Answer:
sorry not sure but i can try help with telling you this...
Explanation:
alls you havs to do is ask the internet and it will tell you
A vibrating mass of 300 kg mounted on a massless support by a spring of stiffness 40,000 N>m and a damper of unknown damping coefficient is observed to vibrate with a 10-mm amplitude while the support vibration has a maximum amplitude of only 2.5 mm (at resonance). Calculate the damping constant and the amplitude of the force on the base.
Answer:
400 N
Explanation:
[tex]\text { Given: } m=300 \mathrm{~kg}, k=40,000 \mathrm{~N} / \mathrm{m}, \omega_{b}=\omega_{n}(r=1), X=10 \mathrm{~mm}, Y=2.5 \mathrm{~mm}[/tex] .
Find damping constant
[tex] \frac{X}{Y}=\left[\frac{1+(2 \zeta r)^{2}}{\left[\left(1-r^{2}\right)^{2}+(2 \zeta r)^{2}\right.}\right]^{1 / 2} \\ \left.\frac{10}{2.5}=\frac{\left\lceil 1+4\zeta^{2}\right]^{1 / 2}}{4 \zeta^{2}}\right] \\ 16=\frac{1+4 \zeta^{-2}}{4 \zeta^{2}} \\ \zeta^{2}=\frac{1}{60}=\frac{c^{2}}{4 k m} \\ c=\sqrt{\frac{4(40,000)(300)}{60}} \\ c=894.4 \mathrm{~kg} / \mathrm{s}[/tex]
Amplitude of force on base:
[tex]F_{T}=k Y r^{2}\left[\frac{1+(2 \zeta r)^{2}}{\left(1-r^{2}\right)^{2}+(2 \zeta r)^{2}}\right]^{1 / 2}[/tex]
substituting the values in above formula we get
F_T = 400 N
Two people pull on the wagon each with a constant 20N force. Both people pull to the left. What is the Net Force on the wagon?
Answer:
I dont understand what you are trying to ask
Explanation:
Two identical positive charges exert a repulsive force of 6.4x10^-9 N when separated by a distance of 3.8x10^10 m. Calculate the charge of each.
Answer:F = kq2/d2 ⇒
q = √(Fd2/k)
q = d √(F/k)
d = 3.8 x 10-10 m
F = 6.4 x 10-9 N
Look up k in your physics book in appropriate units, and plug in the numbers. You should get q in coulombs.
Explanation:
The value of each charge will be 1.64 ×10⁻⁴. The concept of the columb force is used.
What is the charge?When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.
Charges that are similar repel each other, whereas charges that are dissimilar attract each other. The term "neutral" refers to an item that has no net charge.
[tex]\rm F =K \frac{q_1Q_2}{d^2} \\\\ F = K\frac{q^2}{d^2} \\\\ q = \sqrt{\frac{dF}{k} } \\\\ q = \sqrt{\frac{3.8 \times 10^{10}\times 6.4 \times 10^{-9}}{9 \times 10^9} } \\\\ q=1.64 \times 10-4[/tex]
Hence the value of each charge will be 1.64 ×10⁻⁴. The concept of the columb force is used.
To learn more about the charge refer to the link;
https://brainly.com/question/24391667
Which of the following represents a concave lens?
A. -di
B. +di
C. -f
D. +f
Answer:
The answer is option D. +f
Option D represents a concave lens. There are two types of lenses, one is a concave lens while the other is the convex lens.
What is the definition of a concave lens?A concave lens deviates a direct beam from the source into a reduced form. At minimum, one interior face of concave lenses is curved.
Because it is curved round inwards at the center and bulges outwards through the edges, causing the light to diverge, a concave lens is also known as a diverging lens.
It forms an upright , virtual picture both real and virtual pictures are formed from the concave lens.
The image formed from the positive side of the focus of the lens.+f shows the positive side of the focus of the lens.f is the focal length.
Hence, option D represents a concave lens
To learn more about the concave lens, refer to the link;
https://brainly.com/question/2919483
#SPJ2
How can parents help children to gain friends?
Answer:
You could try finding a familiar peer to join the activity with your child. Or ask your child who their friends are at school, or what they look for in a friend at school.
Answer:
Let the parents their Children to play outside
Explanation:
I HOPE I HELP YOU
A child makes a ramp to push his toy dump truck up to his sandbox. If he uses 5 newtons of force to push the 12-newton truck up the ramp, what is the mechanical advantage of his ramp?
