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2 years ago

5

Suppose two equal charges of 0.65 C each are separated by a distance of 2.5 km in air. What is the magnitude of the force acting

between them, in newtons?

1 answer:

mafiozo [28]2 years ago

7 0

Answer:

Force between two equal charges will be 608.4 N

Explanation:

We have given charges $q_1=0.65C\ and\ q_2=0.65C$

Distance between the charges = 2.5 km = 2500 m

According to coulombs law force between two charges is given by

$F=\frac{1}{4\pi \varepsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}$ , here K is constant which value is $9\times 10^9Nm^2/C^2$

So force $F=\frac{9\times 10^9\times 0.65\times 0.65}{2500^2}=608.4N$

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A mass is hung from a spring and set in motion so that it oscillates continually up and down. The velocity v of the weight at ti

otez555 [7]

To solve the problem it is necessary to identify the equation in the manner given above.

This equation corresponds to the displacement of a body under the principle of simple harmonic movement.

Where,

$\xi = Acos(\omega t +\phi)$

PART A) Our equation corresponds to

$y = -5cos(4\pi t)$

Therefore the value of omega is equivalent to that of

$\omega = 4\pi$

From the definition we know that the period as a function of angular velocity is equivalent to

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{4\pi}$

$T = \frac{1}{2}$

This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the $cos\theta$ Is equivalent to . So the maximum speed that the body can reach is,

$y = -5cos(4\pi t)$

$y = -5cos(4\pi (1/2))$

$y = -5*(-1)$

$y = 5$

Therefore the maximum felocity will be 5ft / s

PART B) The period of graph is the time taken to reach from one maximum point to next point maximum point, then

$t = \frac{T}{2} = \frac{1}{2}*\frac{1}{2}$

$t = \frac{1}{4}s$

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3 years ago

The total mass of the wheelbarrow and the road is 80 kg calculate the weight of the wheelbarrow and the road

Alja [10]

Answer:

The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.

Explanation:

Given that,

The total mass of the wheelbarrow and the road is 80 kg.

The weight of an object is given by :

W = mg

where

g is acceleration due to gravity

So,

W = 80 × 9.8

= 784 N

So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.

7 0

3 years ago

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 m/s when the

german

Answer: 4 s

Explanation:

Given

The ball leaves the hand of student with a speed of $u=19\ m/s$

When the hand is $h=2.5\ m$ above the ground

Using the equation of motion we can write

$h=ut+\dfrac{1}{2}at^2$

Substitute the values

$\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]$

Thus, the ball will take 4 s to hit the ground.

5 0

2 years ago

Objects 1 and 2 attract each other with a gravitational force of 12 units. If the mass of Object 2 is tripled, then the new grav

olasank [31]

Explanation:

Fgravity = G*(mass1*mass2)/D².

G is the gravitational constant, which has the same value throughout our universe.

D is the distance between the objects.

so, if you triple one of the masses, what does that do to our equation ?

Fgravitynew = G*(3*mass1*mass2)/D²

due to the commutative property of multiplication

Fgravitynew = 3* G*(mass1*mass2)/D² = 3* Fgravity

so, the right answer is 3×12 = 36 units.

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2 years ago

A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The

nordsb [41]

Answer:

$t=6.96s$

Explanation:

From this exercise, our knowable variables are <u>hight and initial velocity </u>

$v_{oy}=96ft/s$

$y_{o}=112ft$

To find how much time does the <u>ball strike the ground</u>, we need to know that the final position of the ball is y=0ft

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

$0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}$

Solving for t using quadratic formula

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

$a=-\frac{1}{2} (32.2)\\b=96\\c=112$

$t=-0.999s$ or $t=6.96s$

<u><em>Since time can't be negative the answer is t=6.96s</em></u>

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3 years ago