Answer:
Explanation:
For circular motion of stone the formula is
v = ω R where ω is angular velocity , v is linear velocity .
For first motion ,
R = length of sling = .6 m
ω = 2π n , n is no of revolution per second
ω = 2 x 3.14 x 5 = 31.4 rad /s
v = 31.4 x .6 = 18.84 m /s
For second motion ,
R = length of sling = .9 m
ω = 2π n , n is no of revolution per second
ω = 2 x 3.14 x 5 = 31.4 rad /s
v = 31.4 x .9 = 28.26 m /s
The figure below shows a motorcycle leaving the end of a ramp with a speed of 39.0 m/s and following the curved path shown. At the peak of the path, a maximum height h above the top of the ramp, the motorcycle's speed is 37.1 m/s. What is the maximum height h? Ignore friction and air resistance. (Enter your answer in m.)
_____m
The maximum height h of the curved path is 7.38 m.
Maximum height of the curved path
Apply the following kinematic equation;
v² = u² - 2gh
where;
v is the final velocity of the motorcycleu is initial velocity of the motorcycleh is the maximum height(u² - v²)/2g = h
(39² - 37.1²)/(2 x 9.8) = h
7.38 m = h
Thus, the maximum height h of the curved path is 7.38 m.
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According to the "Law of Increasing Opportunity Costs," what would be the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning?
A. Food or Drink.
B. Money or income.
C. Sleep or rest.
C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.
What is law of opportunity cost?
The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.
As the student gives up his sleep or night rest in the place of his exam preparation, we say that the opportunity cost is the sleep or rest.
Thus, the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.
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Define a dipole. Hence, write the expression for calculating the electric moment of a dipole
Answer:
Explanation:
An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is defined as p = qa
please need answers ASAP.
Answer:
overloading can be avoided if two many appliances are not connected to a single socket short circuiting is a name given to a situation in which they live and the natural voice accidentally coming contact
8. The pulley is assumed massless and frictionless and rotates freely about its axle. The blocks have masses m = 40 g and m₂ = 20 g, and block mi is pulled to the right by a horizontal force of magnitude F = 0.03 N. Find the magnitude of the acceleration of block m2 and the tension in the cord if the surface is frictionless. (2pt) m₂ a₂ T₂ T₂ T₁ m₁
The magnitude of the acceleration of block m2 is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.
Acceleration of the blocksThe acceleration of the blocks is calculated from the net force on the blocks.
∑F = ma
a = ∑F/m
a = (F) / (m₁ + m₂)
where;
F is the horizontal force appliedm₁ is mass of first block = 40 g = 0.04 kgm₂ is mass of the second block = 20 g = 0.02 kga = (0.03)/(0.04 + 0.02)
a = 0.09 m/s²
Tension due to block m₂T = m₂a
T = (0.02 x 0.09) = 1.8 x 10⁻³ N
Thus, the magnitude of the acceleration of block m2 is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.
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Compare and contrast visible light, infrared light, and ultraviolet light.
Answer:
NE BİLİM
Explanation:
PUAN İÇİN YAZIYORUM SEN BENİ GÖRMEZDEN GEL ANLDIN MI DOSTUM bu arada Suriye >>>>>
A man exerts a horizontal force of 123 N on a crate with a mass of 40.2 kg.
(A) If the crate doesn't move, what's the magnitude of the static friction force (in N)?
______N
(B) What is the minimum possible value of the coefficient of static friction between the crate and the floor? (Assume the crate remains stationary.)?
The magnitude of the static friction force = 123 N
The coefficient of static friction = 0.31
What is static friction?Static friction is the frictional force that must be overcome in other for a body to that moving over another.
Since the crate does not move, the magnitude of the static frictional force is equal to the applied force.
The magnitude of the static frictional force = 123 N
The coefficient of static friction = frictional force/normal reaction
The coefficient of static friction = 123/(40* 9.8)
The coefficient of static friction = 0.31
In conclusion, the static frictional force on the on the crate is equal to the applied force on the crate.
