Answer:
(a) 0.31
(b) 0.245
Explanation:
(a)
F' = μ'mg.................... Equation 1
Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.
make μ' the subject of the equation above
μ' = F'/mg............. Equation 2
Given: F' = 75 N, m = 25 kg
constant: g = 9.8 m/s²
Substitute these values into equation 2
μ' = 75/(25×9.8)
μ' = 75/245
μ' = 0.31.
(b) Similarly,
F = μmg.................. Equation 3
Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.
make μ the subject of the equation
μ = F/mg.............. Equation 4
Given: F = 60 N, m = 25 kg, g = 9.8 m/s²
Substitute these values into equation 4
μ = 60/(25×9.8)
μ = 60/245
μ = 0.245
An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the speed remained the same, by what factor would its centripetal acceleration change?
Answer:
The centripetal acceleration changed by a factor of 0.5
Explanation:
Given;
first radius of the horizontal circle, r₁ = 500 m
speed of the airplane, v = 150 m/s
second radius of the airplane, r₂ = 1000 m
Centripetal acceleration is given as;
[tex]a = \frac{v^2}{r}[/tex]
At constant speed, we will have;
[tex]v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1[/tex]
a₂ = 0.5a₁
Therefore, the centripetal acceleration changed by a factor of 0.5
At what minimum speed must a roller coaster be traveling when upside down at the top of a 7.4 m radius loop-the-loop circle so the passengers will not fall out?
Answer:
v = 8.5 m/s
Explanation:
In order for the passengers not to fall out of the loop circle, the centripetal force must be equal to the weight of the passenger. Therefore,
Weight = Centripetal Force
but,
Weight = mg
Centripetal Force = mv²/r
Therefore,
mg = mv²/r
g = v²/r
v² = gr
v = √gr
where,
v = minimum speed required = ?
g = 9.8 m/s²
r = radius = 7.4 m
Therefore,
v = √(9.8 m/s²)(7.4 m)
v = 8.5 m/s
Minimum speed for a roller coaster while travelling upside down so that the person will not fall out = 8.5 m/s
For a roller coaster be traveling when upside down the Force balance equation can be written for a person of mass m.
In the given condition the weight of the person must be balanced by the centrifugal force.
and for the person not to fall out centrifugal force must be greater than or equal to the weight of the person
According to the Newton's Second Law of motion we can write force balance
[tex]\rm mv^2/r -mg =0 \\\\mg = mv^2 /r (Same\; mass) \\\\\\g = v^2/r\\\\v = \sqrt {gr}......(1)[/tex]
Given Radius of loop = r = 7.4 m
Putting the value of r = 7.4 m in equation (1) we get
[tex]\sqrt{9.8\times 7.4 } = \sqrt{72.594} = 8.5\; m/s[/tex]
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A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you at 13.0 m/s. The maximum force you apply in catching the water balloon is twice the average force. How long must the interaction time of your catch be to make sure the water balloon doesn't soak you
Answer:
t = 0.029s
Explanation:
In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:
[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}[/tex] (1)
m: mass of the water balloon = 1.20kg
Δv: change in the speed of the balloon = v2 - v1
v2: final speed = 0m/s (the balloon stops in my hands)
v1: initial speed = 13.0m/s
Δt: interaction time = ?
The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:
[tex]|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s[/tex]
The interaction time to avoid that the water balloon breaks is 0.029s
Find the terminal velocity (in m/s) of a spherical bacterium (diameter 1.81 µm) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be 1.10 ✕ 103 kg/m3. (Assume the viscosity of water is 1.002 ✕ 10−3 kg/(m · s).)
Answer:
The terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.
