An object on a horizontal, frictionless surface is attached to a spring, displaced, and then released. If it isdisplaced 0.12m from its equilibrium position and released after 0.8s its displacement is found to be 0.12m onthe opposites side, and passed the equilibrium position once during the interval. Find:
Answers
Given:
The maximum displacement from the equilibrium position is A = 0.12 m.
Half of the time period is
[tex]T_{\frac{1}{2}}=\text{ 0.8 s}[/tex]To find the amplitude, time period, and frequency.
Explanation:
Amplitude is the maximum displacement from the equilibrium position.
Thus, the amplitude is A = 0.12 m.
One time period is the time taken from maximum displacement on one side(say A) to maximum displacement on the opposite side and back to the maximum displacement on the same side(A).
Thus, the time period is
[tex]\begin{gathered} T=2T_{\frac{1}{2}} \\ =2\times0.8 \\ =1.6\text{ s} \end{gathered}[/tex]The frequency will be
[tex]\begin{gathered} f=\frac{1}{T} \\ =\frac{1}{1.6} \\ =0.625\text{ Hz} \end{gathered}[/tex]Related Questions
What is the resistance in a circuit that has a current of 2.5A and a voltage of 40v
Answers
16 ohms
Explanation
Ohm's law relates the strength of a direct current is directly proportional to the potential difference and inversely proportional to the resistance of the circuit,it is given by the expresssion
[tex]\begin{gathered} V=IR \\ if\text{ we isolate R} \\ R=\frac{V}{I} \end{gathered}[/tex]then
Step 1
a) Let
[tex]\begin{gathered} I=2.5\text{ A} \\ V=40\text{ V} \end{gathered}[/tex]b)replace in the formula
[tex]\begin{gathered} R=\frac{V}{I} \\ R=\frac{40V}{2.5A} \\ R=16\text{ ohms} \end{gathered}[/tex]therefore, the answer is 16 ohm
I hope this helps you
40) Climbing the Empire State Building A new record for running the stairs of the Empire State Building was set on February 4, 2003. The 86 flights, with a total of 1576 steps, was run in 9 minutes and 33 seconds. If the height gain of each step was 0.20 m, and the mass of the runner was 70.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower.
Answers
The power is given by:
[tex]P=\frac{W}{t}[/tex]where W is the work and t is the time.
We know that the work done is related to the gravitational potential energy by:
[tex]W=mg(y_f-y_0)[/tex]where yf is the final height of the object and y0 is the initial height.
Now, in this case we have a total of 1576 steps, each of them with a height of 0.2 meters that means that the total height gained is:
[tex](1576m)\cdot0.2=315.2\text{ m}[/tex]This in turns means that the work done is:
[tex]W=(70\operatorname{kg})(9.8\frac{m}{s^2})(315.2m)=216227.2\text{ J}[/tex]Now, the time it takes to achieve this is 9 minutes 33 seconds, this is the same as:
[tex]9(60)+33=573\text{ s}[/tex]Finally we use the power formula:
[tex]P=\frac{216227.2\text{ J}}{573\text{ s}}=377.36\text{ W}[/tex]Now we need to remember that 1 Hp is equal to 745.7 W, then we have that this is the same as:
[tex]377.36\text{ W}\cdot\frac{1\text{ Hp}}{745.7\text{ W}}=0.506[/tex]Therefore the power to make that climb is 377.36 W or 0.506 Hp
Suppose a 345-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.1 m from the ground to a branch.How much work, in joules, did the bird do on the snake? How much work, in joules, did it do to raise its own center of mass to the branch?
Answers
The work done in each case can be calculated with the change in potential energy of the body:
[tex]Work=m\cdot g\cdot h[/tex]The work done by the bird on the snake will use only the mass of the snake (in kg):
[tex]\begin{gathered} Work=0.075\cdot9.8\cdot2.1\\ \\ Work=1.5435\text{ J} \end{gathered}[/tex]The work done by the bird to raise its own center of mass will use only the bird mass (in kg):
[tex]\begin{gathered} Work=0.345\cdot9.8\cdot2.1\\ \\ Work=7.1\text{ J} \end{gathered}[/tex]what is the mass on grams of 0.56 moles of NaCl
Answers
Answer:
1 mole of Na = 23 g
1 mole of Cl = 35 g
1 mole of NaCl = 58 g
.56 * 58 g = 32.5 g
Neptune circles the Sun at a distance of 4.50 × 1012 m once every 164 years. Saturn circles the Sun at a distance of 1.43 × 1012 m. What is the orbital period of Saturn?