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A block with a mass of 33.0 kg is pushed with a horizontal force of 150 N. The block moves at a constant speed across a level, rough surface a distance of 4.95 m.
(a) What is the work done (in J) by the 150 N force?
_________J
(b) What is the coefficient of kinetic friction between the block and the surface?
________
The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.
What is the work done?The work done is given by the use of the formula;
W = F * x
Where;
F = force applied
x = distance covered
W = 150 N * 4.95 m = 742.5 J
Now;
The coefficient of kinetic friction is given by;
μ = F/mg
μ = 150/ 33 * 9.8
μ = 0.46
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200 g of water is heated and its temperature goes from 280 K to 300 K. What was the change in temperature for this process?
A. 280 K
B. 20 F
C. 20 K
D. 300 K
The change in temperature is 20 kelvin
What is Temperature?
Temperature can simply be described as how hot or how cold an object is at a particular period in time. The unit of temperature is Kelvin.
The formula for calculating change in temperature is
Final temperature - Initial temperature
Final temperature = 300 kelvin
Initial temperature= 280 kelvin
Change in temperature= 300-280
= 20 kelvin
Thus the change in temperature for this process is 20 kelvin
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A traffic light hangs from a pole as shown in (Figure 1). The uniform aluminum pole AB is 7.60 m long and has a mass of 10.0 kg . The mass of the traffic light is 23.5 kg. 1.) Determine the tension in the horizontal massless cable CD. 2) Determine the vertical component of the force exerted by the pivot A on the aluminum pole. 3.) Determine the horizontal component of the force exerted by the pivot A on the aluminum pole.
There is a 446N force in the horizontal massless cable CD. The vertical component of the pivot A's force on the aluminum pole, which is 328.3N. The force that the pivot A applies to the aluminum pole has a horizontal component of 446N.
We need to be aware of the force in order to discover the solution.
How can I determine the tension in the CD's horizontal massless cable?The free body diagram of the masses must be drawn in order to determine the tension in the horizontal cable CD.To obtain the tension on CD in the free body diagram, let's balance all the vertical and horizontal forces. [tex]TH-mg\frac{l}{2}cos\alpha -Mglcos\alpha =0\\T=\frac{glcos\alpha (\frac{m}{2}+M )}{h} \\T=446N[/tex]where, H=3.8m, l=7.6m, m=10kg, M=23.5 kg and alpha= 37 degree,
How to calculate the force the pivot A exerted on aluminum's vertical and horizontal components?The overall force acting vertically is,[tex]F_V-mg-Mg=0\\F_v=328.2N\\[/tex]
The overall force acting horizontally is,[tex]F_H=T=446N[/tex]
In light of this, we may say that the tension in the horizontal massless cable CD and the horizontal component of the force applied by the pivot A to the aluminum pole are identical and each exert 466N of force, whereas the vertical component of that force is 328.3N.
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*Look at attachment for photo of object**
An object, whose mass is 0.520 kg, is attached to a spring with a force constant of 106 N/m. The object rests upon a frictionless, horizontal surface (shown in the figure below).
The object is pulled to the right a distance A = 0.150 m from its equilibrium position (the vertical dashed line) and held motionless. The object is then released from rest.
(a) At the instant of release, what is the magnitude of the spring force (in N) acting upon the object?
__________ N
(b) At that very instant, what is the magnitude of the object's acceleration (in m/s2)?
________ m/s2
(c)
In what direction does the acceleration vector point at the instant of release?
- The direction is not defined (i.e., the acceleration is zero).
- Toward the equilibrium position (i.e., to the left in the figure).
- Away from the equilibrium position (i.e., to the right in the figure).
- You cannot tell without more information.
