A cat stalks an unsuspecting mouse. The movement of the cat is shown in the following graph of the horizontal position, x, against time, t. What is the average speed of the cat between t = 2s and t = 6s?

Explanation:

Volume is the amount of space an object takes up, while mass is the amount of matter in an object. ... To find the volume of an irregular sized object, one would use the displacement method for measuring volume and place the object in water and measure the amount of water that is displaced.

Answer:

To measure the volume of an irregularly shaped object, pour some water in a measuring cylinder. Then suspend the irregularly shaped object with a thread. After that , move the object gradually downwards and immerse it in water. The volume of the irregularly shaped object is the difference between the volume of the liquid before and after. After measuring the difference, we come to know about the volume of the irregularly shaped object.

Answer:

PFA

:-)

Explanation:

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.

Answer:

Positive

Explanation:

In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive

Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

[tex]x=4t^3+3t^2-5t+2[/tex]

Where,

x is in meters and t is in sec

We know that,

Velocity,

[tex]v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5[/tex]

(a) i. t = 2 s

[tex]v=12(2)^2+6(2)-5=55\ m/s[/tex]

At t = 4 s

[tex]v=12(4)^2+6(4)-5=211\ m/s[/tex]

(b) Acceleration,

[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6[/tex]

Pu t = 3 s in above equation

So,

[tex]a=24(3)+6\\\\a=78\ m/s^2[/tex]

Hence, (a) (i) v = 55 m/s (ii) v = 211 m/s and (b) 78 m/s²

Answer:

1. direction of propagation of sound

2.medium through which sound trsnsmitted

Answer:

The correct option is energy levels

Explanation:

Rutherford's model of an atom suggests that an atom has a tiny positively charged central mass (now called the nucleus) which is surrounded by electrons (negatively charged) in a cloud-like manner.

Bohr's model went a bit further than the Rutherford's model in describing an atom by suggesting that the electrons which surrounds in the nucleus travel in fixed circular orbits. This description by Bohr was able to describe the energy levels of orbitals which assumes that smallest orbitals have the lowest energy while the largest orbitals have the highest energy.

Answer:

energy levels

hope this helped!

Explanation:

Answer:

The volume of the cavity is 0.013m^3

Explanation:

To find the volume of the cavity, the major parameter missing is the diameter of the cavity itself. we can obtain this using the following steps:

Step one:

Obtain the volume of the cylinder by dividing the mass of the cylinder by the density.

Volume of the cylinder = 2.1 / 11.053 =0.19[tex]m^{3}[/tex]

Step two:

From the volume of the cylinder, we can get the radius of the cylinder.

[tex]radius = \sqrt{\frac{V}{\pi \times h}} = \sqrt{\frac{0.19}{\pi \times 0.32}} =0.44m[/tex]

Step three:

From the cross-sectional area, we can obtain the radius of the cavity.

Let the radius of the cavity be = r, while the radius of the cylinder be = R

CSA of cavity =

[tex]\pi({R^2}-r^2) = CSA\\0.65 = \pi (0.32^2-r^2)\\r= 0.115m[/tex]

Step Four:

calculate the volume of the cavity using volume =[tex]\pi r^2 \times h[/tex]

Recall that the cavity has the same height as the original cylinder

[tex]volume = \pi \times 0.115^2\times 0.32= 0.013m^3[/tex]

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:

4.0

Explanation:

The following data were obtained from the question:

Force (F) = 20 N

Mass (m) = 5 kg

Acceleration (a) =.?

Force is simply defined as the product of mass and acceleration. Mathematically, it is expressed as

Force (F) = mass (m) x acceleration (a)

F = ma

With the above formula, we can obtain th acceleration of the body as follow:

Force (F) = 20 N

Mass (m) = 5 kg

Acceleration (a) =.?

F = ma

20 = 5 x a

Divide both side by 5

a = 20/5

a = 4 m/s²

Therefore, the value that will complete the last cell in the question above is 4.

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.

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

Answer:

E= 600 W

Explanation:

Given that

m = 2 kg

Speed ,  v= 10 m/s

Force , F= 60 N

Given that box is moving with constant velocity, it means that friction force will be 60 N.

f = 60 N

Therefore total energy generated

E= f x v

E= 60 x 10 = 600 W

E= 600 W

Thus the answer will be 600 W.

Answer:

Explanation:

The velocity ratio of  a wheel and axle is the ratio of the radius (R) of the wheel to the radius (r) of the axle. It is expressed as;

VR = R/r

Since radius = diameter/2

VR = (D/2)/(d/2)

VR = D/d

D is the diameter of the wheel and 'd' is the diameter of the axle.

Given VR = 3 and d = 5cm

3 = D/5

D = 15 cm

If the diameter of the wheel is 15cm, the radius of the wheel will be 15/2 = 7.5cm.

b) Workdone by the load = Load * distance moved by load

Given load = 60kg

Distance moved by load = 2π*radius of axle

Distance moved by load  = 2π(0.025) = 0.157

workdone by load = 60* 0.157 = 9.42J

Effort = Workdone by load/distance moved by the wheel

Effort = 9.42/2π(0.075)

Effort = 9.42/0.471

Effort = 20kg

Hence the effort applied is 20kg

c) MA = Load/Effort

MA = 60/20

MA = 3

d) Efficiency = MA/VR * 100%

Efficiency = 3/3 * 100%

Efficiency = 100%

Answer:

I assume its c. Since its talking about testing.

