Answer:
k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
give us some initial conditions
1) friction force fr = 1N when v = 4m / s
2) an initial displacement of x = 0.08 m for t=0 s
Explanation:
In this exercise, you are asked to state the problem you are posing. We are going to find the equation of motion for this exercise. Let's start with Newton's second law
Let's set a reference system with the y-axis in a vertical and positive direction upwards.
We have four forces: an external downward force, negative in sign, the but that goes down and is negative, the Hook force that goes up and is positive and the friction force that opposes the movement, in this case it goes down being negative
let's write Newton's second law
F_e -F -fr - W = m a
where
F_e = -kDy = - k y
fr = - b v = -b dy / dt
W = mg
we substitute for the specific case, that is, using the signs
k y -b [tex]\frac{dy}{dt}[/tex] - m g - F = m [tex]\frac{d^2y}{dt^2}[/tex]
In the initial condition of the problem, before starting the movement, the friction force is zero and the acceleration is also zero
k y - m g - F = 0
from this equation you can find the spring constant, y= 9m and F=2 N
It is not clear if when the movement starts this external force becomes zero, but since it balances the weight we can eliminate the two forces that have the same magnitude and opposite direction, so the equation remains
k y - b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
give us some initial conditions
1) friction force fr = 1N when v = 4m / s
2) an initial displacement of x = 0.08 m for t=0 s
therefore, to initiate the movement, a small external force F 'is applied that moves the system to a new equilibrium position and this small force F' is made zero, thus initiating an oscillatory movement, described by the equation.
k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
This is a differential equation of the second degree, therefore it needs two initial conditions for its complete solution
The initial amount of displacement corresponds to the amplitude of movement A = 0.08 m
PLEASE ANSWER WITH ACTUAL ANSWER AND I WILL MARK BRAINLIEST (IF YOU GIVE ME A SCAMMY ANSWER I WILL REPORT YOU!!!)
A student wants to determine the local value of the gravitational field strength, g , in their classroom. Which of the following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured?
Select TWO answers.
A: Run a lab cart down an inclined plane; measure the length of the ramp and the time it takes the cart to reach the bottom.
B: Hang a known mass from a spring scale; measure the spring scale reading when the mass is at rest.
C: Accelerate a lab cart horizontally; measure the mass of the cart and its acceleration.
D: Drop a heavy metal ball; measure the drop height and the time it takes the ball to hit the ground.
Answer:
Most likely (B)
Explanation:
B in the passage is the most representative out of all your choices and it has evidence from the passage
Hope dis helps Jit!
Sorry i forgot to type C
B and C both measure mass while the others are calculations and are bias
The following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured:
Hang a known mass from a spring scale; measure the spring scale reading when the mass is at rest.Drop a heavy metal ball; measure the drop height and the time it takes the ball to hit the ground.What is gravitational field?A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies.
When a known mass from a spring scale is hung; by e; measuring the spring scale reading when the mass is at rest, the magnitude of the gravitational field strength ( reading/mass) can be calculated.
When a heavy metal ball is dropped, by measuring e the drop height and the time it takes the ball to hit the ground, the magnitude of the gravitational field strength ( h = gt²/2) can be calculated. Hence, option (B) and option (D) is correct.
Learn more about gravitational field here:
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Pls help me mark Brainliest here the answer choices
4.0N
8.0N
12.0N
16.0N
20.0N
Answer:
20.0N
Becuase It's the largest
Answer:
20.0
Explanation:
It's the biggest number
As altitude decreases, what happens to
air pressure?
A. increases
B. decreases
C. stays the same
D. not enough information to tell
Answer:
A. Increases
Explanation:
As altitude decreases, the amount of gas molecules in the air increases - the air becomes less dense. As altitude increases, the amount of gas molecules in the air decreases—the air becomes less dense.
Answer:
It decreases so it is B
Explanation:
As altitude rises, air pressure drops.
Temperature is the measure of the average kinetic energy of the particles in an object. True False
Answer:
true
Explanation:
with increased temperature particles move faster as they gain kinetic energy
Help me please with both questions?
Answer:
question #1 is A
Question #2 is C
Explanation:
A bird travels at a speed of 14.2 m/s for 514 meters. How many seconds did it
fly?
Answer:
0.54 sec
Explanation:
Answer:
Time = 36.19 secondsExplanation:
Speed = 14.2 m/s
Distance = 514 m
Time = Distance / Speed
Time = 514 / 14.2
Time = 36.19 seconds
why is potassium and sodium considered as reactive metals?
