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asambeis [7]
3 years ago
6

I. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maxi

mum compression. Indicate whether the work done by the spring is positive, negative, or zero.
Physics
1 answer:
nalin [4]3 years ago
6 0

Answer:

Negative

Explanation:

If the box is heading right in the positive direction, the work will be negative. The spring has an opposite force to that of the box.

Hope this helped. :)

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Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is
victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = \frac{-kQq}{r^2}

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = \frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2} = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive  x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = \frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2} = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

3 0
3 years ago
How does mass affect the gravitational pull of the Sun, Earth, and the Moon?
madreJ [45]
Larger mass creates a stronger pull
3 0
2 years ago
Read 2 more answers
A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s
Paraphin [41]

Answer:

\omega = 0.016\,\frac{rad}{s}

Explanation:

The rotation rate of the man is:

\omega = \frac{v}{R}

\omega = \frac{0.80\,\frac{m}{s} }{5\,m}

\omega = 0.16\,\frac{rad}{s}

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega

The final angular speed is:

\omega = 0.016\,\frac{rad}{s}

3 0
3 years ago
What is the equivalent resistance of the circuit?
Kaylis [27]

Answer:

80 Ω.

Explanation:

In this circuit the resistances are in series.The equivalent resistance of a series circuit is equal to the sum of the resistances. Req= 60 + 20 = 80 Ω.

8 0
3 years ago
The small ball of mass m and its supporting wire become a simple pendulum when the horizontal cord is severed. Determine the rat
natali 33 [55]

Answer:

See the attached image and the explanation below

Explanation:

We must draw a schematic of the described problem, after the sketch it is necessary to make a free body diagram, at the time before and after cutting the cord.

These free body diagrams can be seen in the attached image.

First we perform a sum of forces on the x & y axes before cutting the cord, to be able to find the T tension of the wire. (This analysis can be seen in the attached image).

In this way we get the T-wire tension equation, before cutting.

Now we make another free body diagram, for the moment when the wire is cut (see in the attached diagram).

It is important to clarify that when the cord is cut, the system will no longer be in statically, therefore newton's second law will be used for summation of forces which will be equal to the product of mass by acceleration.

Finally with equations 1 and 2 we can find the K ratio.

5 0
3 years ago
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