Explanation:
The terminal velocity of the bacterium can be calculated using the following equation:
[tex] F = 6\pi*\eta*rv [/tex] (1)
Where:
F: is drag force equal to the weight
η: is the viscosity = 1.002x10⁻³ kg/(m*s)
r: is the radium of the bacterium = d/2 = 1.81 μm/2 = 0.905 μm
v: is the terminal velocity
Since that F = mg and by solving equation (1) for v we have:
[tex] v = \frac{mg}{6\pi*\eta*r} [/tex]
We can find the mass as follows:
[tex] \rho = \frac{m}{V} \rightarrow m = \rho*V [/tex]
Where:
ρ: is the density of the bacterium = 1.10x10³ kg/m³
V: is the volume of the spherical bacterium
[tex] m = \rho*V = \rho*\frac{4}{3}\pi*r^{3} = 1.10 \cdot 10^{3} kg/m^{3}*\frac{4}{3}\pi*(0.905 \cdot 10^{-6} m)^{3} = 3.42 \cdot 10^{-15} kg [/tex]
Now, the terminal velocity of the bacterium is:
[tex] v = \frac{mg}{6\pi*\eta*r} = \frac{3.42 \cdot 10^{-15} kg*9.81 m/s^{2}}{6\pi*1.002 \cdot 10^{-3} kg/(m*s)*0.905 \cdot 10^{-6} m} = 1.96 \cdot 10^{-6} m/s [/tex]
Therefore, the terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.
I hope it helps you!
What direct current will produce the same amount of thermal energy, in a particular resistor, as an alternating current that has a maximum value of 2.59 A?
Answer:
The direct current that will produce the same amount of thermal energy is 1.83 A
Explanation:
Given;
maximum current, I₀ = 2.59 A
The average power dissipated in a resistor connected in an AC source is given as;
[tex]P_{avg} = I_{rms} ^2R[/tex]
Where;
[tex]I_{rms} = \frac{I_o}{\sqrt{2} }[/tex]
[tex]P_{avg} = (\frac{I_o}{\sqrt{2} } )^2R\\\\P_{avg} = \frac{I_o^2R}{2} ----equation(1)[/tex]
The average power dissipated in a resistor connected in a DC source is given as;
[tex]P_{avg} = I_d^2R --------equation(2)[/tex]
where;
[tex]I_d[/tex] is direct current
Solve equation (1) and (2) together;
[tex]I_d^2R = \frac{I_o^2R}{2} \\\\I_d^2 = \frac{I_o^2}{2} \\\\I_d=\sqrt{\frac{I_o^2}{2} } \\\\I_d = \frac{I_o}{\sqrt{2}} \\\\I_d = \frac{2.59}{\sqrt{2} } \\\\I_d = 1.83 \ A[/tex]
Therefore, the direct current that will produce the same amount of thermal energy is 1.83 A
key points that can be found in the realist philosophical position
Answer:
Key points that can be found in the realist philosophical position are as follows:
The view that we observe or identify is real, truly out there.The objects which are identified are independent of someone's perceptions, linguistic practices, conceptual scheme, and beliefs.Quantum mechanics is an example of philosophical realism that claims world is mind-independent.2. A 2.0-kg block slides down an incline surface from point A to point B. Points A and B are 2.0 m apart. If the coefficient of kinetic friction is 0.26 and the block is starting at rest from point A. What is the work done by friction force
Answer:a
Explanation:
A 2.0-kg object moving 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:
20J
Explanation:
In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;
m₁u₁ + m₂u₂ = (m₁ + m₂)v ---------------(i)
Where;
m₁ and m₂ are the masses of first and second objects respectively
u₁ and u₂ are the initial velocities of the first and second objects respectively
v is the final velocity of the two objects after collision;
From the question;
m₁ = 2.0kg
m₂ = 8.0kg
u₁ = 5.0m/s
u₂ = 0 (since the object is initially at rest)
Substitute these values into equation (i) as follows;
(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v
(10.0) + (0) = (10.0)v
10.0 = 10.0v
v = 1m/s
The two bodies stick together and move off with a velocity of 1m/s after collision.
The kinetic energy(KE₁) of the objects before collision is given by
KE₁ = [tex]\frac{1}{2}[/tex]m₁u₁² + [tex]\frac{1}{2}[/tex]m₂u₂² ---------------(ii)
Substitute the appropriate values into equation (ii)
KE₁ = ([tex]\frac{1}{2}[/tex] x 2.0 x 5.0²) + ([tex]\frac{1}{2}[/tex] x 8.0 x 0²)
KE₁ = 25.0J
Also, the kinetic energy(KE₂) of the objects after collision is given by
KE₂ = [tex]\frac{1}{2}[/tex](m₁ + m₂)v² ---------------(iii)
Substitute the appropriate values into equation (iii)
KE₂ = [tex]\frac{1}{2}[/tex] ( 2.0 + 8.0) x 1²
KE₂ = 5J
The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision
K = KE₂ - KE₁
K = 5 - 25
K = -20J
The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J
The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 3 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 88,000 m/s. What are the masses of the two stars
Answer:
Explanation:
given
T = 3months = 7.9 × 10⁶s
orbital speed = 88 × 10³m/s
V= 2πr÷T
∴ r = (V×T) ÷ 2π
r = (88km × 7.9 × 10⁶s) ÷ 2π
r = 1.10 × 10⁸km
using kepler's 3rd law
mass of both stars = (seperation diatance)³/(orbital speed)²
M₁ + M₂ = (2r)³/([tex]\frac{1}{4}[/tex]year)²
= (1.06 × 10²⁵)/(6.2×10¹³)
1.71×10¹²kg
since M₁ = M₂ =1.71×10¹²kg ÷ 2
M₁ = M₂ = 8.55×10¹¹kg
Blue light (λ = 475 nm) is sent through a single slit with a width of 2.1 µm. What is the maximum possible number of bright fringes, including the central maximum, produced on the screen? (Hint: What is the largest angle that can be used?)