Answer:
m = 2.4
Explanation:
Given that,
Input force, [tex]F_i=5\ N[/tex]
Output force, [tex]F_o=12\ N[/tex]
We need to find the mechanical advantage of the ramp. The ratio of output force to the input force is equal to mechanical advantage. So,
[tex]m=\dfrac{12}{5}\\\\m=2.4[/tex]
So, the mechanical advantage of his ramp is 2.4.
During an experiment, Ellie records a measurement of 0.0034 m. How would
she write her measurement in scientific notation?
A. 3.4 x 10-3 m
B. 3.4 x 10-4 m
O C. 3.4 x 10-5 m
D. 3.4 x 10-2 m
Answer:
(A) She needs to move the decimal point by 3 places
Please help !!!!!!!!!!!!
Answer:
152 Volts
Explanation:
First, the Resistors are in series. So, the net resistance, R = 3 + 4 +3 + 4 +5
= 19 ohms
Using, V = IR;
V (the needed P.D) = 8 x 19 = 152 Volts
will be needed to successfully transport 8 amps of current round the circuit
Answer:
4q
Explanation:
wait nvm i dont know
Which of the following is an example of predation?
Answer:
an owl hunts, catches, and consumes a field mouse for energy
Explanation:
Which planet is least like earth? Mars,Venus, or Jupiter
Answer:
mars, reason why is because they both are diff from the size
Explanation:
The plates of a vacuum-gap parallel plate capacitor have a 100.0 mm2 area, a vacuum gap of 5.00 mm and are connected to a 1.5-volt battery. After the capacitor is charged, the battery is disconnected from the capacitor. After the battery is disconnected, the plates are pulled apart until the vacuum gap is 7.50 mm. a. What are the initial and final energies stored in the capacitor
Answer:
E₀ = 2.0*10⁻¹¹ J = 0.2 pJ
Ef = 3.0*10⁻¹¹ J = 0.3 pJ
Explanation:
The energy stored between the plates of a parallel plate capacitor can be expressed in terms of the capacitance C and the potential difference between plates V as follows:[tex]E = \frac{1}{2} * C * V^{2} (1)[/tex]
When the capacitor is fully charged, the potential difference between plates must be equal to the voltage of the battery, 1.5 V.In a parallel plate capacitor, the value of the capacitance is independent of the applied voltage, and depends only on geometric constants and the dielectric constant of the medium between plates, as follows:[tex]C = \frac{\epsilon_{o}*A}{d} (2)[/tex]
We can find the initial value of C replacing in (2) by the givens below:A = 100.0 mm2d= 5.00 mmε₀ = 8.85*10⁻¹² F/m[tex]C_{o} = \frac{\epsilon_{o}*A}{d} = \frac{(8.85*(10)^{-12} F/m)*(10^{-4} m2)}{5.0*(10)^{-3}m} = 1.77*10^{-13} F (3)[/tex]With this value of C₀, and the value of the initial potential difference between plates (1.5 V), we can find the initial charge on the capacitor, starting from the definition of capacitance:[tex]C =\frac{Q}{V} (4)[/tex]Solving for Q in (4):[tex]Q = C_{o}* V = 1.77*10^{-13} F * 1.5 V = 2.65*10^{-13} C (5)[/tex]Finally, we can find the initial energy stored in the capacitor, replacing (3) and V in (1):[tex]E_{o} = \frac{1}{2} * C_{o} * V_{o} ^{2} = \frac{1}{2} * 1.77*10^{-13}F*(1.5V)^{2} = 0.2 pJ (6)[/tex]
If we pull apart the plates until the vacuum gap is 7.50 mm, we will change the expression of C in (2), decreasing its value due to the expanded gap.Replacing in (2) the new value of the gap (7.50 mm), we can find the new value of C, as follows:[tex]C = \frac{\epsilon_{o}*A}{d} = \frac{(8.85*10^{-12}F/m)*10^{-4} m2}{7.5*10^{-3}m} = 1.18*10^{-13} F (7)[/tex] In order to find the final energy stored in the capacitor, we need also the value of the final potential difference between plates.Once disconnected from the battery, the charge on any of the plates must remain the same, due to the principle of conservation of the charge.So, since we have the value of Q from (5) and the new value of C from (7), we can find the new potential difference between plates as follows:[tex]V_{f} = \frac{Q}{C_{f}} = \frac{2.7*10^{-13}C}{1.18*10^{-13}F} = 2.25 V (8)[/tex]With the values of Vf and Cf, we can find the value of the final energy stored in the capacitor, replacing these values in (1):[tex]E_{f} = \frac{1}{2} * C_{f} * V_{f} ^{2} = \frac{1}{2} * 1.18*10^{-13}F*(2.25V)^{2} = 0.3 pJ (9)[/tex]
The body mass of Asaiah is 70 Kg.