A. The magnitude of the spring force (in N) acting upon the object is 15.9 N
B. The magnitude of the object's acceleration (in m/s²) is 30.58 m/s²
C. The direction of the acceleration vector points toward the equilibrium position (i.e., to the left in the figure).
A. How to determine the force Extension (e) = 0.150 mSpring constant (K) = 106 N/mForce (F) = ?F = Ke
F = 106 × 0.15
F = 15.9 N
B. How to determine the accelerationMass (m) = 0.52 KgForce (F) = 15. 9 NAcceleration (a) =?F = ma
Divide both sides by m
a = F / m
a = 15.9 / 0.52
a = 30.58 m/s²
C. How to determine the direction of the acceleration vectorConsidering the diagram, we can see that the spring was pulled away from the equilibrium point.
Thus, when the spring is released, it will move toward the equilibrium point. This is also true about the acceleration.
Therefore, we can conclude that the direction of the acceleration vector is towards the equilibrium point.
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b. Calculate the total resistance of the circuit below. (4 points)
c. In the circuit diagram above, the meters are labeled 1 and 2. Write 2 - 3 sentences identifying each type of meter and how it is connected with the 12 Ω resistor. (4 points)
d. In the circuit diagram above, predict which resistors (if any) will stop working when the switch is opened. Write 2 - 3 sentences explaining your reasoning. (4 points)
Please answer in complete sentences. Will mark brainliest.
The total resistance in the circuit, R is 4 Ω .
The meters connected to the 12 Ω resistance are:
Voltmeter - connected in parallelAmmeter - connected in seriesIf the switch is opened, only the 4 Ω and 2 Ω resistances will stop working.
What is the equivalent or total resistance in the circuit?The resistances in the circuit are connected both in series and in parallel
The resistances in series are the 4 Ω and the 2 Ω resistances.
Equivalent resistance = 4 + 2 = 6 Ω
The 4 Ω and the 2 Ω resistances are then connected in parallel with the 12 Ω resistance.
Total resistance, R is calculated as follows:
1/R = 1/12 + 1/6
1/R = 3/12
R = 12/3
R = 4Ω
The meters connected to the 12 Ω resistance are:
Voltmeter - connected in parallelAmmeter - connected in seriesIf the switch is opened, only the 4 Ω and 2 Ω resistances will stop working because the circuit connecting them to the cell is broken whereas the circuit to the 12 Ω resistance is continuous.
In conclusion, resistances can be connected in parallel or in series.
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What is the formula that relates current and voltage?
A. V = I/R
B. V = R/I
C. V = IR
D. V = I^2/R
"V = IR" is the equation that relates current to voltage.
Relationship between current and voltage:
Georg Simon Ohm, a German scientist, and mathematician carried out an experiment in 1827 using multiple circuits with variable wire lengths. He took measurements of both the voltage across the electrical component and the current flowing through the circuit.Ohm's law outlines the connection between voltage, current, and resistance. According to the equation,V = IR or,
I = V/R,
the amount of current (I) flowing through a circuit is inversely
related to the amount of resistance (r) and directly proportional to
the voltage (v).
Therefore the correct answer is option C i.e., "V = IR".
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By how much does the gravitational potential energy of a 55- kg pole vaulter change if her center of mass rises about 4.0 m during the jump?
Express your answer to two significant figures and include the appropriate units.
The change in the gravitational potential energy of the pole vaulter is 2,156 J.
Change in the gravitational potential energy
The change in the gravitational potential energy of the pole vaulter is calculated as follows;
ΔP.E = mg(Δh)
where;
m is massΔh is change in heightΔP.E = (55)(9.8)(4)
ΔP.E = 2,156 J
Thus, the change in the gravitational potential energy of the pole vaulter is 2,156 J.
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The four tires of an automobile are inflated to a gauge pressure of 2.02×10^5 Pa. Each tire has an area of 217 cm^2 in contact with the ground. Determine the weight of the automobile.
The weight of the automobile is 17,533.6 N.
