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lora16 [44]
2 years 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]2 years 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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5 0
2 years ago
A 4.0 Ω resistor, an 8.0 Ω resistor, and a 12.0 Ω resistor are connected in parallel across a 24.0 V battery. What is the equiva
dsp73

PART A)

Equivalent resistance in parallel is given as

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

now we have

\frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12}

R = 2.18 ohm

PART B)

since potential difference across all resistance will remain same as all are in parallel

so here we can use ohm's law

V = iR

for 4 ohm resistance we have

24 = 4(i)

i = 6 A

PART C)

since potential difference across all resistance will remain same as all are in parallel

so here we can use ohm's law

V = iR

for 8 ohm resistance we have

24 = 8(i)

i = 3 A

3 0
2 years ago
Read 2 more answers
A vertical wire carries a current straight up in a region where the magnetic field vector points due north. What is the directio
Elanso [62]

Answer:

The direction of the resulting force on this current is due east.

Explanation:

Given;

direction of the magnetic field to be due north

Applying right hand rule which states that: to determine the direction of the magnetic force on a positive moving charge point the thumb of the right hand in the direction of velocity v, the fingers in the direction of magnetic field B, and a perpendicular to the palm points in the direction of magnetic force.

Since the magnetic force must be perpendicular to the magnetic field, and direction of the magnetic field is due north, then the magnetic force must be due East.

Therefore, the direction of the resulting force on this current is due east.

7 0
2 years ago
A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a
pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

120^2=130^2+50^2-2\times130\times50\cos\alpha

\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}

\alpha=\cos^{-1}(0.3846)

\alpha=67.38^{\circ}

We need to calculate the angle β

Using cosine law

50^2=130^2+120^2-2\times130\times120\cos\beta

\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}

\beta=\cos^{-1}(0.923)

\beta=22.63^{\circ}

We need to calculate the force on 130 cm side

Using formula of force

F_{130}=ILB\sin\theta

F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0

F_{130}=0

We need to calculate the force on 120 cm side

Using formula of force

F_{120}=ILB\sin\beta

F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63

F_{120}=0.1385\ N

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

F_{50}=ILB\sin\alpha

F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38

F_{50}=0.1385\ N

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

8 0
2 years ago
A densly wound cylindrical coil has 210 turns per meter, a 5 cm radius, and carries 38 mA. What is the magnitude of the uniform
kaheart [24]

Answer:

(a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

Explanation:

Given that,

Number of turns = 210

Radius = 5 cm

Current = 38 mA

Current in the wire = 500 mA

We need to calculate the magnetic field inside a coil

Using formula of magnetic field

B=\mu_{0} ni

Put the value into the formula

B_{c}=4\pi\times10^{-7}\times210\times38\times10^{-3}

B_{c}=0.0000100\ T

B_{c}=10.02\times10^{-6}\ T

Now a straight wire is inserted into the coil that carries 500 mA. It travels down the central axis of the coil

Distance d=\dfrac{5}{2}

d=2.5\ cm

We need to calculate the magnetic field from the straight wire

Using formula of magnetic field

B_{w}=\dfrac{\mu_{0}I}{2\pi d}

Put the value into the formula

B_{w}=\dfrac{4\pi\times10^{-7}\times500\times10^{-3}}{2\times\pi\times2.5\times10^{-2}}

B_{w}=4.0\times10^{-6}\ T

This field is perpendicular to the wire.

The magnitude of magnetic field is

B=\sqrt{B_{c}^2+B_{w}^2}

B=\sqrt{(10.02\times10^{-6})^2+(4.0\times10^{-6})^2}

B=0.00001078\ T

B=10.78\times10^{-6}\ T

B=10.78\ \mu\ T

Hence, (a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

6 0
2 years ago
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