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3 years ago

1.

Physics

2 answers:

xz_007 [3.2K]3 years ago

5 0

Answer: A. All of the answers are correct.

Anton [14]3 years ago

5 0

Answer:

A. All are correct

Explanation:

When a force is applied on an object, that object is accelerated. Acceleration is a change in the objects speed

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A small, positively charged ball is moved close to a large, positively charged ball. which describes how the small ball likely r

NNADVOKAT [17]

Answer;

-it will move away from the large ball because like charges repel.

Explanation;

-Electric force is the force that pushes apart two like charges, or that pulls together two unlike charges. The basic law of electrostatics Like charges of electricity repel each other, whereas unlike charges attract each other.

When small, positively charged ball is moved close to a large, positively charged ball it would be pushed away from the large positively charged ball since they are both positively charged. One has to put in energy to try to move the small ball closer to the large ball. The closer one try to move it to the large ball, the more energy one has to put in, so the more electrical potential energy the small ball would have.

6 0

3 years ago

Read 2 more answers

If C is 1kg and D is 100kg, and the initial velocities of both balls are 5m/s, how would the magnitude of the forces exerted by

kupik [55]

Answer:

Explanation:

The forces exerted by each mass is best understood in terms of their momentum.

Momentum is a sort of compelling force or impulse. It is given as:

Momentum = mass x velocity

Let us consider the momentum of the balls;

Substance C;

Mass = 1kg

Velocity = 5m/s

Momentum of C = 1 x 5 = 5kgm/s

Substance D:

Mass = 100kg

Velocity = 5m/s

Momentum of D = 100kg x 5m/s = 500kgm/s

Body D has a higher momentum compared to Body C. This suggests that body D will exert a higher force than C when they collide.

The higher the momentum, the more the force of impact it has.

3 0

3 years ago

16. Kinematic equations can only be used if...

Effectus [21]

the acceleration or the next force acting on the body is constant

7 0

3 years ago

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

A large cylindrical tank contains 0.750 cubic meters of nitrogen gas at 27 degrees celsius and 1.5 e5 pa absolute pressure. the

k0ka [10]

<span>3.36x10^5 Pascals The ideal gas law is PV=nRT where P = Pressure V = Volume n = number of moles of gas particles R = Ideal gas constant T = Absolute temperature Since n and R will remain constant, let's divide both sides of the equation by T, getting PV=nRT PV/T=nR Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation P1V1/T1 = P2V2/T2 where P1, V1, T1 = Initial pressure, volume, temperature P2, V2, T2 = Final pressure, volume, temperature Now convert the temperatures to absolute temperature by adding 273.15 to both of them. T1 = 27 + 273.15 = 300.15 T2 = 157 + 273.15 = 430.15 Substitute the known values into the equation 1.5E5*0.75/300.15 = P2*0.48/430.15 And solve for P2 1.5E5*0.75/300.15 = P2*0.48/430.15 430.15 * 1.5E5*0.75/300.15 = P2*0.48 64522500*0.75/300.15 = P2*0.48 48391875/300.15 = P2*0.48 161225.6372 = P2*0.48 161225.6372/0.48 = P2 335886.7441 = P2 Rounding to 3 significant figures gives 3.36x10^5 Pascals. (technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>

8 0

3 years ago