The speed of the second ball after impact is 1.43 m/s at 60.8⁰.
Speed of the second ballApply the principle of conservation of linear momentum;
in y direction0.6(2) = 0.6(1.5 sin30) + 0.6(vy)
1.2 = 0.45 + 0.6vy
0.6vy = 0.75
vy = 0.75/0.6
vy = 1.25 m/s
in x direction0.6(2) = 0.6(1.5 cos30) + 0.6(vx)
1.2 = 0.78 + 0.6vx
0.6vx = 0.42
vx = 0.42/0.6
vx = 0.7 m/s
Resultant speedv = √(vy² + vx²)
v = √(1.25² + 0.7²)
v = 1.43 m/s
Direction of second massθ = arc tan(vy/vx)
θ = arc tan (1.25/0.7)
θ = 60.8⁰
Thus, the speed of the second ball after impact is 1.43 m/s at 60.8⁰.
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mass is the amount of matter in an object what describes the amount of space the object takes up?
A.Color
B. Temperature
C. Volume
D. Texture
answer asap this is a exam
I need help with this question, could you help me? :))) ASAP
A supersonic airplane is flying horizontally at a speed of 2610 km/h.
1. What is the centripetal acceleration of the airplane, if it turns from North to East on a circular path with a radius of 80.5 km?
2. How much time does the turn take?
3. How much distance does the airplane cover during the turn?
Answer:
Centripetal acceleration of this aircraft: approximately [tex]6.52\; {\rm m \cdot s^{-2}}[/tex].
Distance covered during the turn: approximately [tex]63.2\; {\rm km}[/tex].
Time required for the turn: approximately [tex]0.0242\; \text{hours}[/tex] (approximately [tex]87.2\; {\rm s}[/tex].)
Explanation:
Convert velocity and radius to standard units:
[tex]\begin{aligned}v &= 2610\; {\rm km \cdot h^{-1}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &= 725\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
[tex]\begin{aligned} r = 80.5\; {\rm km} \times \frac{1000\; {\rm m}}{1\; {\rm km}} = 8.05 \times 10^{4}\; {\rm m}\end{aligned}[/tex].
Hence, the centripetal acceleration of this aircraft:
[tex]\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(725\; {\rm m\cdot s^{-1}})^{2} }{8.05 \times 10^{4}\; {\rm m}} \\ &\approx 6.53\; {\rm m \cdot s^{-2}}\end{aligned}[/tex].
The trajectory of the turn is an arc with a radius of [tex]r = 80.5\; {\rm km}[/tex] and a central angle of [tex]\theta = 90^{\circ} = (\pi / 4)[/tex]. The length of this arc would be:
[tex]\begin{aligned} s &= r\, \theta \\ &= 80.5\; {\rm km} \times (\pi / 4) \\ &\approx 63.2\; {\rm km}\end{aligned}[/tex].
The time required to travel [tex]63.2\; {\rm km}[/tex] at a speed of [tex]2610\; {\rm km \cdot h^{-1}}[/tex] would be:
[tex]\begin{aligned}t &= \frac{s}{v} \\ &\approx \frac{63.2\; {\rm km}}{2610\; {\rm km \cdot h^{-1}}} \\ &\approx 0.0242\; {\rm h} \\ &\approx 0.0242 \; {\rm h} \times \frac{3600\; {\rm s}}{1\; {\rm h}} \\ &\approx 87.2\; {\rm s} \end{aligned}[/tex].
Find the components of vtot along the x and y axes in Figure 3.25, where = 23.0° and vtot = 7.56 m/s.
vtot, x =
m/s
vtot, y =
m/s
The component of the total velocity in the x - direction is 6.96 m/s.
The component of the total velocity in the y - direction is 2.95 m/s.
Component of the velocity in x direction
The component of the total velocity in the x - direction is calculated as follows;
v(x) = vtot cosθ
where;
vtot is total velocityv(x) is velocity in x directionv(x) = 7.56 x cos(23)
v(x) = 6.96 m/s
Component of the velocity in y - directionv(y) = vtot sinθ
v(y) = 7.56 x sin(23)
v(y) = 2.95 m/s
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A 5.0 kg hammer strikes a 0.25 kg nail with a force of 10.0N, causing the nail to accelerate at 40.0m/s2. What is the acceleration of the hammer?
According to Newton's second law
F=maa=F/ma=10/5.25a=1.9m/s²A gas in a cylinder expands from a volume of 0.110 m³ to 0.320 m³. heat flows into the gas just rapidly enough to keep the pressure constant at 1.65×10⁵ during the expansion. the total heat added is 1.15×10⁵ j. find the change in internal energy of the gas.
