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3 years ago

What is the purpose of each of the components (each shown with leader and line) of the circuit shown is the diagram?

Physics

1 answer:

NemiM [27]3 years ago

5 0

The wires is what is needed to put together the whole thing, kinda like glue when you're gluing a piece of paper on it.

Anyways, the battery is the main source and main energy per say.

That energy that comes from the battery, thanks to the wires, it can transfer that said energy to both the switch and light bulb.

And as you flick the switch, it depends of how you put it together, there's two options, turning the light bulb on or turning it off.

Though it doesn't mean that since the light bulb is connected to the battery makes the bulb turn on no matter what since the switch can cancel the main source's energy.

- Ouma :>

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The temperature measurement scale that begins at absolute zero is

ikadub [295]

<h3>Answer: A. Kelvin</h3>

========================================================

Explanation:

At 0 Kelvin, we're at the coldest temperature possible which is referred to as absolute zero. We cannot get any colder than this. At this temperature, the molecules aren't moving or vibrating to provide heat energy. The higher the temperature, the more the molecules are moving or vibrating.

The 0 degree mark on the Celsius scale is well above absolute zero. So that allows for negative Celsius temperatures to be possible. The same can be said about Fahrenheit as well. We cannot have negative Kelvin temperatures.

4 0

3 years ago

Read 2 more answers

A bus travels 6 km east and then 8 km south. The magnitude of the bus’s resultant displacement is ___km.

Crank

Answer:

10 km

Explanation:

You can draw your problem out. You can use the pathagoran theorem to figure it out. So $6^{2}$ + $8^{2}$ and then square root the answer which will be $\sqrt{100}$ which equals to 10.

3 0

3 years ago

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0

3 years ago

Can any juniors who go to Texas connections academy help me out with physics?

zimovet [89]

I'm not from that school but I can help you.

3 0

3 years ago

spotlight on a boat is 2.5 m above the water, and the light strikes the water at a point that is 8.0 m horizontally displaced fr

ludmilkaskok [199]

Answer:

Explanation:

Let i be the angle of incidence and r be the angle of refraction .

From the figure

Tan ( 90 - i ) = 2.5 / 8

cot i = 2.5 / 8

Tan i = 8 / 2.5 = 3.2

i = 72.65°

From snell's law

sini / sin r = refractive index

sin 72.65 / sinr = 1.333

sin r = .9545 / 1.333

= .72

r = 46⁰

From the figure

Tan r = d / 4

Tan 46 = d /4

d = 4 x Tan 46

= 4 x 1.0355

=4.14 m .

3 0

3 years ago