Explanation:
meter=100cm
cm=10^-2 m
min=60 sec
0.0141666667 m / s
The conversion of 85 cm/min would be 0.0133 m/s
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.
As given in the problem position-time graph of an object.
As we know that slope of the position time curve represents the velocity of the object
As given in the problem we have to convert 85 cm/min to m/s.
Let us first convert the displacement in cm to meters
100 cm = 1 m
1 cm = 1/100 m
85 cm = 85×1/100 m
= 0.85 m
similarly, convert time given in minutes to the seconds
1 min = 60 s
velocity = displacement /time
velocity = 0.85 / 60 m/s
=0.01333 m/s
Thus, we converted 85 cm/m into 0.0133 m/s
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Suppose an experiment is designed to test the durability of batteries in different conditions. All of the batteries tested are double-A (AA) Brand X. All sets of batteries are preconditioned in different environmental conditions for exactly 168 hours (1 week).
Set 1: 0°C (freezing point of water)
Set 2: 24°C (approximately room temperature)
Set 3: 37°C (approximately body temperature)
The batteries are then continuously used to power identical mechanical drummer toys. As long as the toy keeps drumming the battery is considered functional. The drumming time for each toy is measured as an indication of battery durability. In this experiment, which condition is not controlled?
A.) temperature
B.) brand of batteries
C.) test for durability
D.) type of battery (battery size)
Answer:
I assume its c. Since its talking about testing.
Explanation:
Answer:
The answer is test of durability
Explanation:
you walk 6 block east, 2 blocks north, 3 blocks west and then 2 blocks north. the total distance you travel is blocks
Answer:
The answerI travel 13 blocksAn object of mass 25kg is at rest. What is its momentum ?
Answer:
[tex]\boxed{0}[/tex]
Explanation:
Momentum is the measure of mass in motion.
[tex]\sf momentum = mass \times velocity[/tex]
An object at rest has a velocity of 0.
[tex]p=mv[/tex]
[tex]p = 25 \times 0[/tex]
[tex]p=0[/tex]
The momentum of an object at rest is always 0.
THE LENGTH OF A PENDULUM IS (1.5±0.01)m AND THE ACCELERATION DUE TO GRAVITY IS TAKEN AS (9.8±0.1)ms-² calculate the time period of the pendulum with uncertainty in it
Answer:
2.4583 ± 0.0207 seconds
Explanation:
The time period of a pendulum is approximately given by the formula ...
T = 2π√(L/g)
The maximum period will be achieved when length is longest and gravity is smallest:
Tmax = 2π√(1.51/9.7) ≈ 2.47903 . . . seconds
The minimum period will be achieved for the opposite conditions: minimum length and maximum gravity:
Tmin = 2π√(1.49/9.9) ≈ 2.43756 . . . seconds
If we want to express the uncertainty using a symmetrical range, we need to find half their sum and half their difference.
T = (2.47903 +2.43756)/2 ± (2.47903 -2.43756)/2
T ≈ 2.4583 ± 0.0207 . . . seconds
__
We have about 2+ significant digits in the given parameters, so the time might be rounded to 2.46±0.02 seconds.
Describe the motion of water waves.
Answer:
Water waves are an example of waves that involve a combination of both longitudinal and transverse motions. As a wave travels through the waver, the particles travel in clockwise circles. The radius of the circles decreases as the depth into the water increases.
Monochromatic light of wavelength 649 nm is incident on a narrow slit. On a screen 2.25 m away, the distance between the second diffraction minimum and the central maximum is 1.99 cm. (a) Calculate the angle of diffraction θ of the second minimum. (b) Find the width of the slit.
