Please help me ASAP...

A train increases its speed steadily from 10 m/s to 20 m/s in 1 minute.

a What is its average speed during this time, in m/s?

b How far does it travel while increasing its speed?

Answer:

a) Strings of guitar (being played)

b) Heart-beat in a healthy person

c) Motion of an apple falling from a tree

Explanation:

a) The motion of the string of a guitar being played is the only motion involving simple harmonic motion.

b) All the other motions are circular motion except the heart beat in a healthy person, which is periodic.

c) The motion of an apply falling from a tree is the only motion under the complete influence of gravity from the onset till the end.

Answer:

27.34 (no unit)

Explanation:

26.975*82.33%+29.018*17.67%

=27.34

Answer:

The hypothesis may or may not be true and needs to be tested. It might be the answer to the problem. Hence, it must be tested thoroughly. When these predictions are tested again and again in independent scientific experiments and gets verified, the hypothesis is converted into a scientific theory.

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:

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: Bleach is a base

Explanation:  If bleach had a pH level of 12, a number above 7, than it is a base.  Hope this helps!

Answer:

5.3333 sec

Explanation:

initial speed: u = 108km/hr or 30 m/s

final speed: v = 0m/s

distance travelled: s = 80m

time the car took to stop: = t sec

[tex]v^{2} - u^{2}[/tex] = 2as,

a = ([tex]v^{2} - u^{2}[/tex])/2s

a = (0-900)/160

a = -5.625 [tex]ms^{-2}[/tex]

v = u + at,

t = (v - u)/a

t= (0 - 30)/(-5.625)

t = 5.3333 sec

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: React with the skin

Explanation:

lab safety test Safety goggles and an apron must be worn when handling chemicals labeled corrosive because they React with the skin

Answer:

We have the position vector given in terms of time t. r(t) = t^3*i + t^2*j

To find the velocity vector we have to differentiate r(t) with respect to time.

r'(t) = 3t^2*i + 2t*j

The vector representing acceleration is the derivative of the position vector

r''(t) = 6t*i + 2*j

When time t = 2.

The velocity vector is 3*2^2*i + 2*2*j

=> 12*i + 4*j

The speed is the absolute value of the velocity vector or sqrt(12^2 + 4^2) = sqrt (144 + 16) = sqrt 160

The acceleration vector is 6*2*i + 2*j

=> 12*i + 2*j

The required acceleration at t=2 is 12*i + 2*j and the speed is sqrt 160.

Explanation:

Can I have thx and brainliest?

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

Answer:

Well if they playing a game like that

Answer:

The nail exerts an equal force on the hammer in the opposite direction.

Explanation:

The Newtons third law states that there is an equal an opposite reaction for every action. When hammer pushes the nail, the nail will push the hammer back in opposite direction. When the hammer hits a nail then nail will exert the equal and opposite force to the hammer. These both objects will exert force on each other in opposite directions.

Hello, I am BrotherEye

Answer:

Answers are

1. "The nail exerts an equal force on the hammer in the opposite direction."

2. "500 N"

3. "The iron piece exerts a force of 1 N on the magnet in the opposite direction."

4. "When mass moves closer to the point of rotation, rotational inertia decreases."

5. "The skater spins slower because his rotational inertia has increased."

Explanation:

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

As the angle of incidence increases in Figure 2.8, a point is finally reached where the refracted ray does not emerge at the second layer but lie along the interface. This particular angle of incidence at which the angle of refraction is 90° and the refracted ray lies along the interface is known as the critical angle. At and beyond the critical angle, there is no transmitted ray and therefore a very high reflected ray will be recorded .

Therefore,

sinθisin90=Vp1Vp2

But, sin 90 = 1.

At critical angle,

sinθcritical=Vp1Vp2

A critical refracted wave travels along the interface between layers and is refracted back into the upper layer at the critical angle. The waves refracted back into the upper layer are called head waves or first-break refractions because at certain distances from a source, they are the first arriving energy. Recorded first-break refraction is shown in Figure 2.10.

Note that these first-break refractions can give us important information about the shallow velocities on land seismic data.

Note also that seismic data are acquired in such a way that reflections from horizons of interest are in the pre-critical region, even at the farthest offset in the data.

