Match the variables with its corresponding unit
Answer:
1. s
2. m
3. start m/s
4. end m/s
5. m/s^2
Explanation:
1. t is the suffix for the time and it is related to seconds [s]
2. d is the suffix for distance and it is related to meters [m]
3. Vi is the suffix for the initial velocity and it is related to meters per second [m/s]
4. Vf is the suffix for the final velocity and it is related to meters per second [m/s]
5. a is the suffix for the acceleration and it is related to meters per squared second [m/s^2]
What do all elements in a column in the periodic table have in common? (2 points)
Oa
Their atoms have the same number of valence electrons.
Ob
Their atoms have the same atomic mass.
Their atoms have the same number of protons and electrons.
Od
Their atoms have the same number of neutrons.
Answer:
You mean in periods.
They have the same number of elctron shells.
But if they are in the same group then they have the sane number oifValence electrons.
Explanation:
which energy is primarily responsible for phenomena such as movement of large plates and volcanoes?
Answer:
It should be geothermal energy
How long would it take to travel 50 km traveling at a speed of 10 km/hr
Answer:
it would take 5 hours
Explanation:
plz vote this up i promise its correct
What measurement do geologist use to find the absolute age of once living things
Answer:
Radiocarbon dating
Explanation:
Activity Problem 8.3 This problem asks you to compare the finite current element approximation to the infinite straight wire approximation for the isolated segment of wire carrying current of 2A to the right. The wire is 8cm long. Note, the finite current approximation will be a fairly poor approximation to the actual field which would be found by integration; the infinite wire field is a better approximation in this case. (a)Compute the magnetic field at point C a distance 2cm for the center of the wire using the infinite straight wire approximation. (b)Compute the magnetic field at point C using the finite current element approximation. (c)Compute the magnetic field at point E a distance 2cm from the end using the finite current element approximation.
Answer: (a) B = 2 x 10⁻⁵T
(b) B = 1.94 x 10⁻⁵T
(c) B = 1.8 x 10⁻⁴T
Explanation: A magnetic field due to a current passing through a straight wire is calculated using the Biot-Savart Law:
[tex]dB=\frac{\mu_{0}}{4.\pi} \frac{IdLXR}{r^{3}}[/tex]
where
dL is current length element
[tex]\mu_{0}[/tex] is permeability of free space ([tex]4.\pi.10^{-7}[/tex]T.m/A)
(a) For a infinite straight wire:
[tex]B=\frac{\mu_{0}I}{2.\pi.R}[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}}[/tex]
B = 2x10⁻⁵T
For an infinite, long and straight wire, magnetic field is 2x10⁻⁵T.
(b) For a finite wire:
[tex]B=\frac{\mu_{0}I}{2.\pi.R}\frac{L}{\sqrt{L^{2}+R^{2}} }[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}} \frac{8.10^{-2}}{\sqrt{(8.10^{-2})^{2}+(2.10^{-2})^{2}} }[/tex]
B = 1.94x10⁻⁵T
The magnetic field for a finite wire in the same conditionsas infinite wire is 1.94x10⁻⁵T.
(c) For a finite wire at a point distant from the end of the wire:
[tex]B=\frac{\mu_{0}I}{4.\pi.L\sqrt{2} }[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{4.\pi.8.10^{-2}\sqrt{2} }[/tex]
B = 0.18x10⁻⁵T
At a point at the end, magnetic field is 1.8x10⁻⁴T.
1. What are the two (2) classes of matter?
A. Atom and molecule
B. Element and compound
C. Plant and animal cells
D. Substance and mixture
2. Which is true regarding water?
A. Compound of element
B. Element of gases
C. Mixture of
gases
D. Substances of gases
3. What is the difference between a substance and a mixture?
A. Substance can be separated while mixture cannot be separated.
B. Substance has many components while mixture has one only.
C. Substance is heterogeneous while mixture is homogeneous.
D. Substance is pure while mixture is impure.
Answer:
substance and mixture
8. What is the mass of an object if a force of 34 N produces an acceleration of 4 m/s/s?
F = 34 N
F
34
m-
m-
m = 8.5 kg
a
4
a - 4 m/s
Answer:
The answer is 8.5 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
where
f is the force
a is the acceleration
So we have
[tex]m = \frac{34}{4} = \frac{17}{2} \\ [/tex]
We have the final answer as
8.5 kgHope this helps you
How can you tell if hair was forcibly removed?
Answer:Explanation:
A microscopic hair examination can also determine if a hair was forcibly removed, artificially treated or diseased. A comparison microscope can be used to compare a questioned hair to a known hair sample in order to determine if the hairs are similar and if they could have come from a common source.
5 Which has the greatest average speed?
A. a boat sailing 80 km in 2 hours
B. a car driving 90 km in 3 hours
c. a train traveling 120 km in 3 hours
D. a truck moving 50 km in 1 hour
a train cuz its 120km in 3 hours
how much ume does it take to use 300 W of power to do 1800 J of Work?
