The hypothesis could be stated as follows: "Since there is no direct sunlight at night, the temperature of the air blown by the bedroom air conditioner is colder at night than it is during the day."
You may devise an experiment to gauge the temperature of the air being blasted by the air conditioner both at night and during the day in order to verify this theory. This is a general description of the experiment.
Place a thermometer or temperature sensor close to the air conditioner's outlet, which is where the room's cool air is blown. Make sure the sensor is in the same spot for the measurements during the day and at night.
The exact relationship between the air temperature blasted by the air conditioner and the presence or absence of direct sunshine is what the experiment attempts to investigate. According to the theory, cooler air may be blasted by the air conditioner during the daytime when there is sunlight but warmer air during the night when there is no direct sunlight.
Hence, the hypothesis could be stated as follows: "Since there is no direct sunlight at night, the temperature of the air blown by the bedroom air conditioner is colder at night than it is during the day."
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An athlete crosses a 25-m-wide river by swimming perpendicular to the water current at a speed of 0.5 m/s relative to the water. He reaches the opposite side at a distance 40 m downstream from his starting point.
1) How fast is the water in the river flowing with respect to the ground?
2) What is the speed of the swimmer with respect to a friend at rest on the ground?
Answer:
1) Vx=0.8 m/s
2) V=0.94339 m/s
Explanation:
We know that the speed to cross the river is 0.5 m/s. This is our y axis. Vy
And we don't know the speed of the river. This is in our x axis. Vx
Since we know that the shore is 25m away, and we have a 0.5m/s of speed.
We can find with y=v.t
Since we know y=25m and v=0.5m/s
We can find that 50 seconds is the time we take to cross the river.
Now we need to calculate the velocity of the river, for that we use the same equation, x=v.t where x is 40m at downstream, and t now we know is 50 seconds.
We can find that v of the river, or Vx is 0.8 m/s.
With the two components of the velocity we use this equation to calculate the module or the velocity of the swimmer respect a fix point on the ground.
[tex]V=\sqrt{(Vx)^{2}+(Vy)^{2} }[/tex]
We replace the values of Vx and Vy and we find. V=0.94339 m/s.
The river current is flowing at a speed of 0.8 m/s concerning the ground, and the swimmer's speed concerning a friend on the ground is approximately 0.94 m/s.
Explanation:An athlete swims across a 25-m-wide river with a speed of 0.5 m/s perpendicular to the water current. The swimmer is carried 40 m downstream, indicating the presence of a river current. To find the speed of the river current, we use the Pythagorean theorem since the swimmer's motion relative to the river and the river's motion relative to the ground are perpendicular to each other.
The time taken to cross the river is the width of the river divided by the swimmer's speed concerning the water:
Time = 25 m / 0.5 m/s = 50 s. Now, the speed of the river current can be calculated using the downstream distance covered (40 m) and time (50 s): Speed of river = Distance downstream / Time = 40 m / 50 s = 0.8 m/s.
To calculate the swimmer's speed relative to the ground, we consider both the swimmer's speed across the river and the river's current speed. Using the Pythagorean theorem:
Swimmer's speed relative to the ground = √((0.5 m/s)^2 + (0.8 m/s)^2) = √(0.25 + 0.64) m/s = √(0.89) m/s ≈ 0.94 m/s.
A cartridge electrical heater is shaped as a cylinder of length L = 200 mm and outer diameter D = 20 mm. Under normal operating conditions the heater dissipates 2 kW while submerged in a water flow that is at 20°C and provides a convection heat transfer coefficient of h = 5000 W/m2 · K. Neglecting heat transfer from the ends of the heater, determine its surface temperature Ts. If the water flow is inadvertently terminated while the heater continues to operate, the ____
Answer:
Ts=51.83C
Explanation:
First we calculate the surface area of the cylinder, neglecting the top and bottom covers as indicated by the question
Cilinder Area= A=πDL
L=200mm=0.2m
D=20mm=0.02m
A=π(0.02m)(0.2m)=0.012566m^2
we use the equation for heat transfer by convection
q=ha(Ts-T)
q= heat=2Kw=2000W
A=Area=0.012566m^2
Ts=surface temperature
T=water temperature=20C
Solving for ts
Ts=q/(ha)+T
Ts=2000/(5000*0.012566m^2)+20=51.83C
Final answer:
The surface temperature Ts of the cylindrical cartridge electrical heater is approximately 51.85°C under normal operating conditions. If the water flow is stopped, the temperature will rise sharply and could lead to overheating and damage.
Explanation:
To determine the surface temperature Ts of the cylindrical cartridge electrical heater, we apply the concept of steady-state heat transfer. The heater dissipates power P of 2 kW and is submerged in water with a convection heat transfer coefficient h of 5000 W/m2·K. The formula for heat dissipation per unit area due to convection is:
q = h(Ts - Tinf)
where:
q is the heat flux, the rate of heat transfer per unit area (W/m2),
h is the heat transfer coefficient,
Ts is the surface temperature of the heater,
Tinf is the temperature of the water, which is 20°C.
The total heat dissipated by the heater is equal to the heat transfer rate per unit area multiplied by the surface area of the cylinder A, which can be calculated using the formula:
A = πDL
Given that L = 200 mm = 0.2 m and D = 20 mm = 0.02 m, we can calculate the surface area. Substituting the values:
A = π (0.02 m) (0.2 m) = 0.01256 m2
Now, knowing that P = qA, we can solve for q:
q = № / A = 2000 W / 0.01256 m2 = 159235.67 W/m2
Using q, we then calculate Ts:
q = h(Ts - Tinf) → 159235.67 W/m2 = 5000 W/m2·K (Ts - 20°C)
Ts = (159235.67 W/m2 / 5000 W/m2·K) + 20°C
Ts = 31.85°C + 20°C
Ts = 51.85°C
If the water flow is inadvertently terminated while the heater continues to operate, the temperature of the heater surface will rise sharply. This will likely lead to overheating, which can damage the heater and pose safety risks.
