Answer:
D
Explanation:
A studious physics student is interrupted by a swarm of bees and chased off a cliff. Since she has her calculator in hand she quickly punches in numbers to figure out the initial velocity she needs to make it into the lake below. The cliff is 10 m high and the lake is 15 m away from the edge of the cliff. Find the time it takes her to drop. Find her initial velocity,
Answer:
The time is 1.4 sec
The initial velocity is 10.7 m/s.
Explanation:
Given that,
Height = 10 m
Distance = 15 m
We need to calculate the time
Using equation of motion
[tex]s=ut-\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]10=0+\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]t^2=\dfrac{2\times10}{9.8}[/tex]
[tex]t=\sqrt{\dfrac{2\times10}{9.8}}[/tex]
[tex]t=1.4\ sec[/tex]
We need to calculate the initial velocity
Using formula of velocity
[tex]v=\dfrac{d}{t}[/tex]
Put the value into the formula
[tex]v=\dfrac{15}{1.4}[/tex]
[tex]v=10.7\ m/s[/tex]
Hence, The time is 1.4 sec
The initial velocity is 10.7 m/s.
A dog has a mass of 60kg and an acceleration of 2m/s/s. What is the force of the dog?
1. What is Ohm"s law?
2. If you placed a negatively charged hairbrush near your hair, what charge would your hair be?
3. You must change a lightbulb and the new lightbulb has a larger resistance. If the voltage of the battery does not change, what happens to the current going through the flashlight?
HELLPPPP
1. Ohm's law shows the relationship between:
voltagecurrentresistanceFormula: voltage = current x resistance
2. The negative charge on the hairbrush will induce a positive charge on your hair. As a result, your hair is going to be attracted to the hairbrush (and repelled by other strands of hair.)
3. V = IR, so if the resistance of the current increases, and the voltage of the current stays the same, there is as a result, going to be less current.
Best of Regards!
Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion is reversible but not isothermal and that the temperature decreases as the expansion proceeds. (a) Find an expression
Answer:
(a) The work done by the gas on the surroundings is, 17537.016 J
(b) The entropy change of the gas is, 73.0709 J/K
(c) The entropy change of the gas is equal to zero.
Explanation:
(a) The expression used for work done in reversible isothermal expansion will be,
where,
w = work done = ?
n = number of moles of gas = 4 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = 240 K
= initial volume of gas =
= final volume of gas =
Now put all the given values in the above formula, we get:
The work done by the gas on the surroundings is, 17537.016 J
(b) Now we have to calculate the entropy change of the gas.
As per first law of thermodynamic,
where,
= internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
Thus, w = q = 17537.016 J
Formula used for entropy change:
The entropy change of the gas is, 73.0709 J/K
(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.
As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0
So, from this we conclude that the entropy change of the gas must also be equal to zero.
Explanation:
7N
6
→ 2N
Net Force:
Determine the net force acting on the object.
Answer:
When a force is applied to the body, not only is the applied force acting, there are many other forces like gravitational force Fg, frictional force Ff and the normal force that balances the other force. Therefore, the net force formula is given by, FNet = Fa + Fg + Ff + FN.
Explanation:
A soccer ball accelerates from rest and rolls 6.5m down a hill in 3.1 s. It then bumps into a tree. What is the speed of the ball just before it hits the tree.
Answer:
2.096m/s
Explanation:
The speed of this soccer ball can be calculated using the formula;
Speed = distance/time
According to this question, the distance of the ball before it hits the tree is 6.5m, the time it takes is 3.1s, hence;
Speed = 6.5/3.1
Speed of the ball = 2.096m/s
Therefore, the speed of the ball before hitting the tree is 2.096m/s
QUESTION 2
At the end of an investigation, what should be done if the results do NOT support the hypothesis?
O A Repeat your investigation to ensure your results are accurate and then modify your hypothesis if necessary.
OB. Repeat your investigation over and over again until you get the results that will support your original hypothes
O C. Check your measurement tools to ensure they are working.
OD. Change the topic of your investigation to one that will yield results that support a hypothesis.
Answer:
A. Repeat your investigation to ensure your results are accurate and then modify your hypothesis if necessary.
Explanation:
Having results that do not support the hypothesis is a common occurrence.
Hypotheses always depend on the data and experiment. If at the end of an investigation the results do not support the hypothesis, the investigation should be be repeated to further confirm this discovery.
