All categories

3 years ago

What is the current (Amps) of a circuit with a 9V battery and a 18Ω lamp?

1 answer:

8 0

Answer: A standard 9V battery has about 400-600 mAh capacity. In the most basic terms, these batteries can supply about 500 milliamps for one hour before being "dead". With electricity, we measure the amount of charge flowing through the circuit over a period of time. Current is measured in Amperes (usually just referred to as "Amps"). An ampere is defined as 6.241*10^18 electrons (1 Coulomb) per second passing through a point in a circuit. In a series circuit, amperage, or amplitude, of the current remains constant and can be calculated using Ohm's law V = I/R while the voltage drops across each resistor that can be summed up to get the total resistance.

Hope this helps........ Stay safe and have a Merry Christmas!!!!!!!! :D

You might be interested in

Determine the accelerations that result when a 45 N net force is applied to 3kg object

svp [43]

Answer:

acceleration is 18

Explanation:

45N/3kg=18

4 0

3 years ago

Read 2 more answers

When the moon slowly moves away from the earth.How is it affecting the gravitational force between the earth and the moon?

sukhopar [10]

Well, think about how the tides will be affected when the moon moves farther away. If the moon first started off very close the earth, we would have more tsunamis. (Scientists have found that the moon has possibly been closer to earth long ago.) While it moves away, soon there will no longer be many tides.

3 0

3 years ago

Objects A and B both start at rest. They both accelerate at the same rate. However, object A accelerates for 3x the time as obje

mylen [45]

Answer:

v_Object A = 3v_Object B

1/3v_Object A = v_Object B

Explanation:

Let's see what variables we have in this problem.

Since the objects both start at rest, we have an initial velocity of 0 m/s.

The objects accelerate, so we will have acceleration.

Object A accelerates for 3x the time as Object B, so we have time.

The problem wants us to compare their final speeds, so we have final velocity.

Check what constant acceleration kinematic equation has these variables:

v = v₀ + at

Let's create values for the unknown variables and compare the final velocities.

v = ?
v₀ = 0 m/s (objects start at rest)
a = 5 m/s²
t = 10 s

Since Object A accelerates for 3x the time as Object B, we can use t = 30 s for Object A.

Let's write the two equations:

v = (0) + (5)(10)
v = (0) + (5)(30)

Simplify these equations.

v = 50 m/s
v = 150 m/s

Let's use another set of values to compare the final velocities to see if the velocities differ by 100 m/s or Object A has 3x the final velocity of Object B.

v = ?
v₀ = 0 m/s (objects start at rest)
a = 3 m/s²
t = 7 s

Write the two equations:

v = (0) + (3)(7)
v = (0) + (3)(21)

Simplify these equations.

v = 21 m/s
v = 63 m/s

Now we can clearly see that the final velocities are differed by 3x. Object A has 3x the final speed compared to that of Object B.

3 0

3 years ago

At time t=0 a 2150 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. This forc

amm1812

Explanation:

Given that,

Mass of the rocket, m = 2150 kg

At time t=0 a rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. The force is given by equation :

$F=At^2$

Here F = 888.93 N when t = 1.25 s

$F=At^2\\\\A=\dfrac{F}{t^2}\\\\A=\dfrac{888.93}{(1.25)^2}\\\\A=568.91\ N/s^2$

The value of A is $568.91\ N/s^2$ .

(a) Let J is the impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired. It is given in terms of force as :

$J=\int\limits {F{\cdot} dt}$

Limits will be from 2 s to 2+ 4 = 6 s

It implies :

$J=\int\limits^6_2 {At^2{\cdot} dt}\\\\J=A\int\limits^6_2 {t^2{\cdot} dt}\\\\J=A\dfrac{t^3}{3}|_2^6\\\\J=568.91\times \dfrac{1}{3}\times (6^3-2^3)\\\\J=39444.42\ Ns$

(b) Impulse is also equal to the change in momentum as :

$J=m\Delta v\\\\\Delta v=\dfrac{J}{m}\\\\\Delta v=\dfrac{39444.42}{2150}\\\\\Delta v=18.34\ m/s$

Hence, this is the required solution.

5 0

3 years ago

A competitive go-cart driver is traveling 32 m/s. He sees a caution flag go up, so he slows at a rate of -1.5 m/s^2 in 10.8 s. W

Svetllana [295]

Answer:

15.8 m/s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 32 m/s.

Acceleration (a) = – 1.5 m/s²

Time (t) = 10.8 s.

Final velocity (v) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration (a) = [final velocity (v) – initial velocity (u)] / time (t)

a = (v – u) /t

With the above formula, we can obtain the final velocity of go-cart driver as follow:

Initial velocity (u) = 32 m/s.

Acceleration (a) = – 1.5 m/s²

Time (t) = 10.8 s.

Final velocity (v) =?

a = (v – u) /t

– 1.5 = (v – 32) / 10.8

Cross multiply

(v – 32) = –1.5 × 10.8

v – 32 = – 16.2

Collect like terms

v = – 16.2 + 32

v = 15.8 m/s

Therefore, the final velocity of go-cart driver is 15.8 m/s.

5 0

3 years ago