Answer: the amount of mass is oscillating is 34.8 kg
Explanation:
Given that;
amplitude A = 20.0 cm
time t = 10 s
amplitude decreases x = 15.0 cm
damping coefficient b = 2.00 N.s/m
amount of mass is oscillating = ?
we know that; amplitude can be expressed as;
x = Ae^-(∝t)
we substitute
15 = 20e^-∝(10)
∝ = 0.02877 s⁻¹
Hence mass m will be;
m = b/2∝
we substitute
m = (2 N.s/m) / ( 2 × 0.02877 s⁻¹)
m = 34.8 kg
Therefore the amount of mass is oscillating is 34.8 kg
What is the kinetic energy of a 1500 kg vehicle traveling at a velocity of 8 m/s?
Answer:
48,000 JExplanation:
The kinetic energy of an object can be found by using the formula
[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass
v is the velocity
From the question we have
[tex]k = \frac{1}{2} \times 1500 \times {8}^{2} \\ = 750 \times 64[/tex]
We have the final answer as
48,000 JHope this helps you
football or basketball which one is easier to get you offers
Answer:
basketball
Explanation: cause it is
Answer:
It honestly depends on which one you're better at.
Explanation: I mean that's what I think.
Two objects of same material are travelling near you. Object A is a 1.1 kg mass traveling 10.2 m/s; object B is a 2 kg mass traveling 5 m/s. Which object would make you feel worse if you are hit by it
Answer:
Object A
Explanation:
The object that would make you feel worse if you're hit by it is the object possessing the highest momentum. Thus, we need to find the momentum of the two objects.
Momentum of an object is the product of its mass and that of it's velocity. Momentum is given by the formula
P = M * V, where
P = momentum
M = mass of the object
V = velocity of the object
Now, solving for object A, we have
P(a) = 1.1 * 10.2
P(a) = 11.22 kgm/s
And then, solving for object B, we have
P(b) = 2 * 5
P(b) = 10 kgm/s
The object when the highest momentum is object A, and thus would make you feel worse when hit by it
A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is A) The same for all wavelengths B) In the near-ultraviolet C) In the visible D) In the near-infrared E) Indeterminate
Answer:
B) In the near-ultraviolet
Explanation:
The best option to this problem is : In the near-ultraviolet
A car accidently roll of a cliff. As it leaves the cliff it has horzontal velocity of 13 m/s it hits the ground 60m from the shoreline. Calculate the hight of the cliff
Answer: The height of the cliff is 104.59 m
Explanation:
The horizontal speed of the car when it leaves the cliff is 13 m/s, and it hits the ground 60m from the shoreline.
Here we can use the relationship:
Time*Speed = Distance.
To find the time that the car is in the air, we know that:
speed = 13m/s
distance = 60m
time = T
13m/s*T = 60m
T = (60m)/13m/s = 4.62 s
This means that the car is falling for 4.62 seconds.
Now let's analyze the vertical problem.
As the car leaves the cliff, it only has horizontal velocity, this means that the vertical initial velocity will be zero
The only force acting in the vertical axis is the gravitational force, this means that the acceleration will be equal to the gravitational acceleration, which is:
g = 9.8m/s^2
then:
a = -9.8m/s^2
Where the negative sign is because the acceleration is pulling the car downwards.
To get the vertical velocity, we could integrate over time to get:
v(t) = (-9.8m/s^2)*t + v0
Where v0 is the constant of integration and the initial vertical velocity, that we already know that is equal to zero, then the vertical velocity as a function of time can be written as:
v(t) = (-9.8m/s^2)*t
To get the vertical position equation, we need to integrate again over the time:
P(t) = (1/2)*(-9.8m/s^2)*t^2 + H
Where H is the constant of integration and the initial vertical position, then H will be the height of the cliff.
We know that the car needs 4.62 seconds to hit the ground, this means that:
P(4.6s) = 0m
Then:
P(t) = (1/2)*(-9.8m/s^2)*(4.62s)^2 + H = 0
(-4.9m/s^2)*(4.62s)^2 + H = 0
H = (4.9m/s^2)*(4.62s)^2 = 104.59 m
This means that the cliff is 104.59 meters high
A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The mass is released, separating it 5 cm from its equilibrium position. Find the period of the motion
Answer:
m 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 mm 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 m
Explanation:
That is a reason
The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.4 V is across it at a frequency of 1400 Hz. What is the value of the inductance
Answer:
The value of the inductance is 1.364 mH.
