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2 years ago

What are the equations of angular acceleration, angular velocity and theta?.

Physics

1 answer:

soldier1979 [14.2K]2 years ago

3 0

Θ is the angular displacement = ωt
ω is the angular velocity = θ/t
α is the angular acceleration = ω/t

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what is the percent efficiency of a light bulb that emits 14 joules of light for every 60 joules of electric energy thst enters

jasenka [17]

If you're using the bulb as a source of light, then it's. 14/60 = 23.3% efficient.
If you're using it to heat a bird nest or a hamster cage, then it's. 46/60 = 76.7% efficient !
It just depends on your point of view, and what you consider 'useful' output.

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2 years ago

Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum

Alina [70]

Answer:

Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
Uncertainty in momentum (∆P) = ?
Planck's constant (h) = 6.26 × 10⁻³⁴ Js

$\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi}$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} }$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} }$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192}$

$\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21}$

$\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21}$

$\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}$

4 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

Consider two copper wires with circular cross-sections and equal lengths. one wire has 3 times the diameter of the other. How do

Masteriza [31]

The correct option is (C) The longer wire has 3 times the resistance of the shorter wire.

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

Why do circuits need resistance?

Using ohms (Ω) as the unit, resistance acts as a gauge to quantify how easily current will flow through a circuit. When resistance falls, current rises, and when resistance rises, current falls. In order to make sure that current flows in circuits at the proper rate, resistors are crucial.

How can resistance be determined in a circuit?

By applying Ohm's Law, you can get the total resistance if you know the total current and voltage through the entire circuit: R = V / I. A parallel circuit, for instance, has a voltage of 9 volts and a combined current of 3 amps. Total resistance (RT) is equal to 9 volts / 3 amps, or 3.

Learn more about the Resistance in circuit with the help of the given link:

brainly.com/question/1851488

#SPJ4

I understand that the question you are looking for is "Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistances of these two wires compare?

A) Both wires have the same resistance.

B) The longer wire has 1/3 the resistance of the shorter wire.

C) The longer wire has 3 times the resistance of the shorter wire.

D) The longer wire has 9 times times the resistance of the shorter wire.

E) The longer wire has 27 times times the resistance of the shorter wire."

5 0

2 years ago

A road with a radius of 75.0 m is banked so that a car can navigate the curve at a speed of 15.5 m/s without any friction. When

kirza4 [7]

Answer: 0.683

Explanation: The relationship between a car moving along a curve and the frictional force is given below as

us×g = v²/r

Where us = coefficient of static friction =?, g = acceleration due to gravity = 9.8 m/s², v = 22.4 m/s and r = 75m

By substituting the parameters, we have that

us×9.8 = 22.4²/ 75

us ×9.8 = 501.76/75

us ×9.8 = 6.690

us = 6.690/9.8 = 0.683

8 0

2 years ago