Answer:
because they are found freely in nature uncombined so they are highly reactive with other elements
A proton is moved so that its electric potential energy increases from 4.0 × 10-14 J to 9.0 × 10-14 J. The magnitude of the charge on a proton is 1.602 × 10-19 C.
What is the electric potential difference through which the proton moved?
2.5 × 105 V
3.1 × 105 V
5.6 × 105 V
8.1 × 105 V
Answer:
B. 3.1 × 10^5 V
Explanation:
Answer:
B
Explanation:
e2021
A child of mass 51.9 kg sits on the edge of a merry-go-round with radius 2.4 m and moment of inertia 215.24 kg m2 . The merry go-round rotates with an angular velocity of 2.1 rad/s. The child then walks towards the center of the merry-go-round and stops at a distance 0.864 m from the center. Now what is the angular velocity of the merry-go-round
Answer:
4.25 rad/s
Explanation:
Given that.
Mass, m = 51.9 kg
Radius, r1 = 2.4 m
Moment of inertia, I = 215.24 kgm^2
Angular velocity, ω = 2.1 rad/s
Radius, r2 = 0.864 m
To start with, we are going to use the Conservation of angular momentum to solve the question, which is
l(initial) = l(final)
[I₁ + I₂](initial)*ω(initial) = [I₁ + I₂](final)*ω(final)
Making ω(final) the subject of formula, we have
ω(final) = [I₁ + I₂](initial)*ω(initial) / [I₁ + I₂](final)
ω(final) = [215.24 + (51.9)(2.4)²](2.1) / [215.24 + (51.9)(0.864)²]
ω(final) = [215.24 + 298.944]2.1 / [215.24 + 38.74]
ω(final) = 514.184 * 2.1 / 253.98
ω(final) = 1079.786 / 253.98
ω(final) = 4.25 rad/s
= 5.273 rad/s
Mass is the amount of matter in an
object. Which statement is true?
A. A person's mass on the moon is less than it is on
Earth.
B. Whether on the Earth or the moon, a person's mass
is the same.
C. Gravity changes the amount of matter there is in an
object.
D. Mass is the same thing as weight.
If our atmosphere had a uniform density of 1.25 kg/m3 all the way up to a border with empty space above, that border would be Answer km above sea level. The pressure at sea level is 1 atm = 105 N/m2 and g = 10 m/s2. Enter your answer as an integer.
Answer:
The border is 8km above sea level.
Explanation:
We know that:
Density = 1.25 kg/m^3
Pressure = 10^5 N/m^2
g = 10m/s^2
Now, suppose that we have a virtual rectangle, such that its bases have an area of 1m^2 and the rectangle has a height equal to H.
This virtual figure has a volume V = 1m^2*H, and it is filled with air (which we know that has a density 1.25 kg/m^3)
Then the total mass inside that volume is:
M = (1.25 kg/m^3)*V = (1.25 kg/m^3)*(1m^2*H)
The weight of this mass is:
W = g*M = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)
And if we divide the weight in a given surface, let's say 1 m^2, we get the pressure per square meter, which we know is equal to 10^5 N/m^2
then:
P = 10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)
Whit this equation we can find the value of H.
10^5 N/m^2 = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)*(1/m^2)
10^5 N = (10m/s^2)*(1.25 kg/m^3)*(1m^2*H)
(10^5 N)/(10 m/s^2) = (1.25 kg/m^3)*(1m^2*H)
(10^4 kg) = (1.25 kg/m^3)*(1m^2*H)
(10^4 kg)/( 1.25 kg/m^3) = 1m^2*H
8,000 m^3 = 1m^2*H
(8,000 m^3)/(1m^2) =H
8,000 m = H
And we want this answer in km, knowing that 1,000m = 1km
8,000m = 8km = H
The border is 8km above sea level.
Height of boundaries is 8.2 km
Given that:Normal density = 1.25 kg/m³
1 atm = 101325 N/m²
Find:Height of boundaries
Computation:Pressure = Height × Density × Gravitational acceleration
101325 = Height × 1.25 × 9.8
101325 = Height × 12.25
Height of boundaries = 101325 / 12.25
Height of boundaries = 8271.42 m
Height of boundaries = 8.2 km
Learn more:https://brainly.com/question/23358029
Can someone tell me anything useful about energy management in the human body?
Answer:
The human body carries out its main functions by consuming food and turning it into usable energy. Immediate energy is supplied to the body in the form of adenosine triphosphate (ATP). Since ATP is the primary source of energy for every body function, other stored
Explanation:
this what teacher explain to us
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel during 30 seconds of rotation?