Weight of the automobile
The weight of the automobile is calculated as follows;
P = F / A
F = (W/4)
P = (W/4) / A
P = W/4A
W = 4AP
where;
P is pressure A is areaW = (4)(217 x 10⁻⁴ m²)(2.02 x 10⁵ Pa)
W = 17,533.6 N
Thus, the weight of the automobile is 17,533.6 N.
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deriving projectile motion formulas
Answer:
Projectile motion formula or equations derived (In Tabular format)Motion Path equation:
y = (tanθ) x – (1/2) g . x2/(V0 cosθ)2
Explanation:
I need to figure out how to extend a single paragraph's worth of answers that I can make into 3 paragraphs, requesting a LOT of help here.
Question:
Discuss one feature of a spacesuit that protects astronauts outside the spacecraft.
An astronaut's spacesuit is considerably more than just a pair of garments they put on during spacewalks. An entire spacesuit is actually a single-person spacecraft. The Extravehicular Mobility Unit, or EMU, is the official name for the spacesuit used on the International Space Station and Space Shuttle. "Extravehicular" refers to something that is outside of a vehicle or spaceship. To be "mobile" implies to be able to move around while wearing an astronaut suit. The astronaut is shielded by the spacesuit from the perils of being outdoors in space.
The human body cannot survive in space's hostile atmosphere without protection. A spacesuit must shield the astronaut from the vacuum of space, the drastic temperature changes in space, and, if at all possible, it must lessen the astronaut's exposure to radiation. Therefore, the suit's primary function is to act as a pressure vessel. It must keep an air environment close to the skin constant, deliver a constant supply of clean air to the lungs, and expel stale, carbon dioxide-rich air. Every 90 minutes or so, an astronaut in low Earth orbit (LEO) experiences day and night. They can fast chill to -250 F during the night and reach 250 F while the sun is shining on them (-156 C). The body's normal temperature of 98.6 F must be maintained by the suit (37 C).
To do all this, the current NASA suit has 14 layers. The liquid ventilation and cooling garment is the first three layers. It resembles a body-hugging spandex garment that has tubes inside that carry cool water across the body to dissipate extra heat. Air below the next layer cannot escape since it is a pressure vessel made of nylon coated with urethane. The following layer resembles a tent since it is constructed of Dacron. Its goal is to exert pressure on the pressure garment so that it keeps its form and doesn't expand. Neoprene coasted nylon ripstop is the following layer. It is quite resilient to tears. Seven layers of mylar film and foil blanket are used to cover it. These restrict warmth transfer into or out of the suit by acting like a thermos. Due to its several layers, it also provides defense against very small micrometeroids, which drain energy as they pierce and break each layer. The top layer is made of orthofabric, a goretex, nomex, and kevlar mixture. It is very impervious to tearing and thermally reflecting to help regulate temperature. The suits are made by ILC Dover. Incorporating materials that minimize radiation penetration through the suit is something they are always working on.
A rectangle measuring 30.0 cm by 40.0 cm is located inside a region of a spatially uniform magnetic field of 1.65 T, with the field perpendicular to the plane of the coil (Figure 1). The coil is pulled out at a steady rate of 2.00 cm/s traveling perpendicular to the field lines. The region of the field ends abruptly as shown.
a) Find the emf induced in this coil when it is all inside the field.
b) Find the emf induced in this coil when it is partly inside the field.
c) Find the emf induced in this coil when it is all outside the field.
The solution for the questions below is mathematically given as
induced emf=0induced emf=0.0132Vinduced emf=0What is the emf induced in this coil when it is all inside the field.?(A)
Generally, the equation for the flux is mathematically given as
[tex]\Phi _{B}=B.A[/tex]
[tex]\Phi _{B}=B.A\\\\\Phi _{B}=B(40*30*10^{-4}) ,[/tex]
So, the magnetic flux through the coil is constant.
