The displacement, acceleration, energy and velocity of the simple harmonic motion of the mass attached to the spring are as follows;
A) x(t) = 0.05·sin(10·t + π/2)
B) The acceleration is; a(t) = -2.5 m/s²
C) The total energy is 0.0625 J
D) The velocity is ±√3/4 m/s
What is a simple harmonic motion?The restoring force of a body in simple harmonic motion is directly proportional to the displacement of the body from its mean or central position.
Mass of the block, m = 2 kg
The spring constant, k = 200 N/m
The extension of the spring = 5 cm
Time at which the spring is released, t = 0
A. The motion of the spring with the mass is a Simple Harmonic Motion
The angular velocity can be obtained using the formula;
ω = √(k/m)
Therefore;
ω = √(200/2) = 10
The angular velocity of the block on the spring is, ω = 10 rad/s
The period, T = The time to complete 2·π rad
Therefore; T = 2·π rad/(10 rad/s) = π/5 s
The amplitude, A, is the cistance of the mass from the at rest position, which is 5 cm = 0.05 m
The equation of the extension of the spring is therefore;
x(t) = 0.05·sin(10·t + c)
At t = 0, x(t) = 0.05, therefore;
sin(10 × 0 + c) = sin(c) = 1
c = π/2
The equation for the displacement as a function of time is therefore;
x(t) = 0.05·sin(10·t + π/2)B. The acceleration when x(t) = A/2 is obtained as follows;
x(t) = 0.05·sin(10·t + π/2)
A/2 = 0.05·sin(10·t + π/2)
A = 0.05
0.05/2 = 0.05·sin(10·t + π/2)
sin(10·t + π/2) = 1/2
10·t + π/2 = π/6
t = -π/30
cos(10×(-π/30) + π/2) = ±√3/2
v(t) = x'(t) = 0.05 × 10 × cos(10·t + π/2)
a(t) = v'(t) = -5·sin(10·t + π/2)
a(t) = v'(t) = -5·sin(10·t + π/2) = -5 × 1/2 = -2.5
The acceleration when X = + A/2 is -2.5 m/s²C. The energy in a pring = (1/2)·k·x²
When x = A/2, we get;
E = (1/2) × 200 × (0.05/2)² = 0.0625
The energy in the spring when x = A/2 is 0.0625 JD) The velocity when x = A/2 is; v(t) = x'(t) = 0.05 × 10 × cos(10·t + π/2)
v(t) = 0.5 × cos(10·t + π/2)
When x = A/2, sin(10·t + π/2) = 1/2, therefore;
cos(10·t + π/2) = ±√3/2
v(t) = 0.5 × ±√3/2 = ±√3/4
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Question
Stopping Distance
48.0 ft
49.0 ft
50.0 ft
51.0 ft
52.0 ft
2
The stopping distance for 10 cars traveling at 55 miles per hour is shown in the table above. What is the
mean (average) stopping distance for the 10 cars?
Frequency
1
2
3
2
a three tiered birthday cake rests on a table. From the bottom to the top, the cake tiers weigh 15 N, 8N, and 6N, respectively. What is the magnitude and direction of the normal force acting on the base of the seconf tier
The normal force on the second tier of the cake is 14N. It is the sum of the weight of the last and the second tier of the cake.
What is Normal force?The normal force is the force which the surfaces exert to prevent solid objects from passing through each other. Normal force is a contact force. If two objects or surfaces are not in contact, they cannot exert a normal force on each other.
Fₙ = mg
Fₙ = Normal force
Fₙ on the second tier is the sum of the first and second tier of the cake is:
Fₙ = 8N + 6N
Fₙ = 14N
The normal force on the second tier is 14N.
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A single movable pulley is used to lift 4000N load to the height of 50 cm. If the efficiency of the pulley is 75%, Calculate effort distance,effort applied,output work and mechanical advantage.
Since the efficiency of the pulley is 75% and the single movable pulley is used to lift 4000N load to the height of 50 cm, the effort distance, effort applied, output work and mechanical advantage are as follows:
Effort Distance = 50 cm
Effort Applied = 5333.33 N
Output work = 2000 J
Mechanical Advantage = 0.75.
When a single movable pulley is used to lift a load, the effort applied is equal to the weight of the load being lifted, and the effort distance is the distance the effort is applied.