In reality, part of the seismic energy arriving at an interface is transmitted and refracted, and another part of the energy is reflected at that same interface. Given that there are many reflectors in the subsurface, there will be many paths from source to receiver, each of them with a different travel time. The proportion of energy reflected depends on the material properties of the two bounding layers and on the angle of incidence

Answer:

The correct option is;

3. The total mechanical energy is constant as the block moves back and forth

Explanation:

The total mechanical energy is the sum of the potential and kinetic energies of the system

For a system that is isolated from the effects of external forces, but being acted upon by the internal conservative forces within the system, the total mechanical energy is constant

For a black and spring system, we have total mechanical energy, E = 1/2×K×A².

Where;

K = Constant

A = The amplitude of motion

Therefore, where there is no loss to friction, with A, remaining constant, the total mechanical energy will be constant.

Answer:

Explanation:

[tex]m =800kg\\v = 15\\\\K.E = \frac{1}{2}mv^2\\ K.E= \frac{1}{2} \times 800\times 15^2\\= 400 \times 225\\= 90000 joules\\= 90 kilojoules[/tex]

Answer:

K.E of truck > K.E of car

Explanation:

Mass of the truck = 8000Kg

K.E=[tex]\frac{1}{2} mv[/tex]

K.E =[tex]\frac{1}{2}*8000*v\\ 4000v[/tex]

Mass of the car = 1000 Kg

K.E of the car =[tex]\frac{1}{2}*1000*v\\ 500v[/tex]

Therefore Kinetic energy of the truck is greater than that of the car

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:

Explanation:

Assuming the 10 m distance is the vertical displacement, the work done by both people is the same.

Work = force × distance

W = (250 N) (10 m)

W = 2500 J

The power of the first person is:

Power = work / time

P = 2500 J / 120 s

P = 20.83 W

The power of the second person is:

P = 2500 J / 60 s

P = 41.67 W

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:

v1 = −2.201946 m/s ( to the left)

v2 = 0.7780534 m/s ( to the right)

Explanation:

Given the following :

Mass of first glider (m1) = 0.149kg

Initial Speed of first glider (u1) = 0.710 m/s

Mass of second glider (m2) = 0.308kg

Initial Speed of second glider (u2) = 2.27m/s

For elastic collision:

m1u1 + mu2u2 = m1v1 + m2v2

Where V1 and v2 = final velocities if the body after collision.

Taking right as positive ; left as negative

u1 = 0.710m/s ; u2 = - 2.27m/s

u1 - u2 = - (v1 - v2)

0.710 - - 2.27 = - v1 + v2

v2 - v1 = 2.98 - - - - (1)

From:

m1u1 + mu2u2 = m1v1 + m2v2

(0.149 * 0.710) + ( 0.308 * - 2.27) = (0.149 * v1) + (0.308 * v2)

0.10579 + (-0.69916) = 0.149 v1 + 0.308v2

−0.59337 = 0.149 v1 + 0.308v2

Dividing both sides by 0.149

v1 + 2.067v2 = −0.59337 - - - - - (2)

From (1)

v2 = 2.98 + v1

v1 + 2.067(2.98 + v1) = −0.59337

v1 + 6.16 + 2.067v1 = −0.59337

3.067v1 = −0.59337 - 6.16

3.067v1 = −6.75337

v1 = −6.75337 / 3.067

v1 = −2.201946 m/s ( to the left)

From v2 = 2.98 + v1

v2 = 2.98 + (-2.201946)

v2 = 0.7780534 m/s ( to the right)

Answer:

The distance the car travels is 115500 m in S.I units

Explanation:

Distance d = vt where v = speed of the car and t = time taken to travel

Now v = 99 km/h. We now convert it to S.I units. So

v = 99 km/h = 99 × 1000 m/(1 × 3600 s)

v = 99000 m/3600 s

v = 27.5 m/s

The speed of the car is 27.5 m/s in S.I units

We now convert the time t = 70 minutes to seconds by multiplying it by 60.

So, t = 70 min = 70 × 60 s = 4200 s

The time taken to travel is 4200 s in S.I units

Now the distance, d = vt

d = 27.5 m/s × 4200 s

d = 115500 m

So, the distance the car travels is 115500 m in S.I units

Answer:

I assume its c. Since its talking about testing.

Explanation:

Answer:

The answer is test of durability

Explanation:

Answer:

c

Explanation:

The marginal cost curve image  has been attached from which we can clearly, indicate that

ATC = average total cost

AFC = average fixed cost

AVC = average variable cost.

From the graph we can indicate that the marginal cost curve

(c) Lies above the AVC curve when the AVC curve slopes downward.

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:

Answer:

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

Explanation:

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:

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.