Answer:
m = 4
Explanation:
The expression that explains the constructive interference of a diffraction pattern is
a sin θ = m λ
where a is the width of the slit and λ the wavelength
sin θ = m λ / a
The maximum value is for when the sine is 1, let's substitute
1 = m λ/a
m = a /λ
let's reduce the magnitudes to the SI system
a = 2.1 um = 2.1 10⁻⁶
lam = 475 nm = 475 10⁻⁹ m
let's calculate
m = 2.1 10⁻⁶ / 475 10⁻⁹
m = 4.42
with m must be an integer the highest value is
m = 4
A slender rod of length L has a varying mass-per-unit-length from the left end (x=0) according to dm/dx=Cx where C has units kg/m2. Find the total mass in terms of C and L, and then calculate the moment of inertia of the rod for an axis at the left end note: you need the total mass in order to get the answer in terms of ML^2
Answer:
ML²/6
Explanation:
Pls see attached file
The total mass is M = CL²/2, and the moment of inertia is I = ML²/2,
Moment of inertia:The length of the rod is L. It has a non-uniform distribution of mass given by:
dm/dx = Cx
where C has units kg/m²
dm = Cxdx
the total mass M of the rod can be calculated by integrating the above relation over the length:
[tex]M =\int\limits^L_0 {} \, dm\\\\M=\int\limits^L_0 {Cx} \, dx\\\\M=C[x^2/2]^L_0\\\\M=C[L^2/2]\\\\[/tex]
Thus,
C = 2M/L²
Now, the moment of inertia of the small element dx of the rod is given by:
dI = dm.x²
dI = Cx.x²dx
[tex]dI = \frac{2M}{L^2}x^3dx\\\\I= \frac{2M}{L^2}\int\limits^L_0 {x^3} \, dx \\\\I= \frac{2M}{L^2}[\frac{L^4}{4}][/tex]
I = ML²/2
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A trough is filled with a liquid of density 810 kg/m3. The ends of the trough are equilateral triangles with sides 8 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough. (Use 9.8 m/s2 for the acceleration due to gravity.)
Answer:
The hydrostatic force on one end of the trough is 54994.464 N
Explanation:
Given;
liquid density, ρ = 810 kg/m³
side of the equilateral triangle, L = 8m
acceleration due to gravity, g = 9.8 m/s²
Hydrostatic force is given as;
H = ρgh
where;
h is the vertical height of the equilateral triangle
Draw a line to bisect upper end of the trough, to the vertex at the bottom, this line is the height of the equilateral triangle.
let the half side of the triangle = x
x = ⁸/₂ = 4m
The half section of the triangle forms a right angled triangle
h² = 8² - 4²
h² = 48
h = √48
h = 6.928m
F = ρgh
F = 810 x 9.8 x 6.928
F = 54994.464 N
Therefore, the hydrostatic force on one end of the trough is 54994.464 N
An ideal gas in a cubical box having sides of length L exerts a pressure p on the walls of the box. If all of this gas is put into a box having sides of length 0.5L without changing its temperature, the pressure it exerts on the walls of the larger box will be...
p.
2p.
4p.
8p.
12p.
Answer:
2P
Explanation:
See attached file
What happens when you increase the number of slits per millimeter (decrease the spacing between slits)?
