What is the color of visible light of the lowest frequencies? Of the highest frequencies?

How does electric force depend on the amount of charge and the distance between charges

Gennadij [26K]

Explanation:

The attractive or repulsive forces which act between any two charged species is an electric force.
The electric force depends on the distance between the charged species and the amount of charge which can be calculated by the formula given as follows

F = k× $\frac{q1q2}{r2}$

where, K is coulombs constant, which is equal to - 9 x10^9 $Nm^2/C^2.$

The unit for K is newtons square meters per square coulombs.
This is known as Coulomb's Law.

6 0

3 years ago

Calculate the force of gravity on the 1.2-kg mass if it were 1.9×107 m above earth's surface (that is, if it were four earth rad

Anettt [7]

Answer:

Force of gravity, F = 0.74 N

Mass of an object, m = 1.2 kg

Distance above earth's surface, $d=1.9\times 10^7\ m$

Mass of Earth, $M=5.97\times 10^{24}\ kg$

Radius of Earth, $r=6.37\times 10^6\ m$

We need to find the force of gravity above the surface of Earth. It is given by :

$F=G\dfrac{mM}{R^2}$

R = r + d

R = 25370000 m

$F=6.67\times 10^{-11}\times \dfrac{5.97\times 10^{24}\ kg\times 1.2\ kg}{(25370000\ m)^2}$

F = 0.74 N

So, the force of gravity on the object is 0.74 N. Hence, this is the required solution.

5 0

4 years ago

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 375 N

spayn [35]

Answer:

The conservation of energy should be used to answer this question.

a)

At the position where the spring is unstretched, the elastic potential energy of the spring is zero.

$K_1 + U_1 - W_f = K_2 +U_2\\K_1 - W_f = U_2$

since $U_1$ and $K_2$ is equal to zero.

$W_f = F_fx\\\\U_2 = \frac{1}{2}kx^2\\\\19 - (10)x = \frac{1}{2}(375)x^2\\\\375x^2 + 20x - 38 = 0$

The roots of this quadratic equation can be solved by using discriminant.

$\Delta = b^2 - 4ac\\x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}$

$x_1 = -0.346\\x_2 = 0.292$

We should use the positive root, so

x = 0.292 m.

b)

We should use energy conservation between the point where the spring is momentarily at rest, and the point where the spring is unstretched.

$K_2 + U_2 - W_f = K_3 + U_3\\U_2 - W_f = K_3$

since the kinetic energy at point 2 and the potential energy at point 3 is equal to zero.

$\frac{1}{2}kx^2 - F_fx = K_3\\K_3 = 15.987 - 2.92 = 13.067 J$

Explanation:

In questions with springs, the important thing is to figure out the points where kinetic or potential energy terms would be zero. When the spring is unstretched, the elastic potential energy is zero. And when the spring is at rest, naturally the kinetic energy is equal to zero.

In part b) the cookie slides back to its original position, so the distance traveled, x, is equal to the distance in part a). The frictional force is constant in the system, so it is quite simple to solve part b) after solving part a).

8 0

3 years ago

A 15.7 kg block is dragged over a rough, horizontal surface by a constant force of 83.1 N

Kay [80]

Answer:

The work done by the 83.1 N force is 4687.5 J.

The magnitude of the work done by the force of friction is 1187.5 J.

Explanation:

Given:

Mass of the block, $m=15.7$ kg

Force acting on it, $F=83.1$ N

Angle of application of force, $\theta = 25.7$ °

Displacement of the block, $d=62.6$ m

Coefficient of friction, $\mu =0.161$

Acceleration due to gravity, $g=9.8$ m/s²

Work done by a force is given as:

$Work,W=F\times d\times \cos\theta$

So, work done by the constant force is given as:

$W_{force}=83.1\times 62.6\times \cos(25.7)\\W_{force}=4687.5\textrm{ J}$

Now, in order to find work done by friction, we need to evaluate friction.

For the vertical direction, the net force is zero as there is no vertical motion.

Therefore,

$N + F\sin\theta = mg\\ N = mg-F\sin\theta$

Frictional force is given as:

$f=\mu N=\mu (mg-F \sin \theta)=0.161\times ((15.7\times 9.8)-(83.1\times \sin(25.7))=18.97\textrm{ N}$

Now, friction acts in the direction opposite to the displacement. Thus, angle between frictional force and displacement is 180°.

Therefore, work done by friction is:

$W_{fric}=f\times d\times \cos\theta_f\\W_{fric}=12.72\times 62.6\times \cos(-180)\\W_{fric}=18.97\times 62.6\times -1\\W_{fric}=-1187.5\textrm{ J}$

Negative sign indicates that frictional force is acting opposite to motion.

So, the magnitude of the work done by the force of friction is 1187.5 J.

8 0

4 years ago

A mass of 3 slugs (this is the English unit of mass, a pound is a force) is attached to a vertical spring with a spring constant

motikmotik

Answer:

equation of motion for the mass is x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin √{α² - ω²} t )

Explanation:

Given data

mass = 3 slugs = 3 * 32.14 = 96.52 lbs

constant k = 9 lbs/ft

Beta = 6lbs * s/ft

mass is pulled = 1 ft below

to find out

equation of motion for the mass

solution

we know that The mass is pulled 1 ft below so

we will apply here differential equation of free motion i.e

dx²/dt² + 2 α dx/dt + ω² x =0 ........................1

here 2 α = Beta / mass

so 2 α = 6 / 96.52

α = 0.031

α² = 0.000961 ...............2

and

ω² = k/mass

ω² = 9 /96.52

ω² = 0.093 ..................3

we can say that from equation 2 and 3 that α² - ω² = -0.092239

this is less than zero

so differential equation is

x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin √{α² - ω²} t )

equation of motion for the mass is x(t) = e^αt ( C1 cos √{α² - ω²} t + C2 sin √{α² - ω²} t )

3 0

3 years ago