Explanation:

Answer:

The answer is test of durability

Explanation:

Answer:

Answer:

Well if they playing a game like that

Answer:

66.86m

Explanation:

Velocity of ball thrown, u = 23.5 m/s

Initial height of the ball above the water, H = 11.5 m

Angle of projection, θ = 39.5°

Vertical components of veloclty = usinθ

Horizontal components of veloclty = ucosθ

The soft ball hits the water after time 't'

Considering the second equation of motion

S = ut + 1/2at^2........ 1

But since the ball went through motion under gravity ( free fall ) rather than linear motion, then equation 1 can be rewritten as:

H = ut +/- 1/2gt^2

H = - 11.5m

U = usinθ

θ = 39.5°

a = -g = -9.8m/s^2

- 11.5m = 23.5(sin39.5°)t + 1/2(-9.8)t^2

-11.5m = 23.5(0.6360)t - 4.9t^2

-11.5m = 14.946t - 4.9t^2

4.9t^2 -14.946t-11.5m = 0

Since the ball drifted horizontally

D = (Ucosθ)t

Where θ = 39.5°

U = 23.5m/s t=

Alternatively,

horizontal component of the velocity is 23.5 cos 39.5º = 18.1331 m/s

now how long does it take the ball to raise to a peak and fall to the water.

vertical component of velocity = 23.5 sin 39.5º = 14.947m/s

time to reach peak t = v/g = 11.947/9.8 = 1.5252 sec

peak reached above cliff top is

h = ½gt² = ½(9.8)(1.5252)²

= ½×22.797

= 11.3985m

now the ball has to fall 11.3985+ 11.5 = 22.8985m

time to fall from that height is

t = √(2h/g) = √(2• 22.8986/9.8) = 2.1617 sec

add up the two times to get time it is in the air, 2.1617 + 1.5252 = 3.6869

now haw far does the ball travel horizontally in that time

d = vt = 18.1331 ×3.6869= 66.856m

= 66.86m

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:

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:

Answer:

27.34 (no unit)

Explanation:

26.975*82.33%+29.018*17.67%

=27.34

Answer:

3504 m/s

Explanation:

Let x be the horizontal component of distance

y - vertical component of distance

t-time

ax- horizontal component of acceleration

ay-Vertical component of acceleration

Vx-horizontal component of velocity

Vy-Vertical component of velocity

horizontally: x = V_x ×t + ½×a_x×t²  

plugging the values we get

23500× cos 55º = 1350×cos25.0º × 10.20 + ½×a_x× (10.20)²  

⇒ax = 19.2 m/s²  

Moreover,

V'x = V_x + a_x×t = 1350×cos25.0º + 19.2×10.20= 1419 m/s  

similarly in vertical direction:

y = V_y×t + ½×a_y×t²  

23500×sin55º = 1350×sin25.0º×10.20s + ½×a_y×(10.20)²  

⇒a_y = 258 m/s²  

Also,

V'y = V_y + a_y×t = 1350×sin25.0º + 258×10.20 = 3204 m/s  

Therefore

V = √(V'x² + V'y²) = 3504 m/s  

therefore,  magnitude of final velocity of missile=3504 m/s

THANKS  

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Brainliest

Answer:1m

Explanation:

2n=0.4m

5n=?

5n×0.4/2n=1m

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]

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:

b. varies inversely with the square of the distance from the center of Earth.

Explanation:

Comparing the Newton's law of universal gravitation and second law of motion;

from Newton's second law of motion,

F = ma ............. 1

from New ton's law of universal gravitation,

F = [tex]\frac{GMm}{r^{2} }[/tex] ........... 2

Equating 1 and 2, we have;

mg =  [tex]\frac{GMm}{r^{2} }[/tex]

g = [tex]\frac{GM}{r^{2} }[/tex]

Therefore, the acceleration due to gravity near Earth, g, is inversely proportional to the square of the distance from the center of Earth.

Answer:

First, as you may know, the light travels at a given velocity.

In vaccum, this velocity is c = 3x10^8 m/s.

And we know that:

distance = velocity*time

Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.

This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)

Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.

The astronomers could know what was happening inside galaxies way back then by the fact that;

they examine the spectra of galaxies (or the overall colors of galaxies) with the highest redshifts they can find

Astronomers Measure the wavelength of the light that is stretched, so the light is seen as 'shifted' towards the red part of the spectrum by using spectroscopy. This measure is also called redshift.

This invokes a ray of light through a triangular prism that splits the light into various components known as spectrum.

The way the astronomers could use this concept to know what was happening in the galaxies before is by examining the spectra of galaxies that have the highest redshifts.

Read more at; https://brainly.com/question/15995216

Explanation:

To find the average of these numbers, we just have to add the three numbers together and divide by 3.

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.