The game "Whack-a-Mole", in which the player must "whack" randomly appearing moles with a hammer as quickly as possible when they peek their heads out, is best described as a _____ task.
Answer:
reaction time
Explanation:
According to my research on studies conducted by various neurologists, I can say that based on the information provided within the question this game best describes the reaction time task. This is the amount of time it takes for your brain to process the visual information and for your body to react accordingly. This is because the toy moles only pop their heads out for an extremely small amount of time in which you have to hit them with the hammer before they go back down.
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
Alice drops a rock down from a 40 meter tall building (vAlice,i = 0) at the same time that Bob (lying on the ground) throws a rock straight upwards from the ground with vBob,i = 20 m/s. At what height do the rocks pass each other?
Answer:
Rocks will pass each other at 20.4m.
Explanation:
Alice drops the rock from [tex]y_{a,i}=40m[/tex] (y is in the vertical axis) at [tex]v_{a,i}=0[/tex].
Bob throws the rock from [tex]y_{b,i}=0m[/tex] at [tex]v_{b,i}}=20\frac{m}{s}[/tex].
[tex]g[/tex] is gravity's acceleration.
Position equation for alice's rock:
[tex]y_{a}(t)=y_{a,i} +v_{a,i}t - \frac{1}{2}gt^{2} =40m-\frac{1}{2}gt^{2}[/tex]
Position equation for Bob's rock:
[tex]y_{b}(t)=y_{b,i} + v_{b,i}t - \frac{1}{2}gt^{2} =20\frac{m}{s}t -\frac{1}{2}gt^{2}[/tex]
We we'll first find the time [tex]t_{0}[/tex] at which the rocks meet. For this, we will equalize Bob's and Alice's equations:
[tex]y_{a}(t_{0})=y_{b}(t_{0})[/tex]
[tex]40m-\frac{1}{2}gt_{0}^{2}=20\frac{m}{s}t_{0} -\frac{1}{2}gt_{0}^{2}[/tex]
⇒[tex]20\frac{m}{s}t_{0}=40m[/tex] ⇒ [tex]t_{0}=2s[/tex]
So, we can take [tex]t_{0}=2[/tex] and just put it in any of the two laws of motion to see at what height the rocks meet. We will take Alice's equation (using g=9.8m/s):
[tex]y_{a}(2) =40m-\frac{1}{2}g2^{2}=20.4m[/tex]
Rocks will pass each other at 20.4m.
Radioisotopes are unstable forms of isotopes because they contain an excess number of ____________ . Radioisotopes lose nuclear energy as they decay or break down, thus becoming ____________ stable. The ____________ half-life is the amount of time it takes for 50% of the radioisotope to become stable. The amount of time it takes for 50% of radioactive material to leave the body is referred to as the ____________ half-life.
Answer:
neutrons, more, physical, biological
Explanation:
Radioisotopes are those isotopes of an atom which due to excessive energies are unstable, this unstability results from combination of protons and neutrons which are unstable.
Thus to achieve stability, these isotopes loses their energy when they decay or disintegrate.
The time taken by the radioisotope to achieve about 50% stability after disintegration is known as Physical half-life.
The biological half-life refer to the time taken by the radioisotope for the elimination of about 50% of the radioactive substance.
Radioisotopes are unstable due to an excess number of neutrons, and as they decay, they become more stable. The physical half-life is the time for 50% to decay, while the biological half-life refers to the time till 50% leaves the body.
Explanation:Radioisotopes are unstable forms of isotopes because they contain an excess number of neutrons. Radioisotopes lose nuclear energy as they decay or break down, thus becoming more stable. Radioisotopes are unstable forms of isotopes because they contain an excess number of neutrons. Radioisotopes lose nuclear energy as they decay or break down, thus becoming more stable. The physical half-life is the amount of time it takes for 50% of the radioisotope to become stable. The amount of time it takes for 50% of radioactive material to leave the body is referred to as the biological half-life.The physical half-life is the amount of time it takes for 50% of the radioisotope to become stable. The amount of time it takes for 50% of radioactive material to leave the body is referred to as the biological half-li.
A lead ball is dropped into a lake from a diving board 5.0 meters above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 seconds after it is released. How deep is the lake?
Answer:
28.8 meters
Explanation:
We must first determine at which velocity the ball hits the water. To do so we will:
1) Assume no air resistance.
2) Use the Law of conservation of mechanical energy: E=K+P
Where
E is the mechanical energy (which is constant)
K is the kinetic energy.
P is the potential energy.
With this we have [tex]\frac{m}{2} *v^{2} = m*g*h[/tex]
Where:
m is the balls's mass <- we will see that it cancels out and as such we don't need to know it.
v is the speed when it hits the water.
g is the gravitational constant (we will assume g=9.8[tex]\frac{m}{s^{2} }[/tex].
h is the height from which the ball fell.
Because when we initially drop the ball, all its energy is potential (and [tex]P = - m*g*h[/tex]) and when it hits the water, all its energy is kinetic ([tex]K=\frac{m}{2} *v^{2}[/tex]. And all that potential was converted to kinetic energy.
Now, from [tex]\frac{m}{2} *v^{2} = m*g*h[/tex] we can deduce that [tex]v=\sqrt{2*g*h}[/tex]
Therefore v=9.6[tex]\frac{m}{s}[/tex]
Now, to answer how deep is the lake we just need to multiply that speed by the time it took the ball to reach the bottom.
So D=9.6[tex]\frac{m}{s}[/tex]*3[tex]s[/tex]=28.8[tex]m[/tex]
Which is our answer.