Answer:
Increasing the number of slits not only makes the diffraction maximum sharper, but also much more intense. If a 1 mm diameter laser beam strikes a 600 line/mm grating, then it covers 600 slits and the resulting line intensity is 90,000 x that of a double slit. Such a multiple-slit is called a diffraction grating.
What is the work done in stretching a spring by a distance of 0.5 m if the restoring force is 24N?
Answer:
3Nm
Explanation:
work = 0.5 x 12 x 0.5 = 3
The work done in stretching the spring by a distance of 0.5 m, with a restoring force of 24 N, is 6 joules.
To calculate the work done in stretching a spring, we can use the formula for work done by a spring:
Work = (1/2) * k *[tex]x^2[/tex]
where:
k = spring constant
x = distance the spring is stretched
Given that the restoring force (F) acting on the spring is 24 N, and the distance the spring is stretched (x) is 0.5 m, we can find the spring constant (k) using Hooke's law:
F = k * x
k = F / x
k = 24 N / 0.5 m
k = 48 N/m
Now, we can calculate the work:
Work = (1/2) * 48 N/m * [tex](0.5 m)^2[/tex]
Work = (1/2) * 48 N/m * [tex]0.25 m^2[/tex]
Work = 6 joules
Therefore, the work done in stretching the spring by a distance of 0.5 m, with a restoring force of 24 N, is 6 joules.
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6a. A special lamp can produce UV radiation. Which two statements
describe the electromagnetic waves emitted by a UV lamp? *
They have a higher frequency than X-rays.
They have the same wave speed as visible light
They have a longer wavelength than microwaves.
They have a lower frequency than gamma rays.
They have a greater wave speed than radio waves.
Answer:
The correct options are:
B) They have the same wave speed as visible light
D) They have a lower frequency than gamma rays.
Explanation:
B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s
D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).
We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.
Two lenses of focal length 4.5cm and 1.5cm are placed at a certain distance apart, calculate the distance between the lenses if they form an achromatic combination
3.0cm
Explanation:
For lenses in an achromatic combination, the following condition holds, assuming the two lenses are of the same materials;
d = [tex]\frac{f_1 + f_2}{2}[/tex] ---------(i)
Where;
d= distance between lenses
f₁ = focal length of the first lens
f₂ = focal length of the second lens
From the question;
f₁ = 4.5cm
f₂ = 1.5cm
Substitute these values into equation (i) as follows;
d = [tex]\frac{4.5+1.5}{2}[/tex]
d = [tex]\frac{6.0}{2}[/tex]
d = 3.0cm
Therefore, the distance between the two lenses is 3.0cm
An ultrasound machine uses 1.64 × 105 watts of power. If it draws 12.0 amps of current, what is the resistance?
Answer:
R = 1138.9 Ω
Explanation:
Hello,
In this case, for the given power (P) and current (I), we can compute the resistance (R) via:
R = P / I²
Thus, we obtain:
R = 1.64x10⁵ W / (12.0 A)²
R = 1138.9 Ω
Best regards.
When a hydrometer (see Fig. 2) having a stem diameter of 0.30 in. is placed in water, the stem protrudes 3.15 in. above the water surface. If the water is replaced with a liquid having a specific gravity of 1.10, how much of the stem would protrude above the liquid surface
Answer:
5.79 in
Explanation:
We are given that
Diameter,d=0.30 in
Radius,r=[tex]\frac{d}{2}=\frac{0.30}{2}=0.15 in[/tex]
Weight of hydrometer,W=0.042 lb
Specific gravity(SG)=1.10
Height of stem from the water surface=3.15 in
Density of water=[tex]62.4lb/ft^3[/tex]
In water
Volume of water displaced [tex]V=\frac{mass}{density}=\frac{0.042}{62.4}=6.73\times 10^{-4} ft^3[/tex]
Volume of another liquid displaced=[tex]V'=\frac{V}{SG}=\frac{6.73\times 10^{-4}}{1.19}=5.66\times 10^{-4}ft^3[/tex]
Change in volume=V-V'
[tex]V-V'=\pi r^2 l[/tex]
Substitute the values
[tex]6.73\times 10^{-4}-5.66\times 10^{-4}=3.14\times (\frac{0.15}{12})^2l[/tex]
By using
1 ft=12 in
[tex]\pi=3.14[/tex]
[tex]l=\frac{6.73\times 10^{-4}-5.66\times 10^{-4}}{3.14\times (\frac{0.15}{12})^2}[/tex]
l=2.64 in
Total height=h+l=3.15+2.64= 5.79 in
Hence, the height of the stem protrude above the liquid surface=5.79 in
Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The power created is [tex]P_{avg} = F * v_{avg}[/tex]
Explanation:
From the question we are told that
The that the average power is mathematically represented as
[tex]P_{avg} = \frac{W }{\Delta t }[/tex]
Where W is is the Workdone which is mathematically represented as
[tex]W = F * s[/tex]
Where F is the applies force and s is the displacement due to the force
So
[tex]P_{avg} = \frac{F *s }{\Delta t }[/tex]
Now this displacement can be represented mathematically as
[tex]s = v_{avg} * \Delta t[/tex]
Where [tex]v_{avg }[/tex] is the average velocity and [tex]\Delta t[/tex] is the time taken
So
[tex]P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }[/tex]
=> [tex]P_{avg} = F * v_{avg}[/tex]
Answer:
Pavg = Fvavg
Explanation:
Since the P (power) done by the F (force) is:
P = Fs/t
and we are looking for the velocity, so then it would be:
P = Fv
with the average velocity the answer is:
Pavg = Favg
If an object is moving at a constant speed, and the force F is also constant, this formula can be used to find the average power. If v is changing, the formula can be used to find the instantaneous power at any given moment (with the quantity v in this case meaning the instantaneous velocity, of course).
A loop of wire with cross-sectional area 1 m2 is inserted into a uniform magnetic field with initial strength 1 T. The field is parallel to the axis of the loop. The field begins to grow with time at a rate of 2 Teslas per hour. What is the magnitude of the induced EMF in the loop of wire
Answer:
The magnitude of the EMF is 0.00055 volts
Explanation:
The induced EMF is proportional to the change in magnetic flux based on Faraday's law:
[tex]emf\,=-\,N\, \frac{d\Phi}{dt}[/tex]
Since in our case there is only one loop of wire, then N=1 and we get:
[tex]emf\,=-\,N\, \frac{d\Phi}{dt}[/tex]
We need to express the magnetic flux given the geometry of the problem;
[tex]\Phi=B\,\,A[/tex]where A is the area of the coil that remains unchanged with time, and B is the magnetic field that does change with time. Therefore the equation for the EMF becomes:
[tex]emf\,=-\,N\, \frac{d\Phi}{dt} = \frac{d\Phi}{dt} =-\frac{d\,(B\,A)}{dt} =-\,A\,\frac{d\,(B)}{dt}=- 1\,m^2(2\,\,T/h})= -2\,\,m^2\,T/(3600\,\,s)= -0.00055\,Volts[/tex]
A total electric charge of 2.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 26.0 cm . The potential is zero at a point at infinity.
a) Find the value of the potential at 45.0 cm from the center of the sphere.
b) Find the value of the potential at 26.0 cm from the center of the sphere.
c) Find the value of the potential at 16.0 cm from the center of the sphere.
Answer:
a) 40 V
b) 69.23 V
c) 69.23 V
Explanation:
See attachment for solution
A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:
Answer:
The ratio is [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
Explanation:
Generally the Moment of inertia of a spherical object (shell) is mathematically represented as
[tex]I = \frac{2}{3} * m r^2[/tex]
Where m is the mass of the spherical object
and r is the radius
Now the the rotational kinetic energy can be mathematically represented as
[tex]RE = \frac{1}{2}* I * w^2[/tex]
Where [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{v}{r}[/tex]
=> [tex]w^2 = [\frac{v}{r}] ^2[/tex]
So
[tex]RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2[/tex]
[tex]RE = \frac{1}{3} * mv^2[/tex]
Generally the transnational kinetic energy of this motion is mathematically represented as
[tex]TE = \frac{1}{2} mv^2[/tex]
So
[tex]\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}[/tex]
[tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
We observe that a small sample of material placed in a non-uniform magnetic field accelerates toward a region of stronger field. What can we say about the material?
Answer:
C) It is either ferromagnetic or paramagnetic
Explanation:
The complete question is given below
We observe that a small sample of material placed in a non-uniform magnetic field accelerates toward a region of stronger field. What can we say about the material?
A) It must be ferromagnetic.
B) It must be paramagnetic.
C) It is either ferromagnetic or paramagnetic.
D) It must be diamagnetic.