Explanation:
Given;
amplitude current, I₀ = 200 mA = 0.2 A
amplitude voltage, V₀ = 2.4 V
frequency of the wave, f = 1400 Hz
The inductive reactance is calculated;
[tex]X_l = \frac{V_o}{I_o} \\\\X_l = \frac{2.4}{0.2} \\\\X_l =12 \ ohms[/tex]
The inductive reactance is calculated as;
[tex]X_l = \omega L\\\\X_l = 2\pi fL\\\\L = \frac{X_l}{2 \pi f}[/tex]
where;
L is the inductance
[tex]L = \frac{12}{2 \pi \times \ 1400} \\\\L = 1.364 \times \ 10^{-3} \ H\\\\L = 1.364 \ mH[/tex]
Therefore, the value of the inductance is 1.364 mH.
Given values are:
Current, [tex]I_0 = 200 \ mA, or \ 0.2 \ A[/tex]Voltage, [tex]V_0 = 2.4 \ V[/tex]Frequency, [tex]f = 1400 \ Hz[/tex]As we know the formula,
→ [tex]X_l = \frac{V_0}{I_0}[/tex]
By substituting the values, we get
[tex]= \frac{2.4}{0.2}[/tex]
[tex]= 12 \ ohms[/tex]
Now,
The inductive reactance will be:
→ [tex]X_l = \omega L[/tex]
or,
→ [tex]X_l = 2 \pi f L[/tex]
Hence,
The Inductance will be:
→ [tex]L = \frac{X_i}{2 \pi f}[/tex]
By putting the values, we get
[tex]= \frac{12}{2 \pi\times 1400}[/tex]
[tex]= 1.364\times 10^{-3} \ H[/tex]
[tex]= 1.36 \ mH[/tex]
Thus the above answer is right.
Learn more about Inductance here:
https://brainly.com/question/17136897
A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic friction coefficient is 0.1.
Draw a FBD of all the forces acting on the sled.
What is the weight of the sled?
What force is needed to start the sled moving?
What force is needed to keep the sled moving at a constant velocity?
Answer:
a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.
b) The weight of the sled is 490.35 newtons.
c) A force of 147.105 newtons is needed to start the sled moving.
d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.
Explanation:
a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:
[tex]F[/tex] - External force exerted on the sled, measured in newtons.
[tex]f[/tex] - Friction force, measured in newtons.
[tex]N[/tex] - Normal force from the ground on the mass, measured in newtons.
[tex]W[/tex] - Weight, measured in newtons.
b) The weight of the sled is determined by the following formula:
[tex]W = m\cdot g[/tex] (1)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]m = 50\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the weight of the sled is:
[tex]W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]W = 490.35\,N[/tex]
The weight of the sled is 490.35 newtons.
c) The minimum force needed to start the sled moving on the horizontal ground is:
[tex]F_{min,s} = \mu_{s}\cdot W[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]W[/tex] - Weight of the sled, measured in newtons.
If we know that [tex]\mu_{s} = 0.3[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to start the sled moving is:
[tex]F_{min,s} = 0.3\cdot (490.35\,N)[/tex]
[tex]F_{min,s} = 147.105\,N[/tex]
A force of 147.105 newtons is needed to start the sled moving.
d) The minimum force needed to keep the sled moving at constant velocity is:
[tex]F_{min,k} = \mu_{k}\cdot W[/tex] (3)
Where [tex]\mu_{k}[/tex] is the kinetic coefficient of friction, dimensionless.
If we know that [tex]\mu_{k} = 0.1[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to keep the sled moving at a constant velocity is:
[tex]F_{min,k} = 0.1\cdot (490.35\,N)[/tex]
[tex]F_{min,k} = 49.035\,N[/tex]
A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.
