A ) = πr²,
= π( 0.275 )² = 0.2376 m²,
B = ( 10,000 V )( 4.42 [tex]*[/tex] 10⁻³ s ) / ( 500 )( 0.2376 m² ) = ( About ) 0.372 T
The magnetic field strength : 0.372 T
Suppose a 185 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m.
Randomized Variables
m = 185 kg
v = 29 m/s
h = 32 m
A. How high can it coast up the hill. if you neglect friction in m?
B. How much energy is lost to friction if the motorcycle only gains an altitude of 33 m before coming to rest?
Answer:
a) Height reached before coming to rest is 42.86 m
b) Energy lost to friction is 17902.45 J
Explanation:
mass of the motorcycle = 185 kg
speed of the towards the hill = 29 m/s
The wheels weigh 12 kg each
Wheels are annular rings with an inner radius of 0.280 m and outer radius of 0.330 m
a) To go up the hill, the kinetic energy of motion of the motorcycle will be converted to the potential energy it will gain in going up a given height
the kinetic energy of the motorcycle is given as
[tex]KE[/tex] = [tex]\frac{1}{2}mv^{2}[/tex]
where m is the mass of the motorcycle
v is the velocity of the motorcycle
[tex]KE[/tex] = [tex]\frac{1}{2}*185*29^{2}[/tex] = 77792.5 J
This will be converted to potential energy
The potential energy up the hill will be
[tex]PE[/tex] = mgh
where m is the mass
g is acceleration due to gravity 9.81 m/s^2
h is the height reached before coming to rest
[tex]PE[/tex] = 185 x 9.81 x m = 1814.85h
equating the kinetic energy to the potential energy for energy conservation, we'll have
77792.5 = 1814.85h
height reached before coming to rest = 77792.5/1814.85 = 42.86 m
b) if an altitude of 33 m was reached before coming to rest, then the potential energy at this height is
[tex]PE[/tex] = mgh
[tex]PE[/tex] = 185 x 9.81 x 33 = 59890.05 J
The energy lost to friction will be the kinetic energy minus this potential energy.
energy lost = 77792.5 - 59890.05 = 17902.45 J
A) The motorcycle can coast up the hill by ; 42.86m
B) The amount of energy lost to friction : 17902.45 J
A) Determine how high the motorcycle can coast up the hill when friction is neglected
apply the formula for kinetic and potential energies
K.E = 1/2 mv² ---- ( 1 )
P.E = mgH ---- ( 2 )
As the motorcycle goes uphiLl the kinetic energy is converted to potential energy.
∴ K.E = P.E
1/2 * mv² = mgH
∴ H = ( 1/2 * mv² ) / mg ---- ( 3 )
where ; m = 185 kg , v = 29 m/s , g = 9.81
Insert values into equation ( 3 )
H ( height travelled by motorcycle neglecting friction ) = 42.86m
B) Determine how much energy is lost to friction if the motorcycle attains 33m before coming to rest
P.E = mgh = 185 * 9.81 * 33 = 59890.05 J
where : h = 33 m , g = 9.81
K.E = 1/2 * mv² = 77792.5 J ( question A )
∴ Energy lost ( ΔE ) = ( 77792.5 - 59890.05 ) = 17902.45 J
Hence we can conclude that The motorcycle can coast up the hill by ; 42.86m , The amount of energy lost to friction : 17902.45 J.
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A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]f= -75 \ cm = - 0.75 \ m[/tex]
b
[tex]P = -1.33 \ diopters[/tex]
Explanation:
From the question we are told that
The image distance is [tex]d_i = -75 cm[/tex]
The value of the image is negative because it is on the same side with the corrective glasses
The object distance is [tex]d_o = \infty[/tex]
The reason object distance is because the object father than it being picture by the eye
General focal length is mathematically represented as
[tex]\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}[/tex]
substituting values
[tex]\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}[/tex]
=> [tex]f= -75 \ cm = - 0.75 \ m[/tex]
Generally the power of the corrective lens is mathematically represented as
[tex]P = \frac{1}{f}[/tex]
substituting values
[tex]P = \frac{1}{-0.75}[/tex]
[tex]P = -1.33 \ diopters[/tex]
Before you start taking measurements though, we’ll first make sure you understand the underlying concepts involved. By what method is each of the spheres charged?
