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emmasim [6.3K]
3 years ago
8

Which of the following is a definition of acceleration? *

Physics
1 answer:
slavikrds [6]3 years ago
8 0

Answer:

Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing.

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Why is energy required to get an object moving?
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Answer:

this is what popped up when I searched it up:In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.

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Two traveling sinusoidal waves are described by the wave functions y1 = 4.85 sin [(4.35x − 1270t)] y2 = 4.85 sin [(4.35x − 1270t
Tamiku [17]

Answer:

Approximately 9.62.

Explanation:

y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0].

y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)].

Notice that sine waves y_1 and y_2 share the same frequency and wavelength. The only distinction between these two waves is the (-0.250) in y_2\!.

Therefore, the sum (y_1 + y_2) would still be a sine wave. The amplitude of (y_1 + y_2)\! could be found without using calculus.

Consider the sum-of-angle identity for sine:

\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b).

Compare the expression \sin(a + b) to y_2. Let a = (4.35\, x - 1270) and b = (-0.250). Apply the sum-of-angle identity of sine to rewrite y_2\!.

\begin{aligned}y_2 &= 4.85\, \sin[(\underbrace{4.35\, x - 1270\, t}_{a}) + (\underbrace{-0.250}_{b})]\\ &= 4.85 \, [\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}.

Therefore, the sum (y_1 + y_2) would become:

\begin{aligned}& y_1 + y_2\\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t) \\ &\quad \quad \quad\;+\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}.

Consider: would it be possible to find m and c that satisfy the following hypothetical equation?

\begin{aligned}& (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)\\&= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}.

Simplify this hypothetical equation:

\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\&=\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)\end{aligned}.

Apply the sum-of-angle identity of sine to rewrite the left-hand side:

\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\[0.5em]&=m\, \sin(4.35\, x - 1270\, t)\cdot \cos(c) \\ &\quad\quad + m\, \cos(4.35\, x - 1270\, t)\cdot \sin(c) \\[0.5em] &=\sin(4.35\, x - 1270\, t)\cdot (m\, \cos(c)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot (m\, \sin(c)) \end{aligned}.

Compare this expression with the right-hand side. For this hypothetical equation to hold for all real x and t, the following should be satisfied:

\displaystyle 1 + \cos(-0.250) = m\, \cos(c), and

\displaystyle \sin(-0.250) = m\, \sin(c).

Consider the Pythagorean identity. For any real number a:

{\left(\sin(a)\right)}^{2} + {\left(\cos(a)\right)}^{2} = 1^2.

Make use of the Pythagorean identity to solve this system of equations for m. Square both sides of both equations:

\displaystyle 1 + 2\, \cos(-0.250) +  {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2.

\displaystyle {\left(\sin(-0.250)\right)}^{2} = m^2\, {\left(\sin(c)\right)}^2.

Take the sum of these two equations.

Left-hand side:

\begin{aligned}& 1 + 2\, \cos(-0.250) + \underbrace{{\left(\cos(-0.250)\right)}^2 + {\left(\sin(-0.250)\right)}^2}_{1}\\ &= 1 + 2\, \cos(-0.250) + 1 \\ &= 2 + 2\, \cos(-0.250) \end{aligned}.

Right-hand side:

\begin{aligned} &m^2\, {\left(\cos(c)\right)}^2 + m^2\, {\left(\sin(c)\right)}^2 \\ &= m^2\, \left( {\left(\sin(c)\right)}^2 +  {\left(\cos(c)\right)}^2\right)\\ &= m^2\end{aligned}.

Therefore:

m^2 = 2 + 2\, \cos(-0.250).

m = \sqrt{2 + 2\, \cos(-0.250)} \approx 1.98.

Substitute m = \sqrt{2 + 2\, \cos(-0.250)} back to the system to find c. However, notice that the exact value of c\! isn't required for finding the amplitude of (y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c).

(Side note: one possible value of c is \displaystyle \arccos\left(\frac{1 + \cos(0.250)}{\sqrt{2 \times (1 + \cos(0.250))}}\right) \approx 0.125 radians.)

As long as \! c is a real number, the amplitude of (y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c) would be equal to the absolute value of (4.85\, m).

Therefore, the amplitude of (y_1 + y_2) would be:

\begin{aligned}|4.85\, m| &= 4.85 \times \sqrt{2 + 2\, \cos(-0.250)} \\&\approx 9.62 \end{aligned}.

8 0
3 years ago
This is a science Question. Please help.
sattari [20]

Answer:

B

Explanation:

This is a physics question, know that force is equals to mass divided by acceleration (acc.), so if the same force is applied, say 10 Newton and the mass of A is 2 and the mass of B is 4, then the acceleration of A is 0.2 and that of B is 0.4 by equating, and this applies to all cases.

6 0
3 years ago
A mass spectrometer applies a voltage of 2.00 kV to accelerate a singly charged positive ion. A magnetic field of B = 0.400 T th
N76 [4]

The mass of the ion is 5.96 X 10⁻²⁵ kg

<u>Explanation:</u>

The electrical energy given to the ion Vq will be changed into kinetic energy \frac{1}{2}mv^2

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force \frac{mv^2}{r}.

So,

Vq = \frac{1}{2}mv^2

and

Bqv = \frac{mv^2}{r}

Right from these eliminating v, we can derive

m = \frac{B^2r^2q}{2V}

On substituting the value, we get:

m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\

m = 5.96 X 10⁻²⁵ kg.

3 0
3 years ago
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