80000 Joule is the change in the internal energy of the gas.
In Thermodynamics, work done by the gas during expansion at constant pressure:ΔW = -pdV
ΔW = -pd (V₂ -V₁)
ΔW = - 1.65×10⁵ pa (0.320m³ - 0.110m³)
= - 0.35×10⁵ pa.m³
= - 35000 (N/m³)(m³)
= -35000 Nm
ΔW = -35000 Joule
Therefore, work done by the system = -35000 Joule
Change in the internal energy of the gas,ΔV = ΔQ + ΔW
Given:
ΔQ = 1.15×10⁵ Joule
ΔW = -35000 Joule
ΔU = 1.15×10⁵ Joule - 35000 Joule
= 80000 Joule.
Therefore, the change in the internal energy of the gas= 80000 Joule.
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Question:- A planet X moves in an elliptical path as shown below. At which point does the planet moves faster. Assume, Ethiopian institute of space science and technology, allows to launch a space station with you. While you are in orbit, you gets thirsty. The only thing in the space station were a cup of water. you go to the cup of water and pick it up,But unfortunately your hand is twisted and the cup is turned down.Because the cup is turned down, water pour on to the floor.How would you get water?
We can get water from the sweat and exhaled breath of the people present in the space station.
How would you get water?We can get water from the sweat and exhaled breath. The water we drink is recycled from the sweat and exhaled breath of the people present in the space station which was collected through condensation on the Space Station's walls.
So we can conclude we can get water from the sweat and exhaled breath of the people present in the space station.
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A force F parallel to X-axis is starching a very light horizontal spring. Variation of force F vs. X, elongation of the spring is shown in the figure. (a)What is the spring force constant? (b) Calculate the work done by the force F as the spring elongates from x=4.0 cm to x=8.0 m.
The spring constant is obtained from the ratio of applied force to the elongation of the spring. The work done by the force is 2F.
Spring constant
The spring constant is obtained from the ratio of applied force to the elongation of the spring.
k = F/x (N/m)
Work done by the forceThe work done by the force is calculated as follows;
W = ¹/₂FΔx
where;
F is applied forceΔx is elongation of the spring = 8 m - 4 m = 4 mW = ¹/₂F(4)
W = 2F
Thus, the spring constant is obtained from the ratio of applied force to the elongation of the spring. The work done by the force is 2F.
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Calculate the work done by pulling an object through a distance of 30metres by a force of 220 newton
Answer:
6600J
Explanation:
the formula for work is: W= F x d
F- force, d- distance
Many years ago, a scientist discovered an animal P and studied its characteristics. Some of the characteristics of animal P are as follows:
A Lays egg
B has four legs
C Can swim in the water
D Can produce milk to feed the young.
Which of the above characteristics made it difficult to classify animal P as a mammal?
Answer:
A
Explanation:
most mammals are viviparous instead of oviparous. oviparous means laying eggs
The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in
0.60s. The gravitational field intensity is 9.8N/kg (acceleration due gravity). Calculate
the force the escaping gas exerts of the rocket.
a) 12.14 N [up]
b) 41. N [up]
c) 12.9 N [down]
d) 13 N [up]
The force the escaping gas exerts of the rocket is 10.42 N.
Force escaping gas exerts
The force the escaping gas exerts of the rocket is calculated as follows;
F = m(v - u)/t
where;
m is mass of the rocketv is the final velocity of the rocketu is the initial velocity of the rockett is time of motionF = (0.25)(40 - 15)/0.6
F = 10.42 N
Thus, the force the escaping gas exerts of the rocket is 10.42 N.
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An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.
The magnitude of the tension in the lower rope is 12.8 N. Calculate the magnitude of the tension in the upper rope.
24.93 N is incorrect.
The tension in the upper rope is determined as 50.53 N.
Tension in the upper rope
The tension in the upper rope is calculated as follows;
T(u) = T(d)+ mg
where;
T(u) is tension in upper ropeT(d) is tension in lower ropeT(u) = 12.8 N + 3.85(9.8)
T(u) = 50.53 N
Thus, the tension in the upper rope is determined as 50.53 N.
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HELP ASAP PLEASE!!!
In the circuit, the resistance of C is 24-Ohms, the resistance of A is 4-Ohms and the resistance of B is 8-Ohms. If the current through resistor A is 2A, what is the current through the resistor C?
If resistor A, B and C are arranged in series, same current will flow in all the resistors and 2A will flow is resistor C.
What is series circuit?
This is the type of circuit arrangement in which the resistors are arranged in series order.