Load (output force) = 4000 N
Height lifted = 50 cm = 0.5 m
Efficiency of the pulley = 75%
Given the above information, we can calculate the effort applied, output work, and mechanical advantage as follows:
Effort applied (input force) = Load ÷ Efficiency
= 4000 N ÷ 0.75
= 5333.33 N
Output work = Load x Height lifted
= 4000 N x 0.5 m
= 2000 J
Effort distance = Height lifted
= 50 cm
Mechanical advantage = Load ÷ Effort applied
= 4000 N ÷ 5333.33 N
= 0.75
It is worth noting that pulley efficiency is the ratio of output work to the input work. And since the efficiency of the pulley is 75% so the input work is greater than the output work, which means some of the energy is wasted in overcoming friction and other losses.
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The Astronomical Unit (AU) as defined by astronomers is
Answer:
a unit of length effectively equal to the average, or mean, distance between Earth and the Sun
Explanation:
A graph of v(t) is shown for a world-class track sprinter in a 100 -m race. (See figure below. For each answer, enter a number.)
(a) What is his average velocity (in m / s ) for the first 4 s} ?
______ m/ s
(b) What is his instantaneous velocity (in m / s ) at t=6 s ?
______ m / s
(c) What is his average acceleration (in m / s²) between 0 and 4s?
________m / s²
(d) What is his time (in s) for the race?
______ s
a) The average velocity at 4s is 12 m/s
b) The instantaneous velocity at t= 6s is 12 m/s
c) The average acceleration is; 12 - 0/ 4 - 0 n= 3 m/s^2
d) The time for the race is 10 s
What is the velocity time graph?We know that the velocity time graph is the kind of set up that has two axis and we can be able to use it to be read off the velocity of the acceleration of the object. I would want us to recall that the acceleration is the rate of change of the velocity of the object with time.
Now we have that;
The graph for the velocity of the object can be obtained by looking at the image that has been shown and then we read off the points that are on the graph. The vertical axis shows the velocity of the graph while the horizontal axis shows the time that is covered in the graph.
The acceleration would be the ratio of the change in the velocity to the change in the time of the object.
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Six vectors (a through f ) have the magnitudes and directions indicated in the figure. (Figure 1)
1.Rank the vector combinations on the basis of their magnitude.
Rank from largest to smallest. To rank items as equivalent, overlap them.
a+c
f+c
d
a+b
a+e
a+d
2.Rank the vector combinations on the basis of their angle, measured counterclockwise from the positivexaxis. Vectors parallel to the positivexaxis have an angle of0^\circ. All angle measures fall between0^\circand360^\circ.
Rank from largest to smallest. To rank items as equivalent, overlap them.
a+d
f+c
a+b
a+e
a+c
d
1. The rank of the vector combinations on the basis of their magnitude
a+b = a +d > a+c > f +c = d > a + e.
2. The rank of the vector combinations on the basis of their magnitude
a+d = a +b > a+c > f +c =d > a+e.
What is vector quantity?Any physical quantity having both magnitude and direction is called vector quantity.
According to figure:
a = 2 unit b = 2 unit c = 1 unit d = 2 unit e = 1 unit and f = √5 unit.
a + c = 2+1 unit = 3 unit
a + b = 2√2 unit
a + e = 2 -1 = 1 unit
a +d = 2√2 unit
f+ c = 2 unit
a+d = 2√2 unit
f+c = 2 unit
a+b = 2√2 unit
a+e = 1 unit
a+c = 3 unit
Hence, rank of the vector combinations on the basis of their magnitude
a+b = a +d > a+c > f +c = d > a + e.
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A 4.51 kg object is placed upon an inclined plane which has an incline angle of 23.0*. The object slides down the inclined plane with a constant speed. Find the normal force, friction force and the coefficient of sliding friction
To find the normal force, we can use the equation: normal force = weight + friction force * cos(incline angle).
How to find the normal force ?The weight of the object is (4.51 kg) * (9.8 m/s^2) = 44.398 NTo find the friction force, we can use the equation: friction force = coefficient of friction * normal force.We can assume that the friction force is equal to the force of gravity acting against the object because it is moving down the inclined plane at a constant pace. As a result, the friction force is equal to the product of the object's weight and sin (incline angle)Friction force is equal to (9.927 N)*sin(23.0)*(44.398 N)We can use the following equation to determine the coefficient of sliding friction:friction coefficient is calculated as friction force divided by normal force.coefficient of sliding friction = 9.927 N /44.398 N = 0.224Therefore, the normal force is 44.398 N, the friction force is 9.927 N, and the coefficient of sliding friction is 0.224.To know more about normal force , check out :
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A 30kg metal ball is dropped from a height of 12.5 m.
a. Find the final velocity when the ball hits the ground.
b. Find the time it takes for the ball to hit the ground.
