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4 years ago

5

Which of the following is an example of acceleration? I. A car speeds up. II. A car slows down. III. A car travels in a straight

line at a constant speed. IV. A car travels at a constant speed and turns left.

1 answer:

svetlana [45]4 years ago

6 0

I., II., and IV. are examples of acceleration. III. isn't.

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What is the acceleration of a 10kg mass pushed by a 5N force?

GenaCL600 [577]

Answer:

<h2>0.5 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

From the question we have

$a = \frac{5}{10} = \frac{1}{2} \\ = 0.5$

We have the final answer as

<h3>0.5 m/s²</h3>

Hope this helps you

5 0

3 years ago

You have a length of tubing that is closed at one end. You cut the tubing into two pieces of unequal length, giving you a tube o

umka21 [38]

Answer:

The funda mental frequency of the original tube is 182Hz.

Explanation:

See the attachment for the calculation steps.

In order to calculate the fundamental frequency of the original closed tube we need to find the length of the tube which is equal to the sum of the lengths of the two new tubes.

For closed tubes

f = nv/4L (n = 1, 3, 5,...n)

f = nv/2L (n = 1, 2, 3,...n)

The details of calculation can be found below in the attachment.

4 0

3 years ago

A(n) ___ keeps a reaction from occurring by increasing the activation energy required for a reaction to occur. A. inhibitor B. c

skelet666 [1.2K]

Due to requirements i am typing this but the ans is a

8 0

3 years ago

Read 2 more answers

In your research lab, a very thin, flat piece of glass with refractive index 1.50 and uniform thickness covers the opening of a

lara31 [8.8K]

Answer:

The longest wavelength in vacuum for which there is constructive interference for the reflected light, λ = 3472.

Explanation:

Refractive index of Glass (given) = 1.5

For the case of a constructive interference,

2nt = (m + 1/2) λ

For case 1,

2nt = (m + 1/2) 496 nm

For case 2,

2nt = (m +1+ 1/2) 386 nm

2nt = (m+3/2) * 386 nm

(m + 1/2) 496 nm = (m+3/2) * 386 nm

m = 3

Inserting the value of m in 1.

2nt = (m + 1/2) 496 nm

2*1.5t = (3 + 1/2) * 496 nm

t = ((3 + 1/2) * 496 nm)/ 3

t = 578.6 nm

The thickness of the glass, t = 578.6 nm

b)

It is generally known that for constructive interference,

2nt = (m + 1/2) λ

λ = 2nt / ((m + 1/2))

For Longest Wavelength, m = 0

λ = 2*1.5*578.6/ (1/2)

λ = 3472 nm

3 0

3 years ago

Is it possible to have a charge of 5 x 10-20 C? Why?

ruslelena [56]

1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is $2.3\cdot 10^{39}$ times bigger than the gravitational force

Explanation:

1)

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

$e=1.6\cdot 10^{-19}C$

Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than $e$ , because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

$Q=5\cdot 10^{-20}C$

If we compare this value to $e$ , we notice that $Q$ , so no object can have a charge of $Q$ .

2)

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

$Q=ne$

where

Q is the charge of the object

n is an integer multiple

e is the fundamental charge

Here we have

$Q=2.4\cdot 10^{-18}C$

Substituting the value of e, we find n:

$n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15$

n is integer, so this value of the charge is possible.

3)

We now do the same procedure for the new object in this part, which has a charge of

$Q=2.0\cdot 10^{-19}C$

Again, the charge on this object can be written as

$Q=ne$

where

n is the number of electrons in the object

Using the value of the fundamental charge,

$e=1.6\cdot 10^{-19}C$

We find:

$n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25$

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

4)

To solve this part, we use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (Action force), then object B exerts an equal and opposite force on object A (reaction force)".

In this problem, we have two objects:

- A charge Q

- A charge 5Q

Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

5)

The magnitude of the electric force between two single-point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

r = 0.30 cm = 0.003 m is the separation

So, the electric force between the two charges is

$F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-6})(7.2\cdot 10^{-6})}{(0.003)^2}=32400 N$

6)

The electric force between two charged objects has direction as follows:

- If the two objects have charges of opposite signs (+ and -), the force between them is attractive

- If the two objects have charges of same sign (++ or --), the force between them is repulsive

In this problem, the two charges are:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

We see that the two charges have same sign: therefore, the force between them is repulsive.

7)

The electric force between the proton and the electron in the atom can be written as

$F_E=k\frac{q_1 q_2}{r^2}$

where

$q_1 = q_2 = e = 1.6\cdot 10^{-19}C$ is the magnitude of the charge of the proton and of the electron

$r=5.3\cdot 10^{-11} m$ is the separation between them

So the force can be rewritten as

$F_E=\frac{ke^2}{r^2}$

The gravitational force between the proton and the electron can be written as

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p = 1.67\cdot 10^{-27}kg$ is the proton mass

$m_e=9.11\cdot 10^{-27}kg$ is the electron mass

Comparing the 2 forces,

$\frac{F_E}{F_G}=\frac{ke^2}{Gm_p m_e}=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{(6.67\cdot 10^{-11})(1.67\cdot 10^{-27})(9.11\cdot 10^{-31})}=2.3\cdot 10^{39}$

8 0

3 years ago