A ferromagnetic material will respond towards a magnetic field. They are those materials that are attracted to a magnet. Ferromagnetism is associated with our everyday magnets and is the strongest form of magnetism in nature. Iron and its alloys is very good example of a material that readily demonstrate ferromagnetism.
Paramagnetic materials are weakly attracted to an externally applied magnetic field. They usually accelerate towards an electric field, and form internal induced magnetic field in the direction of the external magnetic field.
The difference is that ferromagnetic materials can retain their magnetization when the externally applied magnetic field is removed, unlike paramagnetic materials that do not retain their magnetization.
In contrast, a diamagnetic material is repelled away from an externally applied magnetic field.
A water-balloon launcher with mass 5 kg fires a 1 kg balloon with a velocity of
8 m/s to the east. What is the recoil velocity of the launcher?
Answer:
1.6 m/s west
Explanation:
The recoil velocity of the launcher is 1.6 m/s west.
What is conservation of momentum principle?When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.
A water-balloon launcher with mass 5 kg fires a 1 kg balloon with a velocity of 8 m/s to the east.
Final momentum will be zero, so
m₁u₁ +m₂u₂ =0
Substitute the values for m₁ = 5kg, m₂ =1kg and u₂ =8 m/s, then the recoil velocity will be
5 x v +1x8 = 0
v = - 1.6 m/s
Thus, the recoil velocity of the launcher is 1.6 m/s (West)
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Two identical pendulums have the same period when measured in the factory. While one pendulum swings on earth, the other is taken on a spaceship traveling at 95%% the speed of light. Assume that both pendulums operate under the influence of the same net force and swing through the same angle.
When observed from earth, how many oscillations does the pendulum on the spaceship undergo compared to the pendulum on earth in a given time interval?
a. more oscillations
b. fewer oscillations
c. the same number of oscillations
Answer:
Explanation:
As a result of impact of time widening, a clock moving as for an observer seems to run all the more gradually than a clock that is very still in the observer's casing.
At the point when observed from earth, the pendulum on the spaceship takes more time to finish one oscillation.
Hence, the clock related with that pendulum will run more slow (gives fewer oscillations as observed from the earth) than the clock related with the pendulum on earth.
Ans => B fewer oscillations
A typical arteriole has a diameter of 0.080 mm and carries blood at the rate of 9.6 x10-5 cm3/s. What is the speed of the blood in (cm/s) the arteriole
Answer:
v= 4.823 × 10⁻⁹ cm/s
Explanation:
given
flow rate = 9.6 x10-5 cm³/s, d = 0.080mm
r = d/2= 0.080/2= 0.0040 cm
speed = rate of blood flow × area
v = (9.6 x 10⁻⁵ cm³/s) × (πr²)
v = (9.6 x 10⁻⁵ cm³/s) × π(0.0040 × cm)²
v= 1.536 × 10⁻⁹π cm/s
v= 4.823 × 10⁻⁹ cm/s
An arrow is launched vertically upward at a speed of 50 m/s. What is the arrow’s speed at the highest point? Ignore air resistance
Answer:
depending on how high it goes at 100m it has taken 2 secondes
Explanation:
At the highest point, the arrow is changing from moving up to moving down. At that exact point, its speed AND its velocity are both ZERO.
And air resistance actually makes no difference.
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.5, what minimum force magnitude is required from the rope to start the crate moving? (b) If µk= 0.35, what is the magnitude of the initial acceleration of the crate?
Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
A proton with an initial speed of 400000 m/s is brought to rest by an electric field.
Part A- Did the proton move into a region of higher potential or lower potential?
Part B - What was the potential difference that stopped the proton?
?U = ________V
Part C - What was the initial kinetic energy of the proton, in electron volts?
Ki =_________eV
Answer:
moves into a region of higher potential
Potential difference = 835 V
Ki = 835 eV
Explanation:
given data
initial speed = 400000 m/s
solution
when proton moves against a electric field so that it will move into higher potential region
and
we know Work done by electricfield W is express as
W = KE of proton K
so
q × V = 0.5 × m × v² ......................1
put here va lue
1.6 × [tex]10^{-19}[/tex] × V = 0.5 × 1.67 × [tex]10^{-27}[/tex] × 400000²
Potential difference V = 1.336 × 10-16 / 1.6 × 10-19
Potential difference = 835 V
and
KE of proton in eV is express as
Ki = V numerical
Ki = 835 eV