From faradays law,
[tex]\varepsilon =-\frac{\mathrm{d} \Phi _{B}}{\mathrm{d} t}[/tex]
---(1) for the induced emf. Since magnetic flux is constant, LHS. of (1) =0
induced emf=0
(B)
Let x be the length of the coil's magnetic field area.
[tex]then, \Phi _{B}=B.A=B(40*x*10^{-4}) \\\\\frac{\mathrm{d}\Phi _{B} }{\mathrm{d} t}=40\\B*10^{-4}*\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
induced emf=0.0132V
(C)
In conclusion, Therefore, there is no variation in the magnetic flux across the coil when magnetic flux=0.
induced emf=0
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If the magnetic flux through a loop increases according to the relation; where , is in milliweber (mWb) and t is in seconds. Calculate the magnitude of e.m.f induced in the loop when t = 2s
The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
emf induced in the loopThe magnitude of e.m.f induced in the loop is calculated as follows;
emf = dФ/dt
Ф = 6t² + 7t
dФ/dt = 12t + 7
at t = 2 seconds
emf = dФ/dt = 12(2) + 7 = 31 V
Thus, the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.
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do you think an Electromagnet can be used for separating plastic bags from a garbage heap? explain
Answer:
No
Explanation:
Plastic bags are not magnetic materials, only magnetic materials (such as iron) can be attracted by the magnet.
Hope this helps.
A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and 272.0 N (54.4 lbs) stand on the board so that the board is balanced.
What is the upward force exerted on the board by the support?
The upward force exerted on the board by the support is 530.8 N.
Upward force exerted on the board by the supportThe upward force exerted on the board by the support is calculated as follows;
F(up) = 52.8 N + 206.0 N + 272.0 N
F(up) = 530.8 N
Thus, the upward force exerted on the board by the support is 530.8 N.
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This technique allowed multiple peoples DNA to be compared to look for
similarities and differences in order to solve crime
DNA Examination
DNA Elimination
ODNA Sequencing
ODNA Fingerprinting
*2
DNA sequencing is the technique which allows multiple peoples DNA to be compared to look for similarities and differences in order to solve crime and is denoted as option C.
What is DNA?This is referred to as the deoxyribonucleic acid and contains the genetic components of organisms and is also located in the nucleus. On the other hand, DNA sequencing refers to the process of determining the nucleic acid sequence.
Each organisms has a unique DNA sequence which is why it is used to identify individuals through the use of bodily fluids such as saliva, blood etc.
It is used to compare the similarities and differences in order to solve crime thereby making it the most appropriate choice.
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A roller-coaster car shown in the figure below is pulled up to point 1 where it is released from rest. Take y = 39 m .
(Figure 1)
Assuming no friction, calculate the speed at point 2.
Express your answer to two significant figures and include the appropriate units.
Calculate the speed at point 3.
Express your answer to two significant figures and include the appropriate units.
Calculate the speed at point 4.
Express your answer to two significant figures and include the appropriate units.
The speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.
Speed of the roller coasterThe speed of the roller coaster at any position is calculated by applying the principle of conservation of energy.
K.E = P.E
¹/₂mv² = mgh
v = √2gh
where;
h is vertical displacementSpeed at point 2v(2) = √[(2 x 9.8)(39 - 0)]
v(2) = 28 m/s
Speed at point 3v(3) = √[(2 x 9.8)(39 - 26)]
v(3) = 16 m/s
Speed at point 4v(4) = √[(2 x 9.8)(39 - 14)]
v(4) = 22 m/s
Thus, the speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.
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Help me <3 please
Thank you :)
Answer:
11,890
Explanation:
First we need to know what is considered a significant figure.
A significant figure is a value that is not a zero at the start OR end of a value.
Which means, the 0 in the value of 90 or 0.363 are not considered a significant figure.
The 0 in the value of 3056 is considered a significant figure.
So from the table, we can deduce:
0.275 has 3 significant figures
750 has 2 significant figures
[tex]10.4 \times {10}^{5} = 1040000[/tex]
has 3 significant figures.