How does the force of gravity and the force of earth contribute to africa's poverty?
Answer:
The force of gravity is not the same as being on the earth. when your on the earth there no gravitational pull its all up to the air
Explanation:
No explanation
what is the stress in a steel wire that is 5m long and 0.04cm squared in cross section If the wire bears a load of 20kg?
Answer:
Explanation:
stress = ?
length =5 m
area of cross section = 0.04 cm or 0.0004m
force = 20 × 10 = 200 N ( w = mg) g = 10
formula : stress = force / cross-sectional area
stress = 200 / 0.0004
stress = 500,000 Nm^-2
If a current of 1.10 A flows through a 7.00 Ω resistor of length 3.00 m, what is the electric field strength inside the resistor?
Answer:
the electric field strength inside the resistor is 2.57 V/m
Explanation:
Given;
current flowing through the wire, I = 1.10 A
resistance of the wire, R = 7.00 Ω
length of the wire, L = 3.00 m
The emf created inside the resistor is calculated as;
V = IR
V = 1.10 x 7
V = 7.7 V
The electric field strength inside the resistor is calculated as;
E = V/L
E = 7.7 / 3
E = 2.57 V/m
Therefore, the electric field strength inside the resistor is 2.57 V/m
In which type of circuit does charge move in only one direction?
A. A D.C CIRCUIT
B. AN A.C CIRCUIT
C. A COMBINED CIRCUIT
D. A PARALLEL CIRCUIT
Which of the following would NOT be
considered a pollutant?
A. carbon monoxide
B. sulfur dioxide
C. oxygen
D. smoke
Answer:
Answer: Oxygen
Explanation:
Oxygen would not be considered as a pollutant
Answer:
Hey there the answer is C. Oxygen
Explanation:
Oxygen is the chemical element with the symbol O and atomic number 8. It is a member of the chalcogen group in the periodic table, a highly reactive nonmetal, and an oxidizing agent that readily forms oxides with most elements as well as with other compounds. After hydrogen and helium, oxygen is the third-most abundant element in the universe by mass. At standard temperature and pressure, two atoms of the element bind to form dioxygen, a colorless and odorless diatomic gas with the formula O 2. Diatomic oxygen gas constitutes 20.95% of the Earth's atmosphere. Oxygen makes up almost half of the Earth's crust in the form of oxides. Hope this helps! Have a great day!
Help please. Question about a potential energy.
The water pressure to an apartment is increased by the water company. The water enters the apartment through an entrance valve at the front of the apartment. Where will the increase in the static water pressure be greatest when no water is flowing in the system
Answer:
Option C
Explanation:
Options for the question are as follows -
A. At a faucet close to entrance valve
B. At a faucet away from the entrance valve
C. It will be the same at all faucets
D. There will be no increase in the pressure at the faucets
Solution -
The static force will be the same at all faucets and also the area of the faucets be same.
Thus, the pressure created at all faucets will be the same.
Thus, option C is correct
Guys can you please help me real quick with this
Answer:
1. Wavelength = 3.2 m
2. Amplitude = 0.6 m
Explanation:
1. Determination of the wavelength.
The wavelength of a wave is defined as the distance between two successive crest. This implies that for every complete vibration, there is one wavelength.
From the diagram given above, we can see that the wave makes 2½ vibrations.
This means that there are 2½ equal wavelength of the wave. Therefore, the wavelength can be obtained as follow:
Length (L) = 8 m
Wavelength (λ) =?
2½ λ = L
5/2 λ = 8
5λ / 2 = 8
Cross multiply
5λ = 2 × 8
5λ = 16
Divide both side by 5
λ = 16 / 5
λ = 3.2 m
Therefore, wavelength of the wave is 3.2 m.
2. Determination of the amplitude.
The amplitude of a wave is defined as the maximum displacement of the wave from the origin.
From the diagram given above, the distance between the maximum and minimum displacement is given as 1.2 m. Thus, we can obtain the amplitude of wave as follow:
Distance between the maximum and minimum displacement (D) = 1.2
Amplitude (A) =?
A = ½D
A = ½ × 1.2
A = 0.6 m
Thus, the amplitude of the wave is 0.6 m
Each of the two grinding wheels has a diameter of 6 in., a thickness of 3/4 in., and a specific weight of 425 lb/ft3. When switched on, the machine accelerates from rest to its operating speed of 3450 rev/min in 5 sec. When switched off, it comes to rest in 35 sec. Determine the motor torque and frictional moment, assuming that each is constant. Neglect the effects of the inertia of the rotating motor armature.
