Answer:
315,000 ft·lb
Explanation:
At 300 ft and 5 lb/ft, the weight of the cable is (300 f)(5 lb/ft) = 1500 lb. The work done to raise it is equivalent to the work done to raise the cable's center of mass. Since the cable is of uniform density, its center of mass is half the cable length below the winch.
total work done = work to raise cable + work to raise load
= (1500 lb)(150 ft) +(300 lb)(300 ft) = 315,000 ft·lb
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
A soap bubble of radius R is situated in a uniform electric field of magnitude E. The electric flux through the surface of the soap bubble is
Answer:
The electric flux around the soap bubble will be
Ф = q/ε
Explanation:
The radius of the soap bubble = R
The electric field around the soap bubble = E
The electric flux = ?
The soap can be approximated to be a sphere, so we find the surface area of the sphere
For the soap bubble, the surface area will be
A = [tex]4\pi R^{2}[/tex]
Recall that electric flux is given as
Ф = EA
substituting value of A from above, we'll have
Ф = E[tex]4\pi R^{2}[/tex]..... equ 1
Also recall that the electric field E is given as
E = q/(4πε[tex]R^{2}[/tex])
substitute the value of E into equ 1, to get
Ф = q/(4πε[tex]R^{2}[/tex]) * [tex]4\pi R^{2}[/tex]
The electric flux around the soap bubble reduces to
Ф = q/ε
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.
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 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
3. A body moves along a semicircular path. The ratio of distance to displacement is
Answer:
Ratio of distance to displacement is pi/2
Explanation:
Pls see attached file for diagram and explanation
a person lifts 4.5kg block up a vertical distance of 1.2meters and then carries it horizontally for 7.3meters. Calculate the Total Work done by the person and the block.
Answer:
The total work done by the person is given as = m g h
= 4.5kg x 9.8m/s²x1.2m
= 52.92J
This is the work done in moving the block in a vertical distance
However there is no work done when the block is moved in a horizontal direction since ko work is done against gravity.
Explanation:
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5 × 10−4 mm and a crack length of 2.5 × 10−2 mm when a tensile stress of 170 MPa is applied (in MPa)
Answer:
2404 MPa
Explanation:
See attachment for solution
The maximum stress that exists at the tip of the internal crack is 3,400 Mpa.
The given parameters;
radius of the internal crack, r = 2.5 x 10⁻⁴ mmcrack length, l = 2.5 x 10⁻²tensile stress, σ = 170 MPa = 170 x 10⁶ N/m²The maximum stress that exists at the tip of the internal crack is calculated as follows;
[tex]\sigma _{max} = 2\sigma \times \sqrt{(\frac{l}{r} )} \\\\\sigma _{max} = 2 \times 170 \times 10^6 \times \sqrt{(\frac{2.5\times 10^{-2}}{2.5 \times 10^{-4}})} \\\\\sigma _{max} = 3.4 \times 10^{9} \ Pa\\\\\sigma _{max} = 3,400\ Mpa[/tex]
Thus, the maximum stress that exists at the tip of the internal crack is 3,400 Mpa.
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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
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.
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).
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
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
Two mirrors are touching so they have an angle of 35.4 degrees with one another. A light ray is incident on the first at an angle of 55.7 degrees with respect to the normal to the surface. What is the angle of reflection from the second surface
Answer:
54.6°
Explanation:
From law of reflection i=r.
So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.
Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.
If you consider triangle AOB, one angle is ∠AOB=90°
and ∠OAB is 54.6°
From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°
So, the second incident angle will be 54.6°
Hence, the second reflected angle will be 54.6 degrees.
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]
How can global warming lead to changes to the Earth’s surface? a. Global warming can lead to an increased number of earthquakes, which change the Earth’s surface. b. Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses. c. Global warming leads to a decrease in water levels of coastal wetlands. d. Global warming cannot lead to changes to the Earth’s surface.