(a) What is his weight on Earth?
(b) If he goes to the Moon,
(i) What is his mass?
(ii) What is his weight?
Answer:
A I hope its not wrong I hope u do good
I need help plissss..............
Part one: Multiple choices
1) A person sitting in the compartment of moving train is:
a) in the state of rest with respect to surroundings of the compartment,
b) in the state of motion with respect to surroundings of the compartment.
C) in the state of rest with respect to surroundings outside of the compartment
d)all of them
2) The motion of tuning fork prongs on vibration is:
a) Linear motion
b) periodic motion
c) circular motion
d) projectile motion
3) All the following are periodic motion except
a) moving car in straight line
b) Earth's rotation
c) pendulum
d) Swing
4) The rate of change of displacement is:
a) Acceleration
b) force
c) distance
d) velocity
5) When an object moves at negative acceleration in a straight line its:
a) displacement equals zero
b) velocity decrease
C) velocity increase
d) none of them
6) When the object speeds up its acceleration:
a) decreases
b) increases c) it has no acceleration d) All of them
7) The rate of change of velocity is:
a) force
b) variable velocity
c) instantaneous velocity
d) acceleration
8) If a train is moving in a straight line to cover a distance of 600 m in a minute its
velocity is:
a) 600 m/s
b) 60 m/s c) 100 m/s d) 480 m/s
9) The division between total displacement and total time is the:
a) variable velocity b) average velocity c) speed d) Instantaneous velocity
10) The rate of change of displacement at a given instant is called the
a) average velocity
b) instantaneous velocity
C) average velocity
d) instantaneous acceleration
11) A body completes one circular revolution in a roundabout whose diameter
140 m.
Find its displacement,
a) 439.6 m
b) 440 m
c) zero
d) 879.2 m
Answer:
1)
a) in the state of rest with respect to surroundings of the compartment,
2)
b) periodic motion
3)
a) moving car in a straight line
4)
d) velocity
5)
b) velocity decrease
6)
b) increases
7)
d) acceleration
8)
10 m/s
9)
b) average velocity
10)
b) instantaneous velocity
11)
a) 439.6 m
Explanation:
1)
With respect to the inside surrounding the person will be at rest. Because the person is not moving inside the compartment.
2)
The vibration motion follows periodic motion.
3)
The car moving in a straight line is an example of rectilinear motion and its wheels are in rotational motion. They are not in periodic motion.
4)
Definition of velocity.
5)
Acceleration is the rate of change of velocity. So negative acceleration means a decrease in velocity.
6)
Acceleration is the rate of change of velocity. So an increase in velocity means an increase in acceleration.
7)
Definition of acceleration.
8)
[tex]velocity = \frac{Distance}{Time}\\\\velocty = \frac{600\ m}{1\ min}\frac{1\ min}{60\ s}\\\\velocity = 10\ m/s[/tex]
Hence, none of the options is correct. The correct answer is 10 m/s.
9)
Definition of average velocity.
10)
Definition of instantaneous velocity.
11)
[tex]Displacement = Circumference = \pi d\\Displacement = \pi(140\ m)\\Displacement = 439.6\ m[/tex]
The total kinetic energy of an object depends on two (2) factors. Select those factors from the list below.
Mass
Density
Volume
Velocity
Please help me !!!!!!!
Answer: I believe that it is 35 Joules.
Explanation:
100 - 65 = 35
:)
Suppose a rocket in space is accelerating at 1.5 m/s2. If, at a later time, the rocket quadruples its thrust (i.e., net propelling force), what is the new acceleration?
a man pushed on the side ..
Answer:
B.will increase the maximum static friction between the box and the floor
Explanation:
Because static friction is the force that keeps an object at rest
a potted plant falls from a window sill and is gaining speed. which one of the following statements is true of the plant?