And if there is still no correlation, then the hypothesis is not a reasonable explanation for the investigation and should be modified or rejected if necessary.
g An angry rhino with a mass of 2700 kg charges directly toward you with a speed of 3.70 m/s. Before you start running, as a distraction, you throw a 0.180 kg rubber ball directly at the rhino with a speed of 9.05 m/s. Determine the speed of the ball (in m/s) after it bounces back elastically toward you.
Answer:
9.05m/s
Explanation:
given data
m1= 2700kg
v1=3.7m/s
m2=0.18kg
v2=9.05m/s
v3=?
We know that the velocity of the rhino will remains unchanged after impact as the mass of the rubber ball is negligible
m1v1+m2v2=m1v1+m2v3
2700*3.7+0.18*9.05=2700*3.7+0.18*v3
9990+1.629=9990+0.18v3
9991.629-9990=0.18v3
1.629=0.18v3
v3=1.629/0.18
v3=9.05m/s
Research has shown that this type of interview is the most effective in predicting later job
performance.
Answer:
Situational Interview
Explanation:
A situational interview is about as close to the real job as it gets. During this type of interview, candidates may be presented with a visual or audio simulation of a scenario and asked to respond to it. They are asked to analyze a problem and profer suggestions on how they would handle it.
If the candidate has solved similar problems in the past, it will come to the fore.
If they haven't then the best outcome is that it will tell the interviewers how well the candidate is able to solve similar problems.
An example of a Situational Interview question is this:
How would you handle an angry customer who for no justifiable reason has decided to create a problematic scene to disrupt the business?
Because Situational Interviews are about behavioral responses (present, past, and future), they are powerful tools in determining or predicting future job performance. An interviewing technique that is developed using this methodology is called the S.T.A.R.
This is an acronym for Situation, Task, Action, Result.
Situation: the candidate is asked to present a challenging situation that occurred recently. This tests what the candidate sees as a challenging situation.
Task: The candidate based on the situation is asked to identify what they need to do to remedy the problem. This tells the interviewer(s) whether or not the candidate can think up a solution for the problem.
Action: Here they define the actual steps taken to resolve the problem
Result: The candidate against the above is required to give the result gotten
Action and Result tell the interviewer the quality of the candidate's ability to follow through and achieve the intended results. This also judges the quality of execution in terms of cost and time. The candidate with the lowest cost and time and the highest quality of outcome is considered the best.
Cheers
The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm
Answer:
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
Explanation:
The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:
[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]
[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.
[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.
And we expand the equation above by definitions of elastic potential energy and kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.
[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.
If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:
[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]
[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east. what is the distance and displacement
Explanation:
Let east = E, and, west = opposite to east = - E.
Here, displacement:
=> 2m east + 4m west + 1m east
=> 2E + 4(-E) + 1E
=> 2E - 4E + 1E
=> - 1E
=> 1(-E)
=> 1m west
And, distance,
=> 2m + 4m + 1m = 7m
The distance of a person is 7 m and the displacement of the person is 1m west.
To find the distance and displacement, the given values are,
A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east.
What is the distance and the displacement?Displacement:
The displacement is shortest distance between initial and final position or we can say it is the straight line distance between initial and final position.If object moves in a straight line path without any turn then the path length and the displacement is always same.Distance:
The distance is the total path length of the object while it will move from initial to final position.If the object move on curved path then displacement is smaller than the distance moved by the object.Let us consider East = E and west = opposite to east = - E.
Calculating the displacement:
= 2m east + 4m west + 1m east
= 2E + 4(-E) + 1E
= 2E - 4E + 1E
= - 1E
= 1(-E)
= 1m west.
The displacement is 1m west.
Now calculating the distance,
= 2m + 4m + 1m
= 7m
The distance is 7m.
Thus, the displacement and the distance is found as 1 m west and 7m.
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Silly Goose falls 1.0 m to the floor. How long does the fall take
Answer:You need to give more explanation sorry
Explanation:
Answer:
4.20 seconds
Explanation:
Supposing that silly goose weighs 69 pounds, we need to start on the math.
Simple maths, truly and really. 69/1=69, of course.