Answer:
P=300 W
W=1800 J
P=W/t
t=W/P
t=6 s
What is the resultant velocity of a plane that is traveling at 245 m/s North and encounters a tailwind of 55 m/s North?
Answer:
b
Explanation:
You are studying circular motion by placing pennies on a turntable and then turning it on so that it will spin. You keep increasing the speed until one of the pennies slips off. You repeat this procedure and observe that the penny close to the outer edge always slip off first. What is the best inference?
1.The penny near the edge has a greater tangential velocity than the one in the center, so it experiences more air resistance. It’s the effect of the air “blowing” it off.
2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.
3.The centrifugal force acting on the pennies is stronger on the one near the edge than the one near the center.
Answer:
2.The centripetal force required to keep the pennies in place increases with the distance from the center. Eventually, as the turntable spins faster, the friction force between the turntable and the penny near the edge is not enough to supply the required centripetal force.
Explanation:
centripetal force = m ω² R
here m is mass , ω is angular velocity and R is distance of penny from centre
So this force depends upon R
penny on the outer edge will require greater centripetal force to move in circular path .
The centripetal force will be provided by frictional force of table which is same for both the coin . Hence the penny on the outer edge will slip off first the moment , frictional force reach its maximum value for it . But it will be sufficient to keep in balance the penny nearer to the centre .
If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with speed vo along the same line from the opposite direction toward the source, then the frequency of the sound heard by the observer is fo = c + vo c − vs fs where c is the speed of sound, about 332 m/s. (This is the Doppler effect.) Suppose that, at a particular moment, you are in a train traveling at 45 m/s and accelerating at 1.1 m/s2. A train is approaching you from the opposite direction on the other track at 46 m/s, accelerating at 1.6 m/s2, and sounds its whistle, which has a frequency of 459 Hz. At that instant, what is the perceived frequency that you hear? (Round your answer to one decimal place.) Hz How fast is it changing? (Round your answer to two decimal places.) Hz/s
Answer:
457.81 Hz
Explanation:
From the question, it is stated that it is a question under Doppler effect.
As a result, we use this form
fo = (c + vo) / (c - vs) × fs
fo = observed frequency by observer =?
c = speed of sound = 332 m/s
vo = velocity of observer relative to source = 45 m/s
vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).
fs = frequency of sound wave by source = 459 Hz
By substituting the the values to the equation, we have
fo = (332 + 45) / (332 - (-46)) × 459
fo = (377/ 332 + 46) × 459
fo = (377/ 378) × 459
fo = 0.9974 × 459
fo = 457.81 Hz
Suppose you are standing on a scale in an elevator that is accelerating upward. Will the scale read your weight as larger for smaller than the weight it reads when you are stationary?
Answer:
following are the solution to this question:
Explanation:
When I stand at such a scale in an elevated that's already rising upwards, its scale would appear to also be 0 because of free fall and would often reveal that weight whenever the lift is stable.
In this, the free fall is also known as the object, that is influenced exclusively by gravity, and an object operating only through the influence of gravity is said to be in a free-fall state.
A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.64 s. What was the average acceleration during this run?
Answer:
45.89m/s²
Explanation:
Given
Distance S = 305m
Time t = 3.64s
To get the acceleration during this run, we will apply the equation of motion:
S = ut+1/2at²
Substitute the given parameters into the formula and calculate the value of a
305 = 0+1/2 a(3.64)²
304 = 1/2(13.2496)a
304 = 6.6248a
a = 304/6.6248
a = 45.89m/s²
Hence the average acceleration during this run is 45.89m/s²
If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s than the recoil speed of the cannon will be *
A. 0.2 m/s
B. 2 m/s
C. 4 m/s
D. 10 m/s
Answer:
D. 10 m/s
Explanation:
Using the standard convention "positive is up", what is the sign of the acceleration of a rock you throw up (a) as it is going up, and (b) as it is coming back down?
Answer:
(a) Negative
(b) Positive
Explanation:
Here is a question on sign.
(a) When the rock is going up, it is working against gravity. In this case, the acceleration due to gravity will take a negative value. This is because gravity works or acts downwards
(b) When the rock is coming down, it is actually working with gravity. So in this case, the sign will ne positive
Goal posts at the ends of football fields are padded as a
safety measure for players who might run into them. How
does thick padding around the goal post reduce injuries
to players?
It would be awesome if you could use a sample response
Goal posts at the end of football field are padded as a safety measure to players who might run into them,these thick paddings of the goal posts are done to prevent injuries to the players because the padding of the goal post increases the time of impact and this in-turn decreases the force on the the players body or the force it will have when a player's body hit against it and therefore decreasing the force applied to bring the player to a stop.
We can also say that it increases the time of collision between the player's body and the post.
An elephant pushes with 2000 N on a load of trees. It then
pushes these trees for 150 m. How much work did the
elephant do?