Answers
The orbital time period of the Saturn is 29.6 years
We are given that,
Distance from Sun to Saturn is = a = 1.43 × 10¹²
The mass of the Sun is = M =1.99 × 10³⁰kg
The Gravitational constant = G = 6.67 × 10⁻¹¹N-m²kg⁻²
To find the orbital period of Saturn we can use the equation ,
[tex]T^{2} = \frac{4\pi }{GM}a^{3}[/tex]
Where, T is the orbital time period of the of the Saturn , M is the mass of the sun , G is the gravitational constant.
Therefore, after putting the value in above equation we can get,
[tex]T^{2} = \frac{4(\(3.14)^{2} }{(6.67*10)^{-11} )N-m^{2} kg^{-2}}(1.43*10^{12}) ^{3}m[/tex]
[tex]T^{2} = \sqrt{8.688*10^{17} } s[/tex]
[tex]T = 932094415.818s[/tex]
So that , from above to convert the orbital time period of Saturn from second into year i.e. above seconds divided by seconds (1 sec = 3.154 ×10⁷ Earth years)
Thus, the orbital time period can be ,
[tex]T = 29.6 years[/tex]
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An aluminum rod and a nickel rodare both 5.00 m long at 20.0°C.The temperature of each is raisedto 70.0°C. What is the differencein length between the two rods?AluminumNickela = 23.10-6C+ B = 69.10-6 0-1a = 13.10-6C1 B = 39.10-6-1(Unit = m)Enter
Answers
The difference in length between the two rods = 0.0025m
Explanations:Linear expansivity of a material is given by the formula:
[tex]\alpha\text{ = }\frac{l_2-l_1}{l_1(\theta_2-\theta_1)}[/tex]For the Aluminium rod:
[tex]\begin{gathered} l_{A1}\text{ = 5.0m} \\ \theta_{A1}=20^0C \\ \theta_{A2}=70^0C \\ \alpha_A\text{ = }23\times10^{-6}C^{-1} \\ \alpha_A\text{ = }\frac{l_{A2}-l_{A1}}{l_{A1}(\theta_{A2}-\theta_{A1})} \\ \text{ }23\times10^{-6}\text{ = }\frac{l_{A2}-5}{5(70-20)} \\ 5\times50\times\text{ }23\times10^{-6}=\text{ }l_{A2}-5 \\ l_{A2}=\text{ (}5750\text{ }\times10^{-6})\text{ + 5} \\ l_{A2}=\text{ 0.00575+5} \\ l_{A2}=\text{ 5.00575m} \end{gathered}[/tex]For the Nickel rod:
[tex]\begin{gathered} l_{N1}\text{ = 5.0m} \\ \theta_{N1}=20^0C \\ \theta_{N2}=70^0C \\ \alpha_N=\text{ 13}\times10^{-6}C^{-1} \\ \alpha_N\text{ = }\frac{l_{N2}-l_{N1}}{l_{N1}(\theta_{N2}-\theta_{N1})} \\ \text{ 1}3\times10^{-6}\text{ = }\frac{l_{N2}-5}{5(70-20)} \\ 5\times50\times\text{ 1}3\times10^{-6}=\text{ }l_{A2}-5 \\ l_{N2}=\text{ (32}50\text{ }\times10^{-6})\text{ + 5} \\ l_{N2}=\text{ }0.00325+5 \\ l_{N2}=\text{ 5.00325m} \end{gathered}[/tex]The difference in length between the two rods will be given as:
[tex]\begin{gathered} l_{A2}-l_{N2}=\text{ 5.00575-5.00325} \\ l_{A2}-l_{N2}=0.0025m \end{gathered}[/tex]The difference in length between the two rods = 0.0025m
ine? Bir10. Two people are pulling on opposite ends of a rope so that ithas a tension of 150 newtons. If the rope is not moving, withwhat pulling force is each of the two people pulling?
