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UkoKoshka [18]
3 years ago
9

As a train accelerates uniformly, it passes successive 5-kilometer markers while traveling at velocities 10 m/s and 25 m/s. what

is the velocity in km/h when it passes the next marker? round to nearest tenth of a km/h.
Physics
1 answer:
gregori [183]3 years ago
8 0
<span>122.0 km/hr. First let’s make sure all of our units are in the base meter form: i.e. convert 5km to 5000m. (We will convert back to km later). The first thing to do is look at the equation relating velocity, acceleration, and distance: Vf^2 = Vi^2 + 2*a*d, where Vf is final velocity, Vi is initial velocity, a is acceleration, and d is distance. 25^2 = 10^2 + 2*a*5000 =?> 625 = 100 +10000a => a= 0.0525m/s^2. Now that we have acceleration, we can use the same equation again with different numbers.: Vf^2 = Vi^2 + 2*a*d = 25^2 + 2*0. 0525m*5000 = 625 + 525 =1150 => Vf^2 = 1150 => 33.9m/s. Convert to km/hour: 33.9m/s * 1km/1000m *60s/1min * 60min/ 1 hr = 122.0 km/hr.</span>
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nevsk [136]

Answer:

option a.

Explanation:

We can think of an atom as a nucleus (where the protons and neutrons are) and some electrons orbiting it.

We also know that the mass of an electron is a lot smaller than the mass of a proton or the mass of an electron.

So, if all the protons and electrons of an atom are in the nucleus, we know that most of the mass of an atom is in the nucleus of that atom.

Then we define the mass number, which is the total number of protons and neutrons in an atom. Such that the mass of a proton (or a neutron) is almost equal to 1u

Then if we define A as the total number of protons and neutrons, and each one of these weights about 1u

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option a.

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3 years ago
Help with this physics task pls
cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )  

t=\frac{v}{a} ; d=s*(t-t_{0} )

2)

k=\frac{2*U}{x^{2} }; T_{2}=\frac{P_{2}*V_{2}*T_{1}  }{P_{1}*V_{1}  }  \\

3)

L=\frac{F}{\pi*r*P}; d=\frac{w}{F*cos(o)}

4)

t^{2}=\frac{2*x}{g}  ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} }  \\

5)

h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}

6)

h=\frac{m}{(1/2)*\pi *r^{2} }  ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2}   }

7)

b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}

8)

a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2}  }{M}

9)

v_{o}=\frac{x-\frac{1}{2}*a*t^{2}  }{t}  ; F=\frac{W+uNd}{d*cos(o)}

10)

h=\frac{E-\frac{1}{2}*m*v^{2}  }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2}  }{\frac{1}{2}m }

11)

N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2}   }{\frac{1}{2}*k }

12)

x=x_{o} +\frac{v^{2-v_{o}^{2}  } }{2a}  ;  m=\frac{P*A-F_{1}-F_{2} }{g}

13)

x_{o} = x-\frac{F}{k} ;  u=\frac{cos(o)-\frac{a}{g} }{sin(o)}

14)

t=\frac{d}{v} +t_{o} ; t_{o} = t-(\frac{v-v_{o} }{a} )

15)

F_{2}=\frac{W-F_{1} *d}{d}+F_{3}   ;  v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}

16)

y_{1}=y-\frac{u}{mg}  ; x^{2} = \frac{2W}{k}+x_{o} ^{2}

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Answer:

<em>TRUE</em>

Explanation:

The gravitational force between two objects becomes<u><em> weaker</em></u><em> if the two objects are </em><u><em>moved apart</em></u> and <em>stronger</em><em> if they are brought </em><em>closer</em><em> together</em>; that is, the force depends on the distance between the objects

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