Final answer:
The depth of the lake can be calculated by determining the time it takes for the ball to reach the bottom. Using the equation d = 1/2gt^2, where d is the distance fallen, g is the acceleration due to gravity, and t is the time taken, we can solve for t. The time taken to fall from the diving board is subtracted from the total time taken to sink to the bottom, and then the distance fallen from the diving board to the bottom of the lake is calculated using the equation d = 1/2gt^2. The depth of the lake is found to be 19.42 meters.
Explanation:
To determine the depth of the lake, we first need to calculate how long it takes for the ball to reach the bottom. Since the ball sinks with a constant velocity, the time it takes to reach the bottom is the same as the time it takes to fall from the diving board. Using the equation d = 1/2gt^2, where d is the distance fallen, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken, we can solve for t: 5 = 1/2 * 9.8 * t^2. Rearranging the equation, we have t^2 = 5 / (1/2 * 9.8), which simplifies to t^2 = 1.02. Taking the square root of both sides, we find t = 1.01 s. Since the ball reaches the bottom after 3.0 s, we subtract the time taken to fall from the diving board to get the time taken to sink to the bottom: 3.0 s - 1.01 s = 1.99 s.
Next, we calculate the distance fallen from the diving board to the bottom of the lake using the equation d = 1/2gt^2. The time taken is 1.99 s, and the acceleration due to gravity is 9.8 m/s^2. Plugging in the values, we get d = 1/2 * 9.8 * (1.99)^2. Simplifying the equation, we find d = 19.42 m. Therefore, the depth of the lake is 19.42 meters.
A sound mixer is impressed by the new equipment that was just installed in his recording studio. He says that now he will be able to help singers and musicians produce better CDs because he can eliminate unneeded and undesired wavelengths. What term describes the characteristic of sounds waves that the sound mixer is now able to alter?
a) amplitude
b) volleying
c) frequency
d) saturation
e) wavelength
Answer:
Wavelnght
Explanation:
He says undesired wavelengths can be eliminated with the new sound mixer. Musical sounds are spread over a medium through waves of different characteristics, one of them is wavelenght, that is the distance between begining and the end of an oscilation, normally measured on small distance units such as milimeters or even less. Sound may be higher or deeper depending on its wavelenght, so this sound mixer is able to filter all the sound the user does not want to include in his work, for example, noises from enviroment. Now the producer can make music focused on high notes, like the ones from violins, or focused on low ones like the ones from a bass.
A cubical box with edges of length k centimetres is to be enlarged so that the dimensions of the larger box are k + 2 centimetres, k + 3 centimetres, and k centimetres. The volume of the larger box is how many cubic centimetres greater than the volume of the original box?
Answer:[tex]5k^2+6k[/tex]
Explanation:
Given
initial side of cube is k cm
New dimensions are
k+2 cm
k+3 cm
k cm
[tex]V_{initial}=k^3[/tex]
[tex]V_{Final}=\left ( k+3\right )\left ( k+2\right )\left ( k\right )[/tex]
Now [tex]V_{final}-V{initial}=\left ( k+3\right )\left ( k+2\right )\left ( k\right )-k^3[/tex]
[tex]\Delta V=k^3+5k^2+6k-k^3=5k^2+6k[/tex]
A motorcycle traveling 95.0 km/hr approaches a car traveling in the same direction at 87.0 km/hr. When the motorcycle is 54.0 m behind the car, the rider accelerates and passes the car 17.0 s later. What is the acceleration of the motorcycle (in meters/second^2)?
Answer:
[tex]a = 0.1137 m/s^2[/tex]
Explanation:
Let Vc be the velocity of the car and Vm the velocity of the motorcycle. If we convert their given values, we get:
Vc = 87 km/h * 1000m / 1km * 1h / 3600s = 24.17m/s
Vm = 95 km/h * 1000m / 1km * 1h / 3600s = 26.38m/s
Since their positions are equal after 17s we can stablish that:
[tex]Xc = d + Vc*t = Xm = Vm*t + \frac{a*t^2}{2}[/tex]
Where d is the initial separation distance of 54m. Solving for a, we get:
[tex]a = \frac{d+Vc*t-Vm*t}{t^2}*2[/tex] Repacing the values:
[tex]a = 0.1137 m/s^2[/tex]
Two identical grasshoppers jump into the air with the same initial speed and experience no air resistance. Grasshopper A goes straight up, but grasshopper B goes up at a 66° angle above the horizontal. Which of the following statements about these grasshoppers are correct? (There could be more than one correct choice.
Choices below:
a.) At their highest point, both of them have the same amount of mechanical energy.
b.) At their highest point, grasshopper B is moving faster than grasshopper A.
c.) At their highest point, both of them have the same amount of kinetic energy.
d.) At their highest point, both of them have the same amount of gravitational potential energy.
e.)At their highest point, grasshopper A has more gravitational potential energy than grasshopper B.
At their highest point, grasshoppers A and B have the same mechanical and gravitational potential energy. Grasshopper B has a non-zero total velocity due to its horizontal motion at its highest point, while grasshopper A does not. They both reach the same height, so they have the same gravitational potential energy.
Explanation:To answer your question about the two identical grasshoppers that jump into the air with the same initial speed, experiencing no air resistance: At their highest point, grasshopper A going straight up and grasshopper B going up at a 66° angle, we need to understand some basic concepts of physics first.
At the highest point of their jump, the vertical component of their velocity becomes zero. There is no kinetic energy associated with vertical motion for both of them.
