Answer:
They are:
1) change position
2) distract yourself
3) Get fresh air
4) Face the direction you are going.
5) Drink water.
6) Play music.
7) Put your eyes on horizon.
Explanation:
Hope it helps.
Matter must have two physical properties 1. Have mass, and 2
∆ Must move
∆ Use energy
∆ Take up space
∆ Be measure
able
Answer:
Take up space
Explanation:
Actually we know this by the definition of matter which states that "matter is any substance that has mass and takes up space by having volume."
hope it helped you:)
i)Distinguish between different methods of charging. ii) You are provided with a positively charged gold leaf electroscope. State and explain what happens when a. a glass rod rubbed with silk is brought near the disc of electroscope. b. an ebonite rod rubbed with fur is brought near the disc of electroscope. c. an uncharged metal rod is brought near the disc of electroscope d. a glass rod rubbed with silk is rolled on the disc of electroscope.
Answer:
Explanation:
On rubbing a glass rod with silk, the electrons from the glass rod get transferred to the silk. The silk now has an excess of electrons and so is negatively-charged. On the other hand, the glass rod is deficient in electrons and hence is positively-charged.
In the above case, the silk undergoes negative electrification.
Now, when the positively charged glass rod is touched on the disc of a negatively charged gold leaf electroscope, the electrons shifts towards rod, hence amount of charge on gold leaves decreases and the divergence between the gold leaves decreases as unlike charges attract each other.
Hence, the divergence decreases when a glass rod rubbed with silk is brought near the disc of negatively charged electroscope.
hope it helps pls mark me as brainliest
Question is on the picture. Answers: A. 0.1 J/g*C B. 0.2 J/g*C C. 0.4 J/g*C D. 4 J/g*C
Answer:
B. 0.2 J/g/°C
Explanation:
The solid phase is the first segment (from 0°C to 50°C).
q = mCΔT
200 J = (20 g) C (50°C)
C = 0.2 J/g/°C
The ways to measure the mass and volume of irregular object
Answer:
When we have irregular objects, it may become very hard to calculate the volume of the object, as we actually can not use any simple equation to find it.
The mass is less tricky, just find a scale and wheight it, now we know the mass of the irregular object.
One way to measure the volume of the object is using water... how we do it?
Get some recipient with water, measure the height of the water.
Introduce your object into the water and totally submerge it, now the level of the water will rise. This is because as you introduce the object under the water, you are displacing up a given volume of water that has the same volume as the irregular object.
Now that you know the height of the water before and after you put your object, you can easily calculate the volume of water displaced, and that will be the volume of the object (the tricky part may be totally submerging the object if, for example, is wood and it floats, here you can use a thin wire to push it down but it will affect a little bit the measures.)
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.
the unit of energy is a derived unit
Explanation:
Hi, there!!
Energy is defined as the capacity or ability to do work. It's SI unit is Joule.
here,
Joule = (kg×m×m)/(s×s)
= kg×m^2/s^2.
Therefore, the derived unit is kg.m^2 by s^2.
Hope it helps...
You are the driver of the car in the photos above. You Are traveling at 30 mph when suddenly the car goes from its position in the first photo to the position in the second photo. What is happening
Answer:
the car uses teleportation, to zip to one side of the photo, to the other
Explanation:
Matter is anything that has mass and takes up
space.
Which of the following is an example of
matter?
A. ear phones
B. music
C. sunlight
D. heat
Answer: ear phones
Explanation:
You can physically hold ear phones, but you can't hold music, sunlight, or heat.
Shortly after receiving a traffic ticket for speeding, Fred made numerous comments about the road signs being inadequate and is GPS telling him a different speed limit. This would be an example of:
Answer:
External locus of control
Explanation:
External locus of control is an attitude people possess that makes them attribute their failures or successes to factors other than themselves. The opposite of this type of attitude is the Internal locus of control where the individuals take responsibility for the outcomes of their actions whether good or bad. One good thing about the external locus of control is that when the individuals with this characteristic record successes, they attribute it to others and this presents them as people with team spirit. However, when they record failures, they do not want to take the blame, but rather attribute it to others.
Fred exhibits an external locus of control because he attributed his speeding to other factors like the road signs and GPS instead of fully admitting that it was his fault.
State 1 difference between 1 way rotary motion and reversible rotary motion
Answer:
The difference between One-way as well as reversible rotary motion is described below.