Answer:
a)0.51°
b)1.47×10^-4m
Explanation:
a)for a single slit experiment, the minima that has an angle of θ towards the centre needs to satisfy the expression below.
bsin(θ)= mλ.........................(*)
Where b= width of the slit
The distance on the screen from Central angle can be expressed as
Sin(θ)= y/d............. (**)
d and y is the horizontal distance between slit and screen
If we input eqn(**) into equation (*) we have
y= mλd/b................(z)
In order to find angle (θ) we have
(θ)= sin-(1.99×10^-2)/2.25
= 0.51°
Therefore, angle of diffraction θ of the second minimum is 0.51°
b)to find the width of the sloth using eqn(z) by substitute the values, we have
b= (2)(649×10^-9)(2.25)/1.99×10^-2
b= 1.47×10^-4m
Therefore, the width of the slit is 1.47×10^-4m
When the hydraulic conductivity Ks = 10 mm/hr; effective matrix potential Ns = 20 mm, and rainfall intensity I = 30 mm/hr , determine the amount of runoff generated when the runoff rate reaches 15 mm/hr?( 2.7 mm or 0.21 mm or 18 mm or 0.67 mm)
Answer:
[tex]t = 0.75 \ hr = 0.75 *60 = 45 \ minutes[/tex]
Explanation:
From the question we are told that
The hydraulic conductivity is [tex]Ks = 10\ mm/hr[/tex]
The effective matrix potential [tex]Ns = 20 \ mm[/tex]
The intensity of rainfall is [tex]I = 30 \ mm/hr[/tex]
The runoff rate [tex]R = 15 \ mm/hr[/tex]
Generally the run off rate is mathematically represented as
[tex]R = (I - Ks) t[/tex]
Here t is the amount of runoff generated
=> [tex]15 = (30 - 10) t[/tex]
=> [tex]t = 0.75 \ hr[/tex]
converting to minutes
[tex]t = 0.75 \ hr = 0.75 *60 = 45 \ minutes[/tex]
Hi please, I Have An attachment on Waves, Just two Objective Questions Whoever Answers Will be Marked Brainliest thank you.
Answer:
The first answer is W and Z, since they appear to be a period apart. Dont know the second question. I did what I could, hope someone can answer the second.
1.A particle is moving up an inclined plane. Its velocity changes from 15m/s to 10m/s in two seconds. What is its acceleration? *
[tex] \huge{ \underline{ \boxed{ \bf{ \blue{Solution:}}}}}[/tex]
Provided:-Initial velocity = 15 m/sFinal velocity = 10 m/sTime taken = 2 sTo FinD:-Accleration of the particle....?How to solve?We will solve the above Question by using equations of motion that are:-
v = u + ats = ut + 1/2 at²v² = u² + 2asHere,
v = Final velocity u = Initial velocity a = acceleration t = time taken s = distance travelled Work out:By using first equation of motion,
⇛ v = u + at
⇛ 10 = 15 + a(2)
⇛ -5 = 2a
Flipping it,
⇛ 2a = -5
⇛ a = -2.5 m/s² [ANSWER]
❍ Acclearation is negative because final velocity is less than Initial velocity.
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A student runs 35 m east and then 12m west. What is the distance run by the student
Answer:
The answer is either 47 or 23. Most likely 47
A wire of length 5mm and Diameter 2m extends by 0.25 when a force of 50N was use. calculate the
a. Stress of the wire.
b. Strain of the wire
Answer and Explanation:
Data provided in the question
Force = 50N
Length = 5mm
diameter = 2.0m = [tex]2\times 10^{-3}[/tex]
Extended by = 0.25mm = [tex]0.25\times 10^{-3}[/tex]
Based on the above information, the calculation is as follows
a. The Stress of the wire is
[tex]= \frac{force\ applied}{area\ of \ circle}[/tex]
here area of circle = perpendicular to the are i.e cross-sectional i.e
= [tex]\frac{\pi d^{2}}{4}[/tex]
= [tex]\frac{\pi(2\times 10^{-3})^2}{4}[/tex]
Now place these above values to the above formula
[tex]= \frac{4\times 50}{\pi\times 4 \times 10^{-6}} \\\\ = \frac{50}{\pi}[/tex]
= 15.92 MPa
As 1Pa = 1 by N m^2
So,
MPa = 10^6 N m^2
b. Now the strain of the wire is
[tex]= \frac{Change\ in\ length}{initial\ length} \\\\ = \frac{0.25\times 10^{-3}}{5}[/tex]
= [tex]5 \times 10^{-5}[/tex]
what are some factors that affect the frequency of sound
Answer:
1. direction of propagation of sound
2.medium through which sound trsnsmitted
On the graph of voltage versus current, which line represents a 2.0 resistor?