11,890 has 4 significant figures.
320,050 has 5 significant figures.
So from the above, we can already see the answer.
A 80-kg base runner begins his slide into second base when he is moving at a speed of 3.0 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
(a) How much mechanical energy is lost due to friction acting on the runner?
______J
(b) How far does he slide?
________m
The mechanical energy lost due to friction is 360 J.
The distance the runner slides is, s = 0.655 m
What is the mechanical energy of the runner?
The mechanical energy lost due to friction acting on the runner is equal to the change in kinetic energy.
Change in Kinetic energy = 1/2m(v -u)²
Change in Kinetic energy = 80 * (0 - 3.0)²/2
Change in Kinetic energy = 360 J
Mechanical energy lost = 360 J
Distance he slide is determined using the formula below as follows:
Acceleration of runner = coefficient of friction * acceleration due to gravity
acceleration, a = 0.7 * 9.8 = 6.86
v² = u² - 2as
s = v² + u²/2a
s = 0 + 3³/2 * 6.86
s = 0.655 m
In conclusion, the mechanical energy lost due to friction is the loss in kinetic energy.
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In Milikan’s experiment, a drop of radius of 1.64 μm and density 0.851 g/cm3 is suspended in the lower chamber when a downward-pointing electric field of 1.92 * 105 N/C is applied.
What is the weight of the drop?
Find the charge on the drop, in terms of e.
How many excess or deficit electrons does it have?
(a) The weight of the drop is 1.54 x 10⁻²⁵ N,
(b) The the charge on the drop, in terms of e is 5 x 10⁻¹²e and
(c) The excess electrons is 5 x 10⁻¹² electron.
Weight of the dropThe weight of the drop is calculated as follows;
Volume of the drop; V = ⁴/₃πr³
V = ⁴/₃π(1.64 x 10⁻⁶)³ = 1.845 x 10⁻¹⁷ m³
mass = density x volume
mass = 0.851 g/cm³ x 1.845 x 10⁻²³ cm³ = 1.572 x 10⁻²³ g = 1.57 x 10⁻²⁶ kg.
Weight = 1.57 x 10⁻²⁶ kg x 9.8 m/s² = 1.54 x 10⁻²⁵ N.
Charge on the dropq = F/E
q = (1.54 x 10⁻²⁵ N)/(1.92 x 10⁵)
q = 8.01 x 10⁻³¹ C
1.6 x 10⁻¹⁹ C = 1e
8.01 x 10⁻³¹ C = ?
= 5 x 10⁻¹²e
Excess electron on the drop1.6 x 10⁻¹⁹ C ------- 1 electron
8.01 x 10⁻³¹ C ------- ?
= 5 x 10⁻¹² electron
Thus, the weight of the drop is 1.54 x 10⁻²⁵ N, the the charge on the drop, in terms of e is 5 x 10⁻¹²e and the excess electrons is 5 x 10⁻¹² electron.
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A 0.80-kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m). When the mass is at the highest point of the circle, the speed of the mass is 9.0 m/s. What is the magnitude of the force of the string on the mass at this position?
Answer:
Approximately [tex]25\; {\rm N}[/tex] assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
This mass is in a circular motion of radius [tex]r[/tex]. Hence, when the velocity of the mass is [tex]v[/tex], the acceleration of this mass should be [tex](v^{2} / r)[/tex]. The net force on this mass should be [tex](\text{net force}) = (m\, v^{2}) / r[/tex] towards the center of the circle.
When this [tex]m = 0.80\; {\rm kg}[/tex] mass is at the top of this circle, both gravitational pull and the force of the string (tension) point downwards. Hence, the net force on this mass would be:
[tex](\text{net force}) = (\text{weight}) + (\text{tension})[/tex].