Explanation:
Unless rotary motions occur restricted to single direction exclusively (i.e. whether clockwise as well as anti-clockwise only), it is defined as another rotary 'one-way' motion.This motor establishes that continuously variable movement at 360 ° chemically guided is conceivable. The rotating is regulated, and for a particular direction, the biochemical occurrences guiding rotation become incredibly selective.A 1.00-kg glider attached to a spring with a force constant 25.0 N/m oscillates on a frictionless, horizontal air track. At t = 0, the glider is released from rest at x = -2.80 cm (that is, the spring is compressed by 2.80 cm). (a) Find the period of the glider's motion. How does the period depend on the mass and the spring constant? Does it depend on the amplitude of oscillation? s (b) Find the maximum values of its speed and acceleration. speed m/s acceleration m/s2 (c) Find the position, velocity, and acceleration as functions of time. (Where position is in m, velocity is in m/s, acceleration is in m/s2, and t is in s. Use the following as necessary: t.) x(t) = v(t) = a(t) =
Answer:
a) T = 1.26 s , b) v_max = 0.14 m / s , a_max = 0.7 m / s²
c) x = 0.028 cos (5 t) , v = - 0.14 sin 5t, a = - 0.7 cos 5t
Explanation:
This is a simple harmonic motion exercise that is described by the equation
x = A cos (wt +Ф)
with
w = √ (k / m)
let's apply this expression to our case
a) Angular velocity is related to frequency
w = 2π f
frequency and period are related
f = 1 / T
we substitute
2π / T = √ (k / m)
T = 2π √(m / k)
let's calculate
T = 2π √(1/25)
T = 1.26 s
In the expression for the period, the amplitude does not appear, therefore there is no dependence, as long as Hooke's law is fulfilled, which is correct for small amplitudes.
b) in the initial equation we have the position as a function of time, let's use the definition of speed and acceleration
v = dx / dt
v = - A w sin (wt + Ф)
the speed is maximum when the sine is -1
v_max = A w
w = √ (k / m)
w = √ 25/1
w = 5 rad / s
the amplitude of the movement is equal to the maximum compression of the spring
A = 2.8 cm = 0.028 m
we substitute
v_max = 0.028 5
v_max = 0.14 m / s
acceleration
a = dv / dt
a = - A w² cos (wt + Ф)
the acceleration is maximum when the cosine is -1
a_max = A w²
let's calculate
a_max = 0.028 5²
a_max = 0.7 m / s²
c) let's start by finding the phase constant
v = -A w cos (wt + Ф)
at t = 0 they indicate that the system has v = 0
0 = -A w sin (0 + Ф)
Ф = sin⁻¹ 0
Ф = 0
we write the equation
x = 0.028 cos (5 t)
v = - A w sin (wt + Ф)
v = - 0.028 5 sin (5t + 0)
v = - 0.14 sin 5t
acceleration
a = - A w² cos (wt + Ф)
a = - 0.028 5 2 cos (5 t + 0)
a = - 0.7 cos 5t
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
I need help pls now plleeeeeeeeaaassseeeee
Answer:
[tex]r = \frac{v}{i} = v = ri \\ i = \frac{v}{r} [/tex]
A narrow beam of light containing red (660 nm) and blue (470 nm) wavelengths travels from air through a 1.00 cm thick flat piece of crown glass and back to air again. The beam strikes at an incident angle of 30 degrees. (a) At what angles do the two colors emerge
Answer:
The color blue emerges at 19.16° and the color red emerges at 19.32°.
Explanation:
The angle at which the two colors emerge can be calculated using the Snell's Law:
[tex]n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2})[/tex]
Where:
n₁ is the refractive index of the incident medium (air) = 1.0003
n₂ is the refractive index of the refractive medium:
blue light in crown glass = 1.524
red light in crown glass = 1.512
θ₁ is the angle of the incident light = 30°
θ₂ is the angle of the refracted light
For the red wavelengths we have:
[tex] \theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.0003*sin(30)}{1.512}) = 19.32 ^{\circ} [/tex]
For the blue wavelengths we have:
[tex] \theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.0003*sin(30)}{1.524}) = 19.16 ^{\circ} [/tex]
Therefore, the color blue emerges at 19.16° and the color red emerges at 19.32°.
I hope it helps you!
You are hiking in a canyon and you notice an echo. You decide to let out a yell and notice it took 2 seconds before you heard the echo of your yell. How far away is the canyon wall that reflected your yell
Answer:
d = 343 m
Explanation:
Given that,
You notice that an echo took 2 seconds before you heard the echo of your yell.
We need to find that how far away is the canyon wall that reflected your yell. It means we need to find the distance.
The distance covered by an object is given by :
d = v × t
v is speed of sound in air, v = 343 m/s
The sound took 1 s to reach the wall and 1 s back to you.
It means that,
d = 343 × 1
d = 343 m
So, canyon will reflect your yell at a distance of 343 m
In a Young's double-slit experiment, a set of parallel slits with a separation of 0.102 mm is illuminated by light having a wavelength of 575 nm and the interference pattern observed on a screen 3.50 m from the slits.(a) What is the difference in path lengths from the two slits to the location of a second order bright fringe on the screen?(b) What is the difference in path lengths from the two slits to the location of the second dark fringe on the screen, away from the center of the pattern?
Answer:
Rounded to three significant figures:
(a) [tex]2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
(b) [tex]\displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m[/tex].
Explanation:
Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.