From conservation of linear momentum, the final velocity is 15.7 m/s and the time taken is 1.6 s
What is Velocity ?Velocity can be defined as a distance travel in a specific direction per time taken. It is a vector quantity.
Given that 30kg metal ball is dropped from a height of 12.5 m.
a. Find the final velocity when the ball hits the ground.
The maximum K.E of the ball at it final velocity will be equal to its maximum P.E at height 12.5 m. That is,
K.E = P.E
1/2mv² = mgh
Where
m = 30 Kgg = 9.8 m/s²h = 12.5 mv = ?Substitute all the parameters
1/2 × 30 × v² = 30 × 9.8 × 12.5
v² = 245
v = √245
v = 15.65 m/s
b. The time it takes for the ball to hit the ground can be found through
h = ut + 1/2gt²
but u = 0
h = 1/2gt²
Substitute all the necessary parameters
12.5 = 1/2 × 9.8 × t²
12.5 = 4.9t²
t² = 12.5/4.9
t² = 2.55
t = √2.55
t = 1.6 s
Therefore, the final velocity when the ball hits the ground is 15.65 m/s and the time it takes for the ball to hit the ground is 1.6 s
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the angel between vector a A = 2.00i + 3.00i and vecto B is 45 degree , the scalar product of vectors A and B is 3.00 if the x component of vector B is positive , what is the vector B
I think the answer is b but im not really sure
A box is pushed with a 245 N of force on a horizontal surface as it moves with a
constant velocity. The mass of the box is 104 kg. The box experiences a force of friction
of 45 N.
a. Draw a free body diagram with Fg, FN, F₂ and F₁.
b. Find the weight of the box.
c. Find the coefficient of friction as the box is moving.
d. Find the net force acting the box.
Answer:
Explanation:
b) Fg = W = mg = (104 kg)(9.8 m/s²) = 1019.2 N (weight of the box)
Ff = (coeff. friction)(N)
N = normal force = Fg = 1019.2 N
c) coeff. friction = Ff/N = 45 N/1019.2 N = 0.044
d) Fnet = 245 N - 45 N = 200 N
It's hard to draw a free body diagram with the available symbols, but here's a try at it. Just add a box between the arrows!
W
↓
Ff → ← Fapplied
↑
N
Taylor has set this fitness goal: I will go to the YMCA and lift weights for 30 minutes every other day for the entire month of January. What is MOST likely TRUE about this goal?
Since Taylor has set this fitness goal: I will go to the YMCA and lift weights for 30 minutes every other day for the entire month of January. The option that MOST likely TRUE about this goal is option A:It is a short-term goal.
What is the fitness goal about?A short-term goal is a goal that can be achieved within a relatively short period of time, usually less than a year. In this case, the goal is to go to the YMCA and lift weights for 30 minutes every other day for the entire month of January.
In this case, Taylor's goal is to go to the YMCA and lift weights for 30 minutes every other day for the entire month of January. This goal is specific, as it clearly states what Taylor wants to accomplish, and it is measurable.
This goal can be achieved within the month of January and it's a specific, measurable and time-bound goal, which are characteristics of short-term goals.
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See full question below
Taylor has set this fitness goal: I will go to the YMCA and lift weights for 30 minutes every other day for the entire month of January. What is MOST likely TRUE about this goal?
It is a short-term goal.
It is a long-term goal.
It is an unrealistic goal.
It is a mental health goal.
Answer:it’s a short-term goal
Explanation:
because it’s only for the month of January she will be doing this goal.
At what separation will two charges, each of magnitude 6.0 μC, exert a force of 0.70 N on
each other?
The two charges, each of magnitude 6.0 μC, exert a force of 0.70 N at separation of 1.47 meters.
What is electric force?Electric force is the attracting or repulsive interaction between any two charged things. Similar to any force, Newton's laws of motion describe how it affects the target body and how it does so. One of the many forces that affect objects is the electric force.