Therefore it will take 4.20 seconds for silly goose to hit the ground. if he is going to be a silly goose though, he can just go in the pond, instead of wasting his time.
plzzz helppp
You are pushing a box North in the hallway, at 20n, and a friend gets in front of the box and goes in the opposite direction, at 30n. What direction is the box going at? How much force does the box have going in that direction?
Answer:
the box is going south at 10n
Explanation:
Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s
Answer:
The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]
Explanation:
From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:
[tex]v = r\cdot \omega[/tex] (Eq. 1)
Where:
[tex]r[/tex] - Radius of rotation of the particle, measured in meters.
[tex]\omega[/tex] - Angular velocity, measured in radians per second.
[tex]v[/tex] - Linear velocity of the point, measured in meters per second.
But we know that angular velocity is also equal to:
[tex]\omega = \frac{\theta}{t}[/tex] (Eq. 2)
Where:
[tex]\theta[/tex] - Angular displacement, measured in radians.
[tex]t[/tex] - Time, measured in seconds.
By applying (Eq. 2) in (Eq. 1) we get that:
[tex]v = \frac{r\cdot \theta}{t}[/tex] (Eq. 3)
From Geometry we must remember that circular arc ([tex]s[/tex]), measured in meters, is represented by:
[tex]s = r\cdot \theta[/tex]
[tex]v = \frac{s}{t}[/tex]
The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]
You have made a simple circuit with one bulb. If you wanted to add an
extra bulb without the first bulb dimming. What would you need to
design?
A. A series circuit
B. A complex circuit
C. A parallel circuit
D. An incomplete circuit
Answer:
[tex]A. \: A \: series \: circuit[/tex]
Explanation:
♨Rage♨
Answer:
C. A parallel circuit
Explanation:
Adding a bulb in parallel with the existing bulb will apply the same voltage to both bulbs. They will light equally bright.
You would design a parallel circuit.
_____
In a series circuit the same current would flow in both bulbs, but that current would be at half the original current level. Both bulbs would be dimmer than the first bulb was.
It is difficult to create a "complex" circuit with only two components. An "incomplete" circuit would result in no light at all.
when hydrogen shares electrons with oxygen the outermost shell of the hydrogen atoms are full with how many electrons? and oxygens valence shell is full with how many electrons? because the valence shells of these atoms are full,the atoms are stable.
Answer:
2 and 8
Explanation:
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Magnets are usually made up of which material
A. plastic
B. iron ore
C. copper
D. gold
Answer:
B. iron ore
Explanation:
Hope this helps
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According to Newton's first law, which characteristic of a moving object would remain constant if there were no other
forces acting on it?
O size
c
mass
O shape
O speed
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The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the system is released from rest, what will its acceleration be
This question is incomplete
Complete Question
m1 is 10kg, m2 is 4.0kg. The coefficient of static friction between m1 and the horizontal surface is 0.50. and the Coefficient of kinetic friction is 0.30.
a) if the system is released from rest what will be its acceleration
Answer:
0.7 m/s²
Explanation:
The coefficient of static friction between m1 and the horizontal surface is 0.50. and the coefficient of kinetic friction is 0.30.
(a) if the system is released from rest what will be its acceleration
g = acceleration due to gravity = 9.81 m/s²
Coefficient of Kinetic Friction = μk = 0.30
m1 = 10kg
m2 = 4.0kg
The formula to solve question a is given as:
a = acceleration at rest
m2g- μk m1g = (m1+ m2) a
Making a the subject of the formula:
a = (m2g- μk×m1g )/(m1+ m2)
a = [(4.0 kg × 9.81m/s²) – (0.30 ×9.81 × 10) ]/(10+4)
a = 0.7 m/s²
am I right? be honest
Answer:
I chose c because it is the greater slope at point c
21. A toy car starts from rest and begins to accelerate at 11.0 m/s2. What is the toy
car's final velocity after 6.0 seconds?
Answer:
Explanation:
Given parameters:
Initial velocity = 0
Acceleration = 11m/s²
Time = 6s
Unknown:
Final velocity = ?
Solution:
From the given parameters, we use one of the appropriate equations of motion to solve this problem.
