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3 years ago

a) An electric circuit is rates, 240V, 13Ω calculate the amount of current that can be used by this appliance.

Physics

1 answer:

zaharov [31]3 years ago

5 0

Answer:

18.4615385 amps

Explanation:

The voltage V in volts (V) is equal to the current I in amps (A) times the resistance R in ohms (Ω):

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True or False? Materials that are good conductors of heat are usually poor conductors of electricity.

alexandr402 [8]

False

HOPE THIS WILL HELP YOU

5 0

3 years ago

Read 2 more answers

At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.70 m/s2. At the same inst

kifflom [539]

Answer:

66.85 m

Explanation:

We are given that

Acceleration ,a= $2.7m/s^2$

Speed of truck, v=9.5 m/s

We have to find the distance beyond which the traffic signal will the automobile overtake the truck.

Initial speed of automobile, u=0

We know that

$s=ut+\frac{1}{2}at^2$

Using the formula

$s=0+\frac{1}{2}(27)t^2=\frac{27}{2}t^2$

For constant speed

Acceleration, a=0

Again

$s=vt+0=9.5t$

$9.5t=\frac{27}{2}t^2$

$t=\frac{9.5\times 2}{2.7}=7.037s$

Substitute the value of t

$x=9.5(7.037)=66.85m$

Hence, the distance beyond which the traffic signal will the automobile overtake the truck=66.85 m

3 0

3 years ago

2. In contrast to Venus, the coldest temperature yet measured on the surface of

garik1379 [7]

Answer:

Explanation:

<u>Temperature Scales </u>

There are three temperature scales in the modern sciences: Fahrenheit, Celsius, and Kelvin. Fahrenheit temperature scale assigns the value 32 for the freezing point of water and 212 for the boiling point of water and divides that interval into 180 parts. Celsius scale has a similar reference, giving 0 to the freezing point of water and 100 for the boiling point of water. The conversion between them is as follows

$\displaystyle F=\frac{9}{5}C+32$

$K=C+273.15$

The coldest temperature yet measured on the surface of any body in the solar system is -235°C. Converting to Fahrenheit

$\displaystyle F=\frac{9}{5}(-235)+32=-391^oF$

$K=-235+273.15=38.15^oK$

4 0

3 years ago

A particle moves according to a law of motions=f(t),t≥0 Wheretismeasured in seconds andsin feet.a) Find the velocity at timet.b)

Ray Of Light [21]

Answer:

a) $v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}$

b) 3s

c) t < 3s

d) 1.8ft

e) particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

Explanation:

a)Suppose the equation for motion is:

$f(t) = \frac{9t}{t^2+9}$

Then the velocity is the derivative of the motion function

$v(t) = \frac{df(t)}{dt} = (9t(t^2+9)^{-1})^'$

From here we can apply product rule

$v(t) = (9t)^{'}(t^2+9)^{-1} + (9t)((t^2+9)^{-1})^'$

$v(t) = \frac{9}{t^2+9} - \frac{9t(2t)}{(t^2+9)^2}$

$v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}$

b) The particle is at rest when v(t) = 0:

$\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} = 0$

$\frac{9}{t^2+9} = \frac{18t^2}{(t^2+9)^2}$

Multiply 2 sides by $(t^2+9)^2$ we have:

$t^2 + 9 = 2t^2$

$t^2 = 9$

$t = 3s$

$\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} > 0$

$\frac{9}{t^2+9} > \frac{18t^2}{(t^2+9)^2}$

Multiply 2 sides by $(t^2+9)^2$ we have:

$t^2 + 9 > 2t^2$

$t^2 < 9$

$t < 3s$

(d) As particle is moving in positive direction when t < 3s and negative direction when t > 3s, we can calculate the distance it's moving up to 3s and then after 3s

$f(3) = \frac{9*3}{3^2+9} = \frac{27}{18} = 1.5 ft$

At 3s, particle is changing direction to negative, so its position at 6s is

$f(6) = \frac{9*6}{6^2+9} = \frac{54}{45} = 1.2 ft$

Therefore from 3s to 6s it would have moved a distance of 1.5 - 1.2 = 0.3 ft

Then the total distance it has moved in the first 6 s is 1.5 + 0.3 = 1.8 ft

e) Acceleration is the derivative of velocity function:

$a(t) = \frac{dv(t)}{dt} = (\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2})^'$

$a(t) = -\frac{9(2(t^2+9))(2t)}{(t^2+9)^2} - \frac{36t}{(t^2+9)^2} + \frac{18t^2(2(t^2+9))(2t)}{(t^2+9)^3}$

$a(t) = -\frac{36t}{t^2+9} - \frac{36t}{(t^2+9)^2} + \frac{72t^3}{(t^2+9)^2}$

$a(t) = -\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2}$

Particle is speeding up when a(t) > 0:

$-\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2} > 0$

$\frac{72t^3 - 36t}{(t^2+9)^2} > \frac{54t}{t^2+9}$

as $t \& (t^2 + 9)^2 \geq 0$ we can multiply/divide both sides by it:

$8t^2 - 4 > 6(t^2+9)$

$8t^2 > 6t^2 + 58$

$t^2 > 29$

$t > \sqrt(29) \approx 5.385 s$

so particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

8 0

3 years ago

active and passive secruity measures are employed to identify, detect, classify and analyze possible threats inside of which zon

Juli2301 [7.4K]

Answer:

Assessment zone

Explanation:

It is the assessment zone in various security zones where active and passive security measures are employed to identify, detect, classify and analyze possible threats inside the assessment zones.

8 0

3 years ago

a) An electric circuit is rates, 240V, 13Ω calculate the amount of current that can be used by this appliance.​

a) An electric circuit is rates, 240V, 13Ω calculate the amount of current that can be used by this appliance.