Answer:
89.16pounds
Explanation:
The equation that defines this problem is as follows
W=k/X^2
where
W=Weight
K= proportionality constant
X=distance from the center of Earth
first we must find the constant of proportionality, with the first part of the problem
k=WX^2=131x3960^2=2054289600pounds x miles^2
then we use the equation to calculate the woman's weight with the new distance
W=2054289600/(4800)^2=89.16pounds
Which of the following is NOT part of the kinetic theory of gases?
a. There is very little empty space in a gas.
b. A gas is composed of very small particles.
c. Gas particles do not attract or repel one another.
d. Gas particles move faster when the temperature increases.
Answer:
Option a
Explanation:
The three basic and major points of the kinetic theory of gases are listed below:
1. In the collision between the gas molecules, no loss or gain of energy takes place.
2. According to the theory, the gaseous molecules exhibits constant and linear motion.
3. The molecules of the gas occupies very little space as compared to the container. Thus the empty space in a gas is not very little.
Answer:
The option that does NOT correspond to the kinetic theory of gases is the option a."There is very little empty space in a gas."
Explanation:
The kinetic theory of gases allows to deduce the properties of the ideal gas using a model in which the gas molecules are spheres that comply with the laws of classical mechanics. This theory states that heat and movement are related, that particles of all matter are moving to some extent and that heat is a sign of this movement.
The postulates of this theory are:
A gas is formed by a large number of spherical particles whose size is negligible compared to the distance between the particles. That is, there are no attractive forces between the molecules of a gas .The molecules move in a straight line at high speed and only interact when they collide. Crashes between particles and with the vessel walls are considered perfectly elastic, conserving translational kinetic energy. Expressed in other words, gas molecules possess kinetic energy. In the movement, the gas molecules collide elastically with each other and with the walls of the container that contains them in a perfectly random way, that is, in each shock the energy is delivered from one particle to another, and therefore they can continue in constant motion The frequency of collisions with the walls of the vessel explains the pressure exerted by the gases. The energy of such particles can be converted into heat or another form of energy. but the total kinetic energy of the molecules will remain constant if the volume and temperature of the gas do not vary; Therefore, the pressure of a gas is constant if the temperature and volume do not change. Then, an increase in the temperature of a gas also increases the speed at which the particles move.Taking into account the above, the option that does NOT correspond to the kinetic theory of gases is the option a."There is very little empty space in a gas."
If fe represents the electrostatic force in N that point charge q1 in C exerts on point charge q2 in C, and r represents the distance between the point charges in m, what is the unit for the electrostatic constant k in the formula F = kq1q2/r^2?
The electrostatic constant is also known as Coulomb's constant or the electric constant. Its unit is [tex]\rm \frac{Nm^2}{C^2}[/tex].
It is a fundamental physical constant that appears in Coulomb's law, which describes the interaction between electric charges. It's denoted by the symbol K.
Given information:
Unit of force (F) = Newton
Unit of charge (Q) = Coloumb
Unit of distance (r) = Meters
Now,
On substituting the values of given units:
[tex]\rm F = \frac{Kq_1q_2}{r^2}\\N =\frac{K C\times C}{m^2}\\N = \frac{K C^2}{m^2}\\K = \frac{Nm^2}{C^2}[/tex]
The unit of electrostatic constant (K) is [tex]\rm \frac{Nm^2}{C^2}[/tex]. This constant (K) plays a crucial role in understanding and calculating the forces and interactions between charged particles in the realm of electrostatics.
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Final answer:
The unit for the electrostatic constant k is Newton meters squared per Coulomb squared (N·m²/C²), and k is also known as Coulomb's constant with a value of approximately 8.99 × 10⁹ N·m²/C².
Explanation:
The unit for the electrostatic constant k in the formula F = k[tex]q_{1}[/tex][tex]q_{2}[/tex] /r² can be determined by rearranging the formula to solve for k, which gives us k = Fr² / [tex]q_{1}[/tex][tex]q_{2}[/tex] . Since the unit for the electrostatic force F is Newtons (N), the distance r is in meters (m), and the charge [tex]q_{1}[/tex]and [tex]q_{2}[/tex] are in Coulombs (C), the units for k would be Newton meters squared per Coulomb squared (N·m²/C²). The value of the electrostatic constant k is approximately 8.99 × 10⁹ N·m²/C², also known as Coulomb's constant.
An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.4 s. A passenger in the elevator is holding a 3.3 kg bundle at the end of a vertical cord. What is the tension in the cord as the elevator accelerates?
Answer:
35.71 N
Explanation:
The elevator starts from the rest means its initial velocity is zero.
Given that, the height achieved by the elevator in 1.4 s will be, [tex]S=1m[/tex]
Given that the mass of the bundle which is hold by passenger is, [tex]m=3.3 kg[/tex]
Now according to second equation of motion.
[tex]S=ut+\frac{1}{2}at^{2}[/tex]
Here, S is the height, u is the initial velocity, t is the time taken, and a is the acceleration.
Now initial velocity is zero therefore,
[tex]S=\frac{1}{2}at^{2}\\a=\frac{2S}{t^{2} }[/tex]
According to the free body diagram tension and acceleration in upward direction and weight is in downward direction.
