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Ray Of Light [21]
3 years ago
12

Which of these systems is an oscillator? A. A barrel rolling down the hill B. A child sitting on a swing a skater falling on the

ice C. A skater falling on the ice D. A car going uphill
Physics
1 answer:
Paladinen [302]3 years ago
7 0

Answer:

Option B:

A child sitting on a swing.

Explanation:

When we hear the word oscillator, a good example is the pendulum bob of a grandfather clock. We can picture the motion to get a perfect understanding of its path of motion and relate it to other systems of motion in our everyday life.

An oscillator is a system that moves in such a way that it reverses its direction after a period of time. It can be seen as a "to-and-fro" motion.

From the options, a child sitting on a swing is the perfect example of an oscillating system because the child will be moving forwards and backwards, alternately reversing the direction of motion with time.

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A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f
zalisa [80]

Explanation:

Charge=5.00 E-5

Force=0.751N

F=qE

0.751=5.00 E-5*E

E=1.502*10⁴

option a

8 0
3 years ago
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2CO + 2NO → 2CO2 + N2 is it balannced
natita [175]

If the same atoms appear on both sides, then it's balanced.

In this reaction, there are 4 Oxygens, 2 Carbons, and 2 Nitrogens on each side.  So numerically, <em>it's balanced</em>.  But I don't know enough chemistry to say whether the reaction is possible.

4 0
3 years ago
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At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0
3 years ago
Mr. Fuller assigned his science class a lab comparing the masses of objects to the nearest gram using a balance. The mass of a p
Tomtit [17]
The scale would need 10 aluminum cubes on one side. Figure out how many paper clips would be needed on the other side to balance this. You have to use more than one aluminum cube because you need to have enough cubes so that you get a whole number mass. 10 cubes gives you a total mass of 27 g for the aluminum.
5 0
3 years ago
The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1
Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (1)

Where:

F - Explosive force, measured in newtons.

\Delta s - Barrel length, measured in meters.

m - Mass of the shell, measured in kilograms.

v_{o}, v_{f} - Initial and final speeds of the shell, measured in meters per second.

If we know that m = 1250\,kg, v_{o} = 0\,\frac{m}{s}, v_{f} = 750\,\frac{m}{s} and \Delta s = 15\,m, then the explosive force experienced by the shell inside the barrel is:

F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}

F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}

F = 23437500\,N

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0
3 years ago
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