X- rays with a wavelgnth of 0.0700 nm diffract from a crystal. Two adjacne angels of x-ray diffraciton are 45.6 and 21.0 degrees. What is the distance in nm between the atomic planes responsible for the diffraction? (Use 2dcos(θ)=m(λ))
Answer:
0.15 nm
Explanation:
d = Distance between the atomic planes
m = Order
[tex]\theta_{m}[/tex] = First angle = [tex]45.6^{\circ}[/tex]
[tex]\theta_{m+1}[/tex] = Adjacent angle = [tex]21^{\circ}[/tex]
[tex]\lambda[/tex] = Wavelength = 0.07 nm
From Bragg's relation we know
[tex]2d\cos\theta_{m}=m\lambda[/tex]
[tex]2d\cos45.6^{\circ}=m0.07[/tex]
[tex]2d\cos\theta_{m+1}=(m+1)\lambda[/tex]
[tex]2d\cos21^{\circ}=(m+1)0.07\\\Rightarrow 2d\cos21^{\circ}=m(0.07)+0.07[/tex]
So
[tex]2d\cos21^{\circ}=2d\cos45.6^{\circ}+0.07\\\Rightarrow 2d(\cos21^{\circ}-\cos45.6^{\circ})=0.07\\\Rightarrow d=\dfrac{0.07}{2(\cos21^{\circ}-\cos45.6^{\circ})}\\\Rightarrow d=0.14962\ \text{nm}[/tex]
The distance between the atomic planes is 0.15 nm.
g You want to make simultaneous measurements of the position and momentum of an electron and a proton that are moving along a straight line. An alternate statement of the uncertainty principle involves relationship between position (Δx) and momentum (Δp) uncertainties in the form . If both of them are located with an uncertainty of 1 × 10-10 m, what is the ratio of uncertainty in the velocity of the electron to that of the proton?
Answer:
1832
Explanation:
From;
Δp Δx = h/4π
Δp = uncertainty in momentum
Δx = uncertainty in position
h= Plank's constant
But p =mv hence, Δp= Δmv
m= mass, v= velocity
mass of electron = 9.11 * 10^-31 Kg
Mass of proton = 1.67 * 10^-27 Kg
since m is a constant,
Δv = h/Δxm4π
For proton;
Δv = 6.6 * 10^-34/4 * 3.14 * 1.67 * 10^-27 * 1 * 10^-10
Δv = 315 ms-1
For electron;
Δv = 6.6 * 10^-34/4 * 3.14 * 9.11 * 10^-31 * 1 * 10^-10
Δv = 577000 ms-1
Ratio of uncertainty of electron to that of proton = 577000 ms-1/315 ms-1= 1832
What are various names of tropical storms
based upon the ocean they occur in?
Answer:
hurricanes,Typhoons and cyclones
The aorta carries blood away from the heart at a speed of about 42 cm/s and has a diameter of approximately 1.1 cm. The aorta branches eventually into a large number of tiny capillaries that distribute the blood to the various body organs. In a capillary, the blood speed is approximately 0.08 cm/s, and the diameter is about 0.0005 cm. Treat the blood as an incompressible fluid, and determine N the approximate number of capillaries in the human body.
Answer:
ok fine I'll answer u on comment
Coherent light from a laser shines through two slits 0.002 cm apart, producing an interference pattern on a screen. A bright central fringe is seen on the screen in line with a point directly between the slits. The next bright fringe is seen at an angle of 2 degrees above the central fringe (with a dark fringe between). What is the wavelength of the laser’s light? (1 nm = 10-9 m)
Answer:
λ = 698 nm (Approx)
Explanation:
Given:
Width d = 0.002 cm = 0.002 x 10⁻² nm
θ = 2°
Computation:
d sinθ = mλ
(0.002 x 10⁻²) sin2° = mλ
λ = 698 nm (Approx)
Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. If this body is released from rest in a horizontal position, what is the angular speed of the meter stick as it swings through its lowest position
Answer:
5.35 rad/s
Explanation:
From the question, we are toldthat an Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick.
Solving this question, the potential energy of the particles must equal to the Kinectic energy i.e
P.E=K.E
Mgh= m½Iω²-------------eqn(*)
Where M= mass of the particles
g= acceleration due to gravity= 9.81m/s^2
ω= angular speed =?
h= height of the particles in the stick on the metre stick= ( 50cm + 80cm)= (0.5m + 0.8m)= 130cm=1.3m
If we substitute the values into eqn(*) we have
m×9.81× (1.3m)= 1/2× m×[ (0.5m)² + [(0.8m)²]× ω²
m(12.74m²/s²)= 1/2× m× (0.25+0.64)× ω
m(12.74m²/s²)= 1/2× m× 0.89× ω²
We can cancel out "m"
12.74= 1/2×0.89 × ω²
12.74×2= 0.89ω²
25.48= 0.89ω²
ω²= 28.629
ω= √28.629
ω=5.35 rad/s
Hence, the angular speed of the meter stick as it swings through its lowest position is 5.35 rad/s
Match each method of transferring electric charge with the correct description
friction
transfer of electric charge without direct
contact
induction
transfer of electric charge by rubbing
conduction
transfer of electric charge by direct contact
The matching of each method of transferring electric charge with the correct description should be explained below.