So,
[tex]ma=T-mg\\T=m(g+a)[/tex]
Put the value of a from the above
[tex]T=m(g+\frac{2S}{t^{2} })[/tex]
Put all the variables.
[tex]T=3.3(9.8+\frac{2\times 1}{1.4^{2} })\\T=3.3(9.8+1.02)\\t=35.71N[/tex]
This the required tension.
Final answer:
To determine the tension in the cord as the elevator accelerates, the acceleration of the elevator is first calculated using the kinematic equation and is found to be approximately 1.02 m/s². Then, the tension is calculated using the formula T = mg + ma and is determined to be approximately 36.04 N.
Explanation:
To calculate the tension in the cord as the elevator accelerates, we first need to determine the acceleration of the elevator. Using the distance traveled and the time it took, we can apply the kinematic equation s = ut + 0.5at2 to find the acceleration 'a', where s is the distance (1 m), u is the initial velocity (0 m/s), and t is the time (1.4 s). After finding 'a', we can calculate the tension (T) in the cord using the formula T = mg + ma, where m is the mass of the bundle (3.3 kg), g is the acceleration due to gravity (9.81 m/s2), and a is the acceleration of the elevator.
First, we find the acceleration:
s = u t + 0.5at2
1 = 0 + 0.5a(1.4)2
a ≈ 1.02 m/s2
Next, we calculate the tension:
T = mg + ma
T = 3.3 x 9.81 + 3.3 x 1.02
T ≈ 32.67 + 3.366
T ≈ 36.04 N
An emf is induced in a conducting loop of wire 1.07 m long as its shape is changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 4.36 s and the local 0.115-T magnetic field is perpendicular to the plane of the loop.
Answer:
0.517 mV
Explanation:
Length of wire = 1.07 m
For square:
Perimeter = 1.07 m
Let a be the side of square
So, 4a = 1.07
a = 0.2675 m
Area of square, A 1 = side x side = 0.2675 x 0.2675 = 0.07156 m^2
For circle:
Circumference = 1.07 m
Let r be the radius of circle
So, 2 π r = 1.07
2 x 3.14 x r = 1.07
r = 0.1704 m
Area of circle, A 2 = π r^2 = 3.14 x 0.1704 x 0.1704 = 0.09115 m^2
Change in area, dA = A2 - A1 = 0.09115 - 0.07156 = 0.0196 m^2
Time taken in changing the area, dt = 4.36 s
Magnetic field, B = 0.115 T
According to the Farady's law of electromagnetic induction
[tex]e = \frac{d\phi }{dt}=\frac{dBA }{dt}=B\frac{dA }{dt}[/tex]
[tex]e = 0.115\times \frac{0.0196}{4.36}[/tex]
e = 5.17 x 10^-4 V
e = 0.517 mV
Thus, the induced emf is 0.517 mV.
The initial volume reading in a graduated cylinder is 30 mL. The mass of an irregular shape of an unknown metal piece is 55.3 g. The final volume reading in the graduated cylinder is 37 mL. The density of this unknown metal is _____.
Answer:
7.9[tex]\frac{gr}{cm^3}[/tex]
Explanation:
When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.
So now we know that the volume of that piece of unknown metal is 7mL (which is the same as 7[tex]cm^3[/tex]).
Density is [tex]\frac{mass}{volume}[/tex].
So the density of that piece of metal is [tex]\frac{55.3g}{7cm^3}[/tex]
Which leaves us with a final density of 7.9[tex]\frac{gr}{cm^3}[/tex]
Two cars approach each other; both cars are moving westward, one at 79 km/h, the other at 60 km/h. (a) What is the velocity of the first car relative to (in the reference frame of) the second car
Explanation:
Given that,
There are two cars say A and B. Both are approaching each other moving towards westward.
Speed of car A, [tex]v_A=79\ km/h[/tex]
Speed of car B, [tex]v_B=60\ km/h[/tex]
We need to find the velocity of the first car relative to (in the reference frame of) the second car. As both are moving in same direction, there relative velocity is given by :
[tex]v'=v_A-v_B[/tex]
[tex]v'=79\ km/h-60\ km/h[/tex]
v' = 19 km/h
So, the velocity of car A with respect to car B is 19 km/hr. Hence, this is the required solution.
Jack observed his coworker Jane crying when she was called into the boss's office. Jack thinks that crying at work is a sign of weakness that makes women unfit for employment in his industry. Jack is displaying _____.
... prejudice based on misogyny.
Also a good bit of stupidity.
The magnitude of the electrostatic force between two identical ions that are separated by a distance of 5.0A is 3.7×10^-9N.a) what is the charge of each ion?b) how many electrons are missing from each ion( thus giving the ion its charge imbalance)?
Explanation:
Given that,
Electrostatic force, [tex]F=3.7\times 10^{-9}\ N[/tex]
Distance, [tex]r=5\ A=5\times 10^{-10}\ m[/tex]
(a) [tex]F=\dfrac{kq^2}{r^2}[/tex], q is the charge on the ion
[tex]q=\sqrt{\dfrac{Fr^2}{k}}[/tex]
[tex]q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}[/tex]
[tex]q=3.2\times 10^{-19}\ C[/tex]
(b) Let n is the number of electrons are missing from each ion. It can be calculated as :
[tex]n=\dfrac{q}{e}[/tex]
[tex]n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
n = 2
Hence, this is the required solution.
Identify the situation where work is being done.
A.Carrying a box of crayons across the room.
B.Lifting a backpack off the floor
C.Sitting on a stool
D.Holding a football
Answer:
A. carrying a box of crayons across the room.
Explanation:
work is said to be done when a force moves something over a distance.
Answer:
A.
Explanation:
Carrying a box of crayons across the room.
Creation of a proton gradient by the electron transport chain represents
Answer:
Potential Energy
Explanation:
The electron transport chain is a series of proteins and organic molecules located in the inner membrane of the mitochondria.During electron transfer and proton pumping electrons are moved from a higher energy level to a lower one and in the process release energy.This energy is used to make ATP.It is stored in the electrochemical gradient of protons and it has to be released for electron transport to continue.
Creation of a proton gradient by the electron transport chain represents a pivotal process in cell respiration known as oxidative phosphorylation, in which the energy from electrons is used to form a gradient of protons. This gradient powers ATP production, converting the energy of electrons to a form suitable for cellular work.
Explanation:The creation of a proton gradient by the electron transport chain represents a crucial step in cell respiration called oxidative phosphorylation. In the mitochondria, the electron transport chain uses energy from electrons to pump protons (hydrogen ions) from the mitochondrial matrix into the intermembrane space, forming a gradient. This proton gradient is the potential energy that drives ATP production when the protons flow back across the membrane via ATP synthase. Therefore, the electron transport chain converts the energy of electrons into a useable form for the cell to perform work.
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A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium when the string makes a 30 degrees angle with the vertical, what is the net charge on the ball?
Q = _______ C
Answer:
[tex]Q = \frac{0.068}{E}[/tex]
where E = electric field intensity
Explanation:
As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical
So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field
while weight of the ball is vertically downwards
so here we have
[tex]QE = F_x[/tex]
[tex]mg = F_y[/tex]
since string makes 30 degree angle with the vertical so we will have
[tex]tan\theta = \frac{F_x}{F_y}[/tex]
[tex]tan30 = \frac{QE}{mg}[/tex]
[tex]Q = \frac{mg}{E}tan30[/tex]
[tex]Q = \frac{0.012\times 9.81}{E} tan30[/tex]
[tex]Q = \frac{0.068}{E}[/tex]
where E = electric field intensity
The net charge on the small plastic ball which is suspended by a string in a uniform is 0.068/E C.
What is electric field?The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
A point charge on a string in equilibrium with an electric field can be given as,
[tex]E=\dfrac{mg \tan\theta}{Q}[/tex]
Here, (m) is the mass, (g) is acceleration due to gravity, and (Q) is the charge.
A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field.
The ball is in equilibrium when the string makes a 30 degrees angle with the vertical. Thus, by the above formula,
[tex]E=\dfrac{(0.012)(9.81)\tan(30)}{Q}\\Q=\dfrac{0.068}{E}\rm\; C[/tex]
Thus, the net charge on the small plastic ball which is suspended by a string in a uniform is 0.068/E C.
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A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration of 10 cm/s2 for another 5.0 s. The particle moves in the same direction along a straight line. The average speed over the whole time interval is?
Answer:
V=20cm/s
Explanation:
The average speed is the distance total divided the time total:
[tex]V=X/T[/tex]
First stage:
T1=5s
[tex]v_{f} =v_{o} - at[/tex]
But, [tex]v_{f} =0[/tex] (decelerates to rest)
then: [tex]a =v_{o} /t=0.3/5=0.06m/s^{2}[/tex]
on the other hand:
[tex]x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m[/tex]
X1=75cm
Second stage:
T2=5s
[tex]x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m[/tex]
X2=125cm
Finally:
X=X1+X2=200cm
T=T1+T2=10s
V=X/T=20cm/s
The average speed of the particle is 20 centimetre per second.
Average speed is given as total distance covered by the particle divided by the total time of the journey.
The particle decelerates from 30 cm/s to rest (0 cm/s) in 5.0 s. Hence, the deceleration can be given as,
a = (0 - 30 cm/s) / 5 s = - 6 cm/s²
From the kinematic equation:
s = ut + [tex]\frac{1}{2}at^2[/tex], where s = displacement, a = acceleration, t = time, and u = initial velocity of the particle, we can determine the displacement.
so, s = (30 cm/s) (5 s) - [tex]\frac{1}{2} (6 cm/s^2)(5 s)^2[/tex] = 150 cm - 75 cm = 75 cm
The particle then accelerates uniformly at 10 cm/s² for another 5.0 s. The final velocity after this period is given as:
v = u + at
Here, u = 0 cm/s, a = 10 cm/s². t = 5s. Hence,
v = (10 × 5) m/s = 50 cm/s
Using the kinematic equation:
v² = u² + 2as, we get:
(50 cm/s)² = 2 (10 cm/s²) s
or, s = 125 cm
Total displacement of the particle = 75 cm + 125 cm = 200 cm
total time of journey = 5s + 5s = 10s
so, average velocity = 200 cm / 10 s = 20 cm/s
What is the mass of a dog that weighs 382 N?(unit=kg)
Answer:
The answer to your question is: mass = 38.93 kg
Explanation:
Data
mass = ?
