Simple harmonic motion a loaded spring undergoes simple harmonic motion whose period is proportional to the mass and inversely proportional to the spring constant.
What is simple harmonic motion?Simple harmonic motion is the periodic motion of an object along a straight line, such that the acceleration of the object is directed towards a fixed point on that line and is proportional to its distance from that point.
Example of a simple harmonic motion is the motion of a loaded spiral spring.
The restoring force on the spring is directly proportional to the extension of the spring and the spring constant.
The period of motion, T when a mass, m is attached to a spring with spring constant k,is given as:
T = 2π√(m/k)
Therefore, a loaded spring undergoes simple harmonic motion whose period is proportional to the mass and inversely proportional to the spring constant.
What is periodic motion ?Periodic motion, in physics, motion repeated in equal intervals of time. Periodic motion is performed, for example, by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave
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A car hits a tree with a force of 45 N, the mass of the tree is 65g. What is the resulting acceleration?
a. 0.69 m/s2
b. 692 m/s2
c. 2,925 m/s2
d. 2.93 m/s2
Answer:
i think 692m/s2 is the correct answer
Please help I will fail
Answer:
6. A
7. C
8. B
9. The average speed is the distance (a scalar quantity) per time ratio. Speed is ignorant of direction. The average velocity is the displacement or position change (a vector quantity) per time ratio.
A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angle of θ degrees with the horizontal. Assume the spring has been compressed a distance d from its neutral position.
Required:
a. Set your coordinates to have the x-axis along the surface of the plane, with up the plane as positive, and the y-axis normal to the plane, with out of the plane as positive.
b. Denoting the coefficient of static friction by μs, write an expression for the sum of the forces in the x-direction just before the block begins to slide up the inclined plane. Use defined quantities and g in your expression ΣFx = 25%
c. Assuming the plane is frictionless, what will the angle of the plane be, in degrees, if the spring is compressed by gravity a distance 0.1 m?
d. Assuming θ = 45 degrees and the surface is frictionless, how far will the spring be compressed, d in meters?
Answer:
b) k Δx - W cos θ - μ mg cos θ = m a , c) θ = 86.6º, d) Δx = 1.18 m
Explanation:
a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.
F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring
b) Let's apply Newton's second law for when the spring is compressed
let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Y axis
N - W_y = 0
N = W_y
N = W cos θ
X axis
F -Wₓ -fr = ma
the force applied by the spring is given by hooke's law
F = k Δx
friction force has the expression
fr = μ N
fr = μ W cos θ
we substitute
k Δx - W cos θ - μ mg cos θ = m a ( 1)
c) If the plane has no friction, what is the angle so that Δx = 0.1m
We write the equation 1, with fr = 0 and since the system is still a = 0
k Δx - W cos θ -0 = 0
cos θ = [tex]\frac{k \Delta x}{ m g}[/tex]
cos θ = [tex]\frac{880 \ 0.1}{ 150 \ 9.8}[/tex]
cos θ = 0.0598
θ = cos⁻¹ 0.0598
θ = 86.6º
d) In this part they give the angle θ = 45º and there is no friction, they ask the compression
the acceleration is zero, we substitute in 1
k Δx - W cos θ - 0 = 0
Δx = [tex]\frac{mg \ cos \ \theta}{k}[/tex]
Δx = [tex]\frac{ 150 \ 9.8 \ cos45}{880}[/tex]
Δx = 1.18 m
A hedgehog lives in a backyard in England. Every night, the house owner puts out a bowl of canned cat food and hard-boiled egg. The hungry hedgehog eats some of the food, then stops when it is no longer hungry. This pattern helps the hedgehog to maintain a steady energy level and weight.
What is the name for keeping a stable internal environment despite changes in the outside environment?
A. hormonal control
B. stimulus and response
C. balance
D. homeostasis
Answer: I think it C and B but I am really confident in C
Explanation:
Two very small +3.00-μC charges are at the ends of a meter stick. Find the electric potential (relative to infinity) at the center of the meter stick.
Answer:
The electric potential at the center of the meter stick is 54 KV.
Explanation:
Electric potential (V) is given as:
i.e V = [tex]\frac{kq}{r}[/tex]
Where: k is the Coulomb constant, q is the charge and r is the distance.
Given: q = 3.0 μC = 3.0 x [tex]10^{-6}[/tex] C, r = 0.5 m
So that,
V = [tex]\frac{9*10^{9}*3.0*10^{-6} }{0.5}[/tex]
= [tex]\frac{2.7*10^{4} }{0.5}[/tex]
V = 54000
= 54 000 volts
The electric potential at the center of the meter stick is 54 KV.
what is permittivity
Answer:
Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and electric field intensity.