Matching of transferring electric charge?The friction means the transfer of electric charge via rubbing. The conduction means the transfer of electric charge via the direct contact.
Also, the induction means the transfer of the electric charge without the direct contact.
In this way it should be matched.
Learn more about electric here: https://brainly.com/question/23056096
Answer:
this is the answer that is correct
An Airplane moves 70 m/s as it travels around a vertical circular loop which has a 3 km radius. What is the magnitude of the net force on the 95 kg pilot at the bottom of this loop?
Answer:
155.17N.
Explanation:
The magnitude of the net force is expressed as;
F = mv²/r where;
m is the mass
v is the velocity of the airplane
r is the radius of the loop
Given
m = 95kg
v = 70m/s
r = 3km = 3000m
Required
Magnitude of the net force
F = 95*70²/3000
F = 95*4900/3000
F = 95*49/30
F = 4655/30
F = 155.17N
Hence the magnitude of the net force on the 95 kg pilot at the bottom of this loop is 155.17N.
The net force on the pilot is 1551.67 N.
What is force?Force can be defined as the product of mass and acceleration.
To calculate the magnitude of the net force at the bottom of the loop, we use the formula below.
Formula:
F = mv²/r............. Equation 1Where:
m = mass of the pilotv = velocity of the Airplaner = radius of the loopF = net force on the pilot.From the question,
Given:
m = 95 kgr = 3 km = 3000 mv = 70 m/sSubstitute these values into equation 1
F = 95(70²)/300F = 465500/300F = 1551.67 NHence, the net force on the pilot is 1551.67 N.
Learn more about force here: https://brainly.com/question/13370981
Lisa made the electromagnet shown. (Image above) What can Lisa do to increase the strength of the electromagnet?
O She can use a nail with weaker magnetic properties.
O She can change the direction of the nail.
O She can increase the number of wire loops.
O She can reduce the current in the wire.
If correct first gets Brainliest Please hurry this is a timed test( this is second time just in case I didn't go through)
Answer:
I think C
Explanation:
if im wrong then what is it im taking the exam too
Answer: C. She can increase the number of wire loops.
Explanation:
Do dwarf planets has several other bodies in their path orbiting the Sun just as they do?
Answer:
No they just orbit the sun
Explanation:
What is the significance of Nucleotides in Chromosomes?
Answer:
it comprises of the DNA/RNA bipolymer molecules
How long is a simple pendulum with a period of 1 second?
Answer:
the second pendulum ( also called the royal pendulum) o.994m (39.1 in) long, in which each swing takes one second, became widely used in quality clocks.
A man stands on top of a cliff and shouts.
He hears the echo on the third clap when
he claps his hand at the rate of two claps
per second. What is the distance between
man & the obstruction, if the velocity of
sound is 320 m/s
[tex] \small\bf \: let \: the \: distance \: of \: the \: man \: from \: the \: cliff \: be \: x[/tex]
[tex] \small\bf \: thus \: time \: taken \: by \: sound \: to \: hit \: the \: cilff \: and \: return = \frac{2x}{v} = 1[/tex]
[tex] \bf \to \: x = \frac{320}{2} m = 160m[/tex]
[tex] \small \bf \: thus \: the \: distance \: between \: the \: cliffs \: = 160m \times 2 = 320m[/tex]
if the instantaneous current in the circuit is giveen by I=3 sin theta amperes, the rms value of the current will be
Answer:
[tex]I_{rms}=2.12\ A[/tex]
Explanation:
Given that,
The instantaneous current in the circuit is giveen by :
[tex]I=3\sin\theta\ A[/tex]
We need to find the rms value of the current.
The general equation of current is given by :
[tex]I=I_o\sin\theta[/tex]
It means, [tex]I_o=3\ A[/tex]
We know that,
[tex]I_{rms}=\dfrac{I_o}{\sqrt2}\\\\=\dfrac{3}{\sqrt2}\\\\=2.12\ A[/tex]
So, the rms value of current is 2.12 A.
The charge on an electron is
Answer:
there is a negative charge on electron.
Explanation:
Determine e when I = 0.50 A and R = 12 W.
Answer:
The correct answer is "24 V".