Weight = 382 N
gravity = 9.81 m/s2
Formula
Weight = mass x gravity
mass = weight / gravity
mass = 382 / 9.81 substitution
mass = 38.93 kg result
The mass of a dog that weighs 382 N can be calculated using the weight formula w=mg, rearranged as m=w/g. Substituting the values, we find that the mass of the dog is approximately 39 kg.
Explanation:To calculate the mass of a dog that weighs 382 N, we use the formula w = mg, where w is weight, m is mass, and g is the acceleration due to gravity, which on Earth is approximately 9.8 m/s². So, to find the mass, we rearrange the formula to m = w/g.
Substituting the given values into the equation, m = 382 N / 9.8 m/s² = 39.0 kg. So, the mass of the dog is approximately 39 kg when rounded to the nearest whole number.
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A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.90 m/s at an angle of 23.0° below the horizontal. It strikes the ground 5.00 s later.(a) How far horizontally from the base of the building does the ball strike the ground?1 m(b) Find the height from which the ball was thrown.2Your response differs from the correct answer by more than 10%. Double check your calculations. m(c) How long does it take the ball to reach a point 10.0 m below the level of launching?3 s
Answer:
Part a)
[tex]\Delta x = 36.35 m[/tex]
Part b)
height will be 107.2 m
Part c)
[tex]t = 1.78 s[/tex]
Explanation:
Initial velocity of the ball is given as
v = 7.90 m/s
angle of projection of the ball
[tex]\theta = 23^o[/tex]
now the two components of velocity of ball is given as
[tex]v_x = 7.90 cos23 = 7.27 m/s[/tex]
[tex]v_y = 7.90 sin23 = 3.09 m/s[/tex]
Part a)
Since ball strike the ground after t = 5 s
so the distance moved by the ball in horizontal direction is given as
[tex]\Delta x = v_x \times t[/tex]
[tex]\Delta x = 7.27 (5)[/tex]
[tex]\Delta x = 36.35 m[/tex]
Part b)
Now in order to find the height of the ball we can find the vertical displacement of the ball
[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]
[tex]\Delta y = (3.09)(5) - \frac{1}{2}(9.81)(5^2)[/tex]
[tex]\Delta y = -107.2 m[/tex]
So height will be 107.2 m
Part c)
when ball reaches a point 10 m below the level of launching then the displacement of the ball is given as
[tex]\Delta y = -10 m[/tex]
so we will have
[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]
[tex]-10 = 3.09 t - \frac{1}{2}(9.81)t^2[/tex]
so we will have
[tex]t = 1.78 s[/tex]
For an object in free-fall (ignoring air resistance), which of the following statements is false: You are free to choose the origin and direction of the positive axis. The acceleration due to gravity is constant. The acceleration may point in the opposite direct as the velocity. The acceleration always points in the same direct as the velocity.
Answer:hh
The acceleration may point in the opposite direct as the velocity
Explanation:
By definition free-fall is an uniform accelerate movement so speed always is increasing in fact acelertion and velocity always point in the same direction
Answer:
T h e a c c e l . e r a t i o n m a y p o i n t i n t h e o p p o s i t e d i r e c t a s t h e v e l o c i t y .
Explanation:
You are to drive to an interview in another town, at a distance of 300 km on an expressway. The interview is at 11:15 a.m. You plan to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 120 km, but then construction work forces you to slow to 42.0 km/h for 43.0 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview?
Final answer:
After accounting for a construction zone, the minimum speed needed for the rest of the journey to arrive in time for the interview is approximately 132.67 km/h to cover the remaining 137 km in about 1 hour and 2 minutes.
Explanation:
The question involves calculating the least speed needed for the rest of the trip to arrive in time for an interview after encountering a construction zone. Given that the total distance is 300 km and the interview is at 11:15 am, with a departure time of 8:00 am, we need to calculate the travel time at two different speeds and the remaining distance and time available.
First, let's calculate the time taken for the first part of the journey at 100 km/h for 120 km:
Time = Distance / Speed
Time = 120 km / 100 km/h = 1.2 hours.
Next, for the construction zone at 42 km/h for 43 km:
Time = 43 km / 42 km/h ≈ 1.024 hours.
The total time taken so far is 1.2 + 1.024 ≈ 2.224 hours. Since the student left at 8:00 am, by this time it would be approximately 10:13 am (8:00 am plus approximately 2 hours and 13 minutes).
With the interview at 11:15 am, there's 1.0333 hours (1 hour and 2 minutes) left to travel the remaining distance of 300 km - 120 km - 43 km = 137 km.
To find the minimum speed needed for the remaining distance, we use the formula: Speed = Distance / Time Required
Minimum Speed = 137 km / 1.0333 hours ≈ 132.67 km/h.
The least speed needed for the remainder of the trip is approximately 132.67 km/h.
