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2 years ago

What is Resistance? How is it measured?

Physics

2 answers:

ser-zykov [4K]2 years ago

6 0

Answer:

Resistance is a measure of the opposition to current flow in an electrical circuit.

Explanation:

Amiraneli [1.4K]2 years ago

4 0

Answer:

Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance

Explanation:

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How many protons are in Calcium? (* Hint: The Atomic number of Calcium is 20).

ryzh [129]

Answer:

The answer to your question is: 20

Explanation:

Atomic number is the number of proton and atom has. Each element has a specific number of protons, if the number of protons change, then this is a new element.

Mass number is the number of protons and neutrons and atom has.

Mass number = protons + neutrons

Data

Number of protons = ?

Atomic number = 20

Then,

atomic number = number of protons = 20

3 0

2 years ago

At the circus, a 100.-kilogram clown is fired at 15 meters per second from a 500.- kilogram cannon. What is the recoil speed of

photoshop1234 [79]

The recoil velocity of cannon is (4) 5.0 m/s

Explanation:

We can find the recoil velocity from the law of conservation of momentum.

The recoil velocity is velocity of body 2 after release of body 1, i.e. velocity of cannon after release of clown.

Let v2 be cannon's velocity, v1 be clown's velocity given = 15 m/sec

m1 be clown's mass = 100kg and m2 be cannon's mass given = 500kg.

So recoil velocity of cannon v2 is given by,

v2 = -(m1÷m2)v1

v2 = -(100÷500)15

v2 = -5 m/s

where the minus sign refers to the direction of cannon's recoil velocity being opposite to that of clown.

Hence, option (4)5.0 m/s is the correct answer.

6 0

2 years ago

Read 2 more answers

Select all THREE of the correct response about the diagram of an atom.

STALIN [3.7K]

Answer: HI

Explanation:

5 0

2 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago

imagine that you are on a spaceship that is far from the sun or any planet. Explain why a 500 kg piece of equipment would be dif

Mila [183]

Even though the object is weightless, it would need inertia, I.e, you pushing it or any form of transportation. So you would still have to push that 500kg to just keep it moving in space. Say if it were a planet with less gravitational force, it would be weightless.

8 0

3 years ago

What is Resistance? How is it measured? ​

What is Resistance? How is it measured?