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4 years ago

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Jeff’s father is installing a do-it-yourself security system at his house. He needs to get a device from his workshop that conve

rts electric energy to sound energy. Which device is Jeff’s father looking for? switch motor buzzer bulb battery

2 answers:

Nostrana [21]4 years ago

6 0

Answer:

Buzzer

Explanation:

Lilit [14]4 years ago

4 0

Answer: buzzer.

The working principle of a buzzer is the conversion of electrical energy to sound energy.

The switch just cuts or permits the flow of current, the motor convertes electrical or other kind of energy into mechanical energy, a bulb converts electrical energy into light and a battery converts chemical energy into electrical energy.

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In February 1955, a paratrooper fell 1200 ft from an

-Dominant- [34]

Answer:

$s=1.1107\ m$ is the minimum depth of snow for survivable stopping.

Explanation:

Given:

terminal velocity of fall, $u=56\ m.s^{-1}$

mass of the paratrooper, $m=85\ kg$

force on the paratrooper by the ice to stop him, $F=1.2\times 10^5\ N$

<u>Firstly, we calculate the deceleration caused in the snow:</u>

$a=\frac{F}{m}$

$a=\frac{120000}{85}$

$a=1411.765\ m.s^{-2}$

Now, using equation of motion:

$v^2=u^2+2a.s$ ....................(1)

where:

v = final velocity of the body after stopping

u = initial velocity of the body just before hitting the snow

a = acceleration of the body in the snow

s = distance through in the snow

Putting respective values in eq. (1)

$0^2=56^2+2\times (-1411.765)\times s$

$s=1.1107\ m$

5 0

3 years ago

Archimedes supposedly was asked to determine whether a crown made for the king consisted of pure gold (density of gold is 19.3 ×

Vika [28.1K]

Answer:

Explanation:

First, we can find the mass of the object in air

Since W = mg

m = W/g = (8.9)/(9.8) = 0.908 kg

Then, by Archimedes principle, we can find its volume. The volume is found by the weight of the water displaced by the formula

W = Vρg

The Weight is the difference in scale readings. The density of water is 1000 kg/m3

(8.9- 7.8) = V(1000)(9.8)

Thus V = 1 X 10-4 m3

Then, since density is mass / volume

ρ = 0.908/1 X 10-4

ρ = 8254 kg/m3 which is 8 X 103 kg/m3

The density of gold is 19.3 X 103 kg/m3

Since those densities are not the same, the crown is either hollow or not pure gold

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%7B.5mv%7D%5E%7B2%7D%20%20%3D%20mgh" id="TexFormula1" title=" {.5mv}^{2} = mgh" alt=" {.5m

Katyanochek1 [597]

You said 0.5 · m · v² = m · g · h

Divide each side by 'm' : 0.5 · v² = g · h

Multiply each side by 2 : v² = 2 · g · h

Square root each side : v = √(2 · g · h)

You said that g = 9.8 m/s² and h = 875 units

So v = √(9.8 m/s² · 875 units)

v = √(8,575 m·unit/s²)

v = 92.6 / s² · √(m · unit)

8 0

3 years ago

550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c

Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work

7 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u> $m_{1}$ <u>:</u>

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

<u>For </u> $m_{2}$ <u>:</u>

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

<u>For </u> $m_{3}$ <u>:</u>

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago