A train travels at 100 km/h heading east to reach a town in 4 hours. The train then reversed ad heads west at 50 km/h for 4 hours

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Answer 1

What’s something goes up but never comes down?

Related Questions

Wind is ________ rushing in to fill an area of low pressure.

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I think is local winds

Wind is the stream of air rushing in to fill an area of low pressure. Winds are generated by the differential distribution of temperature resulting in a pressure difference.

What are winds?

The natural movement of air or other gases in relation to a planet's surface is referred to as wind. There are many different sizes of winds, ranging from thunderstorm flows that last only a few minutes to local breezes.

Large-scale atmospheric circulation is primarily brought on by the planet's rotation and the difference in temperature between the equator and the poles (Coriolis effect).

Thermal low circulations over topography and high plateaus can cause monsoon circulations in the tropics and subtropics. Local winds can be defined in coastal regions by the sea breeze/land breeze cycle; in regions with varying terrain, mountain and valley.

Find more on winds:

https://brainly.com/question/616790

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Order the layers, with the oldest at the bottom and most recent at the top.
someone plz help will give brainliest if correct

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Answer:

adbce i think

Explanation:

An object is released from a rocket moving upwards at a speed of 8.9 m / s.
What is the downward acceleration of this object?

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Answer:

9.81 m/s²

Explanation:

It is given that the upward speed of the object is 8.9 m/s.

An object is released.

Now, we know that if the object is released then it will move under the control of gravity which means that the initial velocity of the rocket will not affect the acceleration of the object.

Hence, the downward acceleration of the object will be equal to g i.e. 9.81 m/s².

A railroad car of mass M moving at a speed v1 collides and couples with two coupled railroad cars, each of the same mass M and moving in the same direction at a speed v2. (a) What is the speed vf of the three coupled cars after the collision in terms of v1 and v2

Answers

Answer:

vf = v₁/3 + 2v₂/3

Explanation:

Using the law of conservation of linear momentum,

momentum before impact = momentum after impact

So, Mv₁ + 2Mv₂ = 3Mv (since the railroad cars combine) where v₁ = initial velocity of first railroad car, v₂ = initial velocity of the other two coupled railroad cars, and vf = final velocity of the three railroad cars after impact.

Mv₁ + 2Mv₂ = 3Mvf

dividing through by 3M, we have

v₁/3 + 2v₂/3 = vf

vf = v₁/3 + 2v₂/3

How much work is done by a person who
pushes a cant with a force of 200 newtons if
the cart moves 20 meters in the direction
of the fonce.

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it is 82729cm still

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, this picture is not quite complete! You are on the surface of the Earth, which is rotating. While answering the following questions you may ignore the Earth’s motion around the sun, galactic center, etc. (though that is another interesting question!) (a) What is the magnitude of the acceleration of a person sitting in a chair on the equator? (b) At the equator, is your mass times the gravitational acceleration of the Earth greater than, less than, or equal to the normal force exerted on you by the chair you are sitting on? Explain. (c) A classmate of yours asks you why we have ignored this acceleration for the whole first term of physics. "Is everything we’ve learned a lie?" they ask. Ease their fears by calculating the percentage difference between the normal force from the chair and your weight while sitting on equator. (d) The latitude of Corvallis is 44.4˚. What is your acceleration while sitting in your chair?

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Answer:

a) [tex]a=33.73mm/s^{2}[/tex]

b) mg>N

c) [tex]\%_{change}=0.343\%[/tex]

d) [tex]a=24.07mm/s^{2}[/tex]

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

[tex]a_{c}=\omega ^{2}r[/tex]

where:

[tex]\omega=\frac{2\pi}{T}[/tex]

we know the period of rotation of the earth is about 24 hours, so:

[tex]T=24hr*\frac{3600s}{1hr}=86400s[/tex]

so we can now find the angular speed:

[tex]\omega=\frac{2\pi}{86400s}[/tex]

[tex]\omega=72.72x10^{-6} rad/s^{2}[/tex]

So the centripetal acceleration will be:

[tex]a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)[/tex]

which yields:

[tex]a_{c}=33.73mm/s^{2}[/tex]

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

[tex]\sum F=0[/tex]

so we get that:

[tex]N-mg+ma_{c} = 0[/tex]

and solve for the normal force:

[tex]N=mg-ma_{c}[/tex]

In this case, we can clearly see that:

[tex]mg>mg-ma_{c}[/tex]

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

[tex]mg=(60kg)(9.81m/s^{2})=588.6N[/tex]

and let's calculate the normal force:

[tex]N=m(g-a_{c})[/tex]

[tex]N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})[/tex]

N=586.58N

so now we can calculate the percentage change:

[tex]\%_{change} = \frac{mg-N}{mg}x100\%[/tex]

so we get:

[tex]\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%[/tex]

[tex]\%_{change}=0.343\%[/tex]

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

[tex]cos \theta = \frac{AS}{h}[/tex]

In this case:

[tex]cos \theta = \frac{r}{R_{E}}[/tex]

so we can solve for r, so we get:

[tex]r= R_{E}cos \theta[/tex]

in this case we'll use the average radius of earch which is 6,371 km, so we get:

[tex]r = (6371x10^{3}m)cos (44.4^{o})[/tex]

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

[tex]a=\omega ^{2}r[/tex]

[tex]a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m[/tex]

[tex]a=24.07 mm/s^{2}[/tex]

In this exercise we have to use the knowledge of mechanics to solve the magnitude of the acceleration and the acceleration of gravity, in this way we find that:

a)[tex]a=33.73 mm/s^2[/tex]

b)[tex]mg>N[/tex]

c) [tex]Change=0.343%[/tex]

d) [tex]a=24.07 mm/s^2[/tex]