Your grandmother places a pitcher of iced tea next to a plate of warm, freshly baked cookies so that the pitcher and the plate are touching. You tell your grandmother that the plates are in thermal contact, which means thata. heat flows w/in the warm plate but not w/in the cold pitcherb. heat flows from warm plate to cold pitcher and from cold pitcher to warm platec. heat flows from cold pitcher to warm plated. heat flows from the warm plate to the cold pitcher
Answer:
a) heat flows from the warm plate to the cold one
Explanation:
Heat is an expression of energy, the bodies with more heat mean that they have more energy. In this way, they can share or pass this "excess" of energy to the cold body. Both plates are not in total contact but some part are close enough (they are touching) that the heat can pass from one to the other. Also the heat always need a media to travel, in this case, the air that surrounds the plates is also the media that the heat travels from the hot to the cold one.
An electron with speed of 104 m/s enters a ""forbidden"" region where an electric force tries to push it back along its path with a constant acceleration of 107 m/s2 . How far will the electron go into the ""forbidden"" region? How long will it be in that region?
Answer:
The distance travelled is 151.22m and it took 0.97s
Explanation:
Well, this is an ARM problem, so we will need the following formulas
[tex]x(t)=x_{0} +v_{0} *(t-t_{0} )+0.5*a*(t-t_{0} )^{2}[/tex]
[tex]v(t)=v_{0} +a*(t-t_{0} )[/tex]
where [tex]x_{0}[/tex] is the initial position (we can assume is zero), [tex]v_{0}[/tex] is the initial speed of 104 m/s, [tex]t_{0}[/tex] is the initial time (we also assume is zero), a is the acceleration of 107 m/s2, v is speed, x is position and t is time.
Now that we have the formulas, we know that when the electron stops it has no speed. Then we calculate how much time it takes to stop.
[tex]0=104m/s-107m/s^{2} *t\\t=0.97s[/tex]
Finally, we calculate the distance travelled in this time
[tex]x(0.97s)=104m/s*0.97s+0.5*107m/s^{2}*(0.97s)^{2}=151.22m[/tex]
Answer:
Part a)
[tex]d = 5 m[/tex]
Part b)
[tex]T = 2\times 10^{-3} s[/tex]
Explanation:
Part a)
The electron will move in this forbidden region till its speed will become zero
So here we will have
[tex]v_f = 0[/tex]
[tex]v_i = 10^4 m/s[/tex]
also its deceleration is given as
[tex]a = - 10^7 m/s^2[/tex]
so we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - (10^4)^2 = 2(-10^7) d[/tex]
[tex]d = 5 m[/tex]
Part b)
Now the time till its speed is zero
[tex]v_f - v_i = at[/tex]
[tex]0 - 10^4 = -10^7 t[/tex]
[tex]t = 10^{-3} s[/tex]
so total time that it will be in the region is given as
[tex]T = 2 t[/tex]
[tex]T = 2\times 10^{-3} s[/tex]
does air resistence decrease with speed
Answer:
yes
Explanation:
because when you slow down, the resistance slows with the speed.
Answer:
Yes, air resistance decrease with speed
Explanation:
Air resistance is a kind of fluid friction that acts when objects flow through fluids. It is affected by the velocity of moving objects and the area of the objects. When an object moves with a greater velocity the air resistance acting on them will be high, so the speed decreases.
Fluid friction acts in fluids namely liquids and gases. In liquids the friction is called buoyancy and in gases it is called air resistance or drag. When two objects having the same mass but different area move through air, the object with larger area will have less velocity compared with the other object.
On a boat ride, the skipper gives you an extra-large life preserver filled with lead pellets. When he sees the skeptical look on your face, he says that you'll experience a greater buoyant force if you fall overboard than your friends who wear regular-sized Styrofoam-filled life preservers. True or False
Answer:
True.
Explanation:
He is correct but what he does not tell is that you will be drown. Your life preserver will submerge and displace more water than those of your friends who float at the surface. Although buoyant force on you will be greater , the net downward force on you will still be greater. Hence you will be drown inside the water.
In the sum A→+B→=C→, vector A→ has a magnitude of 13.6 m and is angled 40.2° counterclockwise from the +x direction, and vector C→ has a magnitude of 13.8 m and is angled 20.7° counterclockwise from the -x direction. What are (a) the magnitude and (b) the angle (relative to +x) of B→? State your angle as a positive number.
Answer:
[tex]|B|=27.00425726m[/tex]
[tex]\alpha =210.3781372[/tex]°
Explanation:
Let's use the component method of vector addition:
[tex]A_x=13.6cos(40.2)=10.38762599\\A_y=13.6sin(40.2)=8.778224553\\Cx=13.8cos(20.7+180)=-12.90912763\\Cy=13.8sin(20.7+180)=-4.877952844[/tex]
Now, we know:
[tex]C_x=A_x+B_x\\\\C_y=A_y+b_y[/tex]
So:
[tex]B_x=C_x-A_x=-23.29675362\\B_y=C_y-A_y=-13.6561774[/tex]
Now lets calculate the magnitude of the vector B:
[tex]|B|=\sqrt{(B_x)^{2} +(B_y)^{2} }=27.00425726m[/tex]
Finally its angle is given by:
[tex]\alpha =(arctan(\frac{B_y}{B_x}))+180=30.37813438+180=210.3781344[/tex]°
Keep in mind that I added 180 to the angles of C and B to find the real angles measured from the + x axis counter-clock wise.
Vector is quantity. The magnitude of vector B is 4.644 m while the angle from the positive x-direction is 302.88°.
What is Vector?A Vector is a quantity in physics that has both magnitude and direction.