We know that electric force can be defined as:
Force: F = kQq/r²
0.70 = 9.0 × 10⁹ × (6.0×10⁻⁶)²/r²
r = 1.47 meter.
Hence, the separation between them is 1.47 meters.
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Two students push on a box in the same direction, and one student pushes in the opposite direction. What is the net force on the
box if each student pushes with a force of 50 N?
Net force on the box is 50 N if 2 students push it in the same direction with 50 N of force each and one student pushes it in the other direction with 50 N of force.
Define force.An object with mass experiences a pull or push, which changes its velocity. An agent with the ability to change a body's rest or moving condition is known as an external force. It has both magnitude as well as direction.
The force formula according to Newton is what?F = ma, meaning force is equal to the mass times acceleration, is the second of Newton's three laws of motion.
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A bullet traveling at 5.0 x10^2 meters per is brought to rest by an impulse of 50 Newton*seconds. Find the mass of the bullet.
The bullet stops moving on hitting on a surface. Hence, the impulse here is equal to the momentum. Therefore, the mass of the bullet is 0.1 Kg.
What is impulse?Impulse in physics is the change in momentum. It is the product of the force and change in time.
hence, impulse = f. dt
When the bullet is travelling with a velocity of 500 m/s it has a momentum. When it brought to rest, momentum become zero. Thus, the momentum is equal to the impulse here.
Therefore, f.dt = m. v
f.dt = 50 N s
v = 500 m/s
m = 50 N s/500 m/s = 0.1 Kg
Therefore, the mass of the bullet is 0.1 Kg.
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• Calculate the magnetic field strength inside a solenoid with a radius of 2m and has 2000 loops. Furthermore, it carries a 1600 A current?
The magnetic field strength of the solenoid of radius 2 m is 2.0096 T.
What is a magnetic field?Magnetic field is the region or space around which magnetic force is felt or experienced.
To calculate the magnetic field strength inside a solenoid, we use the formula below.
Formula:
B = μni/r......................... Equation 1Where:
μ = permeability of free spacen = Number of loopsr = Radius of the solenoidB = Magnetic field strengthi = CurrentFrom the question,
Given:
μ = 4π×10⁻⁷T.m/Ai = 1600 An = 2000 loopsr = 2 mSubstitute these values into equation 1
B = (4π×10⁻⁷×1600×2000/2)B = 2.0096 THence, the magnetic field strength is 2.0096 T.
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Earth's gravitational
attraction
vanishes at
force of
force
(a) 6400 km (b) infinity
(c) 42300 km
(d) 1000 km
Answer:
b> infinity
Explanation:
i think b is the correct one
Answer:
(B) Infinity
Explanation:
Gravity can basically never become zero except hypothetically at infinity.
Ball A with a mass of 0.280kg makes an elastic head-on collision with ball B initially at rest. After collision, ball B moves off with half the original speed of ball A. Is the momentum conserved in the collision? Why?
The mass of ball B and the final velocity of ball A can complement the conservation of linear momentum. The answer is yes.
What is Momentum ?Momentum can simply be defined as the product of mass and velocity. It is a vector quantity.
Given that ball A with a mass of 0.280kg makes an elastic head-on collision with ball B initially at rest. After collision, ball B moves off with half the original speed of ball A.
In an elastic head-on collision, momentum is mostly always conserved. That is, the sum of the momentum before collision will be equal to the sum of the momentum after collision.
Mathematically, MaUa = MaVa + MbVb
Is the momentum conserved in the collision?
The answer is yes!
Why?
Because we need to consider the mass of the ball B and the final velocity of the ball A.
Therefore, In consideration of the mass of the ball B and the final velocity of the ball A, we can say that the momentum is conserved.
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if a player kicks a football from ground level with a velocity of 27/ms at an angle of 30 degrees above the horizontal then what is the initial horizantal velocity of the football?
The initial horizontal velocity of the football if a player kicks a football from ground level with a velocity of 27/ms at an angle of 30 degrees above the horizontal is 23.662 m/s.
To find the initial horizontal velocity (Vx) of the football, you can use the horizontal component of the velocity vector. The horizontal component of velocity is found by multiplying the velocity by the cosine of the angle of launch.
[tex]V_{x}[/tex] = V × cos(angle)
where V is the initial velocity, the angle is the angle of launch (in this case, 30 degrees), and cos is the cosine function.