Thus:
[tex]\begin{aligned} (\text{tension}) &= (\text{net force}) -(\text{weight})\\ &= \frac{m\, v^{2}}{r} - m\, g \\ &= m\, \left(\frac{v^{2}}{r} - g\right) \\ &= 0.80\; {\rm kg}\times \left(\frac{(9.0\; {\rm m\cdot s^{-1}})^{2}}{2.0\; {\rm m}} - 9.81\; {\rm m\cdot s^{-2}}\right) \\ &\approx 25\; {\rm kg \cdot m\cdot s^{-2}} \\ &= 25\; {\rm N}\end{aligned}[/tex].
A bowling ball of mass m = 1.1 kg is resting on a spring compressed by a distance d = 0.35 m when the spring is released. At the moment the spring reaches its equilibrium point, the ball is launched from the spring into the air in projectile motion at an angle of θ = 39° measured from the horizontal. It is observed that the ball reaches a maximum height of h = 4.4 m, measured from the initial position of the ball. Let the gravitational potential energy be zero at the initial height of the bowling ball.
a) what is the spring constant, k, in newtons per meter?
(I got that the speed of the ball after the launch is 14.76)
The spring constant, k, in newtons per meter is 1,955.9 N/m.
Speed of the ball after the launchh = v²sin²θ/2g
v = √[(2gh)/sin²θ]
v = √[(2 x 9.8 x 4.4)/ (sin 39)²]
v = 14.76 m/s
Energy of the ball at topE = K.E + P.E
E = ¹/₂m(v cosθ)² + mgh
E = ¹/₂(1.1)(14.76 cos39)² + (1.1 x 9.8 x 4.4)
E = 119.8 J
Spring constantE = ¹/₂kx²
k = 2E/x²
k = (2 x 119.8)/(0.35²)
k = 1,955.9 N/m
Thus, the spring constant, k, in newtons per meter is 1,955.9 N/m.
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Aster walks first 70 m in the direction 37° north of east, and then walks 82 m in the
direction 20° south of east, and finally walks 28 m in the direction 30° west of north.(2pt)
a) How far and at what angle is the Aster's final position from her initial position?
b) In what direction would she has to head to return to her initial position?
The Aster's final position from her initial position is 64 m
The angle is 300° and She has to head in West north direction to return to her initial position
What is Displacement ?The displacement is the distance travelled in a specific direction. Displacement is a vector quantity.
Given that Aster walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.
This can be solved by using bearing method. Cosine formula will be the best to solve for the distance D.
[tex]D^{2}[/tex] = [tex]70^{2}[/tex] + [tex]82^{2}[/tex] - 70 x 82 x Cos (37 + 20)
[tex]D^{2}[/tex] = 4900 + 6724 - 5740Cos57
[tex]D^{2}[/tex] = 11624 - 3126.23
[tex]D^{2}[/tex] = 8497.8
D = [tex]\sqrt{8497.8}[/tex]
D = 92.2 m
a) The Aster's final position from her initial position is 92.2 - 28 = 64 m
The angle = 270° + 30° = 300°
b) She has to head in West north direction to return to her initial position
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For a projectile launched horizontally at 650 m/s and travels for 5.75 seconds, What is the range?
The range of horizontal projectile motion is 3737.5 m.
A projectile's horizontal velocity is constant (a never changing value), Gravity causes a downward vertical acceleration with a magnitude of 9.8 m/s/s. A projectile's vertical velocity fluctuates by 9.8 m/s per second, A projectile's vertical motion is unrelated to its horizontal motion. we know that sinθ is maximum at 90°. Therefore horizontal range will maximum at 45°.
To calculate the range of horizontal projectile motion we use;
Δx = vₓ t
where , Δx = Range
vₓ = Velocity
t = Time
Initial horizontal velocity, vₓ = 650 m/s
Time = 5.75 sec
Δx = 650 × 5.75
= 3737.5 m
Therefore, the range of horizontal projectile motion is 3737.5 m.
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