Ask question

Ask question

All categories

3 years ago

15

Select the correct answer. Brian is repairing an old alarm clock. He needs to replace a device that converts the electric energy

into sound energy. Which of these devices should he use? A. battery B. buzzer C. motor D. switch

2 answers:

Zielflug [23.3K]3 years ago

6 0

A buzzer is the answer

Stella [2.4K]3 years ago

3 0

it shouldbe a buzzer

You might be interested in

Georgie was pulling her brother (of mass 16 kg) in a 9.4 kg sled with a constant force of 39 N for one block (128 m). How much w

Genrish500 [490]

Answer:

4992 J

Explanation:

Force, F = 39 N

Distance, s = 128 m

Work done = force x distance

W = 39 x 128

W = 4992 J

Thus, the work done is 4992 J.

7 0

4 years ago

. Orbit of a satellite It is advantageous to place communications satellites in a circular orbit so that their position is fixed

Alexxx [7]

Answer:

a) r = 4.22 10⁷ m, b) v = 3.07 10³ m / s and c) a = 0.224 m / s²

Explanation:

a) For this exercise we will use Newton's second law where acceleration is centripetal and force is gravitational force

F = m a

a = v² / r

F = G m M / r²

G m M / r² = m v² / r

G M / r = v²

The squared velocity is a scalar and this value is constant, so let's use the uniform motion relationships

v = d / t

As the orbit is circular the distance is the length of the circle in 24 h time

d = 2π r

t = 24 h (3600 s / 1 h) = 86400 s

Let's replace

G M / r = (2π r / t)²

G M = 4 π² r³ / t²

r = ∛(G M t² / (4π²)

r = ∛( 6.67 10⁻¹¹ 5.98 10²⁴ 86400² / (4π²)) = ∛( 75.4 10²¹)

r = 4.22 10⁷ m

b) the speed module is

v = √G M / r

v = √(6.67 10⁻¹¹ 5.98 10²⁴/ 4.22 10⁷

v = 3.07 10³ m / s

c) the acceleration is

a = G M / r²

a = 6.67 10⁻¹¹ 5.98 10²⁴ / (4.22 10⁷)²

a = 0.224 m / s²

5 0

3 years ago

Which has more pressure, FIRST one gets all the points.

SSSSS [86.1K]

<u>Answer:</u>

area of point has more pressure.

<u>Explanation:</u>

formula's : pressure = force ÷ area

1st pressure,

200 ÷ 0.2 = 1000 Pa

2nd pressure,

200 ÷ 0.004 = 50000 Pa

<h3>Therefore the pressure on the area of the point is more.</h3>

8 0

3 years ago

1. Bone has a Young’s modulus of about

Blababa [14]

#1

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

#2

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

#3

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

$\rho = 1185.3 kg/m^3$

#4

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

$x = 3.48 cm$

so it is 3.48 cm below the interface

5 0

4 years ago

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

Pie

The only vertical forces acting on the wall are the weights of the man and the ladder as well as the normal force because the wall has no friction.

The maximal frictional forces are μsN2=(0.40)(900N)=360N, and N1=w1+wm=160N+740N=900N for the vertical forces to balance. At a height of 4.0 meters above the ground, the ladder makes contact with the wall. about the point of contact with the ground, balancing torques.

N2 = 174.0N because (4.0m)2 = (1.5m)(160N) + (1.0m)(35)(740N) =684N.m. It is necessary for the frictional force to counteract this horizontal force. Setting the frictional fore and n1 to a maximum of 360N and solving for the distance x along the ladder (0.4m)(360N)=(1.50m)(160N)+x(35)(740N) yields x=2.70m

<h3>Friction</h3>

When two sliding solid surfaces, fluid layers, or material components come into contact with one another, friction acts as a force to stop them from moving in the same direction. Friction comes in several forms:

The relative lateral motion of two solid surfaces in contact is opposed by dry friction. Between moving surfaces, there is kinetic friction, which is different from static friction. With the exception of atomic or molecule friction, asperities on the surface interact to produce dry friction.

When layers of a viscous fluid move in relation to one another, there is friction between them. This is known as fluid friction.

When a lubricant fluid separates two solid surfaces, it is called lubricated friction.

Learn more about friction here:

brainly.com/question/13030552

#SPJ4

8 0

2 years ago