So, in this case, the initial horizontal velocity of the football would be:
[tex]V_{x}[/tex] = 27 m/s × cos(30)
[tex]V_{x}[/tex] = 27 m/s × 0.866
[tex]V_{x}[/tex] = 23.662 m/s
So the initial horizontal velocity of the football, which moves in the x-direction, is approximately 23.662 m/s.
This means the football will move 23.662 m/s in x direction initially and if no air resistance acts upon the football, it will move at the same velocity in x direction throughout the motion.
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The position-versus-time plot of a boat positioning itself next to a dock is shown in the figure (Figure 1).
Rank the six points indicated in the plot order of increasing value of the velocity v, starting with the most negative. Rank the points in order of increasing value of the velocity. To rank items as equivalent, overlap them.
Incorrect: Try Again; 2 attempts remaining: no points deducted
The points in order of increasing value of the velocity can be ranked as:
B < C < A ≡ F < E < D.
What is velocity?The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.
According to the figure:
The points in order of increasing value of the velocity can be written as:
B < C < A ≡ F < E < D.
As the displacement becomes from positive to negative near point B, it has the lowest velocity and as the displacement becomes from negative to positive near point D, it has the highest velocity
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A Ferris wheel, rotating initially at an angular speed of 0.16 rad/s, accelerates over a 8.0-s interval at a rate of 0.040 rad/s2. What is its angular speed after this 8.0-s interval?
Answer:
0.48 rad/s
Explanation:
We can use this equation to find the final velocity
[tex]w_f=w_0+at[/tex]
We are given
[tex]w_0=0.16[/tex]
[tex]t=8.0[/tex]
[tex]a=0.04[/tex]
Inserting those numbers into our equation gives us
[tex]w_f=0.16+0.04*8[/tex]
Lets solve for [tex]w_f[/tex].
[tex]w_f=0.16+0.32[/tex]
[tex]w_f=0.48[/tex]
A horse pulls a wagon with a force of 1450 Newtons. How many meters sis the wagon travel if the horse exerted 798 watts of power over 3.5 minutes? (convert min. to seconds).
The distance traveled by the horse pulling the wagon with a force of 1450N is 115.57 metres.
How to calculate power?Power is defined as the amount of energy used or transferred in a certain amount of time. This means that power can be calculated by dividing the work done by the time taken.
P = W/t
When a force is applied to move an object, work is done on the object. Work done = force × distance
Power = (Force × distance)/time
According to this question, a horse pulls a wagon with a force of 1450 Newtons.
798 = 1450 × d/210
167,580 = 1450d
Distance = 167,580 ÷ 1450
Distance = 115.57 metres
Therefore, 115.57 metres is the distance traveled by the wagon.
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Exercise 24.28
For the capacitor network shown in (Figure 1), the potential difference across ab is 48 V.
Part A
Find the total charge stored in this network.
Express your answer with the appropriate units.
Q = ___ ____
Part B
Find the charge on the 150nF capacitor.
Express your answer with the appropriate units.
Q₁ = 7.2uC
Part C
Find the charge on the120nF capacitor.
Express your answer with the appropriate units.
Q₂ = 5.76 uC
Part D
Find the total energy stored in the network.
Express your answer with the appropriate units.
U = ____ ____
Part E
Find the energy stored in the 150nF capacitor.
Express your answer with the appropriate units.
U₁ = ______
Part F
Find the energy stored in the 120nF capacitor.
Express your answer with the appropriate units.
U₂= _____
Part G
Find the potential difference across the 150nF capacitor.
Express your answer with the appropriate units.
V₁= ____
Part H
Find the potential difference across the 150nF capacitor.
Express your answer with the appropriate units.
V₂ = ____
The evaluation of the capacitor (in series) network is as follows;
Part A
Q = 3.2 μC
Part B
Q₁ = 3.2 μC
Part C
Q₂ = 3.2 μC
Part D
U = 76.8 μJ
Part E
U₁ = 34 2/15 μJ
Part F
U₂ = 53 1/3 μJ
Part G
V₁ = 21 1/3 V
Part H
V₂ = 26 2/3 V
What is a capacitor?A capacitor consists of pairs of conductors separated by insulators. Capacitors are used to store electric charge.