Answer:
answer is B
Explanation:
Acc. to Ohm's Law:
V = I.R and therefore R=( V/I )
For line A, R = 6/2 = 3
line B, R = 4/2 = 2 ( for all the points plotted on this line the ratio of V and I is 2)
Line C = 2/2 = 1
Line D < 1
Hence, the correct answer is Line B :-)
in how many ways can five basketball players be placed in three postitions?
Answer:
Well if they playing a game like that
If the reading of a linear scale is 4 mm and no of division of the circular scale is 50, then what will be the diameter of the wire in mm? (Least count = 0.01) * 2 points 2.25 3.5 4.5 9.0
Answer:
Diameter of Wire = 4.5 mm
Explanation:
First, we need to find the fractional part of the reading. The fractional part o the reading can be given by the following formula:
Fractional Part = Circular Scale Reading x Least Count
where,
Circular Scale Reading = 50
Least Count = 0.01 mm
Therefore,
Fractional Part = (50)(0.01 mm)
Fractional Part = 0.5 mm
Now, the diameter of the wire can be given by using the following formula:
Diameter of Wire = Linear Scale Reading + Fractional Part
Diameter of Wire = 4 mm + 0.5 mm
Diameter of Wire = 4.5 mm
A butterfly is flying around and its velocity(v) as a function of time(t) is given in the graph below where rightwards is the positive velocity direction. What is the butterfly's displacement x from t=2 to 4s? Answer with two significant digits.
Answer: 19 meters.
Explanation:
We want to find the total displacement between t = 2s and t = 4s.
To do it, we can integrate our function, first write our velocity equation.
for t ≤ 3s, we have a linear equation, let's write it:
A linear relationship can be written as:
y = a*t + b
where a is the slope and b is the y-axis intercept.
For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:
a = (y2 - y1)/(x2 - x1).
Now we can see that our line passes through the points (1, 0) and (0, -2)
then the slope is:
a = (0 -(-2)/(1 - 0) = 2/1 = 2
and knowing that when t = 0s, v(0s) = -2m/s, then our equation is:
v(t) = (2m/s^2)*t - 2m/s for t ≤ 3s
now, for t ≥3s the equation is constant, v(t) = 4m/s.
then we have
v(t) = (2m/s^2)*t - 2m/s -------if t ≤ 3s
v(t) = 4m/s ----- if t ≥ 3s
Now we integrate over time to get the position:
for t ≤ 3s we have:
p(t) = (1/2)*(2m/s^2)*t^2 - 2m/s*t + C
where C is a constant of integration, as we are calculating the displacement this constant actually does not matter, so we can use C = 0m
p(t) = (1m/s^2)*t^2 - 2m/s*t for t ≤ 3s
and p(3s) = (1m/s^2)*3s^2 - 2m/s*3s = 9m - 6m = 3m is the initial position of the other part of the function.
for t ≥ 3s we have:
p(t) = 4m/s*t + p(3s) = 4m/s*t + 3m
then the position equation is:
p(t) = (1m/s^2)*t^2 - 2m/s*t ---- t ≤ 3s
p(t) = 4m/s*t + 3m --- if t ≥ 3s
Now the displacement will be:
p(4s) - p(2s) where for each time, you need to use the correct function:
p(4s) = 4m/s*4s + 3m = 16m + 3m = 19m
p(2s) = (1m/s^2)*2s^2 - 2m/s*2s = 4m - 4m = 0m
p(4s) - p(2s) = 19m - 0m = 19m
The butterfly displacement x from t=2 to 4s is 19 meters.
