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lesantik [10]
3 years ago
14

You have created a circuit in science class. You are using a 9 volt battery. You have generated what type of current?

Physics
2 answers:
gavmur [86]3 years ago
7 0

Batteries always provide Direct Current (DC).

If you're powering your project with batteries and you need some Alternating Current (AC) in parts of the circuit, then you have to provide an extra section of the circuit to take the DC from the battery and make AC with it.

marishachu [46]3 years ago
5 0

Batteries produce direct current.

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What force is needed to accelerate an object 5 m/s if the object has a mass of 10kg?
nevsk [136]
The <span>force that is needed to accelerate an object 5 m/s if the object has a mass of 10kg 50N because you multiply 5 and 10</span>
6 0
3 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
4 years ago
Describe the body's typical state during the first stage of sleep. (Site 1)
Ksju [112]

Answer:

There are four different stages of sleep.

Stage 1 NREM

Explanation:

<em>The process of firmly falling asleep has four stages through which a person goes.</em>

<em>It goes from being awake over light sleep and falling firmly into sleeping.</em>

(STAGE 1)

This is a stage in which there are non-rapid movements of the eyes. In other words, it is a process of dreamless sleep. You enter this stage the moment you decide to sleep and shut your eyes. After several minutes, your body is in fact in the sleeping mode, but not entirely. This means that you can easily be woken up without being aware that you have slept.

Features:

  1. <em>You can easily awake</em>
  2. <em>Your may roll and they may be a little open</em>
  3. <em>The blood pressure and the temperature of the brain start to decrease </em>
  4. <em>You experience the natural human reflexes that the brain sends to assure that the place of your sleep is in a safe environment. By sending twitches to your muscles, your brain may awake your body for several seconds which comes in handy if you are tired and close to sleep on work or some dangerous place like a cliff for example.</em>
  5. <em>Your breading starts to slow down alongside with your pressure and temperature, and your heartbeats slow down.</em>
6 0
3 years ago
Read 2 more answers
Two long, parallel wires separated by 3.50 cm carry currents in opposite directions. The current in one wire is 1.55 A, and the
vaieri [72.5K]

Answer:

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

Explanation:

Given:

Two long, parallel wires separated by a distance,

d = 3.50 cm = 0.035 meter

Currents,

I_{1}=1.55\ A\\I_{2}=3.15\ A

To Find:

Magnitude of the force per unit length that one wire exerts on the other,

\dfrac{F}{l}=?

Solution:

Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,

\dfrac{F}{l}=\dfrac{\mu_{0}\times I_{1}\times I_{2}}{2\pi\times d}

where,

\mu_{0}=permeability\ of\ free\ space =4\pi\times 10^{-7}

Substituting the values we get

\dfrac{F}{l}=\dfrac{4\pi\times 10^{-7}\times 1.55\times 3.15}{2\pi\times 0.035}

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

7 0
3 years ago
Read 2 more answers
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