We know that in order to add to vector we need to divide the vector into two parts, a sine(Vertical) and a cosine(Horizontal), therefore,
The vertical addition of the vectors A and B can be written as,
[tex]\vec A_y +\vec B_y = \vec C_y[/tex]
[tex]\vec A(Sin\ \theta_A) +\vec B(Sin\ \theta_B) = \vec C(Sin\ \theta_C)[/tex]
[tex]13.6(Sin\ 40.2^o) +\vec B(Sin\ \theta_B) = 13.8(Sin\ 20.7^o)\\\\\vec B(Sin\ \theta_B ) =-3.9[/tex]
The Horizontal addition of the vectors A and B can be written as,
[tex]\vec A_x +\vec B_x = \vec C_x[/tex]
[tex]\vec A(Cos\ \theta_A) +\vec B(Cos\ \theta_B) = \vec C(Cos\ \theta_C)[/tex]
[tex]13.6(Cos\ 40.2^o) +\vec B(Cos\ \theta_B) = 13.8(Cos\ 20.7^o)\\\\\vec B(Cos\ \theta_B ) =2.5215[/tex]
As the value of Sin is negative and the value of Cos is positive, therefore, Vector B will lie in the fourth quadrant.
The angle of Vector B,
[tex]\dfrac{\vec B\ Sin\ \theta_B}{\vec B\ Cos\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\\dfrac{\ Sin\ \theta_B}{\ Cos\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\Tan\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\\theta_B = Tan^{-1}\ \dfrac{-3.9}{2.5215}\\\\\theta_B = -57.12^o[/tex]
Thus, the angle of vector B is 57.12° clockwise from the -x direction.
In order to make the angle positive, we can deduct the value from 360°,
[tex]\theta_B = -57.12^o\\\\\theta_B = 360^o -57.12^o\\\\\theta_B = 302.88^o[/tex]
The magnitude of Vector B,
We know the value of the Perpendicular component of vector B,
[tex]\vec B(Sin\ \theta_B ) =-3.9\\\\\vec B(Sin\ -57.12^o ) =-3.9\\\\\vec B= 4.644\rm\ m[/tex]
Hence, the magnitude of vector B is 4.644 m while the angle from the positive x-direction is 302.88°.
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A plane starting from rest (vo = 0 m/s) when t0 = 0s. The plane accelerates down the runway, and at 29 seconds, its velocity is +72.2 m/s. (+ indicates to the right). Determine the average acceleration for the plane.
Answer:
Acceleration, [tex]a=2.48\ m/s^2[/tex]
Explanation:
Given that,
The plane is at rest initially, u = 0
Final speed of the plane, v = 72.2 m/s
Time, t = 29 s
We need to find the average acceleration for the plane. It can be calculated as :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{72.2}{29}[/tex]
[tex]a=2.48\ m/s^2[/tex]
So, the average acceleration for the plane is [tex]2.48\ m/s^2[/tex]. Hence, this is the required solution.
Two objects are thrown vertically upward, first one, and then, a bit later, the other. Is it (a) possible or (b) impossible that both objects reach the same maximum height at the same instant of time?
Answer:
No, it is impossible
Explanation:
Kinematics equation:
[tex]Vf^{2} =Vo^{2} -2gy[/tex]
if height is maximum:
y=H and Vf=0
so:
[tex]Vo^{2} =2gH[/tex][tex]H=Vo^{2} /2g[/tex]Analysis: From the last equation we see that the maximum height depends ONLY on the initial speed. This means that if both objects reach the same maximum height, then they necessarily need to have the SAME initial velocity. If they have the same initial velocity and in order to reach the maximum height at the SAME time the only way is that they are released at the SAME TIME.
An operation has a 20 percent scrap rate. As a result, 80 pieces per hour are produced. What is the potential labor productivity that could be achieved by eliminating the scrap?
Answer:
25%
Explanation:
If there is 20% scrap, then
80% of (production + scrap) = 72 pcs/hr
Now,
[tex]\frac{80}{100} \times (72+scrap)= 72 pcs/hr\\0.80\times (72+Scrap) = 72 \\ 72+Scrap = \frac{72}{0.80}\\72 + Scrap = 90\\Scrap = 18 pcs /hr[/tex]
The percent increase in labor productivity is, [tex]\frac{18}{72} =0.25[/tex].
Or it can be written as 25% of which would give 80 pieces per hour.
Water leaves a fireman’s hose (held near the ground) with an initial velocity v0= 22.5 m/s at an angle θ = 28.5° above horizontal. Assume the water acts as a projectile that moves without air resistance. Use a Cartesian coordinate system with the origin at the hose nozzle position.
a) Using v0, θ, and g, write an expression for the time, tmax, the water travels to reach its maximum vertical height.
b) At what horizontal distance d from the building base, where should the fireman place the hose for the water to reach its maximum height as it strikes the building? Express this distance, d, in terms of v0, θ, and g.