Therefore, the answer to your multiple-choice question is:
a.) At their highest point, both of them have the same amount of mechanical energy. This is true because mechanical energy is the sum of kinetic and potential energy which remains constant due to the conservation of energy. d.) At their highest point, both of them have the same amount of gravitational potential energy. It is true because potential energy depends solely on the height from the ground and both reached the same height. b.) At their highest point, grasshopper B is moving faster than grasshopper A. It's not correct as the vertical velocity component of both grasshoppers is zero at their highest points, but grasshopper B still has horizontal velocity component, making its total velocity non-zero. c.) At their highest point, both of them have the same amount of kinetic energy. It's not correct as at the highest point, grasshopper A has zero kinetic energy but grasshopper B still has kinetic energy due to its horizontal motion. e.) At their highest point, grasshopper A has more gravitational potential energy than grasshopper B. It's not correct because they reached the same height, so they have same potential energy. Learn more about Physics of Motion here:
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A volleyball is hit upward with an initial velocity of 6.0 m/s. Calculate the displacement of the volleyball when its final velocity is 1.1 m/s upward.
Answer:
1.78 m upward
Explanation:
We can find the displacement of the volleyball by using the SUVAT equation:
[tex]v^2 - u^2 = 2ad[/tex]
where, assuming upward as positive direction:
u = 6.0 m/s is the initial velocity
v = 1.1 m/s is the final velocity
a = g = -9.8 m/s^2 is the acceleration of gravity
d is the displacement
Solving the equation for d, we find:
[tex]d=\frac{v^2-u^2}{2a}=\frac{1.1^2-6.0^2}{2(-9.8)}=1.78 m[/tex]
And since it is positive, the displacement is upward.
The integral concepts applied include kinematics and gravitational acceleration. Given initial velocity, final velocity, and knowing gravitational due to acceleration, we can determine the time of flight up to its highest point and the maximal height or displacement through kinematic equations. In this particular case, displacement happens to be approximately 2.275 meters upwards.
Explanation:The subject of this question is Physics, specifically, it deals with the concept of kinematics. The first step to solve the problem is to use the final velocity equation in kinematics, v = u + at, where 'v' is the final velocity, 'u' is the initial velocity, 'a' is the acceleration (which would be gravity in this case, ‐9.8 m/s² since it's acting downward or against the direction of the initial velocity), and 't' time. From the given values, we know that v = 1.1 m/s (these are vector quantities, hence the direction is important, and in this case, both v and u are in the same direction, up the way), u = 6.0 m/s and a = -9.8 m/s². You should set up the equation as follows: 1.1 = 6 + (-9.8)t. By simplifying, we get t = (1.1 - 6) / -9.8 ≈ 0.5s. Next, to get the displacement, you can use another kinematic equation s = ut + (1/2)a*t² ('s' stands for displacement). Once you plug in the known values, you'll get s = 6*0.5 + 1/2*(-9.8)*(0.5)² ≈ 2.275 m.
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A house is losing heat at a rate of 1600 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Express the rate of heat loss from this house per K, °F, and R difference between the indoor and the outdoor temperature.
Answer:
1600 kJ/h per K, 888.88 kJ/h per °F and 888.88kJ/h per R
Explanation:
We make use of relations between temperature scales with respect to degrees celsius:
[tex]1 K= 1^{\circ}C+273\\1^{\circ}F= (1^{\circ}C*1.8)+32\\1 R= (1^{\circ}C*1.8)+491.67[/tex]
This means that a change in one degree celsius is equivalent to a change of one kelvin, while for a degree farenheit and rankine this is equivalent to a change of 1.8 on both scales.
So:
[tex]\frac{Q}{\Delta T(K)}=\frac{Q}{\Delta T(^\circ C)}=1600 \frac{kJ}{h} per K\\\frac{Q}{\Delta T(^\circ F)}=\frac{Q}{\Delta T(^\circ C*1.8)}=888.88 \frac{kJ}{h} per ^\circ F\\\frac{Q}{\Delta T(R)}=\frac{Q}{\Delta T(^\circ C*1.8)}=888.88 \frac{kJ}{h} per R[/tex]
The burner on an electric stove has a power output of 2.0 kW. An 810 g stainless steel tea kettle is filled with water at 20∘ C and placed on the already hot burner. If it takes 2.4 min for the water to reach a boil, what volume of water was in the kettle? Stainless steel is mostly iron, so you can assume its specific heat is that of iron.
Answer:
volume of water in the kettle, V = [tex]774 cm^{3}[/tex]
Given:
Power output of burner, P = 2.0 kW = 2000 W
Mass of kettle, m = 810 g = 0.81 kg
Temperature of water in the kettle, T = [tex]20^{\circ}C[/tex]
Time taken by water to boil, t = 2.4 min = 144 s
Temperaturre at boiling, T' = [tex]100^{\circ}C[/tex]
Solution:
Now, we know that:
Iron's specific heat capacity, [tex]c = 0.45 J/g ^{\circ}C[/tex]
Water's specific heat capacity, [tex]c' = 4.18 J/g ^{\circ}C[/tex]
Now,
Total heat, q = Pt
q = [tex]2000\times 144 = 288 kJ[/tex]
Now,
q = (mc +m'c')(T' - T)
[tex]288\times 10^{3} = (0.81\times 0.45 + m'\times 4.18)(100^{\circ} - 20^{\circ})[/tex]
Solving the above eqn m', we get:
m' = 774 g
Now, the volume of water in the kettle, V:
[tex]V = \frac{m'}{\rho}[/tex]
where
[tex]\rho = density of water = 1 g/cm^{3}[/tex]
Now,
[tex]V = \frac{774}{1}[/tex]
Volume, V = [tex]774 cm^{3}[/tex]
Consider a golf ball with a mass of 45.9 grams traveling at 200 km/hr. If an experiment is designed to measure the position of the golf ball at some instant of time with a precision of 1 mm, then what will be the uncertainty in the speed of the golf ball? What percentage of the speed of the golf ball does this uncertainty represent?
