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lora16 [44]
1 year ago
14

Students use a stretched elastic band to launch carts of known mass horizontally on a track. The elastic bands exert a force F,

which is nonlinear as a function of the distance xx that the band is stretched. A meterstick is used to measure xx , and a motion detector is used to measure the resulting maximum speed of the carts. This procedure is repeated to gather data for several different values of x. Which of the following graphs, if linear, supports the hypothesis that F is a function of x^2?
A. A graph of the cart's maximum speed as a function of x
B. A graph of the cart's maximum speed as a function of x^2
C. A graph of the cart's maximum speed squared as a function of x
D. A graph of the cart's maximum speed squared as a function of x^2
E. A graph of the cart's maximum speed squared as a function of x^3
Physics
1 answer:
rodikova [14]1 year ago
4 0

Answer:

the correct answer is  E

A graph of the cart's maximum speed squared as a function of x^3

Explanation:

For this exercise let's use Newton's second law

        F = m a

force has the form

        F = k x²

and acceleration is related to velocity

        a = dv / dt

Let's use the chain rule or L'Hospital

        a = dv /dx   dx/dt

        a = dv /dx    v

let's substitute

     k x² = m v dv / dx

     k /m x² dx = v dv

we integrate

     k /m    x³ /3 = v² / 2

     v² = (2k /3m)   x³

This is the expression for the variation of the speed as a function of the position, to make a linear graph realism the changes of variable

      y = v²

     x´ = x³

       y = (2k/3m)  x´

if we graph y vs x 'we have a linear graph whose slope is

      m = 2k / 3m

By reviewing the different answers, the correct answer is  E

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erica [24]

Answer:

The answer is "4-5-6 because the bank is growing too quickly in the beginning of its turn".

Explanation:

The S-turns is also known as the reference technique, in which the ground track of the aircraft on both sides of a defined ground-based straight-line distance represents 2 different but equivalent circles.  

Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

5 0
1 year ago
A 4kg bird has 8 joules of kinetic energy, how fast is it flying?
Igoryamba
I believe the answer is 2m/s
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Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o
IRINA_888 [86]

Answer:

a) W=-0.0103125\ J

b) W=0.0059375\ J

c) Compressing is easier

Explanation:

Given:

Expression of force:

F=kx-bx^2+cx^3

where:

k=100\ N.m^{-1}

b=700\ N.m^{-2}

c=12000\ N.m^{-3}

x when the spring is stretched

x when the spring is compressed

hence,

F=100x-700x^2+12000x^3

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, x=-0.05\ m

Now,

The spring constant at this instant:

j=\frac{F}{x}

j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}

j=-8.25\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times -8.25\times (-0.05)^2

W=-0.0103125\ J

b)

When compressing the spring by 0.05 m

we have, x=0.05\ m

<u>The spring constant at this instant:</u>

j=\frac{F}{x}

j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}

j=4.75\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times 4.75\times (0.05)^2

W=0.0059375\ J

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0
1 year ago
A biker travels at an average speed of 18 km/hr along a 0.30 km straight segment of a bike path. How much time (in hours) does t
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Answer: 0.016 h

Explanation:

\text{Average speed} = \frac{\text {Total Distance}}{\text {total time taken}}

It is given that, biker has an average speed = 18 km/h

Total distance traveled = 0.30 km

Therefore, time taken by biker to travel this distance:

\Rightarrow \text{total time taken} = \frac{0.30 km}{18 km/h}=0.016 h

Thus, the biker takes 0.016 hours to travel the segment of 0.30 km at an average speed of 18 km/h.

7 0
2 years ago
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

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Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
2 years ago
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