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vagabundo [1.1K]
3 years ago
13

How does dance use energy to influence motion?

Physics
1 answer:
Goshia [24]3 years ago
8 0
Solar system energy. electoral power engery motion force engery . wind energy. kinetic energy
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Anyone know the answer ?
nata0808 [166]

Answer:

d

Explanation:

4 0
3 years ago
Read 2 more answers
The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T
zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than -0.25 m/s^2

f) The two trains avoid collision if the acceleration of the freight train is at least 0.35 m/s^2

Explanation:

a)

We can describe the position of the passenger train at time t with the equation

x_p(t)=u_p t + \frac{1}{2}at^2

where

u_p = 25.0 m/s is the initial velocity of the passenger train

a=-0.100 m/s^2 is the deceleration of the train

On the other hand, the position of the freight train is given by

x_f(t)=x_0 + v_f t

where

x_0=200 m is the initial position of the freight train

v_f = 15.0 m/s is the constant velocity of the train

The collision occurs if the two trains meet, so

x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

b)

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

x_f(t)=x_0 +v_ft

And substituting t = 22.5 s, we find:

x_f(22.5)=200+(15)(22.5)=537.5 m

We can verify that the passenger train is at the same position at the time of the collision:

x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m

So, the two trains collide at x = 537.5 m.

c)

In the graph in attachment, the position-time graph of each train is represented. We have:

  • The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
  • The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

d)

In order to avoid the collision, the freight train must have a initial position of

x_0'

such that the two trains never meet.

We said that the two trains meet if:

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t

Re-arranging,

\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0

Substituting the values for the acceleration and the velocity,

0.05t^2-10t+x_0'=0

The solution of this equation is given by the formula

t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m

Therefore, the freight train must have a head start of 500 m.

e)

In this case, we want to find the acceleration a' of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t

Re-arranging

\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0

Substituting,

-0.5at^2-10t+200=0

The solution to this equation is

t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}

Again, the two trains never meet if the discriminant is negative, so

10^2-4\cdot (-0.5a') \cdot (200)

So, the deceleration must be smaller (towards negative value) than -0.25 m/s^2

f)

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2

where a_f is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2

Re-arranging,

\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0

And the solution is

t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}

Again, the two trains avoid collision if the discriminant is negative, so

10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2

Learn more about accelerated motion:

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#LearnwithBrainly

8 0
3 years ago
Question 1 of 15
Ierofanga [76]

Answer:

d 75 miles per hour

Explanation:

60 ÷ 20

3 × 25 = 75

hoped that helped!!

8 0
3 years ago
State ohm’s law Q.34.7
defon

Ohm's law states that: V=RI

Explanation:

Ohm's law states that in a conductor, the potential difference across the conductor is directly proportional to the current flowing through it. Mathematically,

V \propto I

where

V is the potential difference

I is the current

The constant of proportionality is called resistance (R), and it gives a measure of "how much the conductor opposes" to the flow of current. Therefore Ohm's law can be rewritten as

V=RI

where R is the resistance. By rewriting the equation as

I=\frac{V}{R}

we see that the larger the resistance, the lower the current in the conductor.

Learn more about Ohm's law:

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4 0
3 years ago
Is it true that the cell theory states that cells are produced by non living cells
sergeinik [125]
The cell theory states that:

1) All living organisms are composed of one or more cells.

2) The cell is the basic unit of structure and organization in organisms.

3) Cells arise from pre-existing cells.

Your statement is not present, so the answer would be no.

♧
5 0
3 years ago
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