Answers
ANSWER
150 N
EXPLANATION
The rope is not moving, so the net force on it is 0. The force that each person exerts on the rope is equal and opposite to the tension on the rope, so the sum of the forces acting on it is zero.
Hence, the force that each of the two people is exerting on the rope while pulling is 150 Newtons.
100 POINTS
A man pulls a crate with a rope. The crate slides along the floor in the horizontal direction (x direction). The man exerts a force of 50 N on the rope, and the rope is at an angle . Describe how the force components change as the angle increases from 0° to 90° and use your graph to explain your answer. Give a detailed explanation of the forces at . Show a sample calculation at one angle for both components.
Answers
The exerted 50 N force at an angle on the crate can be resolved into a horizontal and vertical force component in which the horizontal component, Fₓ, decreases, while the vertical force component, [tex]F_y[/tex], increases as the value of the angle formed by the rope, θ, increases from 0° to 90°.
What is a component of a force?The components of a force are the force parts acting in perpendicular directions, horizontal and vertical, that combine to give the specified force.
The direction the crate is sliding = The horizontal, x-direction
The force exerted by the man = 50 N
The angle of the rope = θ
The components of the force are therefore:
Horizontal component, Fₓ = 50 × cos(θ)
Vertical component, [tex]F_y[/tex] = 50 × sin(θ)
The value of cos(θ) and sin(θ) as the angle the rope makes with the horizontal, θ, increases from 0° to 90° are as follows:
[tex]\begin{center} \begin{tabular}{ |c|c |c |} \theta & cos(\theta) & sin(\theta) \\ 0^{\circ} & 1 & 0 \\ 30^{\circ} & \sqrt{3} /2 & 0.5 \\ 45^{\circ}&\sqrt{2}/2 &\sqrt{2}/2 \\ 60^{\circ}&0.5&\sqrt{3}/2 \\ 90^{\circ} &0&1\\ \end{tabular}[/tex]
Therefore, the horizontal component of the force exerted by the man, Fₓ has a maximum value at θ = 0, and decreases to 0, as θ increases from 0° to 90°.
The vertical component of the force exerted, [tex]F_y[/tex], has a minimum value of 0 at θ = 0°, and the value of sin(θ) and therefore [tex]F_y[/tex], increases to a maximum of (sin(90°) = 1) 50 N as increases to 90°.
Please find attached the graph showing the components of the force, Fₓ, and [tex]F_y[/tex], exerted by the man as the angle formed by the rope increases from 0° to 90°
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If the electric intensity between two parallel plates, placed 1cm apart is 104 NC-1and the direction of the field of this intensity is vertically upward. Find the force on an electron in this field and compare the force on the electron with the electron’s weight
Answers
The force of a charge in an electric field is:
[tex]\vec{F}_e=q\vec{E}[/tex]In this case we know the electric field is:
[tex]\vec{E}=104\hat{j}[/tex]and that the charge is that of the electron, then we have:
[tex]\begin{gathered} \vec{F}_e=-1.6\times10^{-19}(104\hat{j}) \\ \vec{F}_e=-1.664\times10^{-17}\text{ N} \end{gathered}[/tex]Therefore, the magnitude of the force is
[tex]1.664\times10^{-17}\text{ N}[/tex]and in points down.
The weight of the electron is:
[tex]\begin{gathered} W=1.67\times10^{-27}(9.98) \\ W=1.6366\times10^{-26} \end{gathered}[/tex]Making the quotient between the force we have:
[tex]\frac{1.664\times10^{-17}}{1.6366\times10^{-26}}=1.02\times10^9[/tex]Therefore, the electric force is approximately 1e9 times the weight.