a) its kinetic energy is constant
b) its kinetic energy is increasing
c) its kinetic energy is decreasing
a solid disk rotates in the horizontal plane at an angular velocity of 4.9 x 10 rad/s with respect to an axis perpendicular to the disk at its center the moment of inertia of the disk is 0.14 kg from above sand is dropped straight down onto this rotating disk so that a thin unifrom ring of sand is formed at distance of 0.4 m from the axis the sand in the ring has mass of 0.5 kg after all the sand is in place what is the angular velocity of the disk
Answer:
ωf = 3.1*10 rad/sec
Explanation:
Assuming no external torques acting while the sand is being dropped, total angular momentum must keep constant.So we can write the following equality:[tex]L_{o} = L_{f} (1)[/tex]
For a rigid body rotating with respect to an axis, the angular momentum can be written as follows:[tex]L = I* \omega (2)[/tex]
where I = moment of inertiaω = angular velocityReplacing (2) on both sides of (1) we get:[tex]I_{o}* \omega_{o} = I_{f}* \omega_{f} (3)[/tex]
In (3) we know the values of I₀ and ω₀ (since they are givens), but we need to find the value of If first.The final moment of inertia, will be equal to the sum of the initial one, plus the one due to the ring of sand, that also rotates with respect to an axis perpendicular to the disk, as follows:[tex]I_{f} = I_{o} + I_ {ring} (4)[/tex]The moment of inertia of a circular ring is as follows:[tex]I_{ring} = m_{ring} *r^{2} (5)[/tex]
Replacing by the givens in (5) we get:[tex]I_{ring} = m_{ring} *r^{2} = 0.5 kg * (0.4m)^{2} = 0.08 kg*m2 (6)[/tex]
Replacing (6) in (4):[tex]I_{f} = I_{o} + I_ {ring} = 0.14kg*m2 + 0.08 kg*m2 = 0.22 kg*m2 (7)[/tex]Replacing I₀, ω₀ and If in (3), we can solve for ωf, as follows:[tex]\omega_{f} =\frac{I_{o} *\omega_{o} }{I_{f} } = \frac{0.14kg*m2*4.9*10rad/sec}{0.22kg*m2} = 3.1*10 rad/sec (8)[/tex]It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 100m above a raging river and attach a 35-m-long bungee cord to your harness. A bungee cord, for practical purposes, is just a long spring, and this cord has a spring constant of 43N/m . Assume that your mass is 79kg . After a long hesitation, you dive off the bridge. How far are you above the water when the cord reaches its maximum elongation? h=
Answer:
h = 47 m
Explanation:
First, we will calculate the force on the cord due to the weight:
[tex]Force = F = Weight\\F = mg\\F = (79\ kg)(9.81\ m/s^2)\\F = 775\ N[/tex]
Now, we will calculate the elongation by using Hooke's Law:
[tex]F = k \Delta x[/tex]
where,
k = spring constant = 43 N/m
Δx = elongation = ?
Therefore,
[tex]775\ N = (43\ N/m)\Delta x\\\\\Delta x = \frac{775\ N}{43\ N/m}\\\\\Delta x = 18\ m\\[/tex]
So, the final length of the cord will be:
[tex]Final\ Length = Initial\ Length + \Delta x\\Final\ Length = 35\ m + 18\ m\\Final\ Length = 53\ m\\[/tex]
Hence, the height from water (h) can be found using the following formula:
[tex]h = Height\ of\ Bridge - Final\ Length\ of\ cord\\h = 100\ m - 53\ m\\[/tex]
h = 47 m
The modern model of the atom describes electrons in a little less specific detail than earlier models did. Why is it that being less sure about the placement of electrons in an atom is actually an improvement over earlier models?
The plum pudding model of the atom states that
Answer:
It is because one cannot know exactly the position of the electron within the atom.
One formulation of Heisenberg's Uncertainty Principle tells us that one cannot know simultaneously the position and momentum of the electron, so one cannot specify exactly either coordinate because the other would be infinite.
Bohr specified the most probable position of the electron at its lowest energy level in hydrogen and the product of the two would be about the Heisenberg value.
What speed must a 600 kg car have in order to have the same momentum as a 1200 kg truck traveling at a velocity of 10 m/s to the west? 5 m/s west B 20 mls west 5 m/s east 20 m/s east
A) 5m/s west
B) 20m/s west
C) 5m/s east
D) 20m/s east
Answer:
20 m/s
Explanation:
Please help. It’s probably easy
There are two main types of collisions that you will study: elastic and perfectly inelastic. In an elastic collision, kinetic energy is conserved. In a perfectly inelastic collision, the particles stick together and thus retain the same velocity after the collision.
a. True
b. False
Answer:
The first part is right (KE is conserved in an elastic collision).
The second part of the statement is false,.
Since momentum is conserved, let moving mass m strike stationary mass M:
m v = (m + M) V where m v is the momentum in
Obviously, v does not equal V.
Mention two ways in which the effects of friction can be minimised
Answer:
Polishing the rough surface.
Oiling or lubricating with graphite or grease the moving parts of a machine.
Providing all bearings or wheels between the moving parts of a machine or vehicles reduce friction and allow smooth movement as rolling friction is less than sliding friction.