What is displacement?The spacing between two specified points is represented by the one-dimensional quantity of displacement (symbolised as d or s), commonly known as length or distance.
The total displacement between t = 2s and t = 4s.
Integrate our function, the velocity equation.
for t ≤ 3s, we have a linear equation, let's write it:
A linear relationship can be written as:
y = a x t + b
where a is the slope and b is the y-axis intercept.
For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:
a = (y2 - y1)/(x2 - x1).
The line passes through the points (1, 0) and (0, -2)
The slope is:
a = (0 -(-2)/(1 - 0) = 2/1 = 2
When t = 0s, v(0s) = -2m/s, then our equation is:
v(t) = (2m/s²) x t - 2m/s for t ≤ 3s
now, for t ≥3s the equation is constant, v(t) = 4m/s.
v(t) = (2m/s²) x t - 2m/s -------if t ≤ 3s
v(t) = 4m/s ----- if t ≥ 3s
Now we integrate over time to get the position:
for t ≤ 3s we have:
p(t) = (1/2) x (2m/s²) x t^2 - 2m/s x t + C
where C is a constant of integration, to calculate the displacement this constant actually does not matter,
p(t) = (1m/s²)*t^2 - 2m/s x t for t ≤ 3s
and p(3s) = (1m/s^2) x 3s² - 2m/s x 3s = 9m - 6m = 3m is the initial position of the other part of the function.
for t ≥ 3s we have:
p(t) = 4m/s x t + p(3s) = 4m/s x t + 3m
then the position equation is:
p(t) = (1m/s^2) x t² - 2m/s x t ---- t ≤ 3s
p(t) = 4m/s x t + 3m --- if t ≥ 3s
Now the displacement will be:
p(4s) - p(2s) where for each time, you need to use the correct function:
p(4s) = 4m/s x 4s + 3m = 16m + 3m = 19m
p(2s) = (1m/s²) x 2s²- 2m/s x 2s = 4m - 4m = 0m
p(4s) - p(2s) = 19m - 0m = 19m
Thus, the displacement is 19 m.
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A net force of 0.7 N is applied on a body. What happens to the acceleration of the body in a second trial if half of the net force is applied?(1 point) The acceleration is double its original value. The acceleration is half of its original value. The acceleration is the square of its original value. The acceleration remains the same.
Answer:
The answer is The acceleration is double its original value.
Explanation:
It is because of the second trial of accelaration. Because of this, an object's acceleration doubles from its original value.Hope this helps....
Have a nice day!!!!
Answer:
The acceleration is half of its original value
Explanation:
A car accelerates at a rate of 3 m/s^2. If its original speed is 8 m/s, how many seconds will it take the car to reach a final speed of 25 m/s?
Answer:
[tex]\Large \boxed{\mathrm{5.67 \ seconds }}[/tex]
Explanation:
[tex]\displaystyle \mathrm{acceleration \ = \ \frac{final \ velocity - initial \ velocity }{elapsed \ time}}[/tex]
[tex]\displaystyle A \ = \ \frac{V_f - V_i }{t}[/tex]
[tex]\displaystyle 3 \ = \ \frac{25 - 8 }{t}[/tex]
[tex]\displaystyle 3 \ = \ \frac{17 }{t}[/tex]
[tex]\displaystyle t \ = \ \frac{17 }{3} \approx 5.67[/tex]
The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth? AU
Answer:
The answer is 29 AU
Hoped I helped
mark me as brainliest
Answer:
29
Explanation:
At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counselors measure the two "dry" legs of a right triangle. What is the length in meters of the swimming course in the figure below?
Answer:
47.17 m
Explanation:
From the diagram of the question attached, The length of the legs are 25 m and 40 m . This legs form a right angle triangle with the length of the swimming course (L).