Answer:
Option: b. Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses.
Explanation:
Global warming is the reason for the changes in environment and climate on earth. Melting of glaciers leads to an increase in water level and a decrease in landmass. One of the most climactic consequences is the decrease in Arctic sea ice. Melting polar ice along with ice sheets and glaciers across Greenland, North America, Europe, Asia, and South America suspected to increase sea levels slowly. There is an increase in the glacial retreat due to global warming, which leaves rock piles that covered with ice.
Answer:
B: Global warming can lead to glaciers melting, causing flooding to areas and the decrease of glacial land masses.
Explanation:
Global warming is primarily caused by the increase in greenhouse gases, such as carbon dioxide, in the Earth's atmosphere. This leads to a rise in global temperatures, which has various impacts on the Earth's surface. One significant effect is the melting of glaciers and ice caps in polar regions and mountainous areas.
As temperatures increase, glaciers and ice sheets start to melt at a faster rate. This melting results in the release of massive amounts of water into rivers, lakes, and oceans. Consequently, there can be an increase in the frequency and intensity of flooding events in regions downstream from these melting glaciers.
Moreover, the melting of glaciers and ice caps contributes to a rise in sea levels. As the melted ice enters the oceans, it adds to the overall volume of water, leading to a gradual increase in sea levels worldwide. This rise in sea levels poses a threat to coastal areas, as they become more vulnerable to coastal erosion, storm surges, and saltwater intrusion into freshwater sources.
Additionally, the loss of glacial land masses due to melting can have long-term effects on ecosystems. Glaciers act as freshwater reservoirs, releasing water gradually throughout the year. With their decline, the availability of freshwater for agriculture, drinking water, and other human needs can be significantly affected.
Therefore, global warming can indeed lead to changes in the Earth's surface, particularly through the melting of glaciers and subsequent impacts on sea levels, flooding, and glacial land masses.
E23 verified.
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
A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and the tires are 0.70 (static) and 0.40 (kinetic) assuming the car is not skidding while traveling along the curve, what is the magnitude of the centripetal acceleration of the car
Answer:
ac = 3.92 m/s²
Explanation:
In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,
Frictional Force = Centripetal Force
where,
Frictional Force = μ(Normal Force) = μ(weight) = μmg
Centripetal Force = (m)(ac)
Therefore,
μmg = (m)(ac)
ac = μg
where,
ac = magnitude of centripetal acceleration of car = ?
μ = coefficient of friction of tires (kinetic) = 0.4
g = 9.8 m/s²
Therefore,
ac = (0.4)(9.8 m/s²)
ac = 3.92 m/s²
Based on the data provided, the centripetal acceleration is 3.92 m/s²
What is centripetal acceleration?Centripetal acceleration is the acceleration of a body moving in a circular path which is directed toward the center of the circle.
In the given question, the frictional force must balance the centripetal force for the car not to skid.
Frictional Force = Centripetal Forcewhere,
Frictional Force = μR
R = mg
F = μmg
Centripetal Force = m
Then
μmg = ma
a = μg
ac = 0.4 * 9.8 m/s²
ac = 3.92 m/s²
Therefore, the centripetal acceleration is 3.92 m/s².
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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
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
Complete the following statement:
Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different:_____
a. volume.
b. specific heat capacities.
c. expansion coefficients.
d. mass.
e. length.
Answer: c. expansion coefficients.
Explanation: Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different expansion coefficients.
I found the answer on Quizlet. :)
Bimetallic strips used as adjustable switches in electric appliances consist of metallic strips that must have different expansion coefficients. The correct option is c.
What is the expansion coefficient?The coefficient of thermal expansion (CTE) is the rate at which a material expands as its temperature rises. This coefficient is determined at constant pressure and without a phase change, i.e. the material is expected to remain solid or fluid.
Bimetallic strips, which are utilized as adjustable switches in electric appliances, are made up of metallic strips with differing expansion coefficients. The coefficient of thermal expansion indicates how the size of an object varies as temperature changes.