The specified parameters are;
The voltage across ab = 48 V
The capacitance of the first capacitor, C₁ = 150 nF
Capacitance of the second capacitor, C₂ = 120 nF
Part A
The total charge in a capacitor network can be found as follows;
[tex]C_{eq} = \left(\dfrac{1}{150} + \dfrac{1}{120} \right)^{-1} nF = \left(\dfrac{3}{200} \right)^{-1}nF[/tex]
[tex]C_{eq} =\left(\dfrac{3}{200} \right)^{-1}nF=66\frac{2}{3} \, nF[/tex]
[tex]Q_{eq} = C_{eq}\times V_{ab}[/tex]
Therefore;
[tex]Q_{eq}[/tex] = 66 2/3 nF × 48 V = 3,200 × 10⁻⁹ C = 3.2 μC
The total charge in the circuit is 3.2 μCPart B
The charge in the 150 nF capacitor is obtained from the formula for the charge in a capacitor; Q = C × V as follows;
Q = C₁V₁ = C₂V₂
The charge in the capacitors, C₁ and C₂ are the same as the total charge of 3.2 μC
The charge, Q₁ on the 150 nF capacitor, C₁ is therefore, 3.2 nC
Q₁ = 3.2 nCPart C
The capacitors, C₁ and C₂ are in series, therefore, the charge in each capacitor is equivalent to the charge in the circuit, which is 3.2 μC.
Therefore, the charge, Q₂, in the 120 nF capacitor, C₂ is 3.2 μC
Q₂ = 3.2 μF
Part D
The total energy stored in the network can be obtained using the formula;
U = (1/2)·C·V²
Where;
U = The energy in the capacitor
C = The equivalent capacitance of the network = 66 2/3 nF
V = The voltage
Therefore;
[tex]U = \dfrac{1}{2} \times C_{eq}\times V^2[/tex]
[tex]U = \dfrac{1}{2} \times 66\frac{2}{3} \times 10^{-9}\times 48^2 = 76.8[/tex]
The total energy in the circuit, U = 76.8 μJPart E
The energy stored in the 150 nF capacitor is found as follows;
[tex]Q_{eq}[/tex] = Q₁ = C₁ × V₁
V₁ = [tex]Q_{eq}[/tex] ÷ C₁
Therefore;
V₁ = 3.2 μC ÷ 150 nF = [tex]21\frac{1}{3}[/tex] V
U₁ = 0.5×C₁×V₁²
U₁ = 0.5 × 150×10⁻⁹ × [tex]\left(21\frac{1}{3} \right)^2[/tex] = 34[tex]\frac{2}{15}[/tex] μJPart F
The energy stored in the 120 nF capacitor, U₂, can be found as follows;
V₂ = 3.2 μC ÷ 120 nF = [tex]26\frac{2}{3}[/tex] V
U₂ = 0.5 × 150 nF × [tex]\left(26\frac{2}{3} \, V\right)^2[/tex] = [tex]53\frac{1}{3}\, \mathrm{ \mu J}[/tex]
The energy in the 120 nF capacitor is; U₂ = 53 1/3 μJPart G;
The potential difference across the 150 nF, obtained in Part E, is 21 1/3 V
V₁ = 21 1/3 VPart H
The potential difference across the 120 nF, obtained in part F, is 26 2/3 V
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draw diagram of how a reflecting telescope works. show how the angles of reflection would work with mirrors placed at 2 different angles
Lenses, which are pieces of curved, clear glass, were employed in early telescopes to focus light.
What is Telescope?Curved mirrors are used by the majority of telescopes nowadays to collect light from the night sky. Light is focused by a telescope's mirror or lens' shape.
Astronomers use a telescope to observe distant things. Curved mirrors are used by the majority of telescopes, including all large telescopes, to collect and concentrate light from the night sky.
The original telescopes employed lenses, which are simply curved pieces of clear glass, to focus light. The "optics" of a telescope are the mirrors or lenses. Strong telescopes may view objects that are extremely faint and incredibly far away.
Therefore, Lenses, which are pieces of curved, clear glass, were employed in early telescopes to focus light.
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A 3.00kg mass is attached to an ideal spring with k=200N\m if the velocity of body at 0.25m Is 2.3m\s find the amplitude and maximum velocity
To solve this we must be knowing each and every concept related to velocity. Therefore, the amplitude and maximum velocity are 0.23 m and 2.75 m/s respectively.