Explanation:
Unless otherwise stated, all objects are located near the Earth's surface, where g = 9.80 m/s2 .
A force acts on a 1.5 kg , mass, giving it an acceleration of 3.0 m/s2 .
1. If the same force acts on a 3.0 kg mass, what acceleration would be produced?
2. What is the magnitude of the force?
Answer:
1) 1.5 m/s^2
2) 4.5 N
Explanation:
From Newton's Second Law of motion, we know
[tex]F=m*a[/tex]
Which states that to calculate the force acting on an object, you multiply its mass and acceleration.
So, we know an object of mass 1.5 kg has an acceleration of 3 m/s^2, then
[tex]F=m*a=1.5*3=4.5[/tex]
A force of 4.5 N is acting on the object.
If a force of 4.5 N acts on a mass of 3kg we have
[tex]a=\frac{F}{m}=\frac{4.5}{3}=1.5[/tex]
So, it would give it an acceleration of 1.5 m/s^2.
You are comparing the beam waste for two different situations with the goal of using the smallest beam waste possible. A Nd-YAG laser system emits light at 532 nm and the beam is 8 mm in diameter. You also have a Ti-sapphire laser that emits at 855 nm and has a beam diameter of 6 mm. Compare the beam waist for both laser systems using a focusing lens with a focal length of 10 mm. Assume the light fills the lenses in each case
Answer:
comparing the beam waist for both lasers ( ratio of the beam waists )
4.536 μm / 2.117 μm = 2.14
Explanation:
Nd-YAG laser system : emits at 532 nm , beam diameter = 8 mm
Ti-sapphire laser system : emits at 855 nm , Beam diameter = 6mm
Comparing the beam waist for both lase systems using a focusing lens
Focal length = 10 mm
assumption : light fills lenses in each laser system
Beam waist radius ( W ) = [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex]
β = wavelength , D = diameter illuminated , F = focal length
For
Nd-YAG laser system
β = 532 mm , D = 8 mm
hence ( Wn ) = [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex] = ( 2*532 / π ) ( 10 / 8 ) = 2.117 μm
For
Ti-sapphire laser
β = 855 nm , D = 6 mm
hence ( Wt ) [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex] = ( 2* 855 ) / π ) ( 50 / 6 ) = 4.536 μm
comparing the beam waist for both lasers ( ratio of the beam waists )
4.536 μm / 2.117 μm = 2.14
The total charge a battery can supply is rated in mA⋅hmA⋅h, the product of the current (in mA) and the time (in h) that the battery can provide this current. A battery rated at 1000mA⋅h can supply a current of 1000 mA for 1.0 h, 500 mA current for 2.0 h, and so on. A typical AA rechargeable battery has a voltage of 1.2 V and a rating of 1800mA⋅h. For how long could this battery drive current through a long, thin wire of resistance22Ω?
Answer:
118800 seconds
Explanation:
Given :
Voltage, V = 1.2 V
Resistance, R = 22 Ω
Applying Ohm's law, we get
Voltage, V = IR
Current [tex]$I=\frac{V}{R}$[/tex]
[tex]$I=\frac{1.2}{22}$[/tex]
I = 0.0545 A
Rate = 1800 mAh
Time taken, [tex]$t=\frac{1800 \times 10^{-3}}{0.0545}$[/tex]
= 33 hr
= 118800 s
the radius of earth is about 6.38 x10^3 km. A 7.20 x10^3 N spacecraft travels away from earth. What is the weight of the spacecraft at the following distances from Earth's surface? a) 6.38 x 10^3 km
Answer:
[tex]1796.65\ \text{N}[/tex]
Explanation:
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
w = Weight of spracecraft at the surface = [tex]7.2\times10^3\ \text{N}[/tex]
m = Mass of spracecraft
R = Radius of Earth = [tex]6.38\times10^3\ \text{km}[/tex]
h = Elevation = [tex]6.38\times10^3\ \text{km}[/tex]
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
M = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]
[tex]w=mg\\\Rightarrow m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{7.2\times 10^3}{9.81}\\\Rightarrow m=733.94\ \text{kg}[/tex]
From the gravitational law we have
[tex]w'=\dfrac{GMm}{(r+h)^2}\\\Rightarrow w'=\dfrac{6.674\times10^{-11}\times 5.972\times 10^{24}\times 733.94}{(6.38\times10^6+6.38\times10^6)^2}\\\Rightarrow w'=1796.65\ \text{N}[/tex]
The weight of the spacecraft at the given height is [tex]1796.65\ \text{N}[/tex]