Answer:
13.3
Explanation
work= force/area
2000/150
A Carnot engine operates with a cold reservoir at a temperature of 455 K and a hot reservoir at a temperature of 619 K. What is the net entropy change (in J/K) as it goes through a complete cycle? Round your answer to the nearest whole number.
Answer:
Assuming that 4000 J of heat transfer occurs from it the change in entropy will be=2.33j/k
Explanation:
Given data
Th=619 K
Tc= 455K
We are going to assume that 4000 J of heat transfer occurs from it, since it was not specified in the question.
we know that the change in entropy is given as
ΔStotal=ΔShot+ΔScold .
ΔShot=−Q/Th=−4000J/619K=−6.46J/K
For the cold reservoir,
ΔScold=Q/Tc=4000J/455K=8.79J/K
therefore the total is
ΔStotal=ΔShot+ΔScold
=(−6.46+8.79)J/K
=2.33j/k
What is the relation between the weight of a body and acceleration due to gravity?
Answer:
The weight W, or gravitational force, is then just the mass of an object times the gravitational acceleration. Since the gravitational constant (g) depends on the square of the distance from the center of the earth, the weight of an object decreases with altitude.
Explanation:
please mark as brainlist
A 495-kg dragster accelerates from rest to a final speed of 105 m/s in 395 m, during which it encounters an average frictional force of 1400 N. show answer No Attempt 50% Part (a) What is its average power output of the dragster, in watts, if this takes 8.2 s
Answer:
Explanation:
According to energy conservation which states that the workdone is equal to change in the system
Workdone = change in kinetic energy + (frictional force * distance)
Workdone = ΔK + fd
Workdone = kf-Ki + fd
Workdone = = 1/2(m(v-u)^2) + fd
Given
Mass m = 495kg
final velocity v = 105m/s
initial velocity = 0m/s
Force f= 1400N
distance d = 395m
Substitute
Workdone = 1/2(495(105-0)^2) + 1400(395)
Workdone = 2,728,687.5+553000
Workdone = 3,281,687.5 Joules
Time = 8.2secs
Power output = Workdone/Time
Power output = 3,281,687.5/8.2
Power output = 885,766.768
Power output = 8.858 * 10^5 watts
The elements in Groups 3 through 12 of the periodic table are the ______.
Answer:
Transition Metals
Explanation:
__________________________
a simple machine makes our work easier and faster. why?
Answer:
simple machines such as ramps lessen the moment required to do work. if a triangle has a base of 5 and the height is 7, a ramp would make the hypotenuse of this triangle lessoning the total distance. using a²+b²=c² 25+49=c² 74≈8.6 and it is obvious that 8.6 is less than 12 in every unit. other simple machines such as pulleys make it lighter making it simply easier for an object to be lifted.
Explanation:
Answer:
Simple machines are useful because they reduce effort or extend the ability of people to perform tasks beyond their normal capabilities. Simple machines that are widely used are the wheel and axle, pulley, inclined plane, screw, wedge and lever.
as a result, the net electric force experienced by each negatively charged particle is reduced to F/2. The value of q is
Answer:
The value of q is [tex]\dfrac{Q}{8}[/tex]
Explanation:
Given that,
Each charge = -Q
Distance between charges = L
Reduced force = [tex]\dfrac{F}{2}[/tex]
Suppose, Two particles each with a charge -Q are fixed a distance L apart as shown above. Each particle experiences a net electric force F. A particle with a charge +q is now fixed midway between the original two particles.
We know that,
The force on each end is
[tex]F=\dfrac{kQ^2}{L^2}[/tex]...(I)
If the charge q is placed at mid point then
The force on each end charge is
[tex]\dfrac{F}{2}=F+F'[/tex]....(II)
We need to calculate the value of q
Using equation (II)
[tex]\dfrac{F}{2}=F+F'[/tex]
Put the value of F into the formula
[tex]\dfrac{\dfrac{kQ^2}{L^2}}{2}=k\dfrac{Q^2}{L^2}+k\dfrac{q\times(-Q)}{(\dfrac{L}{2})^2}[/tex]
[tex]\dfrac{kq(-Q)}{(\dfrac{L}{2})^2}=-\dfrac{kQ^2}{2L^2}[/tex]
[tex]\dfrac{q}{\dfrac{1}{4}}=\dfrac{Q}{2}[/tex]
[tex]q=\dfrac{Q}{8}[/tex]
Hence, The value of q is [tex]\dfrac{Q}{8}[/tex]
What is psychoanalysis
The higher something is off the ground, the more
potential energy it has.
True
False
Colored lights are called additive colors. Why do you think this is so?
You release a ball from rest at the top of a ramp. 6 s later it is moving at 4.0
m/s. What is the acceleration? (in meters per second squared) *
Your answer
[tex]a = \frac{vf - vi}{t} [/tex]
here initial velocity vi=0 as ball release from rest
the final velocity is vf=4.0
time is t=6
so putting all these values in above equation
[tex]a = \frac{ 4.0- 0}{6} [/tex]
[tex]a = 0.6667m \s {}^{2} [/tex]