In this type of circuit arrangement, same current flows in every component of the circuit.
Thus, if resistor A, B and C are arranged in series, same current will flow in all the resistors and 2A will flow is resistor C.
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A watermelon is blown into three pieces by a large firecracker. Two pieces of equal mass m fly away perpendicular to one another, one in the x direction another in the y direction. Both of these pieces fly away with a speed of V = 23 m/s. The third piece has three times the mass of the other two pieces.
Write an expression for the speed of the larger piece, that is in terms of only the variable V.
What is the numeric value for the speed of the larger piece, in meters per second?
At what angle does the largest piece travel with respect to the -y axis, in degrees?
(a) The expression for the speed of the larger piece is [tex]\frac{v\sqrt{2} }{3}[/tex].
(b) The numeric value for the speed of the larger piece is 10.84 m/s.
(c) The direction of the larger piece is 45⁰ with respect to y - axis.
Expression for the speed of the larger piece
Apply the principle of conservation of linear momentum;
m₁v₁ + m₂v₂ + m₃v₃ = 0
where;
m₁ is mass of first piecem₂ is mass of second piecem₃ is mass of third piece = three times the mass of other piece = 3mmv.i + mv.j + (3m)v₃ = 0
(3m)v₃ = -mv.i - mv.j
3v₃ = - v.i - v.j
[tex]v_3 = \frac{-v_i \ - \ v_j}{3} = \frac{v\ (\sqrt{(-1)^2 + (-1)^2} )}{3} = \frac{v\ \sqrt{2} }{3}[/tex]
Numeric value for the speed of the larger piece[tex]v_3 = \frac{23 \times \sqrt{2} }{3} \\\\v_3 = 10.84 \ m/s[/tex]
Direction of the larger piecetan θ = vj/vi
tan θ = 1/1
tan θ = 1
θ = arc tan(1)
θ = 45⁰
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A 10.0 Ω lightbulb is connected to a 12.0 V battery. (a) What current flows through the bulb? (b) What is the power of the bulb?
The current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.
What current flows through the bulb as well as the power of the bulb?From ohm's law; V = I × R
Where V is the voltage, I is the current and R is the resistance.
Also, Power is expressed as; P = V × I
Where V is voltage and I is current.
Given that;
Resistance R = 10.0 ohmsVoltage V = 12.0VCurrent I = ?Power P = ?First, we determine the current flow through the bulb.
V = I × R
12.0V = I × 10.0 ohms
I = 12.0 ÷ 10.0
I = 1.2A
Next, we determine the power of the bulb.
P = V × I
P = 12.0V × 1.2A
P = 14.4 Watts
Therefore, the current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.
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What is the power of a motor that transfers 550 J of energy in 11s?
Answer:
The answer is 50 watts
Explanation:
The formula of power is work done over elapsed time
P= W/t
P= 550 J / 11 sec
P= 50 watts
Question 9 of 27
While working out, a man performed 500 J of work in 8 seconds. What was
his power?
The power of man performing 500 J of work in 8 seconds is 62.5 J/s.
Power can be defined as the pace at which work is completed in a given amount of time.
Horsepower is sometimes used to describe the power of motor vehicles and other machinery.
The pace at which work is done on an item is defined as its power. Power is a temporal quantity.
Which is connected to how quickly a project is completed.
The power formula is shown below.
Power = Energy / Time
Power = E / T
Because the standard metric unit for labour is the Joule and the standard metric unit for time is the second, the standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W.
Here we have given Energy as 500 J and Time as 8 second.
Power = Energy / Time
Power = 500 / 8 Joule / sec
Power = 250 / 4 Joule / sec
Power = 125 / 2 Joule / sec
Power = 62.5 Joule / sec or 62.5 watt
Power came out to be 62.5 J/s when the man performed 500 Joule of work in 8 seconds.
So we can conclude that the power in the Energy transmitted per unit of time, and can be find out by dividing Energy by time. In our case the Power came out to be 62.5 Joule / Second.
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In Bolt’s fastest 100 meter, he accelerated from the starting block to a speed of 27.8 mi/hr in 9.58 s. What was his acceleration?
O 9.58 mi/hr/s
O 3.60 mi/hr/s
O 2.90 mi/hr/s
O 2780 mi/hr/s
Answer: 2.90mi/hr/s
Explanation:
Acceleration = Final Velocity - Initial over time
Acceleration = 27.8mi/hr over 9.58s.