What is velocity?V is the velocity measurement of an object's rate of motion and direction of motion. As a result, in order to calculate velocity using this definition, we must be familiar with both magnitude and direction.
For example, if an item travels west with 5 meters a second (m/s), its velocity to the west will be 5 m/s. The most frequent and simplest approach to determine velocity is using the formula shown below.
v = √(k / m) ×A
v = velocity of the mass
k= spring constant
m =mass of the object
A= amplitude of the oscillation.
substituting all the given values in the above equation, we get
2.3 m/s = √(200 N/m / 3.00 kg)×A
A = 2.3 m/s / √(200 N/m / 3.00 kg)
= 0.23 m
v =√(200 N/m / 3.00 kg) ×0.23 m
= 2.75 m/s
Therefore, the amplitude and maximum velocity are 0.23 m and 2.75 m/s respectively.
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Calculate the potential energy of a rock of mass 500 g, held at a height of 2 m above ground.
The potential energy of a rock of mass 500 g, held at a height of 2 m above ground is 9.8J.
The formula for the gravitational potential energy of an object is:
PE = mgh
where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2 on the Earth's surface), and h is the height of the object above the reference point.
So, in this case, we can plug in the given values to find the potential energy of the rock:
PE = (500 g) x (9.8 m/s^2) x (2 m)
The weight of the rock is m*g = 500 g * 9.8 m/s^2 = 4.9 N
Therefore,
PE = 4.9 N x 2 m = 9.8 J
Therefore, the potential energy of a rock of mass 500 g, held at a height of 2 m above ground is 9.8J.
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8. Jane rides a sled down a slope of angle θ at constant speed v. Determine the coefficient of kinetic friction between the sled and the slope. Neglect air resistance.
(A) μ=gsinθ (B) μ=mgcosθ (C) μ=tanθ (D) μ=gcosθ
Answer:ssssw
Explanation:
QUESTION 4
A student lifts a 400 N sandbag 2 meters off the ground. How much work, in joules, did the student perform?
Answer:
800J
Explanation:
W = Fs, Work equals force times displacement
in this case, the force is 400N and the displacement is 2 meters.
The regular SI unit for work is joules
Robert Galstyan, from Armenia, pulled two coupled railway wagons a distance of 7 m using his teeth. The total mass of the wagons was about 2.20 X 10^5 kg. Of course, his job was made easier by the fact that the wheels were free to roll. Suppose the wheels are blocked and the coefficient of static friction between the rails and the sliding wheels is 0.220. What would be the magnitude of the minimum force needed to move the wagons from rest? Assume that the track is horizontal.
The magnitude of the minimum force needed to move the wagons from rest is 474320 N
How do I determine the force needed to move the wagons?We have come to know that the force and coefficient of friction have a simple relationship as shown by the equation below:
Frictional force (N) = coefficient of friction (μ) × normal reaction (N)
F = μN
Applying the above formula, we can determine the force needed to move the wagons from rest. Details below:
Mass of wagons (m) = 2.20×10⁵ KgAcceleration due to gravity (g) = 9.8 m/s²Normal reaction (N) = mg = 2.20×10⁵ × 9.8 = 2156000 NCoefficient of static friction (μ) = 0.220Force needed (F) = ?F = μN
F = 0.220 × 2156000
F = 474320 N
Thus, the force needed is 474320 N
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What are the names and number of of atoms in a molecule of nitrous oxide,N2O?
Answer:
2 nitrogen and 1 oxygen,
Explanation:
N2=2 nitrogen
O= a single element
Synthetic pesticides and fertilizers were used extensively during the Green Revolution, but the negative health and environmental impacts were not well-known since the effects added up over time.
True
False
(True) Synthetic pesticides and fertilizers were used extensively during the Green Revolution, but the negative health and environmental impacts were not well-known since the effects added up over time.
what are Synthetic pesticidesChemicals created by humans, synthetic pesticides are meant to kill or deter pests. They are mostly utilized in agriculture, although they are also employed in other fields of endeavor and in household settings. In the 1930s, synthetic pesticides were first used in the United States of America (USA).
Chemicals known as synthetic pesticides are employed to manage and control plant pests like weeds, insects, and/or fungal infections.
Danger synthetic pesticidesHuman health problems associated with exposure to synthetic pesticides include
increased rates of cancer, birth defects and fertility problems.learn more about Synthetic pesticides here
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