Answer:
[tex]0.842\ \text{lb ft}[/tex]
[tex]0.1052\ \text{lb ft}[/tex]
Explanation:
d = Diameter of wheel = 6 in
r = Radius = 3 in = [tex]\dfrac{3}{12}=0.25\ \text{ft}[/tex]
t = Thickness = [tex]\dfrac{3}{4}=0.75\ \text{in}=\dfrac{0.75}{12}\ \text{ft}[/tex]
w = Specific weight = [tex]425\ \text{lb/ft}^3[/tex]
[tex]t_2[/tex] = Time taken to slow down = 35 s
[tex]t_1[/tex] = Time taken to reach operating speed = 5 s
[tex]\omega[/tex] = Angular velocity = [tex]3450\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]
Weight is given by
[tex]W=2\pi r^2tw\\\Rightarrow W=2\pi\times 0.25^2\times \dfrac{0.75}{12}\times 425\\\Rightarrow W=10.43\ \text{lbs}[/tex]
Mass is given by
[tex]m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{10.43}{32}\\\Rightarrow m=0.326\ \text{lb}[/tex]
Moment of inertia is given by
[tex]I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{0.326\times 0.25^2}{2}\\\Rightarrow I=0.01019\ \text{lb ft}^2[/tex]
Angular acceleration while slowing down is given by
[tex]\alpha_f=\dfrac{\omega}{t_2}\\\Rightarrow \alpha_f=\dfrac{3450\times \dfrac{2\pi}{60}}{35}\\\Rightarrow \alpha_f=10.32\ \text{rad/s}^2[/tex]
Frictional moment is given
[tex]\tau_f=I\alpha_f\\\Rightarrow \tau_f=0.01019\times 10.32\\\Rightarrow \tau_f=0.1052\ \text{lb ft}[/tex]
Frictional moment is [tex]0.1052\ \text{lb ft}[/tex]
Angular acceleration while speeding up is given by
[tex]\alpha=\dfrac{\omega}{t_1}\\\Rightarrow \alpha=\dfrac{3450\times \dfrac{2\pi}{60}}{5}\\\Rightarrow \alpha=72.26\ \text{rad/s}^2[/tex]
Motor torque is given by
[tex]\tau_m=\tau_f+I\alpha\\\Rightarrow \tau_m=0.1052+0.01019\times 72.26\\\Rightarrow \tau_m=0.842\ \text{lb ft}[/tex]
Motor torque is [tex]0.842\ \text{lb ft}[/tex].
What is the order of the events for the water cycle on a typical warm day?
А
rain, snow, sleet
B
precipitation, evaporation, rain
с
evaporation, condensation, precipitation
D
condensation, evaporation, precipitation
16. Two electric bulbs marked 100W 220V and 200W 200V have tungsten
filament of same length. Which of the two bulbs will have thicker
filament?
Answer:
The second bulb will have thicker filament
Explanation:
Given;
First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V
Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V
Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m
Resistance of the first bulb:
[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]
Resistance of the second bulb:
[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]
Resistivity of the tungsten filament is given by the following equation;
[tex]\rho = \frac{RA}{L}[/tex]
where;
L is the length of the filament
R is resistance of each filament
A is area of each filament
[tex]A = \pi r^2[/tex]
where;
r is the thickness of each filament
[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]
Therefore, the second bulb will have thicker filament
Night terrors and nightmares are
really the same event.
True
False
CiCi is hiking in the woods after a rainstorm when she sees a single large mass of rock and soil moving quickly downhill.
Which type of mass movement is this?
A. landslide
B. slump
C. creep
D. mudflow
A carnival ride starts at rest and is accelerated from an initial angle of zero to a final angle of 6.3 rad by a rad counterclockwise angular acceleration of 2.0 s2 What is the angular velocity at 6.3 rad?
The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Final angular velocity of the carnival ride
The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;
ωf = ωi + 2αθ
where;
ωf is the final angular velocity of the carnival ride = ?ωi is the initial angular velocity of the carnival ride = 0α is the angular acceleration = 2.0 rad/s²θ is the angular displacement of the carnival ride = 6.3 radωf = 0 + 2(2.0) x 6.3
ωf = 25.2 rad/s
Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Learn more about angular velocity here: https://brainly.com/question/6860269
Answer: 5.0 rad/s
Explanation: Because that’s what khan said so try it out.
I need help with this
Fairly easy question I’ll give extra points help.
1. third law
2. first law
3. third law
4. second law
types of wave interactions include