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jok3333 [9.3K]
2 years ago
9

Electrons move from negative to positive while the voltage flows ________ to _______ .

Physics
1 answer:
Sedaia [141]2 years ago
3 0

Answer:

positive to negative.

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Is metal denting a physical change
Scilla [17]
No it is a chemical change
7 0
3 years ago
How many seconds will it take for a the International Space Station to travel 450 km at a rate of 100 m/s?
SVEN [57.7K]

Time = (distance) / (speed)

<em></em>

Time = (450 km) / (100 m/s)

Time = (450,000 m) / (100 m/s)

Time = <em>4500 seconds </em>(that's 75 minutes)

Note:

This is about HALF the speed of the passenger jet you fly in when you go to visit Grandma for Christmas.

If the International Space Station flew at this speed, it would immediately go ker-PLUNK into the ocean.

The speed of the International Space Station in its orbit is more like 3,100 m/s, not 100 m/s.

8 0
3 years ago
As a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change
Umnica [9.8K]
Refer to the diagram shown below.

m =  the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A =  the amplitude ( the maximum distance) of the mass from the equilibrium
        position

The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω =  the circular frequency of the motion
T =  the period of the motion so that ω = (2π)/T

The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)

In the equilibrium position,
x is zero;
v is maximum;
a is zero.

At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.

In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.

6 0
3 years ago
A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

8 0
3 years ago
During baseball practice, a batter hits a high flyball, and then runs in a straight line and catches it. which has the greater a
Jlenok [28]
They have the same velocity because their displacements (shortest line from point A to point B, which is a straight line) are the same and they meet at the same time.
6 0
3 years ago
Read 2 more answers
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