Then calculating from the information given in the text;

a)In this case, we know the centripetal acceleration is given by the following formula:

[tex]a_c=w^2r\\w=\frac{2\pi}{T} \\T=24*\frac{3600}{60} = 86400 s[/tex]

so we can now find the angular speed:

[tex]w=\frac{2\pi}{86400} \\w=72.72*10^{-6} rad/s^2[/tex]

So the centripetal acceleration will be:

[tex]a_c=(72.72*10^{-6}rad/s^2)^2(6478*10^3m)\\a_c=33.73mm/s^2[/tex]

b)So we can do a sum of forces in equilibrium:

[tex]\sum F=o\\N-mg+ma_c=0\\N-mg+ma_c=0\\N=mg-ma_c\\mg>mg-ma_c\\mg>N[/tex]

This exist cause the centripetal increasing speed happen pulling united states of america upwards, that will create the size of the normal force tinier than the result or goods created of the bulk times the increasing speed of importance.

c) So let's calculate our weight and normal force:

[tex]mg=(60)(9.81)=588.6N\\N=m(g-a_c)\\N=(60)(9.81-33.73*10^{-3})\\N=586.58 N[/tex]

So now we can calculate the percentage change:

[tex]\%change= \frac{mg-N}{mg}*100\% \\=\frac{588.6-586.58}{588.6}*100\% \\=0.343\%[/tex]

d) There, we can see that the radius can be found by using the cos function:

[tex]cos\theta=\Delta S/h\\cos\theta=r/R_E\\r=R_E cos\theta\\r=(6371*10^3)cos(44.4)\\r=4,551.91 km[/tex]

And now we can calculate the acceleration at that point:

[tex]a=w^2r\\a=(72.72*10^{-6})^2(4,551.91*10^3)\\a=24.07 mm/s^2[/tex]

See more about acceleration at brainly.com/question/2437624

Consider two object has mass m1=5.0kg m2=0.5kg and the specific heat c1=c2=4200j/kg oc respectively assume that first object has temperature 20oc and second object has temperature 90oc After they place these two object touch each other determine the mixture temperature of the substance

Answers

Answer: [tex]T=26.36^{\circ}C[/tex]

Explanation:

Given

mass of the first object [tex]m_1=5\ kg[/tex]

mass of the second object [tex]m_2=0.5\ kg[/tex]

specific heat of the two is [tex]c_1=c_2=4200\ J/kg/^{\circ}C[/tex]

The temperature of the first [tex]T_1=20^{\circ}[/tex]

The temperature of the second [tex]T_2=90^{\circ}[/tex]

Heat flows from high temperature to low temperature

Suppose T is the common temperature

[tex]m_1c_1(T-T_1)=m_2c_2(T_2-T)\\5\times 4200\times (T-20)=0.5\times 4200\times (90-T)\\5T-100=45-0.5T\\5.5T=145\\T=26.36^{\circ}C[/tex]

Stoplight parrotfish:

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Is this what you were meaning?

HELP ASAP!
Everything on screenshot.

Answers

Answer:

I believe the answer is sea floor spreading

A 1020 kg car is pulling a 365 kg trailer. Together, the car and trailer have an acceleration of 2.21 m/s2 directly forward. (a) Determine the net force on the car. N forward (b) Determine the net force on the trailer. N forward

Answers

Answer:

a. 2936.2 N b. 773.8 N

Explanation:

Let m = mass of car = 1020 kg and m' = mass of trailer and a = acceleration = 2.12 m/s².

a. Determine the net force on the car. N forward

Since the car pulls both itself and the trailer, the combined mass is m + m' and the net force F on the car is F = (m + m')a

= (1020 kg + 365 kg)2.12 m/s²

= 1385 kg × 2.12 m/s²

= 2936.2 N

b. Determine the net force on the trailer. N forward

The net force F' on the trailer is F = m'a

= 365 kg × 2.12 m/s²

= 773.8 N

Car A is traveling at 18.0 m/s and car B at 25.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 1.80 m/s2. How long does it take car A to overtake car B

Answers

Answer:

car A reaches and immediately overtakes the car B at 22.56 s.

Explanation:

After car A accelerate at 1.8 m/s2, it travels a distance x(A) and car B will have travels a distance x(B), let's recall that the initial distance between them is 300 m, so we have:

[tex]x_{A}=300+x_{B}[/tex]

Now, we can rewrite this equation in terms of speed and time

[tex]V_{iA}t+\frac{1}{2}at^{2}=300+V_{iB}t[/tex]

Where:

V(iA) is the initial speed of car A

V(iB) is the initial speed of car B

t is the time when car A reaches the car B

a is the acceleration

[tex]18t+\frac{1}{2}1.8t^{2}=300+25t[/tex]

[tex]0.9t^{2}-7t-300=0[/tex]

Solving this quadratic equation for t, and taking just the positive value, we will have:

t=22.56 s

Therefore, car A reaches and immediately overtakes the car B at 22.56 s.

I hope it helps you!

A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5 kg and 0.5 kg, respectively. The piston diameter is 10 cm. The local atmospheric pressure is 1 bar. T

Answers

Answer:

The force acting on the shaft is 1324.75 N

Explanation:

Given that,

Cross sectional area of shaft [tex]A_{sh}=0.8\ cm^{2}[/tex]

Gas pressure [tex]P_{gas}=3\ bar=3\times10^5\ Pa[/tex]

Total mass M=24.5+0.5=25 kg

Diameter of piston, d=10 cm

We need to calculate the cross section area of piston

Using formula of area

[tex]A_{p}=\pi\times\dfrac{d^2}{4}[/tex]