All categories

3 years ago

A classmate is working with a source that

1 answer:

3 0

The term cell is incorrect because, the average voltage of a cell is 3.40 V and as already mentioned in question, the volt of source is 18 V, so cell couldn't be the appropriate word, actually the source is battery.

Now, the source is most likely related to cell because a battery which is the source here, is made up of a "group of cell".

You might be interested in

a car traveling at 30m/s notices an accident about 65m up the road. The car's brakes are capable of decelerating at a rate of 6.

uysha [10]

Yes because you would have at least 3 car spaces

4 0

3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

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

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

two cars are travelling side by side down a straight road in opposite directions. both cars are travelling at a speed of 22 m/s.

Nimfa-mama [501]

Answer: 44m/s

Explanation:

Speed of Car A travelling left = 22m/s

Speed of Car B travelling right = 22m/s

Now recall that relative speed of objects moving in opposite directions is equal to the sum of each speed

Hence, Relative speed = (Speed of Car A + Speed of Car B)

= (22m/s + 22m/s)

= 44m/s

3 0

4 years ago

A mass of gas under constant pressure occupies a volume of 0.5 m3 at a temperature of 20°C. Using the formula for cubic expansio

Tanzania [10]

Without a change in pressure

V/T = V/T
Before change = after change - expansion

x = volume at m3

(0.5)/20 = x/45

.025 = x/45

.025 * 45 = x

1 = x

The volume of this gas is 1 m3.

Hope this helps :) also the gas expanded .5 of m3.

4 0

3 years ago

Choose all of the true statements regarding the relationship between voltage, resistance, and current. Current is measured in am

wel

-- Current is measured in amps. (You can use any symbol you want to represent current, but the most common one is " I ", not "Δ".)

-- The relationship between current, voltage, and resistance is mathematically defined by Ohm's Law.

-- Current is the flow of electrons through a circuit.

-- (Ohm's Law is NOT mathematically represented by the equation V=I/R.) It should be V = I · R .

(When solving for Resistance in a circuit and both voltage and current are known values, the equation I =V*R is not true, and not the way to solve it.) If the resistance is what you're looking for, then the equation to use is R = V / I .

-- If the voltage in a circuit is increased, the current will also increase.

6 0

3 years ago

Read 2 more answers