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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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5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by
Anna007 [38]

Answer:

a) W=2.425kJ

b) \Delta E=2.425kJ

c) T_f=20.06^{o}C

d) Q=-2.425kJ

Explanation:

a)

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

W=Fd

Where:

W=work done [J]

F= force applied [N]

d= distance [m]

In this case since it will be a vertical movement, the force is calculated like this:

F=mg

and the distance will be the height

d=h

so the formula gets the following shape:

W=mgh

so now e can substitute:

W=(25kg)(9.7 m/s^{2})(10m)

which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

\Delta E=2.425kJ

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

\Delta Q=mC_{p}(T_{f}-T_{0})

Where:

Q= heat transferred

m=mass

C_{p}=specific heat

T_{f}= Final temperature.

T_{0}= initial temperature.

So we can solve the forula for the final temperature so we get:

T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}

So now we can substitute the data we know:

T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C

Which yields:

T_{f}=20.06^{o}C

d)

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

\Delta Q=-2.425kJ

3 0
2 years ago
Your friend Amanda suffers from a condition that reduces her blood's ability to carry oxygen.which of the following is the name
Anestetic [448]
Amanda might be suffering from a disease called Sickle Cell Anemia. It is an inherited red blood cell disorder.
7 0
2 years ago
A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
ikadub [295]
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
  KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
6 0
2 years ago
1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational
sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = \frac{m v^{2} }{r}

==> 1.1 × 10^{19} = (9.4 × 10^{21} × v^{2} ) ÷ 1.5 × 10^{8}

==> 1.1 × 10^{19} = v^{2} × 6.26×10^{13}

==> v^{2} =  1.1 × 10^{19} ÷ 6.26×10^{13}

==> v^{2} = 0.17571885 × 10^{6}

==> v= 0.419188323 × 10^{3} m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5×10^{8} ÷ 0.419188323 × 10^{3}

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0
2 years ago
Read 2 more answers
A glider of mass m1 on a frictionless horizontal track is connected to an object of mass m2 by a massless string. The glider acc
inysia [295]

Answer:

m2g -> T2 -> T1

Explanation:

m2g -> T2 -> T1

5 0
2 years ago
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