Let [tex]k[/tex] denote a natural number ([tex]k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace[/tex].) In a double-split experiment of a monochromatic light:
A maximum (a bright fringe) is produced when light from the two slits arrive while they were in-phase. That happens when the path difference is an integer multiple of wavelength. That is: [tex]\text{Path difference} = k\, \lambda[/tex].Similarly, a minimum (a dark fringe) is produced when light from the two slits arrive out of phase by exactly one-half of the cycle. For example, The first wave would be at peak while the second would be at a crest when they arrive at the screen. That happens when the path difference is an integer multiple of wavelength plus one-half of the wavelength: [tex]\displaystyle \text{Path difference} = \left(k + \frac{1}{2}\right)\cdot \lambda[/tex].MaximaThe path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where [tex]k = 0[/tex] in [tex]\text{Path difference} = 0[/tex].
The path difference increases while moving on the screen away from the center. The first order maximum is at [tex]k = 1[/tex] where [tex]\text{Path difference} = \lambda[/tex].
Similarly, the second order maximum is at [tex]k = 2[/tex] where [tex]\text{Path difference} = 2\, \lambda[/tex]. For the light in this question, at the second order maximum: [tex]\text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
Central maximum: [tex]k = 0[/tex], such that [tex]\text{Path difference} = 0[/tex].First maximum: [tex]k = 1[/tex], such that [tex]\text{Path difference} = \lambda[/tex].Second maximum: [tex]k = 2[/tex], such that [tex]\text{Path difference} = 2\, \lambda[/tex].MinimaThe dark fringe closest to the center of the screen is the first minimum. [tex]\displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda[/tex] at that point.
Add one wavelength to that path difference gives another dark fringe- the second minimum. [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex] at that point.
First minimum: [tex]k =0[/tex], such that [tex]\displaystyle \text{Path difference} = \frac{1}{2}\, \lambda[/tex].Second minimum: [tex]k =1[/tex], such that [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex].For the light in this question, at the second order minimum: [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda = \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m[/tex].
An object accelerates to a velocity of 230 m/s over a time of 2.5 s. The acceleration it experienced was 42 m/s2. What was its initial velocity?
Answer:
230 = x + 105
x= 125
Explanation:
v = v0 + at
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:
Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text later. We can ignore air resistance. What was the volleyball's velocity at the moment it was tossed into the air?
Answer:
If max height = 1.1 meters, then initial velocity is 3.28 m/s
If max height is 1.1 feet, then the initial velocity is 5.93 ft/s
Explanation:
Recall the formulas for vertical motion under the acceleration of gravity;
for the vertical velocity of the object we have
[tex]v=v_0-g \,t[/tex]
for the object's vertical displacement we have
[tex]y-y_0=v_0\,t - \frac{g}{2} \,t^2[/tex]
If the maximum height reached by the object is given in meters, we use the value for g in [tex]m/s^2[/tex] which is: [tex]9.8\,\,m/s^2[/tex]
If the maximum height of the object is given in feet, we use the value for g in [tex]ft/s^2[/tex] which is : [tex]32\,\,ft/s^2[/tex]
Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:
[tex]v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}[/tex]
and now we use this to express the maximum height in the second equation we typed:
[tex]y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}[/tex]
Then if the max height is 1.1 meters, we use the following formula to solve for [tex]v_0[/tex]:
[tex]1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s[/tex]
If the max height is 1.1 feet, we use the following formula to solve for [tex]v_0[/tex]:
[tex]1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s[/tex]
Answer:
11
Explanation:
for khan academy, this is the answer
whats suface tension
Answer: "Surface tension is a film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimize surface area."
Hope this helps!
Answer:
Explanation:
Surface tension is the property of a liquid surface. It is an effect where the surface of the liquid is strong.
example - small insects can walk on water as they do not have enough weight to penetrate it.
This image might help you
Hope it helps
plz mark as brainliest!!!!!!!
Question 1 (2 points)
(01.01 LC)
Which of the following is a characteristic of science? (2 points)
QU
Reproducible by other scientists
Ob
The personal opinion of the scientist
С
Using variable conditions for each test
d
Including only the data that supports a hypothesis
Answer:
Reproducible by other scientists
Explanation:
I just took the test
Which of the following illustrates an increase in potential energy? Group of answer choices a wind-up toy winding down a person climbs a set of stairs an apple dropping from a tree a firecracker explodes
Answer:
A person climbs a set of stairs
Explanation:
Potential energy is said to be possessed by an object due to its position. As the height from the ground level increase, the potential energy increases. It is calculated by the below formula as :
P = mgh
Out of the given options, the option that illustrates an increase in potential energy is option (b) i.e. a person climbs a set of stairs. As he steps one stair, its position from ground increases. It means its potential energy increases.
please help ASAP.
these are examination questions ..
no nonsense answers .
i will mark as brainliest if you got it correct .
Answer:
1st question c part
2nd question c part
What two factors determine how much potential energy an object has?
Answer:
The mass of the object and its height in the gravitational field of the Earth.
Explanation:
If we are talking about gravitational potential energy which is defined as:
[tex]U=m\,*\,g\,*\,h[/tex]
being "m" the object's mass, "g" the acceleration due to gravity, and "h" the height at which the object is located relative to the conventionally picked level for zero of potential energy.
As long as the value of "g" is constant, the only two variables that determine the gravitational potential energy are the mass (m) of the object and its relative height (h).
Answer:
The objects weight and height above Earth's surface
Explanation:
K12 :)