Explanation:
The given values are:
Current,
I = 0.50 A
Resistance,
R = 12 W
As we know,
⇒ [tex]I = 0.5\times (\frac{E}{2R})[/tex]
On substituting the given values, we get
⇒ [tex]0.5= (\frac{E}{4\times 12} )[/tex]
⇒ [tex]0.5= (\frac{E}{48} )[/tex]
⇒ [tex]E=24 \ V[/tex]
If a paratroopers legs have enough strength to support five times her weight, and they can do so over a travel distance of 0.5 m as they flex, from what height can she free fall to a solid surface without injury
Answer: the required height is 2.5 m
Explanation:
given the given data in question
Lime to come in rest;
using S = ((U1 + U2)/2) t
⇒ 0.5 = ((√(2gh) + 0)/2) t
t = (0.5 × 2) / √(2gh)
now fat = impulse = change in momentum dp
F = 5mg = dp/dt = [(m√(2gh))/(0.5 × 2)] × √(2gh)
5mg = 2mgh/(0.5 × 2)
5 = h / 0.5
h = 5 × 0.5
h = 2.5 m
Therefore the required height is 2.5 m
HELP PLZ!!!!!!!!!!!!!!
Answer:
Since binary is only 1 and 0, you can use a flashlight to display something similar to Morse code (see explanation below)
Explanation:
In binary, 1 means "on" and 0 means "off". A way you can use visible light is through turning on and off a flashlight. If the flashlight is turned on, it would represent a 1. If the flashlight is turned off, it would represent a 0. To make the message easier and more accurately understood for the receiver make sure to flash the lights in a consistent pattern (ex. each flash lasts no longer than half a second, one second between each digit, etc.)
For example, let's say you're trying to send the message "11001"
on on off off on
0 1 2 3 4 5 Numbers represent seconds
As you can see above the message starts at 0 seconds. Between 0 and 1 seconds the flashlight is turned on once. Between 1 and 2 seconds the flashlight is turned on again, Between 2 and 3 seconds as well as 3 and 4 seconds the flashlight is not turned on at all. And finally between 4 and 5 seconds the flashlight is turned on.
Which of the following best illustrates Lewin's interactionist perspective?
Answer:
The options are
a. Sally is a very creative kind of person who likes to build things.
b. Jerry only works because he receives a very large income.
c. Rikki is usually shy, but at work she appears to be quite outgoing.
d. Maury gives money to charities because he wants other people to think he is very generous.
The answer is c. Rikki is usually shy, but at work she appears to be quite outgoing.
Lewin's interactionist perspective explains that an individual’s behavior is usually dependent on his personal behavior/ trait and the environment. The best option is that Rikki is usually shy which is her personal behavior but at work she appears to be quite outgoing due to her environment.
My tiger trap needs 735 N of force on top of it to activate. What is the lightest tiger I can trap?
7203 kg
0.013 kg
75 kg
725 kg
Answer:
75kg
Explanation:
[tex]F=mg[/tex]
[tex]m=\frac{F}{g}[/tex]
[tex]m=\frac{735}{9.8}[/tex]
[tex]m=75kg[/tex]
Therefore, the answer is the third option 75 kg
The energy released in the fission of one 235U92 is 206.6 MeV. A Nuclear reactor that uses this element has an output of 28.7 Mega Watts. What is the mass of 235U92 that is consumed in one day in this reactor
Answer: the required mass is 1.7628 × 10²⁵ μ
Explanation:
Given that;
energy released in the fission of one ²³⁵U₉₂ is 206.6 MeV
power p = 28.7 Mega Watts = 28.7 × 10⁶ W = 28.7 × 10⁶/ 1.6× 10⁻¹³ = 17.9375 × 10¹⁹ MeV/s
now fission need per second will be;
⇒ power / energy released i fission
= 17.9375 × 10¹⁹ MeV / 206.6 MeV = 8.68 × 10¹⁷ per second
now fission need per day will be;
⇒ ( 8.68 × 10¹⁷ × 360 × 24 ) = 7.5 × 10²² per day
hence mass of ²³⁵U₉₂ that is consumed in one day in this reactor will be;
⇒ (235 × 7.5 × 10²²)μ
= 1.7628 × 10²⁵ μ
Therefore the required mass is 1.7628 × 10²⁵ μ
Investigator Campbell has bullets that were collected from the crime scene and puts them under her comparison microscope. What other item would she need to examine?
The victim's clothing with the gunshot hole
Test bullets fired from the suspected gun
A bullet that has been fired into a wall or hard surface
The barrel of the suspected gun
Answer:
Test bullets fired from the suspected gun
Explanation:
This is the correct answer!! I took the test!!