(20%) Problem 5: Two identical springs, A and B, each with spring constant k = 54.5 N/m, support an object with a weight W = 11.6 N. Each spring makes an angle of 0 = 20.6 degrees to the vertical, as shown in the diagram. Create an expression for the tension in spring A
Answers
The tension in spring A is T = W/(2cosθ)
What is tension?Tension is the stretching force in a spring.
How to find the expression for the tension in the spring?Let
T = the tension in the each spring, W = weight andθ = angle each spring makes with the verticalResolving the tension in each spring vertically, so we can have that
for spring A, the tension is Tcosθ and for spring B, the tension is TcosθNow the vertical component of the tension in each equals the weight. So, we have that
Tcosθ + Tcosθ = W
Adding them together, we have that
2Tcosθ = W
Dividing both sides by 2cosθ, so, we can have that
T = W/2cosθ
Thus, the tension in each spring is T = W/(2cosθ)
So, the tension in spring A is T = W/(2cosθ)
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How much work is done on a medicine ball with a force of 29 newtons when you lift it 5 meters?
Answers
Given data
*The given force is F = 29 N
*The given distance is s = 5 m
The formula for the work is done on a medicine ball is given as
[tex]W=F\mathrm{}s[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} W=(29)(5) \\ =145\text{ J} \end{gathered}[/tex]Hence, the work is done on a medicine ball is W = 145 J
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Which of the following best represents R= A - B ?
Please help, it’s due soon!
Answers
The option (C) best represents the R = A- B
What is vector in mathematics?A vector in mathematics is a quantity that not only expresses magnitude but also motion or position of an object in relation to another point or object. Euclidean vector, geometric vector, and spatial vector are other names for it.
In mathematics, a vector's magnitude is defined as the length of a segment of a directed line, and the vector's direction is indicated by the angle at which the vector is inclined.
What are the components in vector quantity?A vector primarily consists of two elements, the horizontal component and the vertical component. The horizontal component's value is cosθ, and the vertical component's value is sinθ.
There are two types of vector multiplication, they are dot products and cross products.
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What is the mass of 12 m3 of methylated spirit whose relative density is 0.8 ? ( Hint the density of water = 1000 kgm-3 ) A. 9600 kg B.9400 kg C. 8600 kg
Answers
ANSWER
A. 9600 kg
EXPLANATION
Given:
• The volume of the substance, V = 12m³
,• The relative density of the substance, 0.8
,• The density of water, 1000 kg/m³
Unknown:
• The mass of the substance, m
The relative density of a substance is defined as,
[tex]\rho_{relative}=\frac{\rho_{substance}}{\rho_{water}}[/tex]And the density of a substance is,
[tex]\rho=\frac{m}{V}[/tex]Let's solve the first formula for the density of the substance,
[tex]\rho_{substance}=\rho_{relative}\cdot\rho_{water}=0.8\cdot1000\operatorname{kg}/m^3=800\operatorname{kg}/m^3[/tex]Then, solve the second formula for m,
[tex]m=\rho\cdot V=800\operatorname{kg}/m^3\cdot12m^3=9600\operatorname{kg}[/tex]Hence, the mass of this substance is 9600 kg
carts, bricks, and bands
10. Predict the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks.
a. Approximately 0.16 m/s2
b. Approximately 0.50 m/s2
c. Approximately 0.64 m/s2
d. Approximately 1.00 m/s2
Answers
D. The acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
What is acceleration?The acceleration of an object is the rate of change of velocity of the object with time.
The acceleration that would occur when four rubber bands were used to pull a cart loaded with two bricks, is determined by applying Newton's second law of motion as follows.
a = F/m
where;
a is accelerationF is the applied forcem is the massLet the mass of a brick = mass of a band = m
the mass of a cart = 2 bricks = 2m
a = (force applied by 4 rubber) / (mass of 1 cart + mass of 2 brick)
a = (4m) / (2m + 2m)
a = (4m)/(4m)
a = 1 m/s²
Thus, the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
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If a car has a momentum of 2315Ns and its mass is 382kg, how fast is it moving?