Water leaves a fireman’s hose (held near the ground) with an initial velocity v0= 22.5 m/s at an angle θ = 28.5° above horizontal, the horizontal distance d from the building base where the fireman should place the hose for the water to reach its maximum height as it strikes the building is given by [tex]\( \frac{v_0^2 \cdot \sin(2\theta)}{g} \)[/tex]
a) To find the time [tex]\( t_{\text{max}} \)[/tex] that the water travels to reach its maximum vertical height, we can use the following kinematic equation for vertical motion:
[tex]\[ v_y = v_{0y} - g \cdot t \][/tex]
Where:
[tex]\( v_y \)[/tex] is the vertical component of velocity at time t.
[tex]\( v_{0y} \)[/tex] is the initial vertical component of velocity (which is [tex]\( v_0 \cdot \sin(\theta) \)[/tex])
g is the acceleration due to gravity
At the maximum height, the vertical component of velocity becomes zero, so we can set [tex]\( v_y = 0 \)[/tex] and solve for [tex]\( t_{\text{max}} \)[/tex]:
[tex]\[ 0 = v_{0y} - g \cdot t_{\text{max}} \][/tex]
[tex]\[ t_{\text{max}} = \frac{v_{0y}}{g} \][/tex]
Substituting [tex]\( v_{0y} = v_0 \cdot \sin(\theta) \)[/tex], we get:
[tex]\[ t_{\text{max}} = \frac{v_0 \cdot \sin(\theta)}{g} \][/tex]
b) To determine the horizontal distance d from the building base where the fireman should place the hose for the water to reach its maximum height as it strikes the building, we need to consider the horizontal motion of the water.
The horizontal distance d can be calculated using the following equation for horizontal motion:
[tex]\[ d = v_{0x} \cdot t_{\text{max}} \][/tex]
Where:
[tex]\( v_{0x} \)[/tex] is the initial horizontal component of velocity (which is [tex]\( v_0 \cdot \cos(\theta) \)[/tex])
[tex]\( t_{\text{max}} \)[/tex] is the time it takes to reach the maximum height (calculated in part a)
Substituting [tex]\( v_{0x} = v_0 \cdot \cos(\theta) \)[/tex] and [tex]\( t_{\text{max}} = \frac{v_0 \cdot \sin(\theta)}{g} \)[/tex], we get:
[tex]\[ d = (v_0 \cdot \cos(\theta)) \cdot \left(\frac{v_0 \cdot \sin(\theta)}{g}\right) \][/tex]
Simplifying:
[tex]\[ d = \frac{v_0^2 \cdot \sin(\theta) \cdot \cos(\theta)}{g} \][/tex]
This can be further simplified using the trigonometric identity [tex]\( \sin(2\theta) = 2 \sin(\theta) \cdot \cos(\theta) \)[/tex]:
[tex]\[ d = \frac{v_0^2 \cdot \sin(2\theta)}{g} \][/tex]
Thus, the horizontal distance d from the building base where the fireman should place the hose for the water to reach its maximum height as it strikes the building is given by [tex]\( \frac{v_0^2 \cdot \sin(2\theta)}{g} \)[/tex].
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The time, tmax, the water takes to reach its maximum vertical height can be calculated using the formula tmax = v0sin(θ) / g. The horizontal distance, d, from the building base where the fireman should place the hose can be found using the formula d = v0cos(θ)tmax.
Explanation:To find the time, tmax, the water travels to reach its maximum vertical height, we can use the formulatmax = v0sin(θ) / g
Where v0 is the initial velocity (22.5 m/s), θ is the angle above horizontal (28.5°), and g is the acceleration due to gravity (9.8 m/s^2).
Substituting the values into the formula, we get:
tmax = (22.5 m/s)sin(28.5°) / 9.8 m/s^2
Solving this equation will give us the value for tmax.
To determine the horizontal distance, d, from the building base where the fireman should place the hose, we can use the formula:
d = v0cos(θ)tmax
Where v0 is the initial velocity, θ is the angle above horizontal, and tmax is the time calculated in part (a).
Substituting the values into the formula, we get:
d = (22.5 m/s)cos(28.5°)tmax
This equation will give us the value for d in terms of v0, θ, and g.
A car traveling at a speed of v can brake to an emergency stop in a distance x , Assuming all other driving conditions are all similar , if the traveling speed of the car double, the stopping distance will be (1) √2x,(2)2x, or(3) 4x:(b) A driver traveling at 40.0km/h in school zone can brake to an emergency stop in 3.00m. What would be braking distance if the car were traveling at 60.0 km/h?
Answer:
D = 6.74 m
Explanation:
Let the f newtons be the braking force
distance to stop = x meters
speed = v m/s
we know that work done is given as
work done W = fx joules
[tex]fx = \frac{1}{2} mv^2[/tex]
If the speed is doubled ,
[tex]fx' = \frac{1}{2} m(2v)^2[/tex]
[tex] = 4[\frac{1}{2}mv^2] = 4fx[/tex]
stooping distance is D = 4x
[tex]40km/h = \frac{40*1000}{60*60} = 11.11 m/s[/tex]
60 km/hr = 16.66 m/s
braking force = f
[tex]f*3 = [\frac{1}{2}mv^2] = [\frac{1}{2}m*11.11^2][/tex]
[tex]f*D = [\frac{1}{2}m*16.66^2][/tex]
[tex]\frac{D}{3} = \frac{16.66^2}{11.11^2}[/tex]
[tex]\frac{D}{3} = 2.2489[/tex]
D = 6.74 m
An object is dropped from the top of a cliff 640 meters high. Its height above the ground t seconds after it is dropped is 640−4.9t^2. Determine its speed 44 seconds after it is dropped.
