Answer:
Option b. it has the same position and the same acceleration as A
Explanation:
Let's analyze every statement:
a. it has the same position and the same velocity as A
In the instant where B passes A, they Do have the same position. Velocity however, cannot be the same because if they were, ball B would never pass ball A. So, this is false.
b. it has the same position and the same acceleration as A
As we said in the previous option, the position is the same. The acceleration is gravity for both balls, so this is true.
c. it has the same velocity and the same acceleration as A
Acceleration is the same but velocities are not, so this is false.
d. it has the same displacement and the same velocity as A
The distance they have traveled is the same, so the displacement is the same, but the velocity is not, so this is false.
e. it has the same position, displacement and velocity as A
The position and displacement is the same but not velocity, so this is false.
Only option b is true.
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
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
An oil may be converted into a substance that is solid at room temperature by
Answer:
A) adding hydrogens, decreasing the number of double bonds in the molecules.
Explanation:
The complete Question is:
An oil may be converted into a substance that is solid at room temperature by
A) adding hydrogens, decreasing the number of double bonds in the molecules.
B) removing water, causing a dehydration synthesis reaction to occur.
C) removing hydrogens, increasing the number of double bonds.
D) cooling it, so that double bonds form and the fats solidify.
Hydrogenation is a process that is commonly used in food industry to reduce the vegetable oil to solid or semi-soild state which is suitable for preserving. Oil is an unsaturated fatty acid having Carbon - Carbon double bonds. Hydrogenation i.e. addition of hydrogen in presence of catalysts to these oils reduce these double bonds to single bonds. This reduction of double bonds also change some physical properties of the oil, the most obvious of which is the melting point. The melting point increases and the oil is converted to a substance that is sold at room temperature.
Therefore, the correct answer to this question is given by option A.
Final answer:
An oil can be converted into a solid fat by hydrogenation, which involves adding hydrogen to unsaturated fatty acids to raise the melting point and create a solid at room temperature, commonly used in making margarine or shortening.
Explanation:
An oil can be converted into a solid fat through a process known as hydrogenation. During this chemical reaction, hydrogen is added to the unsaturated fatty acids present in the oil. This transformation is achieved using a catalyst such as nickel (Ni), platinum (Pt), or palladium (Pd), which facilitates the addition of hydrogen atoms to the oil's fat molecules. The result of these additions changes the molecular structure, thus converting unsaturated fatty acids into saturated fatty acids. This increase in saturation raises the melting point of the triglycerides found in the oil, often turning a liquid oil, like vegetable oil, into a semi-solid or solid form similar to margarine or shortening.
Saturated fats tend to be solid at room temperature due to their chemical structure, which lacks double bonds, allowing them to pack tightly together. Unsaturated fats, which have one or more double bonds in their hydrocarbon chains, are typically liquid at room temperature. The hydrogenation process effectively reduces or eliminates double bonds, increasing the degree of saturation and thereby raising the melting point of the fat. This is a common practice in the food industry to alter the texture and stability of edible fats for various applications.
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]
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]
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]
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 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.
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).
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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.
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°
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.
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]
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:
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.
Which of the following is a unit of speed?
A) h
B) nm
C) cm/y
D) km/s^2
Answer:
Option c that it means cm/year.
Explanation:
Nm means nanometer, 1x10^9m
Km/s^2 is acceleration
h.. it may be height
Answer:
D, km/s^2
Explanation:
This is a unit of speed because it is saying kilometers over seconds squared, and that is a unit of speed similar to miles per hour or mph or m/h.
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.
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.
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.
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]
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]
A cart is propelled over an xy plane with acceleration compo- nents ax 4.0 m/s2 and ay 2.0 m/s2. Its initial velocity has com- ponents v0x 8.0 m/s and v0y 12 m/s. In unit-vector notation, what is the velocity of the cart when it reaches its greatest y coordinate?
To find the velocity of the cart when it reaches its greatest y coordinate, we can combine the x and y components of the velocity. The x-component can be found using Ux = v0x + ax*t, and the y-component can be found using Uy = v0y + ay*t. The total velocity, v, can then be found using v = sqrt(Ux^2 + Uy^2).
Explanation:The velocity of the cart can be found by combining its x and y components. The x-component of the velocity, Ux, can be found using the equation Ux = v0x + ax*t. The y-component of the velocity, Uy, can be found using the equation Uy = v0y + ay*t. The total velocity, v, can be found using the equation v = sqrt(Ux^2 + Uy^2).
Learn more about velocity of a cart here:https://brainly.com/question/25405393
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"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.
reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking 0.916 km due east, 0.918 km due south, and 3.52 km in a direction 49.7 ° north of west. The time required for this trip is 1.750 h. Determine the magnitudes of the bird watcher's (a) displacement and (b) average velocity.
Answer:
Displacement: 2.230 km Average velocity: 1.274[tex]\frac{km}{h}[/tex]
Explanation:
Let's represent displacement by the letter S and the displacement in direction 49.7° as A. Displaement is a vector, so we need to decompose all the bird's displacement into their X-Y compoments. Let's go one by one:
0.916 km due east is an horizontal direction and cane be seen as direction towards the negative side of X-axis.0.928 km due south is a vertical direction and can be seen as a direction towards the negative side of Y-axis.3.52 km in a direction of 49.7° has components on X and Y axes. It is necessary to break it down using trigonometry,First of all. We need to sum all the X components and all the Y componets.
∑[tex]Sx = Ax -0.916[/tex] ⇒ ∑[tex]Sx = [tex]3.52cos(49.7) - 0.916[/tex]
∑[tex]Sx = 1.361 km[/tex]
∑[tex]Sy = Ay - 0.918[/tex] ⇒ ∑[tex]Sy = 3.52sin(49.7) - 0.918[/tex]
∑[tex]Sy = 1.767[/tex]
The total displacement is calculated using Pythagoeran therorem:
[tex]S_{total} =\sqrt{Sx^{2}+ Sy^{2} }[/tex] ⇒
[tex]S_{total} = 2.230 km[/tex]
With displacement calculated, we can find the average speed as follows:
[tex]V = S/t[/tex] ⇒ [tex]V = \frac{2.230}{1.750}[/tex]
[tex]V = 1.274\frac{km}{h}[/tex]
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.
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.
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.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]