Answers
ANSWER:
6.06 m/s
STEP-BY-STEP EXPLANATION:
Given:
Momentum = 2315 N*s
Mass = 382 kg
We can calculate the speed as follows:
[tex]\begin{gathered} p=m\cdot v \\ v=\frac{p}{m} \\ \text{ we replacing} \\ v=\frac{2315}{382} \\ v=6.06\text{ m/s} \end{gathered}[/tex]It moves at a speed of 6.06 m/s
what happens to the state of a variable if it goes through a two series connected NOT gate.
Answers
A 60.0 kg skier with an initial speed of 14 m/s coasts up a 2.50 m high rise as shown in the figure.
Find her final speed right at the top, in meters per second, given that the coefficient of friction between her skis and the snow is 0.38?
Answers
The final speed of the skier at the top mountain is determined as 9.27 m/s.
What is the change in the energy of the skier?
The change in the energy of the skier due to frictional force is calculated as follows;
ΔP.E = Pi + Ef
where;
Pi is the initial potential at the topEf is the energy lost to frictionThe distance of the plane travelled is calculated as;
sin35 = 2.5/L
L = 2.5 / sin35
L = 4.36 m
ΔP.E = mghi - μmgcosθ(L)
where;
m is the masshi is the initial heightg is acceleration due to gravityμ is coefficient of frictionΔP.E = (60 x 9.8 x 2.5) - (0.38)(60)(9.8) cos(35) x (4.36)
ΔP.E = 671.98 1 J
The final speed of the skier at the top of the plane;
P.E = K.E
P.E = ¹/₂mv²
v² = 2P.E /m
v = √(2P.E /m)
v = √(2 x 671.98) / 60)
v = 4.73 m/s
Total speed = -4.73 m/s + 14 m/s = 9.27 m/s
Thus, due to frictional force opposing the upward motion of the skier, the final speed at the top will be smaller than the initial speed.
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Two equal charges q1=q2= -6uC are on the y-axis at y1=3cm and y2= -3cm. What is the magnitude and direction of the electric field on the x-axis at x=4cm. If a test charge q0=2uC is placed at x =4cm find the force the test charge experiences?
Answers
The electric field charges, q₁, and q₂, which are each -6×10⁻⁶ μC, gives:
First part:
The magnitude of the electric field at x = 4 cm is -3.456×10⁷ N/CThe direction of the electric field is towards the origin, along the x-axisSecond part:
The force experienced by the charge is 69.12 NWhat is an electric field?An electric field is the field around a particle that is electrically charged and which exerts a force on charged particles within the field.
The given information are:
The electric charges, q₁ = q₂ = -6 μC
The location of the charge q₁ = y₁ = 3 cm on the y-axis
Location of the charge q₂ = y₂ = -3 cm
First part:
The required location of the point where the electric field magnitude and direction is required is x = 4 cm
The electric field formula is: [tex]\displaystyle{E = \frac{k\cdot q}{r^2}[/tex]
Where:
k = The electrostatic constant ≈ 9 × 10⁹ N·m²/C²
The distances, r, of the charges from the required point are therefore obtained using Pythagorean theorem as follows:
r = √(3² + 4²) = 5
r = 5 cm = 0.05 m
Which gives;
[tex]\displaystyle{E = \frac{9 \times 10^9\times (-6) \times 10^{-6}}{(0.05)^2} = -2.16\times 10^{7}[/tex]
Given that the magnitude of the electric field along the y-axis cancel out, the magnitude of the electric field along the x-axis is found as follows:
[tex]E_x = 2 \times -2.16\times 10^{7}\times \dfrac{4}{5} = -3.456 \times 10^7[/tex]
The magnitude of the electric field at x = 4 is -3.456 × 10⁷ N/C
Second part: The magnitude of the test charge is q₀ = 2 × 10⁻⁶ μC
The force of an electric field, F = E × q
The force experienced by the test charge is therefore:
F = -3.456 × 10⁷ × 2 × 10⁻⁶ = -69.12
The force the test charge experiences is 69.12 N acting towards the origin from the point x = 4 cm.
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A circular loop of wire with a diameter of 13.478 cm is in the horizontal plane and carries a current of 1.607 A counterclockwise, as viewed from above. What is the magnetic field, in microTeslas, at the center of the loop?