Pythagoras theorem states that for a right angle triangle with hypotenuse a and legs b and c, then:
a² = b² + c²
In the triangle, the length of the swimming course (L) is the hypotenuse and the two legs are 25 m and 40 m. Using Pythagoras:
L² = 25² + 40²
L² = 625 + 1600
L² = 2225
L = √2225
L = 47.17 m
Help please answer the two questions
Explanation:
this is it. hope u understand
Question 14 of 30
A bundle of roofing shingles slides off a roof and is falling to the ground. As it
falls, what kind of energy does it possess?
O A. Kinetic only
O B. Potential only
O C. Radiant
D. Kinetic and potential
Answer:
kinetic and potential
Calculate the current passing through a conductor of resistance 4ohms. If a potential difference of 15V its ends______
Explanation:
current = velocity/resistance
I = V/R
15/4
current = 3.75A
hope this helps...
an object weights 0.250 kgf in air 0.150 in water and 0.125 in an oil.find out the density of the object and the oil
Answer:
1) The density of the object = 2500 kg/m³
2) The density of the oil = 1250 kg/m³
Explanation:
1) The information relating to the question are;
The mass of the object in air = 0.250 kgf
The mass of the object in water = 0.150 kgf
The mass of the object in the oil = 0.125 kgf
By Archimedes's principle, we have;
The upthrust on the object in water = Mass in air - mass in water = The weight of the water displaced
The upthrust on the object in water = 0.250 - 0.150 = 0.1 kgf
∴ The weight of the water displaced = 0.1 kgf
Given that the object is completely immersed in the water, we have;
The volume of the water displaced = The volume of the object
The volume of 0.1 kg of water water displaced = Mass of the water/(Density of water)
The volume of 0.1 kg of water = 0.1/1000 = 0.0001 m³
The density of the object = (Mass in air)/ volume = 0.250/0.0001 = 2500 kg/m³
The density of the object = 2500 kg/m³
2) Whereby the mass of the object in the oil = 0.125 kgf
The upthrust of the oil = The weight of the oil displaced
The upthrust of the oil on the object = Mass of the object in air - mass of the object in the oil
The upthrust of the oil on the object = 0.250 - 0.125 = 0.125 kgf
The weight of the oil displaced = The upthrust of the oil
Given that the volume of the oil displaced = The volume of the oil, we have;
The volume of the oil displaced = 0.0001 m³
The mass of the 0.0001 m³ = 0.125 kg
Therefore the density of the oil = 0.125/0.0001 = 1250 kg/m³.
The density of the oil = 1250 kg/m³.
observe the virual skateboarder coming down the hill and over the ramp describe how each of newton’s laws of motion can be observed in this action you can choose the dry wet or muddy conditions or some combination of these
Answer:
first part the skater goes down a constant slope ramp, initially he has Newton's second law
pply Newton's third law, the normal is the reaction to the support of the body on the surface
the ramp shoots off. axis becomes zero and therefore with Newton's first law its speed
Explanation:
It is the description of this movement let's write Newton's laws.
* The first law that a body goes at constant speed or zero if the sum of the external forces is zero
* the second law is F = m a
* The third law states that the forces act in pairs of equal magnitude and opposite direction, one applied to each body.
Let's apply these laws to our case
In the first part the skater goes down a constant slope ramp, initially he has Newton's second law when he accelerates from the initial velocity of zero to a terminal velocity.
The expression for this is
Wₓ - fr = ma
W sin θ - μ W cos θ = m a
W = mg
g (sin θ - μ cos) = a
the value of the coefficient of kinetic friction depends on the condition of the surface, dry, wet or muddy
This is Newton's second law
On the Y axis, which is perpendicular to the ramp we have
N- [tex]W_{y}[/tex] = 0
If we apply Newton's third law, the normal is the reaction to the support of the body on the surface, note that it can be different from the weight.
In the second part when he is on the ramp.
In the ramp the skater enters with a speed v, suppose that the ramp has an incline so that the skater can jump, in this case the angle is positive with respect to the axis x
In this case the analysis is similar to the previous one
Newton's second law gives the acceleration of the skater, who when he reaches the end of the ramp shoots off.