Which example is correctly matched with its type of friction?
A. Pushing a car that isn't moving is an example of slkiding friction.
B. A plane flying through the air is an example of static friction.
OC.
A skateboard wheels on cement is an example of rolling friction.
OD.
A sled sliding down a grassy hill is an example of fluid friction.
Answer:
A. pushing a car that isn't moving
What is net force?
OA. a push or a pull
B. A measure of how fast an object is moving
OC. The amount of energy an object has
D. The combination of all forces acting on an object.
Answer:
D
Explanation:
The net force is the combination of all forces acting on an object.
Find the GCF of each set of numbers.
12, 21, 30
Math
Answer:
3 is the GCF for all these numbers if thats what you're asking
An ideal gas in a 50.0 L tank has a
pressure of 2.45 atm at 22.5°C.
How many moles of gas are in
the tank?
Answer:
5.05225 moles
Explanation:
The computation of the number of moles of gas in the tank is shown below:
Given that
Volume = V = 50 L = 50.0 × 10^-3m^3
Pressure = P = 2.45 atm = 2.45 × 101325
Temperature = T = 22.5°C = (22.5 + 273)k = 295.5 K
As we know thta the value of gas constant R is 8.314 J/mol.K
Now
PV = nRT
n = PV ÷ RT
= ((2.45 × 101325) (50.0 × 10^-3)) ÷ ((8.314) (295.5))
= 5.05225 moles
Need help on another homework question
Rick places the blue lens of one pair of 3D glasses over the red lens of another pair. He then looks through both lenses at the same time. What color will he see?
A. blue
B. black
C. red
D. white
Answer:
a
Explanation:
m
A 3.0 kg model train going right at 2.8 bumps into another 2.0 kg model train car moving in the same
S
m
direction at 1.6 . The heavier train car has a final speed of 2.2 to the right.
S
S
What is the final speed of the lighter 2.0 kg train car?
Answer:
it’s 2.5 m/s
Explanation:
i’m too lazy but trust
This question can be solved by using the law of conservation of momentum.
The final speed of the lighter 2 kg train is " 2.5 ".
When two moving objects collide with each other, the law of conservation of momentum can be applied to them as follows:
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
where,
m₁ = mass of heavier train = 3 kg
m₂ = mass of lighter train = 2 kg
u₁ = initial speed of heavier train = 2.8
u₂ = initial speed of lighter train = 1.6
v₁ = final speed of heavier train = 2.2
v₂ = final speed of lighter train = ?
Therefore,
(3 kg)(2.8) + (2 kg)(1.6) = (3 kg)(2.2) + (2 kg)(v₂)
[tex]v_2 = \frac{5 kg}{2 kg}[/tex]
v₂ = 2.5
Learn more about the law of conservation of momentum here:
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The attached picture illustrates the law of conservation of momentum.
Which statement describes what will most likely occur when warm air cools and the temperature drops to the
point?
A.air will contain more water vapor. B. Dew will form on leaves C.clouds will disappear. D. Water vapor in the air will evaporate
Answer:
The person who did the comment is correct, it B - Dew will form on leaves
Explanation:
Answer:
B
Explanation:
Valeriie.07 tap in g
A car travels 150 kilometers west in 3 hours. What is its average velocity?
Your answer:
150 km/hr
50 km/hr
50 km/hr west
150 km/hr west
Answer:
C= 50km/hr west
Explanation:
150/3= 50
Because it asks for velocity, make sure to include the direction as well.
When rubbing a balloon against your head, you notice the balloon pulling your hair away from your head. What best explains why the balloon and your hair are attracted to each other?
They become oppositely charged, which causes them to be attracted to each other.
They become similarly charged, which causes them to be attracted to each other.
They stick together because of the friction between the two objects.
They are made of different materials, which is why they attract each other.
Answer:
They become oppositely charged, which causes them to be attracted to each other.
Answer: number A
Explanation:
If a positive charge and a negative charge interact, their forces act in the same direction, from the positive to the negative charge. As a result opposite charges attract each other: The electric field and resulting forces produced by two electrical charges of opposite polarity. Have a nice day <3
a passenger on cruise between San Juan, Puerto Rico and Miami, Florida accidentally drops a souvenir metal cube over the side of the boat, into the water. Each side of the metal cube measures 1 meter. The cube promptly sinks to the deepest part of the Puero Rico Trench. Once at the bottom, what pressure does the cube experience? Neglect Atmospheric Pressure. Use wikipedia to see depth of Trench!
