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3 years ago

The lower the value of the coefficient of friction, the___ the resistance to sliding.

Physics

1 answer:

Nuetrik [128]3 years ago

4 0

The lower the value of the coefficient of friction, the lower the resistance to sliding.

<u>Explanation:</u>

The coefficient of friction defines as directly proportionate with the resisting force, which is the frictional force. Hence, if there seems a decrease at coefficient of friction, then it is sure that the frictional force decreases.

We know that the frictional force on a body, is the product of coefficient of frictions and the normal forces acting on the body. Note that friction acts only, if a body is in contact, and it is of three types, static, kinetic and rolling.

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An eagle flying at 15 m/s has 600 J of kinetic energy. About how much is the eagles mass?

Eduardwww [97]

Answer:

5.33kg

Explanation:

Given parameters:

Velocity of eagle = 15m/s

Kinetic energy of the eagle = 600J

Unknown:

Mass of the eagle = ?

Solution:

The kinetic energy of any body is the energy due to the motion of a body. There are different forms of kinetic energy some of which are thermal, mechanical, electrical energy.

The formula of kinetic energy is given as;

Kinetic energy = $\frac{1}{2}$ m v²

where m is the mass, V is the velocity

substitute the parameters in the equation;

600 = $\frac{1}{2}$ x m x 15²

225m = 1200

m = $\frac{1200}{225}$ = 5.33kg

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:

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 $Q$ .

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.

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.

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.

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

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

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.

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

How will heat flow between objects?

alukav5142 [94]

Without an external agent doing work, heat will always flow from a hotter to a cooler object. Two objects of different temperature always interact. There are three different ways for heat to flow from one object to another. They are conduction, convection, and radiation.

5 0

3 years ago

Read 2 more answers

A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police

Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. $x=x_{0}+vt$

2. $x=x_{0}+v_{0}t+\frac{at^{2}}{2}$

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

$vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}$

We simplify the fraction present and rearrange for our formula so that it equals 0:

$0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0$

In the very last step we factored a common factor t. There is two possible solutions to the equation at $t=0$ and:

$0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s$

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at $t=0 s$ (when the SUV passed the police car) and $t=15.56s$ (when the police car catches up to the SUV)

8 0

3 years ago

Most earthquakes happen:

umka2103 [35]

Along plate edges, at points where oceanic or continental plates meet ot at the edges of the plates

4 0

3 years ago

Read 2 more answers