All categories

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)

You might be interested in

Mike is cutting the grass using a human-powered lawn mower. He pushes the mower with a force of 45 n directed at an angle of 41Â

Andrej [43]

Answer:

work done will be equal to 305.05 J

Explanation:

We have given force exerted F = 45 N

Angle with the horizontal $\Theta =41^{\circ}$

Distance moved due to exerted force d = 9.1 m

Work done is equal to $W=Fcos\Theta \times d$ , here F is force $\Theta$ is angle with horizontal and d is distance moved due to force

So work done $W=45\times cos41^{\circ}\times 9.1=309.05J$

So work done will be equal to 305.05 J

7 0

3 years ago

Frank has a sample of steel that has a mass of 80 grams if the density is 8g/cm3 what is the volume

nikklg [1K]

Density is mass divided by volume. rho=m/v. So, v=m/rho. In frank's case this is 80/8 = 10 cm^3.

7 0

3 years ago

A grinding wheel is a uniform cylinder with a radius of 8.65 cm and a mass of 0.400 kg . Calculate its moment of inertia about i

anyanavicka [17]

Answer:

I = 1.5*10⁻³ kg*m²

Explanation:

It can be showed that the moment of inertia (or rotational inertia) for a uniform cylinder of mass m and radius r, respect an longitudinal axis going through its center (parallel to the height of the cylinder) can be written as follows:

$I = \frac{1}{2}*m*r^{2} = \frac{1}{2}*0.400 kg*(0.0865m)^{2} = 1.5e-3 kg*m2$

3 0

3 years ago

A 50.0 kg object rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and

MariettaO [177]

Answer:

$f=140\ N$

Explanation:

Given:

mass of the object on a horizontal surface, $m=50\ kg$

coefficient of static friction, $\mu_s=0.3$

coefficient of kinetic friction, $\mu_k=0.2$

horizontal force on the object, $F=140\ N$

<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

$F_s=\mu_s.N$

where:

$N=$ normal force of reaction acting on the body= weight of the body

$F_s=0.3\times (50\times 9.8)$

$F_s=147\ N$

As we know that the frictional force acting on the body is always in the opposite direction:

So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.

so, the frictional force will be:

$f=140\ N$

8 0

3 years ago

A goal should NOT be:

Korvikt [17]

Answer: unspecific

Explanation:

you should always be specific with your goals, say someones goal is to make it to state for swimming, he/she should be specific to the event in which they want to go to state in.

6 0

2 years ago