Answer:
P = 84.1 MPa
Explanation:
The pressure at the bottom of column of of salt water of height h, is given by the following expression:[tex]P = \rho * g * h (1)[/tex]
where ρ = density of salt water (in Kg/m³),
g = acceleration due to gravity (in m/s²)
h = height of the column of water.
Replacing by their values in (1):[tex]P = \rho * g * h = 1023.6kg/m3*9.8m/s2*8380m = 84.1 MPa (2)[/tex]
Neglecting the atmospheric pressure, the pressure on the cube at the bottom of Puerto Rico Trench is given by (2):P = 84.1 MPa.A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near side a distance d = 0.45 m from the origin. A charge Q2 = 10.4 μC is uniformly spread over the length of the rod.Part (a) Consider a thin slice of the rod, of thickness dx, located a distance x away from the origin. What is the direction of the force on the charge located at the origin due to the charge on this thin slice of the rod? Part (b) Write an expression for the magnitude of the force on the point charge, |dF|, due to the thin slice of the rod. Give your answer in terms of the variables Q1, Q2, L, x, dx, and the Coulomb constant, k. Part (c) Integrate the force from each slice over the length of the rod, and write an expression for the magnitude of the electric force on the charge at the origin. Part (d) Calculate the magnitude of the force |F|, in newtons, that the rod exerts on the point charge at the origin.
Answer:
a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c) F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N
Explanation:
a) q1 is negative and the charge of the bar is positive therefore the force is attractive
b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x
dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]
where k is a constant, Q₁ the charge at the origin, x the distance
c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L
∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]
as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density
λ = dQ₂ / dx
DQ₂ = λ dx
we substitute
F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]
F = k Q1 λ ([tex]-\frac{1}{x}[/tex])
we evaluate the integral
F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]
F = k Q₁ λ [tex]( \frac{L}{d \ (d+L)})[/tex]
we change the linear density by its value
λ = Q2 / L
F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]
d) we calculate the magnitude of F
F =9 10⁹ (-4.2 10⁻⁶) [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]
F = -1.09 N
the sign indicates that the force is attractive
Answer:
a)Toward the rod
b)|dF| = k|Q1|Q2(dx/L)/x^2
c)|F| = k|Q1|Q2/(d(d+L))
d)Plug in for answer c and solve
Explanation:
A)
Q1 is negative and Q2 is positive so it is an attractive force to where the rod is located.
B)
The formula for Force due to electric charges is F=kQ1Q2/r^2
In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.
The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.
The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.
The final formula is |dF| = k|Q1|Q2(dx/L)/x^2
C)
Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:
F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2
factor out constants
F = kQ1Q2/L * integral d to d+L(1/x^2)dx
F = kQ1Q2/L * (-1/x)| from d to d+L
F = kQ1Q2/L * (-1/d+L - -1/d)
F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))
F = kQ1Q2/L * (L)/(d(d+L))
F = kQ1Q2/(d(d+L))
D)
Plug in the given values into c and you have your answer.
If a total 50 J of work are done on an object, it's energy...
Answer:
0.0119502868 kilocalorie
Explanation:
Answer:
increases by 50
Explanation:
It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used as colonies. Inhabitants of the space colonies would live on the inside surface of the cylinder. Inertial effects would resemble gravity's influence and keep them 'plastered to the surface.' Suppose that you are an inhabitant of a space colony which is 1070 miles in length and 4.86 miles in diameter. How many revolutions per hour must the cylinder have in order for the occupants to experience a centripetal acceleration equal to the acceleration of gravity
Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
[tex]a_{c}[/tex] = rω²
ω² = [tex]a_{c}[/tex] / r
ω = √( [tex]a_{c}[/tex] / r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a [tex]a_{c}[/tex] = g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849
The required revolution per hour is 28.6849
Calculation of revolution per hour:
The expression for the linear acceleration with respect to the angular velocity is
= rω²
So,
ω² = / r
ω = √( / r )
Here r represent the radius of the cylinder
ω represent the angular velocity
Now
diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
And,
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
So,
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
Now
1 rad/s = 9.5493 revolution per minute
So,
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
Now
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
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A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 46.1 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding
Answer:
The minimum coefficient of friction required is 0.35.