Answer:
If they are metallic spheres they are connected to earth and a charged body approaches
non- metallic (insulating) spheres in this case are charged by rubbing
Explanation:
For fillers, there are two fundamental methods, depending on the type of material.
If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.
If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.
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
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
A person with normal vision can focus on objects as close as a few centimeters from the eye up to objects infinitely far away. There exist, however, certain conditions under which the range of vision is not so extended. For example, a nearsighted person cannot focus on objects farther than a certain point (the far point), while a farsighted person cannot focus on objects closer than a certain point (the near point). Note that even though the presence of a near point is common to everyone, a farsighted person has a near point that is much farther from the eye than the near point of a person with normal vision.
Both nearsightedness and farsightedness can be corrected with the use of glasses or contact lenses. In this case, the eye converges the light coming from the image formed by the corrective lens rather than from the object itself.
Required:
a. If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that the person would need to see an object at infinity clearly?
b. If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350 m from the person's eyes?
Answer:
a) f₁ = 3.50 m , b) f₂ = 0.84 m
Explanation:
For this exercise we must use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image
a) the distance where the image should be placed is q = 3.50 m and the object is located at infinity p = ∞
1 / f₁ = 1 /∞ + 1 / 3.50
f₁ = 3.50 m
b) in this case the image is at q = -0.600 m and the object p = 0.350 m
1 / f₂ = 1 / 0.350 -1 / 0.600
the negative sign, is because the image is in front of the object
1 / f₂ = 1,1905
f₂ = 1 / 1,1905
f₂ = 0.84 m
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
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.
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current of 26.0 A in the +x direction. The second wire lies along the x-axis. The wires exert attractive forces on each other, and the force per unit length on each wire is 295 µN/m. What is the y-value (in m) of the line in the xy-plane where the total magnetic field is zero?
Answer:
The y-value is z = 0.759 m
Explanation:
From the question we are told that
The position of the first y-axis is [tex]y_1 = 0.300 \ m[/tex]
The current on the first wire is [tex]I_ 1 = 26.0 \ A[/tex]
The force per unit length on each wire is [tex]\frac{F}{l} = 295 \mu N/m = 295 * 10^{-6} \ N/m[/tex]
Generally the force per unit length on first wire is mathematically represented as
[tex]\frac{F}{l} = \frac{\mu_o * I_1 * I_2 }{2*\pi* y_1}[/tex]
Where [tex]\mu _o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting values
[tex]295 *10^{-6} = \frac{ 4\pi * 10^{-7} * 26.0 * I_2 }{2 *3.142* 0.300}[/tex]
[tex]I_2 = \frac{295 *10^{-6 } * 0.300 * 2* 3.142 }{ 4\pi * 10^{-7} * 26 }[/tex]
[tex]I_2 = 17.0 \ A[/tex]
Now the at the point where the magnetic field is zero the magnetic field of each wire are equal , let that point by z meters from the second wire on the y-axis so
[tex]\frac{\mu_o I_2}{2 * \pi * y_1} = \frac{\mu_o I_1}{2 * \pi * (y_1-z)}[/tex]
[tex]I_2 (y_1 - z) = I_1 * y_1[/tex]
substituting values
[tex]17.0 ( 0.300 - z) = 26 * 0.300[/tex]
z = 0.759 m
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 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
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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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]
BIO A trap-jaw ant snaps its mandibles shut at very high speed, a good trait for catching small prey. But an ant can also slam its mandibles into the ground; the resulting force can launch the ant into the air for a quick escape. A 12 mg ant hits the ground with an average force of 47 mN for a time of 0.13 ms; these are all typical values. At what speed does it leave the ground
Answer:
Final velocity (v) = 0.509 m/s (Approx)
Explanation:
Ant use impulse power
Given:
Mass of ant = 12 mg = 12 × 10⁻⁶ kg
Average force = 47 mN = 47 × 10⁻³ N
Initial velocity(u) = 0
Time taken = 0.13 ms = 0.13 × 10⁻³ s
Find:
Final velocity (v)
Computation:
Force × Time = change in momentum
(47 × 10⁻³ N)(0.13 × 10⁻³ s) = mv - mu
(47 × 10⁻³ N)(0.13 × 10⁻³ s) = m(v - u)
6.11 × 10⁻⁶ = 12 × 10⁻⁶(v - 0)
6.11 = 12 v
Final velocity (v) = 0.509 m/s (Approx)
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
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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