Answers
Given:
The number of the loops, n = 1
The diameter of the loop is d = 2r = 13.478 cm
The current in the loop is I = 1.607 A
To find the magnetic field in micro Tesla
Explanation:
The magnetic field can be calculated by the formula
[tex]B\text{ =}\frac{n\mu_0I}{2r}[/tex]Here, the value of the constant is
[tex]\mu_0=\text{ 12.57}\times10^{-7}\text{ H/m}[/tex]On substituting the values, the magnetic field will be
[tex]\begin{gathered} B=\frac{1\times12.57\times10^{-7}\times1.607}{13.478\times10^{-2}} \\ =1.499\text{ }\times10^{-5}\text{ T} \\ =14.99\times10^{-6}\text{ T} \\ =14.99\text{ }\mu T \end{gathered}[/tex]The magnetic field is 14.99 micro Tesla
Timothy wants to know how far his math class is from the orange tree across the street from the school. His feet are ideal feet (meaning they are 1 foot long. 1 foot is 12 inches). Timothy finds that the orange tree is 159 feet from the door of the math classroom. He wants to know that distance in kilometers (km).a. Convert from feet to inches (1 ft =12 in)b. Convert from inches to centimeters (1 in =2.54c. Conver from centimeters to meters (1m = 100cm)d. Convert from meters to kilometers (1km=1000m)
Answers
a) 1 foot = 12 inches
159 feet = 159 x 12 = 1908
The distance in inches is 1908 inches
b) 1 inch = 2.54 cm
1908 inches = 1908 x 2.54 = 4846.32
The distance in centimeters is 4846.32 cm
c) 100 cm = 1 m
4846.32 cm = 4846.32/100 = 48.4632
The distance in meters is 48.4632 m
d) 1000m = 1 km
48.4632 m = 48.4632/1000 = 0.0484632
The distance in kilometers is 0.0484632 km
A small mailbag is released from a helicopter that is descending steadily at 2.82 m/s.(a) After 3.00 s, what is the speed of the mailbag?v = m/s(b) How far is it below the helicopter?d = m(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.82 m/s?v = m/sd = m
Answers
a)
When the package is released from the moving helicopter, the package and the helicopter has a common speed. The package is in freefall. We would calculate the speed of the helicopter after a given time t by applying the formula,
v = vo + gt
where
vo is the initial velocity of of the package and it is equal to the speed of the helicopter
v is the final velocity of the package after time t
g is th acceleration due to gravity
From the information given
vo = 2.82
t = 3
g = 9.8
Thus,
v = 2.82 + 9.8 * 3 = 2.82 + 29.4
v = 32.22 m/s
After 3.00 s, the speed of the mailbag is 32.22 m/s
b) We want to calculate the distance covered by the mailbag in 3 s. We would apply the formula which is expressed as
s = vot + 1/2 x g x t^2
where
s is the distance
vo = 2.82
g = 9.8
t = 3
Thus,
s = 2.82 x 3 + 1/2 x 9.8 x 3^2 = 8.46 + 44.1
s = 52.56 m
Since we want to calculate the distance from the helicopter, we would calculate the diatance that the helicopter also travelled downwards in 3 s by applying the formula for calculating distance which is expressed as
distance = speed x time
Thus
distance = 2.82 x 3 = 8.46 m
Difference in distance = 52.56 - 8.46 = 44.1
The package is 44.1 m from the helicopter
c) If the helicopter is moving upwards, it would be thrown out and it would attain a certain height before it starts descending. The height is calculated by the formula,
h = vo^2/2g
By substituting the values,
h = 2.82^2/2 x 9.8 = 0.406 m
When the mail bag attains this height, it will start moving downwards. At this height, the final velocity is zero. We would calculate the time taken to attain this height by applying the formula,
v = vo - gt
v = 0
Thus,
0 = 2.82 - 9.8 x t
9.8t = 2.82
t = 2.82/9.8 = 0.288
The time left for freefall within the first 3 seconds is
3 - 0.288 = 2.712 s
The height attained by the mailbag in 2.712s is calculated by the formula,
h = gt^2/2
h = 9.8 x 2.712^2/2 = 36.04 m
Distance travelled by helicopter by ascending upward in 3 s is
distance = 2.82 x 3 = 8.46
Height of mailbag from final position after 3 seconds is
36.04 - 0.406 = 35.634
Difference in distance = 35.634 + 8.46 = 44.094
The package is 44.094 m from the helicopter
For the velocity of the mailbag after 3 seconds,
v = - vo + gt
v = - 2.82 + 9.8 x 3 = - 2.82 + 29.4
v = 26.54 m/s
Writing Simple ExpressionsChoose all of the TRUE statement(s).Add 6 and 5, then multiply by 4 is the same as 4(6 + 5).4 times greater than 80 + 25 is the same as 4 x (80 + 25).Subtract 15 from 42 is the same as 15 − 42.9 times greater than 11 + 12 is the same as 9 + 11 + 12.8 times greater than 21 + 15 is the same as 8(21 + 15).