Explanation:
The minimum coefficient of friction required to keep the crate from sliding can be found as follows:
[tex] -F_{f} + F = 0 [/tex]
[tex] -F_{f} + ma = 0 [/tex]
[tex] \mu mg = ma [/tex]
[tex] \mu = \frac{a}{g} [/tex]
Where:
μ: is the coefficient of friction
m: is the mass of the crate
g: is the gravity
a: is the acceleration of the truck
The acceleration of the truck can be found by using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} [/tex]
Where:
d: is the distance traveled = 46.1 m
[tex] v_{f}[/tex]: is the final speed of the truck = 0 (it stops)
[tex]v_{0}[/tex]: is the initial speed of the truck = 17.9 m/s
[tex] a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2} [/tex]
If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.
[tex] \mu = \frac{a}{g} [/tex]
[tex] \mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}} [/tex]
[tex] \mu = 0.35 [/tex]
Therefore, the minimum coefficient of friction required is 0.35.
I hope it helps you!
help me pls it’s a usa test prep pretty easy
Answer:
Im 99.99999% sure its c
Explanation:
i cant see the pictures too well
Velocity time graph and how to draw it
Answer:
Velocity time graph
Explanation:
Draw on graph paper two straight lines originating at the same point and perpendicular to each other. This is the x-y axis. The x-axis is the horizontal line and the y-axis is the vertical line.
Mark appropriate equally-spaced time intervals on the x-axis so that you can easily graph the time values from the table.
Mark appropriate velocity increments on the y-axis so that you can easily graph the velocity values from the table. If you have negative velocity values, extend the y-axis downward.
Find the first time value from the table and locate it on the x-axis. Look at the corresponding velocity value and find it on the y-axis.
Put a dot where a straight line vertically drawn up through the x-axis value and a straight line horizontally drawn through the y-axis value intersect.
Plot in similar fashion for all other velocity-time pairs in your table.
Draw a straight line with a pencil, connecting each dot you have put down on the graph paper, going from left to right
What is the correct definition of amplitude
Answer:
In my textbook's words-
Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path.
Explanation:
The correct definition of amplitude is that it is a maximum displacement
that occurs on a vibrating body from one point to the other.
The initial point of the wave is regarded as its equilibrium position which is
equal to one-half the length of the vibration path.
Amplitude helps to calculate the peak value of different types of waves
such as water waves and in electrical appliances so as to know the peak
current suitable for it.
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An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.
Answer:
The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
Explanation:
We can find the drift speed by using the following equation:
[tex] v = \frac{I}{nqA} [/tex]
Where:
I: is the current = 4.50 A
n: is the number of electrons
q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C
A: is the cross-sectional area = 2.20x10⁻⁶ m²
We need to find the number of electrons:
[tex] n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3} [/tex]
Now, we can find the drift speed:
[tex]v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s[/tex]
Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
I hope it helps you!
Potential energy is energy due to motion.
True or False?
Answer:
true
Explanation:
Answer:
true
Explanation:
please give brainlest need 1
125 cm of gas are collected at 15 °C and
755 mmHg pressure. Calculate the volume of
the gas at s.t.p.
The force of gravity on a person or object on the surface of a planet is called
A. gravity
ОВ.
B. free fall
OC
c. terminal velocity
D. weight
Answer:
D. Weight
Explanation:
Hope that helps:)
The new springs will be identical to the original springs, except the force constant will be 5655.00 N/m smaller. When James removes the original springs, he discovers that the length of each spring expands from 8.55 cm (its length when installed) to 12.00 cm (its length with no load placed on it). If the mass of the car body is 1355.00 kg, by how much will the body be lowered with the new springs installed, compared to its original height
Answer:
Explanation:
For original spring , compression in spring due to a load of 1355 kg is
x = 12 - 8.55 = 3.45 cm = .0345 m
spring constant = W / x
= 1355 x 9.8 / .0345
= 384898.55 N /m
Spring constant of new spring
k = 384898.55 - 5655 = 379243.55 N /m
New compression for new spring
= W / k
= 1355 x 9.8 / 379243.55
= .035 m
= 3.50 cm
Difference of compression = 3.50 - 3.45
= .05 cm .
In later case , car will be more lowered by .05 cm .
PLEASE HELP! WILL GIVE BRANILIEST TO FIRST REAL ANSWER If one marble is rolling three times as fast as a second marble of the same mass, the kinetic energy of the first marble is how many times larger when compared to the kinetic energy of the second marble?
a) 4
b) 9
c) 6
d) 3
(i already know its not 3)
Answer:
9
Explanation:
Two people each do 100 joules of work by pushing a crate to the right. During this process, 50 joules of heat is generated from the
friction between the floor and the crate. How much energy is gained by the crate during this process?