Final answer:
The percent uncertainty in the distance is 0.0593%, the uncertainty in the elapsed time is 1 second, the average speed is 391 m/s, and the uncertainty in the average speed is 0.231 m/s.
Explanation:
To calculate the percent uncertainty in the distance, we divide the uncertainty in distance by the measured distance and then multiply by 100%. In this case, the percent uncertainty is (25 m / 42188 m) × 100% = 0.0593%.The uncertainty in the elapsed time is simply the given uncertainty of 1 second.To calculate the average speed, we divide the measured distance by the measured time. The average speed is 42188 m / (2 h + (30 min × 60 s/min) + 12 s) = 391 m/s.The uncertainty in the average speed can be calculated by taking the product of the percent uncertainties in distance and time. The uncertainty in the average speed is (0.0593% + 0%) × 391 m/s = 0.231 m/s.why is the use of international units important in medicine?
Answer:IU (international unit): An international unit (IU) is an internationally accepted amount of a substance.
This type of measure is used for the fat-soluble vitamins (such as vitamins AD and E) and certain hormones, enzymes, and biologicals (such as vaccines).
Explanation:
The engine on a fighter airplane can exert a force of 105,840 N (24,000 pounds). The take-off mass of the plane is 16,875 kg. (It weighs 37,500 pounds.) If you mounted this aircraft engine on your car, what acceleration would you get? (Please use metric units. The data in pounds are given for comparison. Use a reasonable estimate for the mass of your car. A kilogram mass weighs 2.2 pounds.)
Answer:
The acceleration you can get with that engine in your car is around 70,56 [tex](\frac{m}{s^{2} })[/tex] or 7,26 [tex](\frac{ft}{s^{2} } )[/tex] using 1500kg of mass or 3306 pounds
Explanation:
Using the equation of the force that is:
[tex]F=m*a[/tex]
So, you notice that you know the force that give the engine, so changing the equation and using a mass of a car in 1500 kg or 3306 pounds
[tex]a=\frac{F}{m} =\frac{105840 N }{1500 (kg) }[/tex]
[tex]a=\frac{105840 (\frac{kg*m}{s^{2} } )} {1500 kg }[/tex]
Note: N or Newton units are: [tex]\frac{kg * m}{s^{2} }[/tex]
[tex]a= 70,54 \frac{m}{s^{2} }[/tex]
Also in pounds you can compared
[tex]a= \frac{2400 lf }{3 306 lf}[/tex]
Note: lf in force units are: [tex]\frac{lf*ft}{s^{2} }[/tex]
[tex]a=7,26 \frac{ft}{s^{2} }[/tex]
Final answer:
The acceleration provided by a fighter airplane engine mounted on an average 1,500 kg car would be 70.56 m/s², calculated using Newton's second law of motion by dividing the force of 105,840 N by the mass of the car.
Explanation:
To calculate the acceleration that a fighter airplane engine could give to a car, we must use Newton's second law of motion, which states that the force exerted on an object equals the mass of the object multiplied by the acceleration of the object (F = m*a). In this case, the force given is 105,840 N from the airplane engine. Assuming an average car mass of about 1,500 kg (a reasonable estimate for a personal vehicle), the formula to find the acceleration (a) is as follows:
a = F / m
Substituting in the values provided:
a = 105,840 N / 1,500 kg
When the calculation is performed:
a = 70.56 m/s²
So, the acceleration that would result from mounting this aircraft engine into an average car would be 70.56 meters per second squared.
A force causes a mass of 4 kg to have an acceleration of 8 m/s2. Suppose something causes the mass to be one-quarter of its original amount. What will happen to the acceleration if the amount of force used remains the same?
Explanation:
From Newton's second law:
F = ma
Given that m = 4 kg and a = 8 m/s²:
F = (4 kg) (8 m/s²)
F = 32 N
If m is reduced to 1 kg and F stays at 32 N:
32 N = (1 kg) a
a = 32 m/s²
So the acceleration increases by a factor of 4.
for the displacement vectors a=(3.0m)i+(4.0m)j and b=(5.0m)i+(-2.0m)j, gave a+b in(a) unit vector notation and as a magnitude and an angle (related to i) now give b-a in unit-vector notation and as magnitude and an angle
Answer:
Part (a) a + b = (8.0i - 2.0j) and [tex]\theta = 14.03^o[/tex] from the x-axis
Part (b) b - a = (2.0i - 6.0j) and [tex]\theta = -71.06^o[/tex] from the x- axis
Explanation:
Given,
[tex]\vec{a}\ =\ (3.0m)i\ +\ (4.0m)j[/tex][tex]\vec{b}\ =\ (5.0m)i\ +\ (-2.0m)j[/tex]From the addition of the two vectors,
[tex]\vec{a}\ +\ \vec{b}\ =\ (3.0i\ +\ 4.0j)\ +\ (5.0i\ -\ 2.0j)\\\Rightarrow \vec{a}\ +\ \vec{b}\ =\ (3.0\ +\ 5.0)i\ +\(4.0\ -\ 2.0)j\\\Rightarrow \vec{a}\ +\ \vec{b}\ =\ 8.0i\ +\ 2.0j[/tex]
Let [tex]\theta[/tex] be the angle of the resultant vector of the addition of the vectors with the x-axis (i).
[tex]\therefore Tan\theta\ =\ \dfrac{2.0}{8.0}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{2.0}{8.0}\ \right )\\\Rightarrow \theta\ =\ 14.03^o[/tex]
Part (b)
From the subtraction of the two vectors,
[tex]\vec{b}\ -\ \vec{a}\ =\ (5.0i\ -\ 2.0j)\ -\ (3.0i\ +\ 4.0j)\\\Rightarrow \vec{b}\ -\ \vec{a}\ =\ (5.0\ +\ 3.0)i\ +\(-2.0\ -\ 4.0)j\\\Rightarrow \vec{b}\ -\ \vec{a}\ =\ 2.0i\ -\ 6.0j[/tex]
Let [tex]\theta[/tex] be the angle of the resultant vector of the addition of the vectors with the x-axis (i).