V = U + at
V is the final velocity
U is the initial velocity
a is the acceleration due to gravity
t is the time taken
Input the parameters and solve;
V = 0 + 11 x6
V = 66m/s
The final velocity is 66m/s
If 0.5 kg of this material is used in a transformer core, how long would it have to operate at a frequency of 60 cps to heat up 1
Answer:
Hello your question is incomplete attached below is the complete question
answer : 49 seconds
Explanation:
considering only Hysteresis loss
we have to calculate the Area affected/under the Hysteresis loss
= volume * area
= 4 * ( 1.5 * 20 ) = 120 tesla. A/m
next we calculate the volume of the material
= mass of material / density
= 500 grams / 7.9 g/cm^3 = 6.33 * 10^-5 m^3
next we calculate the heat lost per cycle
= 6.33 * 10^-5 m^3 * 120 = 0.007596 joules
The total heat required to raise temperature by 1°c
= Cp * ΔT * n
= 3R * n * ΔT = 3(8.314) * 8.95 * 1 = 223.23 Joules
where n = number of moles = 500grams / 55.85 = 8.95moles
ΔT = 1
Therefore the time required to have to operate at a frequency of 60 cps
= Total heat required / heat lost per cycle
=( 223.23 / 0.007596 ) 60 cps
= 489.796 seconds ≈ 49 seconds
When electrical energy is used what type of energy is also produced and considered to be waste energy?
Radiant energy
Thermal energy
Mechanical energy
Nuclear energy
Answer:
Thermal Energy
Explanation:
A designer is creating an obstacle for an obstacle course where a person starts on a moveable platform of height H from the ground. The person grabs a rope of length L and swings downward. At the instant the rope is vertical, the person lets go of the rope and attempts to reach the far side of a water-filled moat. The left side of the moat is directly below the position where the person will let go of the rope. The designer runs several tests in which the rope has different lengths and moves the platform such that the rope is always initially horizontal. The designer notices that the person cannot land on the other side if the length L is very short. The designer also notices that the person also cannot land on the other side if the length L is very close to the height H.
Assume the size of the person is much smaller than the lengths L and H. Let D represent the horizontal distance from below the release point to where the person lands.
Required:
a. Why does the person land in the moat if the rope's length is very short?
b. Why does the person land in the moat if the length is nearly the same as the height of the platform?
Answer:
* when L → H chord too long
in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross
* when L → 0 very short string
the speed of the platform is very small, so we do not have the minimum required value
vox = √ (g / (2 (H)) D
Explanation:
For this exercise we are going to solve it using conservation of energy to find the velocity of the body and the launch of projectiles to find the velocity to cross the well.
Let's start with the projectile launch
as the body leaves the vertical its velocity must be horizontal
x = v₀ₓ t
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
when reaching the ground its height of zero (y = 0) and the initial vertical velocity is zero
t = √ 2 y₀ / g
we substitute
x = vox √2y₀ / g
v₀ₓ = √(g / 2y₀) x
In the exercise, it tells us that the width of the well is D (x = D) and the initial height is the height of the platform minus the length of the rope (I = H - L)
v₀ₓ = √(g /(2 (H -L)) D
this is the minimum speed to cross the well.
Now let's use conservation of energy
starting point. On the platform
[tex]Em_{o}[/tex] = U = m g H
final point. At the bottom of the swing
Em_{f} = K + U = 1 / 2m v² + m g (H -L)
as there is no friction the mechanical energy is conserved
Em_{o} = Em_{f}
m g H = 1 / 2m v² + m g (H -L)
v = √ (2gL)
let's write our two equations
the minimum speed to cross the well
v₀ₓ = √ (g /(2 (H -L)) D
the speed at the bottom of the oscillatory motion
v = √ (2g L)
we analyze the extreme cases
* when L → H chord too long
in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross
* when L → 0 very short string
the speed of the platform is very small, so we do not have the minimum required value
vox = √ (g / (2 (H)) D
From this analysis we see that there is a range of lengths that allows us to have the necessary speeds to cross the well
V₀ₓ = v
g / (2 (H -L) D² = 2g L
4 L (H- L) = D²
4 H L - 4 L2 - D² = 0
L² - H L - D² / 4 = 0
let's solve the quadratic equation
L = [H ± √ (H2-D2)] / 2
we assume that H> D
L = ½ H [1 + - RA (1 - (D / H) 2)]
The two values of La give the range of values for which the two speeds are equal
A) The person lands in the moat if the rope's length is very short because :
The speed of the platform is less than the required minimum speedB) The person lands in the moat if the rope length is similar to the height of the platform because :
The speed required to cross the moat approaches infinityFollowing the assumptions;
size of the person is much smaller than L and H
D = horizontal distance
The conditions that will cause the person to land on the moatThe person will land in the moat when the rope's length is very short because as the rope reduces in length the speed reduces as well such that the speed of the platform goes below the required minimum speed which will enable the person cross over. while As the magnitude of the length tends towards the same magnitude of the height the speed required to cross the moat increases towards infinity and this speed cannot be attained by the person hence he will land in the moat.Hence we can conclude that The person lands in the moat if the rope's length is very short because The speed of the platform is less than the required minimum speed and The person lands in the moat if the rope length is similar to the height of the platform because,the speed required to cross the moat approaches infinity.
