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3 years ago

When an electric circuit operates, a buzzer makes a sound. What happens to the electric charges in this circuit?

Physics

2 answers:

Lapatulllka [165]3 years ago

6 0

The best option for the answer of this question is D) The electric charges move from the energy source through the buzzer and back to the energy source.

uranmaximum [27]3 years ago

3 0

Answer: Option (d) is the correct answer.

Explanation:

In an electric circuit, when we press the switch or button then the circuit becomes complete. Then first of all current starts to flow from the energy source which might be a battery or cell etc.

Then the current reaches to the buzzer and hence, it makes a sound. After that, the current moves towards the battery or cell by completing its path.

Thus, we can conclude that the electric charges move from the energy source through the buzzer and back to the energy source.

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Which one of the following types of microscope gives the highest amount of magnification?

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4 years ago

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Which of the following is an example of a long-term climatic change?

Vaselesa [24]

Winter and Summer happen every year, pretty much on schedule.
When they happen, the 'weather' changes, but the 'climate' doesn't.

If you live in the right place, then you might get several blizzards every
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If the temperature where you live stays well below zero for a hundred
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4 years ago

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Which most likely could cause acceleration?

Anastaziya [24]

Answer:

this may seem wierd but a hill

Explanation:

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3 years ago

For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with d

Sedaia [141]

Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material ( $\sigma$ ) = 200 M Pa

Diameter of the bar = 4.5 cm = 45 mm

We know that yield stress of the bar is given by the formula

Yield Stress = $\frac{Maximum load}{Area of the bar}$

⇒ $\sigma$ = $\frac{P_{max} }{A}$ ---------------- (1)

⇒ Area of the bar (A) = $\frac{\pi}{4}$ × $D^{2}$

⇒ A = $\frac{\pi}{4}$ × $45^{2}$

⇒ A = 1589.625 $mm^{2}$

Put all the values in equation (1) we get

⇒ $P_{max}$ = 200 × 1589.625

⇒ $P_{max}$ = 317925 N

In this bar the $P_{max}$ is equal to the weight of the bar.

⇒ $P_{max}$ = $M_{max}$ × g

Where $M_{max}$ is the maximum mass the bar can support.

⇒ $M_{max}$ = $\frac{P_{max} }{g}$

Put all the values in the above formula we get

⇒ $M_{max}$ = $\frac{317925}{9.81}$

⇒ $M_{max}$ = 32408.26 Kg

There fore the maximum mass the bar can support without yielding = 32408.26 kg

3 0

4 years ago

A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a

vredina [299]

(a) 4.0 s

The acceleration of the car is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

$a=-5.0 m/s^2$ is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

$t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s$

(b) 40 m

The stopping distance of the car can be calculated by using the equation

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m$

(c) $-5.0 m/s^2$

The deceleration is given by the problem, and its value is $-5.0 m/s^2$ .

(d) 5000 N

The net force applied on the car is given by

$F=ma$

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

$a=5.0 m/s^2$ is the magnitude of the acceleration

Solving the formula, we find

$F=(1000 kg)(5.0 m/s^2)=5000 N$

(e) $2.0\cdot 10^5 J$

The work done by the force applied by the car is

$W=Fd$

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

$W=(5000 N)(40 m)=2.0\cdot 10^5 J$

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0

4 years ago