[tex]\therefore Tan\theta\ =\ \dfrac{-6.0}{2.0}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{-6.0}{2.0}\ \right )\\\Rightarrow \theta\ =\ -71.06^o[/tex]
What is the Primacy Effect? Question 1 options:
(a) The tendency of people to like winning more than losing
(b)The tendency for items at the beginning of a list to be recalled more easily than other items.
(c)The tendency for people to blame their misdeeds on the first person they see (d)The tendency for people to respect gold medal winners more than silver medal winners
Answer: The primacy effect occurs when you're more likely to remember words at the beginning of a list. A suggested reason for the primacy effect is that the initial items presented are most effectively stored in long-term memory because of the greater amount of processing devoted to them.
So B would be your answer
What is the gravitational potential energy of a two-particle system with masses 4.5 kg and 6.3 kg, if they are separated by 1.7 m? If you triple the separation between the particles, how much work is done (b) by the gravitational force between the particles and (c) by you?
The gravitational potential energy of a two-particle system can be calculated with a specific formula that uses the gravitational constant and both masses. The work done by the gravitational force and an external force when separation is tripled can be found by calculating the change in gravitational potential energy.
Explanation:The gravitational potential energy between two particles can be calculated with the formula: U = -G (M₁M₂/R) , where U is the gravitational potential energy, G is the gravitational constant (6.67 × 10^-11 Nm²/kg²), M₁ and M₂ are the masses of the two bodies, and R is the distance of separation between them.
To find the work done by gravitational force and by you when the separation is tripled, we will need to calculate the change in gravitational potential energy which is given by ΔU = U_final - U_initial. The negative of this value gives the work done by the gravitational force, while the magnitude of ΔU gives the work done by an external force (you in this case).
Learn more about Gravitational Potential Energy here:https://brainly.com/question/23134321
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Find x, the angle of depression from the top of a lighthouse that is 170 ft above water level to the waterline of a ship 1084 ft off shore. Round your answer to the nearest tenth of a degree.
Answer:
The answer to your question is: Ф = 81.1° = 278.9°
Explanation:
Data
angle, x, = ?
height = 170 ft
distance 1084 ft
Process
tan Ф = opposite side / adjacent side
tan Ф = 1084 / 170 = 6.37
Ф = tan ⁻¹ (6.37)
Ф = 81.1° or 360 - 81.1 = 278.9°
A Cessna aircraft has a liftoff speed of v = km/h = 33.3 m/s.
a) What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of S = 240 m?
b) How long does it take the aircraft to become airborne?
Answer:
a) Minimum acceleration is [tex]a=2.31\frac{m}{s^{2} }[/tex].
b) It will take [tex]t_{f}=14.41s[/tex].
Explanation:
Let's order the information.
Initial velocity: [tex]v_{i}=0m/s[/tex]
Final velocity: [tex]v_{f}=33.3m/s[/tex]
Initial position: [tex]x_{i}=0m[/tex]
Final position: [tex]x_{f}=240m[/tex]
a) We can use velocity's equation:
[tex]v_{f}^{2} = v_{i}^{2} +2a(x_{f}-x_{i})[/tex]
⇒ [tex]a=\frac{v_{f}^{2}-v_{i}^{2}}{2(x_{f}-x_{i})}[/tex]
⇒ [tex]a=2.31\frac{m}{s^{2} }[/tex].
b) For this, equation for average acceleration will be helpful. Taking [tex]t_{i}=0[/tex] and having [tex]t_{f}[/tex] as the unknown time it becomes airborne:
[tex]a=\frac{v_{f}-v_{i}}{t_{f}-t_{i}} =\frac{v_{f} }{t_{f}}[/tex]
⇒ [tex]t_{f}=\frac{v_{f}}{a}=\frac{33.3\frac{m}{s}}{2.31\frac{m}{s^{2}}}[/tex]
⇒ [tex]t_{f}=14.41s[/tex].
To take off, the Cessna aircraft needs a minimum acceleration of 2.31 m/s² and it will take approximately 14.4 seconds to become airborne after a run of 240 meters.
To find the minimum constant acceleration required by a Cessna aircraft to become airborne after a takeoff run of 240 meters, and the time it takes to become airborne, follow these steps:
(a) Minimum Constant Acceleration
We use the kinematic equation: v² = u² + 2as, where:
v is the final velocity (33.3 m/s)u is the initial velocity (0 m/s, starting from rest)a is the accelerations is the distance (240 m)Substitute the known values:
[tex](33.3 m/s)^2 = 0 + 2a(240 m)\\1108.89 = 480a\\a = 2.31 m/s^2[/tex]
(b) Time to Become Airborne
We use the kinematic equation: v = u + at, where:
t is the timeOther variables are as previously definedSubstitute the known values:
[tex]33.3 m/s = 0 + (2.31 m/s^2)t\\t = 33.3 / 2.31\\t = 14.4 seconds[/tex]
Big cockroaches can run as fast as 1.50 m/s over short distances. Suppose you turn on the light in an inexpensive motel and see one run directly away from you at a constant speed of 1.50 m/s. If you run after it to squish it beginning 0.80 m behind it with an initial velocity of 0.70 m/s, what minimum acceleration must you have to catch it before it reaches safety 1.30 m away from where it starts running?