At this point the force in the x (horizontal) axis becomes zero and therefore with Newton's first law its speed this axis remains constant and the force in the y axis is the force of gravity and has an acceleration that changes if velocity according to Newton's second law
Answer:look at explanations
Explanation:
You need to repair a broken fence in your yard. The hole in your fence is around 3 meters in length and for whatever reason, the store you go to has oddly specific width 20cm wood. Each plank of wood costs $16.20, how much will it cost to repair your fence?
The correct answer is $243
Explanation:
The hole in the fence is 3 meters, this means it is necessary to buy wood that covers this distance. Now, each meter is equal to 100 centimeters, this means 3 meters is equivalent to 300 centimeters ( 100 cm in each meter x 3). Besides this, it is known each plank covers 20cm and costs $16.20. In this context, the next step is to find how many planks are needed. The process is shown below:
300 cm (total width) ÷ 20 cm (width of 1 plank) = 15 planks
This means 15 planks are needed. Finally, fin the total cost
15 planks x $16.20 (cost of 1 plan) = $243
element X has two isotopes: X-27 and x-29. x-27 has an atomic mass of 26.975 and a relative abundance of 82.33%, and X-29 has an atomic mass of 29.018 and a relative abundance of 17.67%. calculate the atomic mass of element X. show your work
Answer:
27.34 (no unit)
Explanation:
26.975*82.33%+29.018*17.67%
=27.34
A copper-nickel alloy of composition 50 wt% Ni-50 wt% Cu is slowly heated from a temperature of 1200°C (2190 °F). (a) At what temperature does the first liquid phase form? (b) What is the composition of this liquid phase? (c) At what temperature does complete melting of the alloy occur? (d) What is the composition of the last solid remaining prior to complete melting?
Answer:
HELLO BELOW IS THE ATTACHED DIAGRAM USED TO ANSWER YOUR QUESTION AS IT WAS MISSING
A) 1270⁰c
B) 65%
C) 1320⁰c
D) 62%
Explanation:
Nickel alloy composition : 50 wt% Ni - 50 wt%
initial temperature = 1200⁰c = 2190⁰F
A) The temperature at which the first liquid phase form
from the attached diagram the temperature at which the first liquid if formed is 1270⁰c ( at point 2 )
B ) The composition of this liquid phase ( THE FIRST LIQUID )
the composition is found at point 3
wt % of Nickel = 35%, wt% of copper = 100 - 35 = 65%
C ) The temperature at which the alloy melts completely
from the attached diagram the temperature = 1320⁰c ( point 4 )
D) The composition of the last solid remaining prior to complete melting
this can be gotten at point 5 and calculated as
wt % of Ni = 62%
wt % of Cu = 100 - 62 = 38%
A 50.5-turn circular coil of radius 4.75 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.455 T. If the coil carries a current of 22.5 mA, find the magnitude of the maximum possible torque exerted on the coil.
Answer:
The maximum torque, τ = 3.67 × 10⁻³ Nm
Explanation:
The torque τ = NiABsinθ where N = number of turns of circular coil = 50.5, i = current in circular coil = 22.5 mA = 0.0225 A, A = area of circular coil = πr² where r = radius of circular coil = 4.75 cm = 0.0475 m, B = magnetic field strength = 0.455 T and θ = 90° for maximum torque.
So, τ = NiABsinθ
τ = Niπr²Bsinθ
τ = 50.5 × 0.0225 A × π × (0.0475 m)² × 0.455 T × sin90°
τ = 0.003665 Nm
τ = 3.665 × 10⁻³ Nm
τ ≅ 3.67 × 10⁻³ Nm
So the maximum torque, τ = 3.67 × 10⁻³ Nm
A train travels southwest from point A to point B through the Arizona desert at 55 mi/h. How far will the train travel six-and-a-half hours?
Explanation:
Distance = speed × time
d = (55 mi/hr) (6.5 hr)
d = 357.5 mi
