Is there a way for us to control motion

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.

Answer:

The difference between One-way as well as reversible rotary motion​ is described below.

Explanation:

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.)

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

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

Answer:

The answer  is The acceleration is double its original value.

Explanation:

Hope this helps....

Have a nice day!!!!

Answer:

The acceleration is half of its original value

Explanation:

Answer:

1st question c part

2nd question c part

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

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...

Answer:

230 = x + 105

x= 125

Explanation:

v = v0 + at

Answer:

27.34 (no unit)

Explanation:

26.975*82.33%+29.018*17.67%

=27.34

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!!!!!!!

Answer:

the car uses teleportation, to zip to one side of the photo, to the other

Explanation:

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.

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

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 :)

Answer: ear phones

Explanation:

You can physically hold ear phones, but you can't hold music, sunlight, or heat.

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.

Answer:

[tex]r = \frac{v}{i} = v = ri \\ i = \frac{v}{r} [/tex]

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

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!

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:)

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:

The 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].

The 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.

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].

Answer:

Reproducible by other scientists

Explanation:

I just took the test