Answer:
v = -431.2 m/s
Explanation:
Given that,
Initial position of the object, [tex]x=640-4.9t^2[/tex]
Let v is its speed 44 second after it is dropped. The relation between the speed and the position is given by :
[tex]v=\dfrac{dx}{dt}[/tex]
[tex]v=\dfrac{d(640-4.9t^2)}{dt}[/tex]
[tex]v=-9.8t[/tex]
Put t = 44 seconds in above equation. So,
[tex]v=-9.8\times 44[/tex]
v = -431.2 m/s
So, the speed of the ball 44 seconds after it is dropped is 431.2 m/s and it is in moving downwards.
Final answer:
The speed of the object 44 seconds after being dropped from a 640-meter-high cliff is 431.2 m/s. This is found by differentiating the height function to get the velocity function and substituting the time into the velocity equation.
Explanation:
To determine the speed of an object 44 seconds after it is dropped from the top of a 640-meter-high cliff, you can differentiate the height function to find the velocity function.
Given the height formula h(t) = 640 - 4.9t², we get the velocity function v(t) = h'(t) = -9.8t by differentiation. At t = 44 seconds, the speed is simply the absolute value of the velocity. So, we calculate: v(44) = -9.8 * 44
This yields a velocity of -431.2 m/s, and since speed is the magnitude of velocity, the speed is 431.2 m/s.
Which of the following are examples of 1-D motion?
a) A person pacing back and forth down a hallway in a straight line
b) An airplane traveling from Boise to Seattle
c) A student walking from Boone to KAIC
d) A car driving on a straight road
Answer: a, d
Explanation:
A- eaven if the direction changes it's still 1D
B- an airplane always needs more than 1D for take off and landing
C- it isn't likely to get there by an straight way without any land relief
D- a car in a straight road its 1D if the load has no land relief.
A 500-kg object is accelerating to the right at 10 cm/s2. What is the magnitude of the rightward net force acting on it (in Newtons)?
Answer:
F = 50 N
Explanation:
As we know by Newton's II law
{tex]F = ma[/tex]
force is product of mass and acceleration
so we will have
[tex]m = 500 kg[/tex]
[tex]a = 10 cm/s^2[/tex]
[tex]a = 0.10 m/s^2[/tex]
now we have
[tex]F = 500 \times 0.10 [/tex]
[tex]F = 50 N[/tex]
The magnitude of the rightward net force acting on the object is 50 N.
What is force?Force can be defined as the product of mass and acceleration.
To calculate the magnitude of the rightward net force, we use the formula below.
Formula:
F = ma............. Equation 1Where:
F = Force on the objectm = mass of the objecta = acceleration due to gravityFrom the question,
Given:
m = 500 kga = 10 cm/s² = 0.1 m/s²Substitute these values into equation 1
F = 500(0.1)F = 50 NHence, The magnitude of the rightward net force acting on the object is 50 N.
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An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 14.0 m from time t = 1.00 s to time t = 2.00 s. What is the acceleration of the object?a) 5.20 m/s^2 b) 10.4 m/s^2 c) 8.67 m/s^2 d) 6.93 m/s^2 e) 12.1 m/s^2
Answer:
The acceleration of the object is 9.3 m/s²
Explanation:
For a straight movement with constant acceleration, this equation for the position applies:
x = x0 + v0 t + 1/2 a t²
where
x = position at time t
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
we have two positions: one at time t = 1 s and one at time t = 2 s. We know that the difference between these positions is 14.0 m. These are the equations we can use to obtain the acceleration:
x₁ = x0 + v0 t + 1/2 a (1 s)²
x₂ = x0 + v0 t + 1/2 a (2 s)²
x₂ - x₁ = 14 m
we know that the object starts from rest, so v0 = 0
substracting both equations of position we will get:
x₂ - x₁ = 14
x0 + v0 t + 1/2 a (2 s)² - (x0 + v0 t + 1/2 a (1 s)²) = 14 m
x0 + v0 t + 2 a s² - x0 -v0 t - 1/2 a s² = 14 m
2 a s² - 1/2 a s² = 14 m
3/2 a s² = 14 m
a = 14 m / (3/2 s²) = 9.3 m/s²
The constant acceleration of the object is 9.33 m/s².
The given parameters;
distance traveled by the object, s = 14 minitial time of motion of the object, t = 1 sfinal time of motion of the object, t = 2.0sThe acceleration of the object is calculated by applying the second kinematic equation as follows;
s = ut + ¹/₂at²
where;
u is the initial velocity of the object = 0at t = 1.0 s
x₁ = 0 + ¹/₂(1²)a
x₁ = ¹/₂(a)
at t = 2.0 s
x₂ = 0 + ¹/₂(2²)a
x₂ = 2a
The change in position;
Δx = x₂ - x₁ = 14
14 = 2a - ¹/₂(a)
[tex]14 = \frac{3a}{2} \\\\3a = 2(14)\\\\3a = 28\\\\a = \frac{28}{3} = 9.33 \ m/s^2[/tex]
Thus, the constant acceleration of the object is 9.33 m/s².
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