Identify 3 preys and their preys

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

Tpy6a [65]

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)

3 0

3 years ago

If objects at the same distance suddenly decreased in mass the gravitational force between them would

inn [45]

Answer:

The Gravitational force would decrease

Explanation:

To calculate FG the formula is:

Fg=mg

if mass decreases then the gravitational force would decrease

5 0

3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0

3 years ago

Helpon

statuscvo [17]

Answer:

huh ?ion understand

Explanation:

have a bad day <3

7 0

3 years ago

The ash produced when solid waste is incinerated is ______ than the original waste.

11111nata11111 [884]

The ash is more toxic

5 0

3 years ago