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3 years ago

In the equation F = Kq1 q2/r2 solve for q2. Solve for r.

Physics

1 answer:

Tpy6a [65]3 years ago

3 0

Answer:

r = √(k q₁ q₂ / F)

Explanation:

F = k q₁ q₂ / r²

Multiply both sides by r²:

F r² = k q₁ q₂

Divide both sides by F:

r² = k q₁ q₂ / F

Take the square root of both sides:

r = √(k q₁ q₂ / F)

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Help and show work thanks

Zepler [3.9K]

Well its C, cant show you the work its in my head sorry.

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3 years ago

SECTION B<br>THEORY QUESTIONS<br>(1a) Define the following<br>(a) Work:

Vesna [10]

Work is the amount of energy transferred

Explanation:

In physics, work is a measure of the energy transfer occurring in a process. Typically, we talk about work when energy is converted from one form into another.

For instance, work is done when a force is applied on an object. The work done on the object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

We notice the following:

No work is done when the force is perpendicular to the displacement ( $cos 90^{\circ}=0$ )
The work is maximum when the force is parallel to the displacement

Whenever work is done, there is also an energy transfer taking place. For instance, in the previous example, when the force is applied to the object, the object will accelerate (assume there is no friction), and will gain kinetic energy: therefore, there is a transfer of energy to the object.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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3 years ago

A projectile fired up into the air at an angle has a range of 235 m and a flight time of 47 s.

madam [21]

<h2>Answer:5 $ms^{-1}$ ,133.6 $m$ ,51.18 $ms^{-1}$ </h2>

Explanation:

Let $v_{x}$ , $v_{y}$ be the horizontal and vertical components of velocity.

Question a:

Horizontal component of velocity is the ratio of range and time of flight.

So,horizontal component of velocity is $\frac{235}{47}=5ms^{-1}$

So, $v_{x}=5ms^{-1}$

Question b:

Time of flight= $\frac{2v_{y}}{g}$

So, $v_{y}=\frac{47\times 9.8}{2}=51.18ms^{-1}$

Maximum height is given by $\frac{v_{y}^{2}}{2g}$

So,maximum height is $\frac{51.18^{2}}{2\times 9.8}=133.6m$

Question c:

The vertical velocity is already calculated in Question b.

$v_{y}=51.18ms^{-1}$

7 0

3 years ago

A rod of length 0.82 m, rotating with an angular speed, 4.2 rad/s, about axes that pass perpendicularly through one end, has a m

mariarad [96]

Answer:

Explanation:

KE = ½Iω²

ΚΕ = ½(mL²/3)ω²

ΚΕ = ½(0.63(0.82²)/3)4.2²

ΚΕ = 1.24541928

KE = 1.2 J

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2 years ago

a 3 kg piece of putty that is moving with a velocity of 10 m/s collides and sticks to an 8 kg bowling ball that was at rest. wha

defon

The final velocity is 2.7 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.

Therefore we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$