Answer:
a = 4.054 m/ s2
Explanation:
[tex]safety\ Time\ for\ cockroach = \frac{Distance\ travelled\ by\ cockroach}{ speed of cockroach}[/tex]
[tex]t = \frac{1.3}{1.5}[/tex]
t = 0.86 seconds
total distance travelled by you = 0.8 + 1.3 = 2.1 m
Initial velocity u = 0.7 m/s
let a is acceleration,
from equation of motion we have
s = ut + 0.5 at^2
[tex]2.1 = 0.7 \frac{1.3}{1.5} + 0.5 a [\frac{1.3}{1.5}]^2[/tex]
2.1 = 0.60 + 0.37 a
1. 5 = 0.37a
a = 4.054 m/ s2
If y(x, t) = (4.7 mm)sin(kx + (675 rad/s)t + ϕ) describes a wave traveling along a string, how much time does any given point on the string take to move between displacements y = +2.0 mm and y = −2.0 mm?
Answer:
It takes 0.00127 seconds.
Explanation:
The equation its
[tex]y(x,t) = 4.7 \ mm \ sin ( kx + 675 \frac{rad}{s} t + \phi)[/tex].
We want the time for ANY POINT, so, for convenience, lets take x=0
[tex]y(0,t) = 4.7 \ mm \ sin ( k*0 + 675 \frac{rad}{s} t + \phi)[/tex].
[tex]y(0,t) = 4.7 \ mm \ sin ( 675 \frac{rad}{s} t + \phi)[/tex].
Now, we want the positions
[tex]y(0,t_1)=2 \ mm[/tex]
[tex]y(0,t_2)=-2 \ mm[/tex]
so, for the first position:
[tex]y(0,t_1)=2 \ mm[/tex]
[tex]2 \ mm = 4.7 \ mm \ sin ( 675 \frac{rad}{s} t_1 + \phi)[/tex].
[tex]\frac{2 \ mm} {4.7 \ mm }= sin ( 675 \frac{rad}{s} t_1 + \phi)[/tex].
[tex] 0.42 = sin ( 675 \frac{rad}{s} t_1 + \phi)[/tex].
[tex] sin^-1(0.42) = 675 \frac{rad}{s} t_1 + \phi)[/tex].
and for the second one:
[tex]y(0,t_2)=-2 \ mm[/tex]
[tex]-2 \ mm = 4.7 \ mm \ sin ( 675 \frac{rad}{s} t_2 + \phi)[/tex].
[tex]\frac{-2 \ mm} {4.7 \ mm }= sin ( 675 \frac{rad}{s} t_2 + \phi)[/tex].
[tex] -0.42 = sin ( 675 \frac{rad}{s} t_2 + \phi)[/tex].
[tex] sin^-1(0.42) = 675 \frac{rad}{s} t_2 + \phi)[/tex].
Now, we can subtract both:
[tex]sin^-1(0.42) -sin^-1(-0.42) = 675 \frac{rad}{s} t_1 + \phi - 675 \frac{rad}{s} t_2 + \phi[/tex]
[tex]sin^-1(0.42) -sin^-1(-0.42) = 675 \frac{rad}{s} t_1 - 675 \frac{rad}{s} t_2[/tex]
[tex]sin^-1(0.42) -sin^-1(-0.42) = 675 \frac{rad}{s} (t_1 - t_2)[/tex]
[tex]\frac{sin^-1(0.42) -sin^-1(-0.42)}{ 675 \frac{rad}{s}} = (t_1 - t_2)[/tex]
[tex]\frac{0.43 - (-0.43)}{ 675 \frac{rad}{s}} = (t_1 - t_2)[/tex]
[tex]\frac{0.86}{ 675 \frac{rad}{s}} = (t_1 - t_2)[/tex]
[tex] 0.00127 s = (t_1 - t_2)[/tex]
The strings take 0.00127 seconds.
Final answer:
To find the time for a point on a sinusoidal wave to move from y = +2.0 mm to y = -2.0 mm, we calculate the time for half a cycle of the wave using the angular frequency.
Explanation:
The question involves determining the time it takes for a point on a string to move between two displacements in a sinusoidal wave motion. Given that the wave equation is y(x, t) = (4.7 mm)sin(kx + (675 rad/s)t + φ) and the displacements to consider are +2.0 mm and -2.0 mm, we can utilize the properties of the sine function and the given angular frequency to solve for the time interval. First, we find the phase angle (θ) corresponding to y = 2.0 mm by using the inverse sine function and then find the difference in time (t) to reach y = -2.0 mm, considering the periodic nature of the sine wave. Assuming we start from y = +2.0 mm at time t = 0, as the wave progresses to y = -2.0 mm, it completes half a cycle, so δt = π/ω, where ω = 675 rad/s is the angular frequency. Therefore, δt = π/675 s.
The horn on a commuter train emits sound with a frequency of 342.5 Hz when the train is at rest. The train sounds its horn as it passes a passenger platform at a constant speed of 41 m/s. What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding? Remember that change of frequency means the final frequency minus the initial frequency. Be sure to include the correct sign on the answer. (Take the speed of sound to be 345 m/s.)
Answer:
Δf = -82.57 Hz
Explanation:
For this problem we will use Doppler effect formula:
[tex]f = \frac{C+Vr}{C+Vs}*fo[/tex] where:
C is the speed of sound
Vr is the velocity of the person = 0m/s
Vs is the velocity of the source (train): When approaching Vs = -41m/s and when receding Vs = 41m/s
fo = 342.5 Hz
When the train is approaching:
[tex]fa = \frac{C}{C+Vs}*fo = \frac{345}{345-41}*342.5=388.69Hz[/tex]
And when the train is receding:
[tex]fr = \frac{C}{C+Vs}*fo = \frac{345}{345+41}*342.5=306.12Hz[/tex]
So, the change of frequency will be:
Δf = fr - fa = 306.12 - 388.69 = -82.57 Hz
A worker on the roof of a house drops his 0.46 kg hammer, which slides down the roof at constant speed of 9.88 m/s. The roof makes an angle of 27 ◦ with the horizontal, and its lowest point is 17.1 m from the ground. what is the the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground?