Answer is 2.90mi/hr/s
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The larger the mass of a star, the higher the internal pressures. Higher internal pressures causes higher temperatures and it is temperature that determines the types of fusion that can occur deep in a stars interior. Discuss the types of fusion that can occur in a star, the temperatures at which they occur, and the mass required to produce them.
The types of fusion that occurs in stars are:
proton-proton fusion - stars with temperatures less than 15 million Kelvin and masses of 0.08 solar mass or more.carbon cycle fusion - occurs in stars with temperatures of 15 million Kelvin or more and masses of 4.0 solar mass or morehelium fusion - occurs in stars with temperatures of 100 million Kelvin or more and masses of 0.5 solar mass or moreWhat is a nuclear fusion reaction?A nuclear fusion is reaction in which nucleus of small atoms combine together to produces atoms of larger nucleus.
The types of nuclear fusion that can occur in a star include;
Proton-proton fusion - occur in stars with core temperatures less than 15 million Kelvin and masses of 0.08 solar mass or more.Carbon cycle fusion occurs in stars with core temperatures greater than 15 million Kelvin having masses of 4.0 solar mass or more.Helium fusion - occurs in stars with core temperatures greater than 100 million Kelvin and masses of 0.5 solar mass or moreIn conclusion, the types of fusion that occur in stars are proton-proton fusion, carbon cycle fusion and helium fusion.
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A big rock falls from a height towards water below, and hits water at a speed of 12 m/s. Rock was at a height of 15 m before falling and weighs 65 kg.
a) What was the average force of air resistance (e.g., friction) acting on the rock?
b) What is the force of friction underwater if the rock reaches a depth of 2.5 m before stopping and there is a buoyant force of 650 N [upward] acting on the rock once underwater.
In a resonance tube experiment, a tuning fork of frequency 256Hz gave a resonance when the water level was 30cm below the open end of the tube. If the next position of resonance was 100cm, what is the velocity of sound in air?
The velocity of sound in air is 358m/s
Calculations and ParametersGiven;
Frequency= 256 HzWater level= 30cmPosition of resonance= 100cmThe velocity of sound in air= ?v= 2f(L2-L1)
= 2.2.56.(1- 0.3)
=358 m/s
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A hockey puck of mass 150 g is sliding south along the ice and slows at a rate of 1.2 m/s². What is the net force acting on the puck?
Answer:
0.18 N or 180 N
Explanation:
Mass,m = 150g
acceleration,a = 1.2m/s2
CHANGE FROM g TO kg
1kg =1000g
so 150g =150/1000 = 0.15kg
But Force, F = mass,m × acceleration,a
F=ma
F=0.15×1.2
F= 0.18 N
A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capacitor take to loose half its charge?
B) How long does it take the capacitor to loose half of its stored energy?
(a) The time for the capacitor to loose half its charge is 2.2 ms.
(b) The time for the capacitor to loose half its energy is 1.59 ms.
Time taken to loose half of its charge
q(t) = q₀e-^(t/RC)
q(t)/q₀ = e-^(t/RC)
0.5q₀/q₀ = e-^(t/RC)
0.5 = e-^(t/RC)
1/2 = e-^(t/RC)
t/RC = ln(2)
t = RC x ln(2)
t = (12 x 10⁻⁶ x 265) x ln(2)
t = 2.2 x 10⁻³ s
t = 2.2 ms
Time taken to loose half of its stored energyU(t) = Ue-^(t/RC)
U = ¹/₂Q²/C
(Ue-^(t/RC))²/2C = Q₀²/2Ce
e^(2t/RC) = e
2t/RC = 1
t = RC/2
t = (265 x 12 x 10⁻⁶)/2
t = 1.59 x 10⁻³ s
t = 1.59 ms
Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.
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How does the increasing mass effect the force of an object in motion?
Answer:
According to second law of motion,the acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
So simply, it can be affected due to increasing force as there is close relationship between momentum.
Explanation:
The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.
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A set of four capacitors are attached to a 12V battery in the circuit shown below. All capacitances are measured in milli-Farads. Find the amount of electric charge that resides on each capacitor once it is fully charged.
The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
Total capacitance of the circuit
The total capacitance of the circuit is calculated as follows;
Capacitors in series;
1/Ct = 1/8 + 1/7.5
1/Ct = 0.25833
Ct = 3.87 mF
Capacitors is parallel;
Ct = 3.87 mF + 12 mF + 15 mF
Ct = 30.87 mF
Ct = 0.03087 F
Charge in each capacitorQ = CV
Q = 0.03087 x 12
Q = 0.37 C
Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
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FRICTION INVESTIGATION
Summarize your findings in a short report of 150 words. Include your hypothesis, observations, data, calculations, and conclusion. The answers to the questions below should be included in your conclusion.