Answers
Add 6 and 5, then multiply by 4 is the same as 4(6 + 5). TRUE
4 times greater than 80 + 25 is the same as 4 x (80 + 25). TRUE
Subtract 15 from 42 is the same as 15 − 42. TRUE
9 times greater than 11 + 12 is the same as 9 + 11 + 12. FALSE
9 (11+12)
8 times greater than 21 + 15 is the same as 8(21 + 15). TRUE
Yea thanks thank you for the info thanks
Answers
To help us solve this problem let's plot the points given in the table:
From the graph we notice that this the position can be modeled by a sine function, we also notice that the period of this function is 8. We know that a sine function can be modeled by:
[tex]A\sin(B(x+C))+D[/tex]where A is the amplitude, C is the horizontal shift, D is the vertical shift and
[tex]\frac{2\pi}{B}[/tex]is the period.
From the graph we have we notice that we don't have any horizontal or vertical shift, then C=0 and D=0. We also notice that the amplitude is 15, then A=15. Finally, as we said, the period is 8, then:
[tex]\begin{gathered} 8=\frac{2\pi}{B} \\ B=\frac{2\pi}{8} \\ B=\frac{\pi}{4} \end{gathered}[/tex]Plugging these values in the sine function we have:
[tex]x(t)=15\sin(\frac{\pi}{4}t)[/tex]If we graph this function along the points on the table we get the following graph:
We notice that we don't get an exact fit but we get a close one.
Now, that we have a function that describes the position we can find the velocity by taking the derivative:
[tex]\begin{gathered} x^{\prime}(t)=\frac{d}{dt}\lbrack15\sin(\frac{\pi}{4}t)\rbrack \\ =\frac{15\pi}{4}\cos(\frac{\pi}{4}t) \end{gathered}[/tex]Therefore, the velocity is:
[tex]x^{\prime}(t)=\frac{15\pi}{4}\cos(\frac{\pi}{4}t)[/tex]Once we have the expression for the velocity we can find values for the times we need, they are shown in the table below:
From the table we have that:
[tex]x^{\prime}(0.5)=10.884199\text{ cm/s}[/tex]And that:
• The earliest time when the velocity is zero is 2 s.
,• The second time when the velocity is zero is 6 s.
,• The minimum velocity happens at 4 s.
,• The minimum velocity is -11.780972 cm/s
What letter from the picture below represents the position of the maximum kinetic energy?
Answers
Given that a pendulum has a mean position as C, and two extreme points A and E.
We have to find the position of kinetic energy.
Here, the total energy is conserved. So, the sum of potential energy and kinetic energy is constant.
Potential energy increases with the increase in height.
At the extreme positions, A and E, potential energy are maximum and potential energy is zero at point C.
Also, Kinetic energy is zero is at points A and E.
As energy is conserved, Kinetic energy is maximum at point C and potential energy is zero.
