What is the process of conduction in terms of particle movement

A 1500 W electric heater is plugged into the outlet of a 120 V circuit that has a 20 A circuit breaker. You plug an electric hai

Mila [183]

Answer:

Explanation:

Current drawn by electric heater = power/volt =1500/120 = 12.5 A.

current drawn by hair drier at 600 watt = 600/120 =5 A

current drawn by hair drier at 900 watt = 900/120 = 7.5 A.

Total current drawn by heater and hair drier used at 900 watt

= 12.5 + 7.5 = 20 A

Breaking current =20 A

So fuse will trip at this point .

3 0

3 years ago

a ball of diameter 10 cm and mass 10 grams is dropped in a container of water. the cross sectional area of the container is 100

Anastaziya [24]

Answer:

h = 9.83 cm

Explanation:

Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water

let's reduce the magnitudes to the SI system

r = 10 cm = 0.10 m

m = 10 g = 0.010 kg

A = 100 cm² = 0.01 m²

the definition of density is

ρ = m / V

the volume of a sphere

V = $\frac{4}{3} \ \pi r^{3}$

V = $\frac{4}{3}$ π 0.1³

V = 4.189 10⁻³ m³

let's calculate the density of the ball

ρ = $\frac{0.010}{4.189 \ 10^{-3} }$

ρ = 2.387 kg / m³

the tabulated density of water is

ρ_water = 997 kg / m³

we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation

B - W = 0

B = W

where B is the thrust that is given by Archimedes' principle

ρ_liquid g V_submerged = m g

V_submerged = m / ρ_liquid

we calculate

V _submerged = 0.10 9.8 / 997

V_submerged = 9.83 10⁻⁴ m³

The volume increassed of the water container

V = A h

h = V / A

let's calculate

h = 9.83 10⁻⁴ / 0.01

h = 0.0983 m

this is equal to h = 9.83 cm

8 0

3 years ago

The asteroid 2007 VK184 has one chance in 2940 of hitting the Earth between the years 2048 and 2057. The asteroid is 130 meters

N76 [4]

Answer:

The average death rate for this type of event is closest to 350 people per event.

Explanation:

Between the years 2048 and 2057

Is 10 years. Since there is only one chance of event between these years, there is no point to consider it.

The total number of events = 2940 chances.

average death rate per event = total number of dead people divided by total number of events of occurrence.

1000000/2940 = 340.134

The average death rate for this type of event is therefore closest to 350 people per event

4 0

3 years ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago

Can someone help asap with this

kondaur [170]

Answer:

B

Explanation:

i had a test on this and got it correct

7 0

3 years ago