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sweet [91]
3 years ago
11

If the coefficient of kinetic friction between tires and dry pavement is 0.98, what is the shortest distance in which you can st

op an automobile by locking the brakes when traveling at 34.7 m/s
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
7 0

Answer:

The shortest distance is 62.7 m

Explanation:

Given;

coefficient of kinetic friction, μk = 0.98

initial velocity, u = 34.7 m/s

Frictional force on the tire;

Fk = -μkN

where;

N is normal reaction = mg

ma = -μkN

ma = -μkmg

a = -μkg

a = - 0.98 x 9.8 = -9.604 m/s²

The shortest distance in which you can stop an automobile by locking the brakes:

Apply equation of motion;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration of the  automobile

0 = 34.7² + 2(-9.604)x

0 = 1204.09 - 19.208x

19.208x = 1204.09

x = 1204.09/19.208

x = 62.7 m

Therefore, the shortest distance in which you can stop an automobile by locking the brakes when traveling at 34.7 m/s is 62.7 m

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A comet is a
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The answer is D. Small object made of ice and dust that orbits the Sun and forms a coma as it approaches the Sun.
7 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
elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extensi
Citrus2011 [14]

Answer:

40g

Explanation:

20g range > 1.0cm

Therefore,

40g range > 2.0cm

6 0
2 years ago
Describe a situation when you might travel at a high velocity bit with low acceleration
iren [92.7K]

I am sitting in my seat.
I am listening to my mp3 and reading my book.
My eyes are getting heavy.  They start to close.
I try to stay awake, but it's no use.
I am so warm and comfortable and sleepy,
and I have just finished my dinner.
Finally I can't help it.  Resistance is futile.
I give up, and fall deep asleep.
My head rests back against my soft, comfy seat.

My seat is in row 26 on the airplane I'm flying in
to visit my grandmother on the coast.
We are cruising at 560 miles an hour, bearing 280°,
at flight level 320 .
The temperature outside my window is  -60°F .

5 0
3 years ago
A car is travelling at 15 m/s on a horizontal road and stopped after 4 s. The coefficient of kinetic friction between the tires
givi [52]

Answer:

Fr,= umg

umg= ma

a= v/t

umg= mv/t

u= v/gt= 0.38

3 0
3 years ago
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