Whats the percent of 10 of 20
Answers
In order to determine the associated percent of 10 related to 20, proceed as follow:
If x is the percentage, then, you can write:
[tex]\frac{x}{100}\cdot20=10[/tex]which means that x percentage of 20 is equal to 10. By solving for x, you get:
[tex]\begin{gathered} x=\frac{10}{20}\cdot100 \\ x=50 \end{gathered}[/tex]Hence, 10 is the 50% of 20
Answer:2
Explanation:multiply 0.20 times 10 you get it
Choose all the answers that apply.Asteroids:only orbit planetsonly orbit the sunare made of frozen gas and icetake millions of years to orbit the sun sometimes come very close to Earth
Answers
ANSWER:
5th option: sometimes come very close to Earth
STEP-BY-STEP EXPLANATION:
Asteroids are small rocky objects that orbit the Sun. Although asteroids orbit the Sun like planets, they are much smaller.
Therefore, of the options the only totally true option is "sometimes come very close to Earth"
So the correct answer is 5th option: sometimes come very close to Earth
What sequence of two displacements moves from (5, 5) m to (- 5, - 5) * m while traveling a distance of exactly 20 meters? How does this distance compare to the single displacement that connects the same starting and ending point?
Answers
The two displacements that move from (5,5) to (-5,-5)
(5,5) → ( 5,-5) [10 units down]
(5,-5) → (-5,5) [ 10 units left ]
The single displacement that connects the 2 points is the hypotenuse of the formed triangle where each side is 10 m long.
Apply Pythagorean theorem:
c2 = 10^2+10^2
c^2 = 100 + 100
c^2 = 200
c =√200
c= 14.14
Compared to the simple displacement (14.14) that connects both points, it is greater.
20m > 14.14 m
The block of a mass 10.2 kg is sliding at an initial velocity of 3.40 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.153.
Answers
m = mass = 10.2 kg
vo = initial velocity = 3.40 m/s
u = coefficient of kinetic friction = 0.153
g= gravity = 9.8m/s^2
a)
Fr = force of kinetic friction = u m g
Fr = 0.153 x 10.2 x 9.8 = -15.30N
b) Block's acceleration
Newton's second law of motion:
F = m*a
a = F/m = -15.30 / 10.2 = -1.5 m/s^2
c) USe the third equation of motion:
2as = vf^2 - vo^2
Where:
Vf= final velocity= 0 m/s
s = displacement
2 * -1.5 * s = 0^2 - 3.40^2
-3s = -11.56
s= -11.56/-3
s= 3.85 m
Calculate the force between charges of 5.0x10^-8c and 1.0x10^-7c if they are .15m apart
Answers
Given data:
[tex]\begin{gathered} Q_1=5.0\times10^{-8}\text{ C} \\ Q_2=1.0\times10^{-7}\text{ C} \\ r=0.15\text{ m} \end{gathered}[/tex]The electrostatic force is given as,
[tex]F=\frac{KQ_1Q_2}{r^2}[/tex]Here, K is electrostatic force constant.
Substituting all known values,
[tex]\begin{gathered} F=\frac{9\times10^9\times5\times10^{-8}\times1\times10^{-7}^{}}{(0.15)^2} \\ =2\times10^{-3}\text{ N} \end{gathered}[/tex]Therefore, the force between the charges is 2×10^(-3) N. Hence, option (d) is the correct choice.
A mass of 0.520 kilograms is attached to a spring and lowered into a resting position, causing the spring to stretch 18.7 centimeters. The mass is then lifted up and released. a. What is the spring constant of the spring? Include units in your answer.b. What is the frequency of its oscillation? Include units in your answer.Answer must be in 3 significant digits.
Answers
Given data
*The given mass is m = 0.520 kg
*The spring stretches at a distance is x = 18.7 cm = 0.187 m
*The value of the acceleration due to gravity is g = 9.8 m/s^2
(a)
The formula for the spring constant of the spring is given as
[tex]\begin{gathered} F=kx \\ mg=kx \\ k=\frac{mg}{x} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} k=\frac{(0.520)(9.8)}{(0.187)} \\ =27.2\text{ N/m} \end{gathered}[/tex]Hence, the spring constant of the spring is k = 27.2 N/m
(b)
The formula for the frequency of its