Why didn't the box slide off the table before you put enough coins in the bag?
What was the coefficient for the smooth surface? For the rough surface?
What was the difference between the two coefficients? What caused this difference?
What factors caused a margin for human error in this investigation?
Would the coefficient vary if you tied the string to a similar object of different masses, but on the same surface? Explain.
What did you learn from this investigation? Be thoughtful in your answer.
Based on the nature of friction:
the box did not slide off the table before you put enough coins in the bag due to friction.The coefficient of the friction of the rough surface is greater than that of the smooth surfaceThe difference is due to the nature of the surfaces.Human errors could occur if the coins are not added carefullythe coefficient of friction is a constant for a given surface and independent of the mass of the object.What is friction?Friction is the opposition to the relative motion of an object over another at their surfaces of contact.
The coefficient of friction is a ratio of the frictional force and the normal reaction.
Coefficient of friction = frictional force/normal reactionThe coefficient of friction of rough surfaces is greater than that of smooth surfaces.
In the given experiment:
the box did not slide off the table before you put enough coins in the bag because of friction between the box and the tableThe coefficient of the friction of the rough surface is greater than the coefficient of friction of the smooth surfaceThe difference between the two values of coefficient of friction is due to the nature of the surfaces.Human errors in the experiment could occur when the coins are not added carefullyThe coefficient of friction of each surface is constant and independent of the massIn conclusion, the experiment demonstrates that the coefficient of friction is a constant for given surfaces and does not depend on the mass of the object.
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Set local ground level to 700 ft, and record the inches of mercury.
What is the altimeter reading in inches of mercury (from the Kollsman window)?
The main way in which an altimeter measures the altitude of an object is by calculating the location's air pressure.
What is an Altimeter?This refers to the instrument that is used to measure the altitude of an object when it is at a fixed level.
Hence, we can see that an altimeter should NOT be confused with a barometer, because although they both measure pressure, a barometer calculates the change in air pressure and elevation based on available weather.
Please note that your question is incomplete so I gave you a general overview to help you get a better understanding of the concept.
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Calculate the reversible work done in compressing 1 ft3 of mercury at a constant temp. of 0 o C from 1 atm to 3000 atm. The isothermal compressibility of mercury at 0 o C is : κ = 3.9 x 10-6 – 0.1 x 10-9 P , where P (atm) and κ (atm-1 ).
Compressibility of a substance is defined as the ratio of volumetric strain to the change in pressure. It is also equal to the reciprocal of the bulk modulus of the substance. It is denoted by K (kappa).
, for an isothermal process.
As we know that, work done for isothermal reversible expansion is given by-
A device that uses electricity and magnetism to create motion is called a
Answer:
electric motors is the answer
Discuss how oxygen is used in spacecraft's air supplies in at least 3 paragraphs.
Short-duration spacecraft typically have one backup system and carry their own supply of oxygen. A large portion of the required oxygen is produced on long-duration missions, such as the International Space Station (ISS), which has been in orbit since 1998. Different sources provide the oxygen utilized on the ISS. The water electrolyzer is the primary source of metabolic oxygen. As an alternative to the electrolyzer, oxygen candles (also known as SFOGs) can produce metabolic oxygen. Additionally, oxygen is carried up whenever a cargo ship docks and stored in two tanks on the ISS Airlock. The electrolyzer electrolyzes water to create oxygen by running an electric current through it. Since water is a poor electrical conductor by itself, a little quantity of common salt is dissolved in the water to improve its electrical conductivity. Water is split into hydrogen and oxygen throughout the process.
We must keep in mind that oxygen by itself cannot be inhaled; it must be combined in the proper ratio with nitrogen to make it breathable. Two tanks aboard the ISS are used to store nitrogen, and the cargo ships that travel by from time to time also transport nitrogen cylinders. Through the electrical grid of the station, the solar panels on the station supply the necessary electricity for the oxygen generators. The majority of the required water is transported to the station by cargo supply ships. Condensers, which draw water vapor even from the station's air, ensure that not a drop of water is wasted. Using the proper equipment, water is also recycled from the astronauts' urine.
Through a suitable vent, the hydrogen gas produced during the electrolysis process is released into space. Pressurized tanks at the airlock nodes at the space station are pumped with oxygen when the cargo vehicles arrive there. Pressurized tanks there are also pumped with nitrogen. It goes without saying that the station's atmospheric controls combine the gases in the right amounts for the atmosphere of Earth and then distribute the combination throughout the cabin. The production of oxygen in space is impossible.