Therefore, the correct option is c. expansion coefficients.
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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!
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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A 40.0 kg ballet dancer stands on her toes during a performance with 25.0 cm2 in contact with the floor. What is the pressure exerted by the floor over the area of contact if the dancer is stationary
Explanation:
40×10ms^-2
400N.
25/100m^2
0.25m^2
1.P=F/A
=400N/0.25m^2
=100Nm^-2
=100Pa
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.
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
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
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos 5t + π 8 where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos (5t + π / 8) where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm
Answer:(a) 7.392cm
(b) -15.32 cm/s
(c) -184cm/s²
(d) 0.4πs and 8.00cm
Explanation:The general equation of a simple harmonic motion (SHM) is given by;
x(t) = A cos (wt + Φ) --------------(i)
Where;
x(t) = position of the body at a given time t
A = amplitude or maximum displacement during oscillation
w = angular velocity
t = time
Φ = phase constant.
Given from question:
x(t) = 8.00 cos (5t + π / 8) ---------------(ii)
(a) At time t = 0;
The position, x(t), of the body (piston) is given by substituting the value of t = 0 into equation (ii) as follows;
x(0) = 8.00 cos (5(0) + π / 8)
x(0) = 8.00 cos (π /8)
x(0) = 8.00 x 0.924
x(0) = 7.392 cm
Therefore, the position of the piston at time t = 0 is 7.392cm
(b) To get the velocity, v(t), of the piston at t = 0, first differentiate equation (ii) with respect to t as follows;
v(t) = [tex]\frac{dx(t)}{dt}[/tex]
v(t) = [tex]\frac{d(8.00cos(5t + \pi / 8 ))}{dt}[/tex]
v(t) = 8 (-5 sin (5t + π / 8))
v(t) = -40sin(5t + π / 8) --------------------(iii)
Now, substitute t=0 into the equation as follows;
v(0) = -40 sin(5(0) + π / 8)
v(0) = -40 sin(π / 8)
v(0) = -40 x 0.383
v(0) = -15.32 cm/s
Therefore, the velocity of the piston at time t = 0 is -15.32 cm/s
(c) To find the acceleration a(t) of the piston at t = 0, first differentiate equation (iii), which is the velocity equation, with respect to t as follows;
a(t) = [tex]\frac{dv(t)}{dt}[/tex]
a(t) = [tex]\frac{d(-40sin (5t + \pi /8))}{dt}[/tex]
a(t) = -200 cos (5t + π / 8)
Now, substitute t = 0 into the equation as follows;
a(0) = -200 cos (5(0) + π / 8)
a(0) = -200 cos (π / 8)
a(0) = -200 x 0.924
a(0) = -184.8 cm/s²
Therefore, the acceleration of the piston at time t = 0 is -184cm/s²
(d) To find the period, T, first, let's compare equations (i) and (ii) as follows;
x(t) = A cos (wt + Φ) --------------(i)
x(t) = 8.00 cos (5t + π / 8) ---------------(ii)
From these equations it can be deduced that;
Amplitude, A = 8.00cm
Angular velocity, w = 5 rads/s
But;
w = [tex]\frac{2\pi }{T}[/tex] [Where T = period of oscillation]
=> T = [tex]\frac{2\pi }{w}[/tex]
=> T = [tex]\frac{2\pi }{5}[/tex]
=> T = 0.4π s
Therefore, the period and amplitude of the piston's motion are respectively 0.4πs and 8.00cm
define limitations in the operation conditions of a pn junction
Answer:
Such limitations are given below.
Explanation:
Each pn junction provides limited measurements of maximum forwarding current, highest possible inversion voltage as well as the maximum output level.If controlled within certain adsorption conditions, the pn junction could very well offer satisfying performance. In connector operation, the maximum inversion voltage seems to be of significant importance.