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nirvana33 [79]
3 years ago
11

What is meant by the expression action at a distance? (physics)

Physics
1 answer:
Tju [1.3M]3 years ago
4 0
Two things acting on each other without contact.
You might be interested in
A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A
Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

     Δp / p = 1

In percentage form it is 100%

3 0
3 years ago
A runner starts from rest and in 3 s reaches a speed of 8 m/s. If we assume that the speed changed at a constant rate (constant
Stells [14]

Answer:

The average speed of the runner is 4 m/s.

Explanation:

Hi there!

The average speed (a.s) is calculated by dividing the traveled distance (d) over the time needed to travel that distance (t):

a.s = d / t

So, let´s find the distance traveled in those 3 s. For that, we can use the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at the point where the runner starts, then, x0 = 0. Since the runner starts from rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

We have the time (3 s), so let´s find the acceleration. For that, we can use the equation of velocity of an object moving in a straight line with constant acceleration:

v = v0 + a · t

Where "v" is the velocity at a time "t". Since v0 = 0, then:

v = a · t

At t = 3 s, v = 8 m/s

8 m/s = a · 3 s

8/3 m/s² = a

So let´s find the position of the runner at t = 3 s (In this case, the position of the runner will be equal to the traveled distance):

x = 1/2 · a · t²

x = 1/2 · 8/3 m/s² · (3 s)²

x = 12 m

Now, we can calcualte the average speed:

a.s = d/t

a.s = 12 m / 3 s

a.s = 4 m/s

The average speed of the runner is 4 m/s.

4 0
3 years ago
40 Points!!!!!!!!!!!!!
Firlakuza [10]
The answer is A for sure
3 0
3 years ago
A car begins at rest (0 velocity), 5 seconds later it is travelling at 20 meters/per second. What was the acceleration of the ve
klasskru [66]

use the formula

v= u+ at

v is final velocity , u is initial velocity , a is acceleration and t is time

put the values

20 = 0+ a×5

a = 4 m/s²

4 0
2 years ago
Read 2 more answers
How do you do this problem?
mojhsa [17]

Answer:

C

Explanation:

When A and B come in contact with each other, +12 - 12 = 0 so their changes cancel.

Now C has a charge of +12

When A and C come together they each have an equal share of that 12, so each of them has 6

So the answer is

A       B       C

6        0       6

which is C

4 0
3 years ago
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