Answer:
The horizontal distance travelled in that time lapse is 12.94 m
Explanation:
In order to solve this problem, we'll need:
The horizontal speedthe time the hammer takes to fall from the roof to the groundAt the lowest point of the roof, the hammer has a 9.88 m/s speed that makes an angle of 27° with the horizontal, so we can calculate the horizontal and vertical speed with trigonometry. If we take right as x positive and down as y positive we get
[tex]v_{x}=v*cos(27)=9.88 m/s *cos(27)=8.80 m/s \\v_{y}=v*sen(27)=9.88 m/s *sen(27)=4.49 m/s[/tex]
Now, we make two movement equation as we have a URM (no acceleration) in x and an ARM (gravity as acceleration) in y. We will wisely pick the lowest point of the roof as the origin of coordinates
[tex]x(t)=8.8 m/s *t[/tex]
[tex]y(t)=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}[/tex]
Now we calculate the time the hammer takes to get to the floor
[tex]17.1m=4.49m/s*t+\frac{1}{2}*9.8m/s^{2}*t^{2}\\t=1.47s[/tex] or [tex]t=-2.38s[/tex]
Now, we keep the positive time result and calculate the horizontal distance travelled
[tex]x(1.47s)=8.8m/s*1.47s=12.94m[/tex]
The anode heel affect 1. should be considered when the structure of interest is more than 43 inches long. 2. is a variation in photon intensity across the exposure field. 3. occurs because the heel of the anode absorbs more photons than the toe. 4. is best used with the thinner body toward the cathode.
Answer:
Option 2. and 3.
Explanation:
Anode heel effect or simply heel effect is observed in X-ray tubes.
It is the change in the intensity of the emitted X-rays from the anode and this intensity is direction dependent and depends on the direction of the emitted X-rays along the axis of anode-cathode.
This effect results from the absorption of X-rays prior to leaving the anode, their production source.
The absorption of photons is more in the anode heel than the toe of the anode which results in the anode heel effect.
Which of the following quantities have the dimensions of an acceleration? Which of the following quantities have the dimensions of an acceleration?
a. v²/x
b. xt²
c. x/t²
d. v/t
Explanation:
The acceleration of an object is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
Dimension of velocity is, [tex][v]=[LT^{-1}][/tex]
Dimension of time is, [tex][t]=[T][/tex]
Dimension of acceleration is, [tex][a]=[LT^{-2}][/tex]
Option 1.
[tex]\dfrac{v^2}{x}=[LT^{-2}][/tex]
Option 2.
[tex]xt^2=[LT^2][/tex]
Option 3.
[tex]\dfrac{x}{t^2}=[LT^{-2}][/tex]
Option 4.
[tex]\dfrac{v}{t}=[LT^{-2}][/tex]
So, from above calculations, it is clear that option (1),(3) and (4) have the dimensions of acceleration. Hence, this is the required solution.
Answer:
a)[tex]\dfrac{v^2}{x}[/tex]
c)[tex]\dfrac{x}{t^2}[/tex]
d)[tex]\dfrac{v}{t}[/tex]
Explanation:
Acceleration :
Acceleration is the rate of change of velocity of the particle.
The unit of acceleration is m/s².
In mathematical form,
[tex]a=\dfrac{dv}{dt}[/tex]
We will check the all option on base of unit
(a). [tex]\dfrac{v^2}{x}[/tex]
Where, v = velocity
x = position
The unit of velocity and position are m/s and m.
The dimension formula of velocity and position
[tex]v = LT^{-1}[/tex]
[tex]x=L[/tex]
Put the unit in the given equations
[tex]\dfrac{v^2}{x}=\dfrac{L^2T^{-2}}{L} =\dfrac{L}{T^2}[/tex]
(b). [tex]xt^2=L\times T^2[/tex]
(c). [tex] \dfrac{x}{t^2}=\dfrac{L}{T^2}[/tex]
(d). [tex]\dfrac{v}{t}=\dfrac{LT^{-1}}{T}=\dfrac{L}{T^2}[/tex]
Hence, This is the required solution.
"Sequence the kinetic energy, temperature, and density of most solids, liquids, and gases. Use 1 to represent the lowest amount and 3 to represent the highest."
Answer:
Kinetic energy
Solids 1, liquids 2, gases 3
Temperature
Solids 1, liquids 2, gases 3
Density
Solids 3, liquids 2, gases 1
Explanation:
Most of elements can be in the three different states at different temperatures, solid in lower temperatures then liquid and gas in higher temperatures.
When it's gaseous it is more expanded, it's volume is bigger then it's density is lower. As the molecules are more separated they can move more easily.
On the opposite hand in solid state, it's more compressed, the volume is smaller and the density is higher. The molecules are more compressed an they can't move easily.
Suppose f (t) = −11t2 is the position at time t of an object moving along the x axis. Use the limit definition to find the velocity of the object at time t0 = −5.
The velocity of the object at time t0 = -5 can be found by taking the derivative of the position function, which results in v(t) = -22t. Substituting t = -5 into this velocity function yields v(-5) = 110 m/s.
Explanation:The student asks for the velocity of an object at time t0 = -5 given the position function f(t) = −11t2. To find the velocity, we use the limit definition of the derivative. The derivative of the position function with respect to time gives the velocity function:
v(t) = f'(t) = d/dt (-11t2)
This results in:
v(t) = -22t
Now, we can find the velocity at t0 = -5 by plugging the value into the velocity function:
v(-5) = -22(-5) = 110 m/s
Therefore, the velocity of the object at t0 = -5 is 110 m/s.