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Which two types of energy does a book have as it falls to the floor
Answer:
kinetic and potential energy
Explanation:
A person has a mass of 1000g and an acceleration of 20 m/s/s. What is the force on the person
Answer:
20000
Explanation:
Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000
Two particles are separated by 0.38 m and have charges of -6.25x 10 C and 2.91 x 10 C. Use Coulomb's law to predict the force between the particles if the distance is doubled. The equation for Coulomb's law is Fe = g, and the constant, k, equals 9.00 x 10° Nm/C A. -1.13 x 10-6 N OB. 1.13x 106N O C. 2.83 x 10-7 N OD.-2.83x 10N sUBMIT
Answer:
I do not understand what you are asking
One airplane is approaching an airport from the north at 181 kn/hr. A second airplane approaches from the east at 278 km/hr. Find the rate at which the distance between the planes changes when the southbound plane is 30 km away from the airport and the westbound plane is 15 km from airport.
Answer:
The value is [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]
Explanation:
From the question we are told that
The speed of the airplane from the north is [tex]\frac{dN}{dt} = -181 \ km /hr[/tex]
The negative sign is because the direction is towards the south
The speed of the airplane from the east is [tex]\frac{dE}{dt} = -278 \ km/hr[/tex]
The negative sign is because the direction is towards the west
The distance of the southbound plane from the airport is [tex]N = 30 \ km[/tex]
The distance of the westbound plane is [tex]E = 15 \ km[/tex]
Generally the distance between the plane is mathematically represented using Pythagoras theorem as
[tex]R^2 = N^2 + E^2[/tex]
Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes
So
[tex]2R\frac{dR}{dt} = 2N \frac{dN}{dt} + 2E\frac{dE}{dt}[/tex]
Here
[tex]R = \sqrt{N^2 + E^2}[/tex]
=> [tex]R = \sqrt{30^2 + 15^2}[/tex]
=> [tex]R = \sqrt{30^2 + 15^2}[/tex]
=> [tex]R =33.54 \ m [/tex]
[tex]2(33.54) * \frac{dR}{dt} = 2( 30)*(-181) + 2*15*(-278)[/tex]
=> [tex] 67.08 * \frac{dR}{dt} = -19200[/tex]
=> [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]
The rate of change of the distance between the planes is 286.23 km/hr.
The given parameters;
speed of the airplane from North, dn/dt = 181 Km/hspeed of the airplane from the East, de/dt = 278 km/hnorth distance, n = 30 kmeast distance, e= 15 kmThe distance between the two planes is calculated by applying Pythagoras theorem as shown below;
[tex]d^2 = n^2 + e^2\\\\d = \sqrt{n^2 + e^2} \\\\d = \sqrt{30^2 + 15^2} \\\\d = 33.54 \ km[/tex]
The rate of change of the distance between the planes is calculated as follows;
[tex]d^2 = e^2 + n^2\\\\2\frac{dd}{dt} = 2e\frac{de}{dt} + 2n\frac{dn}{dt} \\\\d\frac{dd}{dt} = e\frac{de}{dt} + n\frac{dn}{dt}\\\\(33.54) \frac{dd}{dt} = (15)(278) \ + (30)(181)\\\\(33.54) \frac{dd}{dt} = 9600\\\\\frac{dd}{dt} = \frac{9600}{33.54} \\\\\frac{dd}{dt} = 286.23 \ km/hr[/tex]
Thus, the rate of change of the distance between the planes is 286.23 km/hr.
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which two types of information are found in an elements box in the periodic table
Answer:
Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name.
Explanation:
Answer:
An element's period and group
Logan is a runner he in 60 seconds he can run 360 m what speed did he travel at
Answer:
hhhhhhhh
Explanation: